File "markSceme-SL-paper2.html"
Path: /IB QUESTIONBANKS/4 Fourth Edition - PAPER/HTML/Further Mathematics/Topic 2/markSceme-SL-paper2html
File size: 295.12 KB
MIME-type: application/octet-stream
Charset: utf-8
<!DOCTYPE html>
<html>
<meta http-equiv="content-type" content="text/html;charset=utf-8">
<head>
<meta charset="utf-8">
<title>IB Questionbank</title>
<link rel="stylesheet" media="all" href="css/application-212ef6a30de2a281f3295db168f85ac1c6eb97815f52f785535f1adfaee1ef4f.css">
<link rel="stylesheet" media="print" href="css/print-6da094505524acaa25ea39a4dd5d6130a436fc43336c0bb89199951b860e98e9.css">
<script src="js/application-13d27c3a5846e837c0ce48b604dc257852658574909702fa21f9891f7bb643ed.js"></script>
<script type="text/javascript" async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-MML-AM_CHTML-full"></script>
<!--[if lt IE 9]>
<script src='https://cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv.min.js'></script>
<![endif]-->
<meta name="csrf-param" content="authenticity_token">
<meta name="csrf-token" content="iHF+M3VlRFlNEehLVICYgYgqiF8jIFlzjGNjIwqOK9cFH3ZNdavBJrv/YQpz8vcspoICfQcFHW8kSsHnJsBwfg==">
<link href="favicon.ico" rel="shortcut icon">
</head>
<body class="teacher questions-show">
<div class="navbar navbar-fixed-top">
<div class="navbar-inner">
<div class="container">
<div class="brand">
<div class="inner"><a href="http://ibo.org/">ibo.org</a></div>
</div>
<ul class="nav">
<li>
<a href="../../index.html">Home</a>
</li>
<li class="active dropdown">
<a class="dropdown-toggle" data-toggle="dropdown" href="#">
Questionbanks
<b class="caret"></b>
</a><ul class="dropdown-menu">
<li>
<a href="../../geography.html" target="_blank">DP Geography</a>
</li>
<li>
<a href="../../physics.html" target="_blank">DP Physics</a>
</li>
<li>
<a href="../../chemistry.html" target="_blank">DP Chemistry</a>
</li>
<li>
<a href="../../biology.html" target="_blank">DP Biology</a>
</li>
<li>
<a href="../../furtherMath.html" target="_blank">DP Further Mathematics HL</a>
</li>
<li>
<a href="../../mathHL.html" target="_blank">DP Mathematics HL</a>
</li>
<li>
<a href="../../mathSL.html" target="_blank">DP Mathematics SL</a>
</li>
<li>
<a href="../../mathStudies.html" target="_blank">DP Mathematical Studies</a>
</li>
</ul></li>
<!-- - if current_user.is_language_services? && !current_user.is_publishing? -->
<!-- %li= link_to "Language services", tolk_path -->
</ul>
<ul class="nav pull-right">
<li>
<a href="https://06082010.xyz">IB Documents (2) Team</a>
</li></ul>
</div>
</div>
</div>
<div class="page-content container">
<div class="row">
<div class="span24">
</div>
</div>
<div class="page-header">
<div class="row">
<div class="span16">
<p class="back-to-list">
</p>
</div>
<div class="span8" style="margin: 0 0 -19px 0;">
<img style="width: 100%;" class="qb_logo" src="images/logo.jpg" alt="Ib qb 46 logo">
</div>
</div>
</div><h2>SL Paper 2</h2><div class="specification">
<p><img style="display: block; margin-left: auto; margin-right: auto;" src="images/pig.png" alt></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The diagram shows the line \(l\) meeting the sides of the triangle ABC at the points </span><span style="font-family: times new roman,times; font-size: medium;">D, E and F. The perpendiculars to \(l\) from A, B and C meet \(l\) at G, H and I.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) State why \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} = \frac{{{\rm{AG}}}}{{{\rm{HB}}}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Hence prove Menelaus’ theorem for the triangle ABC.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) State and prove the converse of Menelaus’ theorem.</span></p>
<div class="marks">[13]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A straight line meets the sides (PQ), (QR), (RS), (SP) of a quadrilateral PQRS at </span><span style="font-family: times new roman,times; font-size: medium;">the points U, V, W, X respectively. Use Menelaus’ theorem to show that\[\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = 1.\]</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"> (i) Because the triangles AGF and BHF are similar. <em><strong>R1</strong></em></span></p>
<p> </p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) It follows (by cyclic rotation or considering similar triangles) that</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{BH}}}}{{{\rm{IC}}}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> and \(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{{{\rm{CI}}}}{{{\rm{GA}}}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> Multiplying these three results gives Menelaus’ Theorem, i.e.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{{{\rm{AG}}}}{{{\rm{HB}}}} \times \frac{{{\rm{BH}}}}{{{\rm{IC}}}} \times \frac{{{\rm{CI}}}}{{{\rm{GA}}}}\) <em><strong>M1A1</strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> \( = \frac{{{\rm{AG}}}}{{{\rm{GA}}}} \times \frac{{{\rm{BH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{CI}}}}{{{\rm{IC}}}} = - 1\) <em><strong>M1A1</strong></em></span></p>
<p> </p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) The converse states that if D, E, F are points on the sides (BC), (CA), (AB) of a triangle such that</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = - 1\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> then D, E, F are collinear. <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> To prove this result, let D, E, F′ be collinear points on the three sides so that, using the above theorem, <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\frac{{{\rm{AF'}}}}{{{\rm{F'B}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = - 1\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> Since \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = - 1\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\frac{{{\rm{AF'}}}}{{{\rm{F'B}}}} = \frac{{{\rm{AF}}}}{{{\rm{FB}}}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> and \({\rm{F = F'}}\) which proves the converse. <strong><em>R1</em></strong></span></p>
<p> </p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[13 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><img src="images/beach2.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Draw the diagonal PR and let it cut the line at the point Y. <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Apply Menelaus’ Theorem to the triangle PQR. Then,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YP}}}} = - 1\) <em><strong>M1A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Now apply the theorem to triangle PRS.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{PY}}}}{{{\rm{YR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = - 1\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YP}}}} \times \frac{{{\rm{PY}}}}{{{\rm{YR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = - 1 \times - 1\) <strong><em> M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} \times \frac{{{\rm{PY}}}}{{{\rm{YP}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YR}}}} = 1\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} \times ( - 1) \times ( - 1) = 1\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = 1\) <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The circle \(C\) has centre O. The point Q is fixed in the plane of the circle and outside </span><span style="font-family: times new roman,times; font-size: medium;">the circle. The point P is constrained to move on the circle.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that the opposite angles of a cyclic quadrilateral add up to \({180^ \circ }\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A quadrilateral ABCD is inscribed in a circle \(S\) . The four tangents to \(S\) at the </span><span style="font-family: times new roman,times; font-size: medium;">vertices A, B, C and D form the edges of a quadrilateral EFGH. Given that </span><span style="font-family: times new roman,times; font-size: medium;">EFGH is cyclic, show that AC and BD intersect at right angles.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Show that the locus of a point \({\rm{P'}}\) , which satisfies \(\overrightarrow {{\rm{QP'}}} = k\overrightarrow {{\rm{QP}}} \)</span><span style="font-family: times new roman,times; font-size: medium;"> , is a circle \(C'\) , </span><span style="font-family: times new roman,times; font-size: medium;">where <em><strong>k</strong></em> is a constant and \(0 < k < 1\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that the two tangents to \(C\) from Q are also tangents to \({\rm{C'}}\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/fof.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">recognition of relevant theorem <em><strong> (M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \({\rm{D}}\hat {\rm{O}}{\rm{B}} = 2 \times {\rm{D}}\hat {\rm{A}}{\rm{B}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({360^ \circ } - {\rm{D}}\hat {\rm{O}}{\rm{B}} = 2 \times {\rm{D}}\hat {\rm{C}}{\rm{B}}\) <em><strong>A1</strong></em><br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \({\rm{D}}\hat {\rm{A}}{\rm{B}} + {\rm{D}}\hat {\rm{C}}{\rm{B}} = {\rm{18}}{0^ \circ }\) <strong><em>AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks] </span></em></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><img src="images/runner.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">diagram showing tangents EAF, FBG, GCH and HDE; diagonals cross at M. <strong><em> M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">let \(x = {\rm{E}}\hat {\rm{D}}{\rm{A}} = {\rm{E}}\hat {\rm{A}}{\rm{D}}\) ; \(y = {\rm{B}}\hat {\rm{C}}{\rm{G}} = {\rm{C}}\hat {\rm{B}}{\rm{G}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{D}}\hat {\rm{E}}{\rm{A}} + {\rm{H}}\hat {\rm{G}}{\rm{F}} = 180 - 2x + 180 - 2y = 360 - 2(x + y)\) <em><strong>M1A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{C}}\hat {\rm{D}}{\rm{B}} = y\) and \({\rm{A}}\hat {\rm{C}}{\rm{D}} = x\) , as angles in alternate segments <em><strong>M1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{D}}\hat {\rm{M}}{\rm{C}} = 180 - (x + y) = \left( {\frac{1}{2}} \right)({\rm{D}}\hat {\rm{E}}{\rm{A}} + {\rm{H}}\hat {\rm{G}}{\rm{F}}) = {90^ \circ }\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so the diagonals cross at right angles <strong><em>AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/frollo.png" alt></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong> M1</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">let \({\rm{O'}}\) be the point on OQ such \({\rm{O'P'}}\) is parallel to OP <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">using similar triangles \({\rm{O'Q}} = k{\rm{OQ}}\) , so \({\rm{O'}}\) is a fixed point <strong><em> M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">and \({\rm{O'P}} = k{\rm{OP}}\) which is constant <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \({\rm{P'}}\) lies on a circle centre \({\rm{O'}}\) <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so the locus of \({\rm{P'}}\) is a circle <em><strong>AG </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks] </span></strong></em></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let one of the two tangents to \(C\) from Q touch \(C\) at T</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the image of T lies on TQ <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">and is a unique point \({{\rm{T'}}}\) on \({C'}\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so \({\rm{TT'}}\) is a common tangent and passes through Q <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the same is true for the other tangent <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so the two tangents to \(C\) from Q are also tangents to \({C'}\) <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></p>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(a) Most candidates produced a valid answer, although a small minority used a circular argument.</span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(b) A few candidates went straight to the core of this question. However, many other candidates produced incoherent answers containing some true statements, some irrelevancies and some incorrect statements, based on a messy diagram.</span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(a) This was poorly answered. Many candidates failed to note that the points Q, P and its image were defined to be collinear, and tried to invoke the notion of the Apollonius Circle theory. Others tried a coordinate approach – in principle this could work, but is actually quite tricky without a sensible choice of axes and the origin.</span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">A circle \(C\) passes through the point \((1,{\text{ }}2)\) and has the line \(3x - y = 5\) as the tangent at the point \((3,{\text{ }}4)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the coordinates of the centre of \(C\) <span class="s1">and its radius.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the equation of \(C\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Find the coordinates of the second point on \(C\) </span>on the chord through \((1,{\text{ }}2)\) parallel to the tangent at \((3,{\text{ }}4)\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">attempt to exploit the fact that the normal to a tangent passes through the centre \((a,{\text{ }}b)\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><strong>EITHER</strong></p>
<p class="p2">equation of normal is \(y - 4 = - \frac{1}{3}(x - 3)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1">obtain \(a + 3b = 15\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">attempt to exploit the fact that a circle has a constant radius: <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2">obtain \({(1 - a)^2} + {(2 - b)^2} = {(3 - a)^2} + {(4 - b)^2}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">leading to \(a + b = 5\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2">centre is \((0,{\text{ }}5)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p1">radius \( = \sqrt {{1^2} + {3^2}} = \sqrt {10} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p2">gradient of normal \( = - \frac{1}{3}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">general point on normal \( = (3 - 3\lambda ,{\text{ }}4 + \lambda )\) <span class="Apple-converted-space"> </span><strong><em>(M1)A1</em></strong></p>
<p class="p2">this point is equidistant from \((1,{\text{ }}2)\) and \((3,{\text{ }}4)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2">if \(10{\lambda ^2} = {(2 - 3\lambda )^2} + {(2 + \lambda )^2}\)</p>
<p class="p2"><span class="Apple-converted-space">\(10{\lambda ^2} = 4 - 12\lambda + 9{\lambda ^2} + 4 + 4\lambda + {\lambda ^2}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p3"><span class="Apple-converted-space">\(\lambda = 1\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">centre is \((0,{\text{ }}5)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">radius \( = \sqrt {10\lambda } = \sqrt {10} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">attempt to substitute two points in the equation of a circle <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p4"><span class="Apple-converted-space">\({(1 - h)^2} + {(2 - k)^2} = {r^2},{\text{ }}{(3 - h)^2} + {(4 - k)^2} = {r^2}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p5"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>The <strong><em>A1 </em></strong>is for the two LHSs, which may be seen equated.</p>
<p class="p5"> </p>
<p class="p1">equate or subtract the equations</p>
<p class="p1">obtain \(h + k = 5\) or equivalent <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">attempt to differentiate the circle equation implicitly <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2">obtain \(2(x - h) + 2(y - k)\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p5"> </p>
<p class="p2"><span class="s1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Similarly, <strong><em>M1A1 </em></strong></span>if direct differentiation is used.</p>
<p class="p6"> </p>
<p class="p1"><span class="s2">substitute \((3,{\text{ }}4)\) </span>and gradient \( = 3\) to obtain \(h + 3k = 15\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">obtain centre \( = (0,{\text{ }}5)\) <span class="Apple-converted-space"> </span><strong><em>(M1)A1</em></strong></p>
<p class="p1">radius \( = \sqrt {10} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">equation of circle is \({x^2} + {(y - 5)^2} = 10\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">the equation of the chord is \(3x - y = 1\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">attempt to solve the equation for the chord and that for the circle simultaneously <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2">for example \({x^2} + {(3x - 1 - 5)^2} = 10\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">coordinates of the second point are \(\left( {\frac{{13}}{5},{\text{ }}\frac{{34}}{5}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p2"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was usually well done, using a variety of valid approaches.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was usually well done, using a variety of valid approaches.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was usually well done, using a variety of valid approaches.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The area of an equilateral triangle is \(1\) cm<sup>2</sup>. Determine the area of:</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The points A, B have coordinates (\(1\), \(0\)), (\(0\), \(1\)) respectively. The point P(\(x\), \(y\)) </span><span style="font-family: times new roman,times; font-size: medium;">moves in such a way that \({\rm{AP}} = k{\rm{BP}}\) where \(k \in {\mathbb{R}^ + }\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">the circumscribed circle.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">the inscribed circle.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">When \(k = 1\) , show that the locus of P is a straight line.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">When \(k \ne 1\) , the locus of P is a circle.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Find, in terms of \(k\) , the coordinates of C, the centre of this circle.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Find the equation of the locus of C as \(k\) varies.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><img style="display: block; margin-left: auto; margin-right: auto;" src="images/debra.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">consider the above diagram – [AD] and [BE] are the medians and O is </span><span style="font-family: times new roman,times; font-size: medium;">therefore both the incentre and the circumcentre <em><strong>(R1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">let \({\rm{AB}} = d\) and let \(R\) denote the radius of the circumcircle</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">then,</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(R = {\rm{AO}} = {\rm{AE}}\sec {30^ \circ }\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{d}{2} \times \frac{2}{{\sqrt 3 }} = \frac{d}{{\sqrt 3 }}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area of circumcircle \( = \pi {R^2} = \frac{{\pi {d^2}}}{3}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area of triangle \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{AB}}{\rm{.AC}}\sin {\rm{BAC}}\) <em><strong>M1</strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{\sqrt 3 {d^2}}}{4}\) <strong><em> (A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\sqrt 3 {d^2}}}{4} = 1 \Rightarrow {d^2} = \frac{4}{{\sqrt 3 }}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area of circumcircle \( = \frac{{4\pi }}{{3\sqrt 3 }}\) (\(2.42\)) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [8 marks]</span></em></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">let \(r\) denote the radius of the incircle</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">then</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(r = {\rm{OE}} = {\rm{AE}}\tan {30^ \circ }\) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{d}{{2\sqrt 3 }}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">(A1)</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area of incircle \( = \pi {r^2} = \frac{{\pi {d^2}}}{{12}}\)</span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{\pi }{{3\sqrt 3 }}\) (\(0.605\)) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [3 marks]</span></strong></em></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{A}}{{\rm{P}}^2} = {(x - 1)^2} + {y^2}\) and \({\rm{B}}{{\rm{P}}^2} = {x^2} + {(y - 1)^2}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({x^2} - 2x + 1 + {y^2} = {x^2} + {y^2} - 2y + 1\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(y = x\) which is the equation of a straight line <em><strong>A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({x^2} - 2x + 1 + {y^2} = {k^2}({x^2} + {y^2} - 2y + 1)\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(({k^2} - 1){x^2} + ({k^2} - 1){y^2} + 2x - 2{k^2}y + {k^2} - 1 = 0\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({x^2} + {y^2} + \frac{{2x}}{{{k^2} - 1}} - \frac{{2{k^2}y}}{{{k^2} - 1}} + 1 = 0\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">by completing the squares or quoting the standard result, <strong><em> M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">coordinates of C are</span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\(\left( { - \frac{1}{{{k^2} - 1}},\frac{{{k^2}}}{{{k^2} - 1}}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) let (\(x\), \(y\)) be the coordinates of C</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempting to find \(k\) or \({{k^2}}\) , <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \({k^2} = 1 - \frac{1}{x}\) <strong><em> (A1)</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(y = \frac{{1 - \frac{1}{x}}}{{ - \frac{1}{x}}}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y = 1 - x\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong> </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[9 marks]</span></strong></em></p>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates attempted (a) with many different methods seen although some candidates made algebraic errors in applying the appropriate trigonometrical formulae. </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Very few candidates realised that, because the centre of the triangle divides an altitude in the ratio \(2:1\), the area of the inscribed circle is one quarter the area of the circumscribed circle. </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates solved (a) correctly. </span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Solutions to (b) were often disappointing. Those candidates who used coordinate geometry often made algebraic errors in obtaining the equation of the circle and then finding the coordinates of its centre. Some candidates tried to use Apollonius’ theorem, using the fact that the centre divides the line AB in a ratio dependent upon \(k\), but this approach almost invariably led to algebraic errors. </span></p>
<div class="question_part_label">B.b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In the acute angled triangle ABC, the points E, F lie on [AC], [AB] respectively such </span><span style="font-family: times new roman,times; font-size: medium;">that [BE] is perpendicular to [AC] and [CF] is perpendicular to [AB]. The lines (BE) </span><span style="font-family: times new roman,times; font-size: medium;">and (CF) meet at H. The line (BE) meets the circumcircle of the triangle ABC at P. </span><span style="font-family: times new roman,times; font-size: medium;">This is shown in the following diagram.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/slayer.png" alt></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that CEFB is a cyclic quadrilateral.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Show that \({\rm{HE}} = {\rm{EP}}\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The line (AH) meets [BC] at D.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) By considering cyclic quadrilaterals show that \({\rm{C}}\widehat {\rm{A}}{\rm{D}} = {\rm{E}}\widehat {\rm{F}}{\rm{H}} = {\rm{E}}\widehat {\rm{B}}{\rm{C}}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence show that [AD] is perpendicular to [BC].</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) CEFB is cyclic because \({\rm{B}}\widehat {\rm{E}}{\rm{C}} = {\rm{B}}\widehat {\rm{F}}{\rm{C}} = {90^ \circ }\) <em><strong> R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">([BC] is actually the diameter)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) consider the triangles CHE, CPE <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">[CE] is common <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{H}}\widehat {\rm{E}}{\rm{C}} = {\rm{P}}\widehat {\rm{E}}{\rm{C}} = {90^ \circ }\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}\widehat {\rm{C}}{\rm{E}} = {\rm{P}}\widehat {\rm{B}}{\rm{A}}\) (subtended by chord [AP]) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}\widehat {\rm{B}}{\rm{A}} = {\rm{F}}\widehat {\rm{C}}{\rm{E}}\) (subtended by chord [FE]) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">triangles CHE and CPE are congruent <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore \({\rm{HE}} = {\rm{EP}}\) <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) EAFH is a cyclic quad because \({\rm{A}}\widehat {\rm{E}}{\rm{B}} = {\rm{C}}\widehat {\rm{F}}{\rm{A}} = {90^ \circ }\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{C}}\widehat {\rm{A}}{\rm{D}} = {\rm{E}}\widehat {\rm{F}}{\rm{H}}\) subtended by the chord [HE] <strong><em>R1AG</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">CEFB is a cyclic quad from part (a) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{E}}\widehat {\rm{F}}{\rm{H}} = {\rm{E}}\widehat {\rm{B}}{\rm{C}}\) subtended by the chord [EC] <strong><em>R1AG</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\rm{A}}\widehat {\rm{D}}{\rm{C}} = {180^ \circ } - {\rm{C}}\widehat {\rm{A}}{\rm{D}} - {\rm{D}}\widehat {\rm{C}}{\rm{A}}\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {180^ \circ } - {\rm{C}}\widehat {\rm{A}}{\rm{D}} - (90 - {\rm{E}}\widehat {\rm{B}}{\rm{C)}}\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {90^ \circ } - {\rm{C}}\widehat {\rm{A}}{\rm{D}} + {\rm{E}}\widehat {\rm{B}}{\rm{C}}\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {90^ \circ }\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">hence [AD] is perpendicular to [BC] <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Given that the elements of a \(2 \times 2\) symmetric matrix are real, show that</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) the eigenvalues are real;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) the eigenvectors are orthogonal if the eigenvalues are distinct.</span></p>
<div class="marks">[11]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The matrix \(\boldsymbol{A}\) is given by\[\boldsymbol{A} = \left( {\begin{array}{*{20}{c}}<br>{11}&{\sqrt 3 }\\<br>{\sqrt 3 }&9<br>\end{array}} \right) .\]Find the eigenvalues and eigenvectors of \(\boldsymbol{A}\).</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The ellipse \(E\) has equation \({{\boldsymbol{X}}^T}{\boldsymbol{AX}} = 24\) where \(\boldsymbol{X} = \left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right)\) and \(\boldsymbol{A}\) is as defined in </span><span style="font-family: times new roman,times; font-size: medium;">part (b).</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) Show that \(E\) can be rotated about the origin onto the ellipse \(E'\) having </span><span style="font-family: times new roman,times; font-size: medium;">equation \(2{x^2} + 3{y^2} = 6\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Find the acute angle through which \(E\) has to be rotated to coincide with \(E'\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) let </span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{M} = \left( {\begin{array}{*{20}{c}}<br>a&b\\<br>b&c<br>\end{array}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the eigenvalues satisfy</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\det (\boldsymbol{M} - \lambda \boldsymbol{I}) = 0\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((a - \lambda )(c - \lambda ) - {b^2} = 0\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda ^2} - \lambda (a + c) + ac - {b^2} = 0\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">discriminant \( = {(a + c)^2} - 4(ac - {b^2})\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> \( = {(a - c)^2} + 4{b^2} \ge 0\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that the eigenvalues are real <em><strong>AG</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) let the distinct eigenvalues be \({\lambda _1},{\lambda _2}\) , with eigenvectors </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({{\boldsymbol{X}}_1}\)</span>, \({{\boldsymbol{X}}_2}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _1}{{\boldsymbol{X}}_1} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}\) and </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _2}{{\boldsymbol{X}}_2} = {\boldsymbol{M}}{{\boldsymbol{X}}_1}\)</span> <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">transpose the first equation and postmultiply by \({{\boldsymbol{X}}_2}\) to give</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _1}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">premultiply the second equation by \({\boldsymbol{X}}_1^T\) <br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\({\lambda _2}{\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = {\boldsymbol{X}}_1^T{\boldsymbol{M}}{{\boldsymbol{X}}_2}\)</span> </span><span style="font-family: times new roman,times; font-size: medium;"><em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">it follows that</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((\lambda 1 - {\lambda _2}){\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since \(\lambda 1 \ne {\lambda _2}\) </span><span style="font-family: times new roman,times; font-size: medium;">, it follows that \({\boldsymbol{X}}_1^T{{\boldsymbol{X}}_2} = 0\)</span><span style="font-family: times new roman,times; font-size: medium;"> so that the eigenvectors </span><span style="font-family: times new roman,times; font-size: medium;">are orthogonal <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[11 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">the eigenvalues satisfy \(\left| \begin{array}{l}<br>11 - \lambda \\<br>\sqrt 3 <br>\end{array} \right.\left. \begin{array}{l}<br>\sqrt 3 \\<br>9 - \lambda <br>\end{array} \right| = 0\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\lambda ^2} - 20\lambda + 96 = 0\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\lambda = 8,12\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">first eigenvector satisfies</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>3&{\sqrt 3 }\\<br>{\sqrt 3 }&1<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \left( \begin{array}{l}<br>0\\<br>0<br>\end{array} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \) (any multiple of) \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - \sqrt 3 }<br>\end{array}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">second eigenvector satisfies</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\left( {\begin{array}{*{20}{c}}<br>{ - 1}&{\sqrt 3 }\\<br>{\sqrt 3 }&{ - 3}<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \left( \begin{array}{l}<br>0\\<br>0<br>\end{array} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \) (any multiple of ) \(\left( {\begin{array}{*{20}{c}}<br>{\sqrt 3 }\\<br>1<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) consider the rotation in which \((x,y)\) is transformed onto \((x',y')\) defined by</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right) = \left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\<br>{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">so that \(\left( \begin{array}{l}<br>x\\<br>y<br>\end{array} \right) = \left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\<br>{ - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the ellipse \(E\) becomes</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{x'}&{y'}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\<br>{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{11}&{\sqrt 3 }\\<br>{\sqrt 3 }&9<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\<br>{ - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}<br>\end{array}} \right)\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right) = 24\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{x'}&{y'}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>8&0\\<br>0&{12}<br>\end{array}} \right)\left( \begin{array}{l}<br>{x'}\\<br>{y'}<br>\end{array} \right) = 24\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2{(x')^2} + 3{(y')^2} = 6\) <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) the angle of rotation is given by \(\cos \theta = \frac{1}{2},\sin \theta = \frac{{\sqrt 3 }}{2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since a rotational matrix has the form \(\left( {\begin{array}{*{20}{c}}<br>{\cos \theta }&{ - \sin \theta }\\<br>{\sin \theta }&{\cos \theta }<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \(\theta = {60^ \circ }\) (anticlockwise) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p style="text-align: center;"><img src="images/Schermafbeelding_2017-08-18_om_12.38.09.png" alt="M17/5/FURMA/HP2/ENG/TZ0/06.a"></p>
<p><strong>Figure 1 </strong>shows a tangent [PQ] at the point Q of a circle and a line [PS] meeting the circle at the points R , S and passing through the centre O of the circle.</p>
</div>
<div class="specification">
<p style="text-align: center;"><img src="images/Schermafbeelding_2017-08-18_om_12.40.43.png" alt="M17/5/FURMA/HP2/ENG/TZ0/06.b"></p>
<p><strong>Figure 2 </strong>shows a triangle ABC inscribed in a circle. The tangents at the points A , B , C meet the opposite sides of the triangle externally at the points D , E , F respectively.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \({\text{P}}{{\text{Q}}^2} = {\text{PR}} \times {\text{PS}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>State briefly how this result can be generalized to give the tangent-secant theorem.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\frac{{{\text{A}}{{\text{D}}^2}}}{{{\text{B}}{{\text{D}}^2}}} = \frac{{{\text{CD}}}}{{{\text{BD}}}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By considering a pair of similar triangles, show that</p>
<p style="text-align: center;">\(\frac{{{\text{AD}}}}{{{\text{BD}}}} = \frac{{{\text{AC}}}}{{{\text{AB}}}}\) and hence that \(\frac{{{\text{CD}}}}{{{\text{BD}}}} = \frac{{{\text{A}}{{\text{C}}^2}}}{{{\text{A}}{{\text{B}}^2}}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By writing down and using two further similar expressions, show that the points D, E, F are collinear.</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.iii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>let \(r = \) radius of circle. Consider</p>
<p>\({\text{PR}} \times {\text{PS}} = ({\text{PO}} - r)({\text{PO}} + r)\) <strong><em>M1</em></strong></p>
<p>\( = {\text{P}}{{\text{O}}^2} - {\text{O}}{{\text{Q}}^2}\) <strong><em>A1</em></strong></p>
<p>\( = {\text{P}}{{\text{Q}}^2}\) because POQ is a right angled triangle <strong><em>R1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the result is true even if PS does not pass through O <strong><em>A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>using the tangent-secant theorem, <strong><em>M1</em></strong></p>
<p>\({\text{A}}{{\text{D}}^2} = {\text{BD}} \times {\text{CD}}\) <strong><em>A1</em></strong></p>
<p>so \(\frac{{{\text{A}}{{\text{D}}^2}}}{{{\text{B}}{{\text{D}}^2}}} = \frac{{{\text{CD}}}}{{{\text{BD}}}} \ldots {\text{ (1)}}\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider the triangles CAD and ABD. They are similar because <strong><em>M1</em></strong></p>
<p>\({\rm{D\hat AB}} = {\rm{A\hat CD}}\), angle \({\rm{\hat D}}\) is common therefore the third angles must be equal <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Beware of the assumption that AC is a diameter of the circle.</p>
<p> </p>
<p>therefore</p>
<p>\(\frac{{{\text{AD}}}}{{{\text{BD}}}} = \frac{{{\text{AC}}}}{{{\text{AB}}}} \ldots {\text{ (2)}}\) <strong><em>AG</em></strong></p>
<p>it follows from (1) and (2) that</p>
<p>\(\frac{{{\text{CD}}}}{{{\text{BD}}}} = \frac{{{\text{A}}{{\text{C}}^2}}}{{{\text{A}}{{\text{B}}^2}}}\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>two similar expressions are</p>
<p>\(\frac{{{\text{AE}}}}{{{\text{CE}}}} = \frac{{{\text{B}}{{\text{A}}^2}}}{{{\text{B}}{{\text{C}}^2}}}\) <strong><em>M1A1</em></strong></p>
<p>\(\frac{{{\text{BF}}}}{{{\text{AF}}}} = \frac{{{\text{C}}{{\text{B}}^2}}}{{{\text{C}}{{\text{A}}^2}}}\) <strong><em>A1</em></strong></p>
<p>multiplying the three expressions,</p>
<p>\(\frac{{{\text{CD}}}}{{{\text{BD}}}} \times \frac{{{\text{AE}}}}{{{\text{CE}}}} \times \frac{{{\text{BF}}}}{{{\text{AF}}}} = \frac{{{\text{A}}{{\text{C}}^2}}}{{{\text{A}}{{\text{B}}^2}}} \times \frac{{{\text{B}}{{\text{A}}^2}}}{{{\text{B}}{{\text{C}}^2}}} \times \frac{{{\text{C}}{{\text{B}}^2}}}{{{\text{C}}{{\text{A}}^2}}}\) <strong><em>M1</em></strong></p>
<p>\(\frac{{{\text{CD}}}}{{{\text{BD}}}} \times \frac{{{\text{AE}}}}{{{\text{CE}}}} \times \frac{{{\text{BF}}}}{{{\text{AF}}}} = 1\) <strong><em>A1</em></strong></p>
<p>it follows from the converse of Menelaus’ theorem (ignoring signs) <strong><em>R1</em></strong></p>
<p>that D, E, F are collinear <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iii.</div>
</div>
<br><hr><br><div class="specification">
<p>Consider the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\).</p>
</div>
<div class="specification">
<p>The area enclosed by the ellipse is \(8\pi \) and \(b = 2\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the area enclosed by the ellipse is \(\pi ab\).</p>
<div class="marks">[9]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine which coordinate axis the major axis of the ellipse lies along.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find the eccentricity.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the coordinates of the foci.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the equations of the directrices.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.iv.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The centre of another ellipse is now given as the point (2, 1). The minor and major axes are of lengths 3 and 5 and are parallel to the \(x\) and \(y\) axes respectively. Find the equation of the ellipse.</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(A = 4\int {y{\text{d}}x} \) <em><strong>(M1)</strong></em></p>
<p>\(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \Rightarrow \)</p>
<p>\(y = \frac{{b\sqrt {{a^2} - {x^2}} }}{a}\) <em><strong>(A1)</strong></em></p>
<p>let \(x = a\,{\text{cos}}\,\theta \Rightarrow y = b\,{\text{sin}}\,\theta \) <em><strong>M1</strong></em></p>
<p>\(\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = - a\,{\text{sin}}\,\theta \) <em><strong>A1</strong></em></p>
<p>when \(x = 0,\,\,\theta = \frac{\pi }{2}\). When \(x = a,\,\,\theta = 0\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow A = 4\int_{\frac{\pi }{2}}^0 {b\,{\text{sin}}\,\theta } \left( { - a\,{\text{sin}}\,\theta } \right){\text{d}}\theta \) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow A = - 4ab\int_{\frac{\pi }{2}}^0 {\,{\text{si}}{{\text{n}}^2}\,\theta } \,{\text{d}}\theta \)</p>
<p>\( \Rightarrow A = - 2ab\int_{\frac{\pi }{2}}^0 {\,\left( {1 - \,{\text{cos}}\,2\theta } \right)} \,{\text{d}}\theta \) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow A = - 2ab\left[ {\theta - \frac{{{\text{sin}}\,2\theta }}{2}} \right]_{\frac{\pi }{2}}^0\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow A = - 2ab\left[ {0 - 0 - \left( {\frac{\pi }{2} - 0} \right)} \right]\) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow A = \pi ab\) <em><strong>AG</strong></em></p>
<p><em><strong>[9 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(b = 2\)</p>
<p>hence \(2\pi a = 8\pi \Rightarrow a = 4\) <em><strong>A1</strong></em></p>
<p>hence major axis lies along the <em>x</em>-axis <em><strong>A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({b^2} = {a^2}\left( {1 - {e^2}} \right)\) <em><strong>(M1)</strong></em></p>
<p>\(4 = 16\left( {1 - {e^2}} \right) \Rightarrow e = \frac{{\sqrt 3 }}{2}\) <em><strong>A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>coordinates of foci are \(\left( { \pm ae,\,0} \right) = \left( {2\sqrt 3 ,\,0} \right),\,\left( { - 2\sqrt 3 ,\,0} \right)\) <em><strong>A1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>equations of directrices are \(x = \pm \frac{a}{e} = \frac{8}{{\sqrt 3 }},\, - \frac{8}{{\sqrt 3 }}\) <em><strong>A1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(a = \frac{3}{2},\,b = \frac{5}{2}\) <em><strong>(A1)</strong></em></p>
<p>hence equation is \(\frac{4}{9}{\left( {x - 2} \right)^2} + \frac{4}{{25}}{\left( {y - 1} \right)^2} = 1\) <em><strong>M1A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the ellipse having equation \({x^2} + 3{y^2} = 2\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find the equation of the tangent to the ellipse at the point \(\left( {1,{\text{ }}\frac{1}{{\sqrt 3 }}} \right)\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the equation of the normal to the ellipse at the point \(\left( {1,{\text{ }}\frac{1}{{\sqrt 3 }}} \right)\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Given that the tangent crosses the <span class="s1">\(x\)-axis at P </span>and the normal crosses the <span class="s1">\(y\)-axis at Q, find the equation of (PQ)</span>.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence show that <span class="s1">(PQ) </span>touches the ellipse.</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">State the coordinates of the point where <span class="s1">(PQ) </span>touches the ellipse.</p>
<div class="marks">[1]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the coordinates of the foci of the ellipse.</p>
<div class="marks">[4]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the equations of the directrices of the ellipse.</p>
<div class="marks">[1]</div>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>(i) \(2x + 6y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) <strong><em>M1</em></strong></p>
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{x}{{3y}}\) <strong><em>A1</em></strong></p>
<p>gradient of tangent is \( - \frac{{\sqrt 3 }}{3}\) <strong><em>A1</em></strong></p>
<p>equation of tangent is \(y - \frac{1}{{\sqrt 3 }} = - \frac{{\sqrt 3 }}{3}(x - 1)\) <strong><em>M1A1</em></strong></p>
<p>\(\left( { \Rightarrow y = - \frac{{\sqrt 3 }}{3}x + \frac{{\sqrt 3 }}{3} + \frac{1}{{\sqrt 3 }} \Rightarrow y = - \frac{{\sqrt 3 }}{3}x + \frac{2}{{\sqrt 3 }}} \right)\)</p>
<p> </p>
<p>(ii) gradient of normal is \(\sqrt 3 \) <strong><em>A1</em></strong></p>
<p>equation of normal is \(y - \frac{1}{{\sqrt 3 }} = \sqrt 3 (x - 1)\) <strong><em>A1</em></strong></p>
<p>\(\left( { \Rightarrow y = x\sqrt 3 - \sqrt 3 + \frac{1}{{\sqrt 3 }} \Rightarrow y = \sqrt 3 x - \frac{2}{{\sqrt 3 }}} \right)\)</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">coordinates of <span class="s1">P </span>are \((2,{\text{ }}0)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">coordinates of <span class="s1">Q </span>are \(\left( {0,{\text{ }} - \frac{2}{{\sqrt 3 }}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">equation of <span class="s1">(PQ) </span>is \(\frac{{y - 0}}{{x - 2}} = \frac{{\frac{2}{{\sqrt 3 }}}}{2}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( \Rightarrow y = \frac{1}{{\sqrt 3 }}(x - 2)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">substitute equation of <span class="s1">(PQ) </span>into equation of ellipse</p>
<p class="p1">\({x^2} + 3{\left( {\frac{{x - 2}}{{\sqrt 3 }}} \right)^2} = 2\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( \Rightarrow {x^2} + {x^2} - 4x + 4 = 2\)</p>
<p class="p1">\( \Rightarrow {(x - 1)^2} = 0\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">since the equation has two equal roots <span class="s1">(PQ) </span>touches the ellipse <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\left( {1,{\text{ }} - \frac{1}{{\sqrt 3 }}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({x^2} + 3{y^2} = 2\)</p>
<p class="p1">\(\frac{{{x^2}}}{2} + \frac{{{y^2}}}{{\frac{2}{3}}} = 1\)</p>
<p class="p1">\( \Rightarrow a = \sqrt 2 ,{\text{ }}b = \sqrt {\frac{2}{3}} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>EITHER</strong></p>
<p class="p1">\({b^2} = {a^2}(1 - {e^2})\)</p>
<p class="p1">\(\frac{2}{3} = 2(1 - {e^2})\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( \Rightarrow e = \sqrt {\frac{2}{3}} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">coordinates of foci are \(( \pm ae,{\text{ }}0) \Rightarrow \left( {\frac{2}{{\sqrt 3 }},{\text{ }}0} \right),{\text{ }}\left( { - \frac{2}{{\sqrt 3 }},{\text{ }}0} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">\({f^2} = {a^2} - {b^2}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({f^2} = 4 - \frac{2}{3}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">coordinates of foci are \(\left( {\frac{2}{{\sqrt 3 }},{\text{ }}0} \right),{\text{ }}\left( { - \frac{2}{{\sqrt 3 }},{\text{ }}0} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: </strong>Award accuracy marks if \({a^2}\), \({b^2}\) and \({e^2}\) are given.</p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>equations of directrices are \(x = \pm \frac{a}{e} \Rightarrow x = \sqrt 3 ,{\text{ }}x = - \sqrt 3 \) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>OR</strong></p>
<p>\(d = \frac{{{a^2}}}{f} \Rightarrow x = \sqrt 3 ,{\text{ }}x = - \sqrt 3 \) <strong><em>A1</em></strong></p>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts a) and b) were well done by most candidates, but surprisingly many candidates lost marks on part c). Parts e) and f) were only completed successfully by a small number of candidates and it was common to see parts a) and b) fully correct, parts c) and d) attempted but not fully correct and parts e) and f) not attempted at all.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts a) and b) were well done by most candidates, but surprisingly many candidates lost marks on part c). Parts e) and f) were only completed successfully by a small number of candidates and it was common to see parts a) and b) fully correct, parts c) and d) attempted but not fully correct and parts e) and f) not attempted at all.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts a) and b) were well done by most candidates, but surprisingly many candidates lost marks on part c). Parts e) and f) were only completed successfully by a small number of candidates and it was common to see parts a) and b) fully correct, parts c) and d) attempted but not fully correct and parts e) and f) not attempted at all.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts a) and b) were well done by most candidates, but surprisingly many candidates lost marks on part c). Parts e) and f) were only completed successfully by a small number of candidates and it was common to see parts a) and b) fully correct, parts c) and d) attempted but not fully correct and parts e) and f) not attempted at all.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts a) and b) were well done by most candidates, but surprisingly many candidates lost marks on part c). Parts e) and f) were only completed successfully by a small number of candidates and it was common to see parts a) and b) fully correct, parts c) and d) attempted but not fully correct and parts e) and f) not attempted at all.</p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts a) and b) were well done by most candidates, but surprisingly many candidates lost marks on part c). Parts e) and f) were only completed successfully by a small number of candidates and it was common to see parts a) and b) fully correct, parts c) and d) attempted but not fully correct and parts e) and f) not attempted at all.</p>
<div class="question_part_label">f.</div>
</div>
<br><hr><br><div class="specification">
<p>The hyperbola with equation \({x^2} - 4xy - 2{y^2} = 3\) is rotated through an acute anticlockwise angle \(\alpha \) about the origin.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The point \((x,{\text{ }}y)\) is rotated through an anticlockwise angle \(\alpha \) about the origin to become the point \((X,{\text{ }}Y)\). Assume that the rotation can be represented by</p>
<p>\[\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right].\]</p>
<p>Show, by considering the images of the points \((1,{\text{ }}0)\) and \((0,{\text{ }}1)\) under this rotation that</p>
<p>\[\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right].\]</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By expressing \((x,{\text{ }}y)\) in terms of \((X,{\text{ }}Y)\), determine the equation of the rotated hyperbola in terms of \(X\) and \(Y\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Verify that the coefficient of \(XY\) in the equation is zero when \(\tan \alpha = \frac{1}{2}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the equation of the rotated hyperbola in this case, giving your answer in the form \(\frac{{{X^2}}}{{{A^2}}} - \frac{{{Y^2}}}{{{B^2}}} = 1\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find the coordinates of the foci of the hyperbola prior to rotation.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.iv.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>consider \(\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a \\ c \end{array}} \right]\) <strong><em>(M1)</em></strong></p>
<p>the image of \((1,{\text{ }}0)\) is \((\cos \alpha ,{\text{ }}\sin \alpha )\) <strong><em>A1</em></strong></p>
<p>therefore \(a = \cos \alpha ,{\text{ }}c = \sin \alpha \) <strong><em>AG</em></strong></p>
<p>consider \(\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} b \\ d \end{array}} \right]\)</p>
<p>the image of \((0,{\text{ }}1)\) is \(( - \sin \alpha ,{\text{ }}\cos \alpha )\) <strong><em>A1</em></strong></p>
<p>therefore \(b = - \sin \alpha ,{\text{ }}d = \cos \alpha \) <strong><em>AG</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right]\)</p>
<p>or \(x = X\cos \alpha + Y\sin \alpha ,{\text{ }}y = - X\sin \alpha + Y\cos \alpha \) <strong><em>A1</em></strong></p>
<p>substituting in the equation of the hyperbola, <strong><em>M1</em></strong></p>
<p>\({(X\cos \alpha + Y\sin \alpha )^2} - 4(X\cos \alpha + Y\sin \alpha )( - X\sin \alpha + Y\cos \alpha )\)</p>
<p>\( - 2{( - X\sin \alpha + Y\cos \alpha )^2} = 3\) <strong><em>A1</em></strong></p>
<p>\({X^2}({\cos ^2}\alpha - 2{\sin ^2}\alpha + 4\sin \alpha \cos \alpha ) + \)</p>
<p>\(XY(2\sin \alpha \cos \alpha - 4{\cos ^2}\alpha + 4{\sin ^2}\alpha + 4\sin \alpha \cos \alpha ) + \)</p>
<p>\({Y^2}({\sin ^2}\alpha - 2{\cos ^2}\alpha - 4\sin \alpha \cos \alpha ) = 3\)</p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>when \(\tan \alpha = \frac{1}{2},{\text{ }}\sin \alpha = \frac{1}{{\sqrt 5 }}\) and \(\cos \alpha = \frac{2}{{\sqrt 5 }}\) <strong><em>A1</em></strong></p>
<p>the \(XY{\text{ term}} = 6\sin \alpha \cos \alpha - 4{\cos ^2}\alpha + 4{\sin ^2}\alpha \) <strong><em>M1</em></strong></p>
<p>\( = 6 \times \frac{1}{{\sqrt 5 }} \times \frac{2}{{\sqrt 5 }} - 4 \times \frac{4}{5} + 4 \times \frac{1}{5}\left( {\frac{{12}}{5} - \frac{{16}}{5} + \frac{4}{5}} \right)\) <strong><em>A1</em></strong></p>
<p>\( = 0\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the equation of the rotated hyperbola is</p>
<p>\(2{X^2} - 3{Y^2} = 3\) <strong><em>M1A1</em></strong></p>
<p>\(\frac{{{X^2}}}{{{{\left( {\sqrt {\frac{3}{2}} } \right)}^2}}} - \frac{{{Y^2}}}{{{{(1)}^2}}} = 1\) <strong><em>A1</em></strong></p>
<p>\(\left( {{\text{accept }}\frac{{{X^2}}}{{\frac{3}{2}}} - \frac{{{Y^2}}}{1} = 1} \right)\)</p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the coordinates of the foci of the rotated hyperbola</p>
<p>are \(\left( { \pm \sqrt {\frac{3}{2} + 1} ,{\text{ }}0} \right) = \left( { \pm \sqrt {\frac{5}{2}} ,{\text{ }}0} \right)\) <strong><em>M1A1</em></strong></p>
<p>the coordinates of the foci prior to rotation were given by</p>
<p>\(\left[ {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}} \\ { - \frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { \pm \sqrt {\frac{5}{2}} } \\ 0 \end{array}} \right]\)</p>
<p><strong><em>M1A1</em></strong></p>
<p>\(\left[ {\begin{array}{*{20}{c}} { \pm \sqrt 2 } \\ { \mp \frac{1}{{\sqrt 2 }}} \end{array}} \right]\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iv.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iv.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The points D, E, F lie on the sides [BC], [CA], [AB] of the triangle ABC and [AD], </span><span style="font-family: times new roman,times; font-size: medium;">[BE], [CF] intersect at the point G. You are given that CD \( = 2\)BD and AG \( = 2\)GD .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">By considering (BE) as a transversal to the triangle ACD, show that</span></p>
<p style="text-align: center;"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{3}{2}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Determine the ratios</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}}\) ;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) \(\frac{{{\rm{BG}}}}{{{\rm{GE}}}}\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><img style="display: block; margin-left: auto; margin-right: auto;" src="images/jason.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The diagram shows a hexagon ABCDEF inscribed in a circle. All the sides of the </span><span style="font-family: times new roman,times; font-size: medium;">hexagon are equal in length. The point P lies on the minor arc AB of the circle. </span><span style="font-family: times new roman,times; font-size: medium;">Using Ptolemy’s theorem, show that\[{\rm{PE}} + {\rm{PD}} = {\rm{PA}} + {\rm{PB}} + {\rm{PC}} + {\rm{PF}}\]</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">B.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><img src="images/pizza.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">using Menelaus’ theorem in \(\Delta {\rm{ACD}}\) , </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} \bullet \frac{{{\rm{AG}}}}{{{\rm{GD}}}} \bullet \frac{{{\rm{DB}}}}{{{\rm{BC}}}} = - 1\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} \bullet \frac{2}{1} \bullet \frac{1}{3} = 1\) <strong><em> A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{3}{2}\) <strong><em>AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) using Ceva’s theorem in \(\Delta {\rm{ABC}}\) ,</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} \bullet \frac{{{\rm{AF}}}}{{{\rm{FB}}}} \bullet \frac{{{\rm{BD}}}}{{{\rm{DC}}}} = 1\) <strong><em> M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{3}{2} \bullet \frac{{{\rm{AF}}}}{{{\rm{FB}}}} \bullet \frac{1}{2} = 1\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} = \frac{4}{3}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) using Menelaus’ theorem in \(\Delta {\rm{ABE}}\) , with traversal (FC), <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \bullet \frac{{{\rm{BG}}}}{{{\rm{GE}}}} \bullet \frac{{{\rm{EC}}}}{{{\rm{CA}}}} = - 1\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{4}{3} \bullet \frac{{{\rm{BG}}}}{{{\rm{GE}}}} \bullet \frac{3}{5} = 1\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{BG}}}}{{{\rm{GE}}}} = \frac{5}{4}\) <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">using Ptolemy’s theorem in PAEC, <em><strong>M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{PA}} \bullet {\rm{EC}} + {\rm{AE}} \bullet {\rm{PC = PE}} \bullet {\rm{AC}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since \({\rm{EC = AE = AC}}\) , <strong><em>M1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{PE = PA + PC}}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">similarly for PBDF, <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{PB \bullet DF + BD \bullet PF = PD \bullet BF}}\) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{PD = PB + PF}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">adding these results, </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{PE + PD = PA + PB + PC + PF}}\) <strong><em>AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></em></strong></p>
<div class="question_part_label">B.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part A was reasonably well done by many candidates. It would appear from the scripts that, in general, candidates find Menelaus’ Theorem more difficult to apply than Ceva’s Theorem, probably because the choice of transversal is not always obvious. </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part A was reasonably well done by many candidates. It would appear from the scripts that, in general, candidates find Menelaus’ Theorem more difficult to apply than Ceva’s Theorem, probably because the choice of transversal is not always obvious. </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Few fully correct answers were seen to part B with most candidates unable to identify which cyclic quadrilaterals should be used. </span></p>
<div class="question_part_label">B.</div>
</div>
<br><hr><br><div class="specification">
<p><img style="display: block; margin-left: auto; margin-right: auto;" src="images/12k.png" alt></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The diagram shows triangle ABC together with its inscribed circle. Show that </span><span style="font-family: times new roman,times; font-size: medium;">[AD], [BE] and [CF] are concurrent.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">PQRS is a parallelogram and T is a point inside the parallelogram such that the </span><span style="font-family: times new roman,times; font-size: medium;">sum of \({\rm{P}}\hat {\rm{T}}{\rm{Q}}\) and \({\rm{R}}\hat {\rm{T}}{\rm{S}}\) is \({180^ \circ }\) . Show that \({\rm{TP}} \times {\rm{TR}} + {\rm{ST}} \times {\rm{TQ}} = {\rm{PQ}} \times {\rm{QR}}\) .</span></p>
<div class="marks">[13]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Since the lengths of the two tangents from a point to a circle are equal <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{AF = AE, BF = BD, CD = CE}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Consider </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = 1\) (signed lengths are not relevant here) <em><strong>M2A2 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">It follows by the converse to Ceva’s Theorem that [AD], [BE], [CF] are concurrent. <em><strong>R2 </strong></em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[8 marks]</strong> </span></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/amy.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Draw the \(\Delta {\rm{PQU}}\) congruent to \(\Delta {\rm{SRT}}\) (or translate \(\Delta {\rm{SRT}}\) to form \(\Delta {\rm{PQU}}\) ). <em><strong>R2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Since \({\rm{P}}\hat {\rm{U}}{\rm{Q}} = {\rm{S}}\hat {\rm{T}}{\rm{R}}\) , and \({\rm{S}}\hat {\rm{T}}{\rm{R + P}}\hat {\rm{T}}{\rm{Q}} = {180^ \circ }\) it follows that <em><strong>R2 </strong></em></span><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}\hat {\rm{U}}{\rm{Q + P}}\hat {\rm{T}}{\rm{Q}} = {180^ \circ }\) <strong><em>R1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The quadrilateral PUQT is therefore cyclic <strong><em>R2</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Using Ptolemy’s Theorem, <em><strong>M2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{UQ}} \times {\rm{PT}} + {\rm{PU}} \times {\rm{QT}} = {\rm{PQ}} \times {\rm{UT}}\) <strong><em>A2</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Since \({\rm{UQ = TR}}\) , \({\rm{PU = ST}}\) and \({\rm{UT = QR}}\) <strong><em>R2</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Then \({\rm{TP}} \times {\rm{TR}} + {\rm{ST}} \times {\rm{TQ}} = {\rm{PQ}} \times {\rm{QR}}\) <strong><em>AG </em></strong></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[13 marks]</em> </span></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><img src="images/9.png" alt></p>
<p>The diagram above shows a point \({\text{O}}\) inside a triangle \({\text{ABC}}\). The lines \({\text{(AO), (BO), (CO)}}\) meet the lines \({\text{(BC), (CA), (AB)}}\) at the points \({\text{D, E, F}}\) respectively. The lines \({\text{(EF), (BC)}}\) meet at the point \({\text{G}}\).</p>
</div>
<div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that, with the usual convention for the signs of lengths in a triangle, \(\frac{{{\text{BD}}}}{{{\text{DC}}}} = - \frac{{{\text{BG}}}}{{{\text{GC}}}}\).</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">(b) The lines \({\text{(FD), (CA)}}\) meet at the point \({\text{H}}\) and the lines \({\text{(DE), (AB)}}\) meet at the point \({\text{I}}\). Show that the points \({\text{G, H, I}}\) are collinear.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) applying Ceva’s theorem to triangle </span><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{ABC}}\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{\text{CD}}}}{{{\text{DB}}}} \times \frac{{{\text{AE}}}}{{{\text{EC}}}} \times \frac{{{\text{BF}}}}{{{\text{FA}}}} = 1\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">applying Menelaus’ theorem to triangle </span><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{ABC}}\) with transversal \({\text{(GFE)}}\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CE}}}}{{{\text{EA}}}} \times \frac{{{\text{AF}}}}{{{\text{FB}}}} - = 1\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">multiplying the two equations, <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{\text{CD}}}}{{{\text{DB}}}} \times \frac{{{\text{BG}}}}{{{\text{GC}}}} = - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so that \(\frac{{{\text{BD}}}}{{{\text{DC}}}} = - \frac{{{\text{BG}}}}{{{\text{GC}}}}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) similarly</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{\text{CE}}}}{{{\text{EA}}}} = - \frac{{{\text{CH}}}}{{{\text{HA}}}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">and \(\frac{{{\text{AF}}}}{{{\text{FB}}}} = - \frac{{{\text{AI}}}}{{{\text{IB}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">multiplying the three results,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{\text{BD}}}}{{{\text{DC}}}} \times \frac{{{\text{CE}}}}{{{\text{EA}}}} \times \frac{{{\text{AF}}}}{{{\text{FB}}}} = - \frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CH}}}}{{{\text{HA}}}} \times \frac{{{\text{AI}}}}{{{\text{IB}}}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">by Ceva’s theorem, as shown previously, the left hand side is equal to \(1\), therefore so is the right hand side <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">that is \(\frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CH}}}}{{{\text{HA}}}} \times \frac{{{\text{AI}}}}{{{\text{IB}}}} = - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">it follows from the converse to Menelaus’ theorem that </span><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{G, H, I}}\) are collinear <em><strong>R1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Prove that the interior bisectors of two of the angles of a non-isosceles triangle </span><span style="font-family: times new roman,times; font-size: medium;">and the exterior bisector of the third angle, meet the sides of the triangle in three </span><span style="font-family: times new roman,times; font-size: medium;">collinear points.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">A.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">An equilateral triangle QRT is inscribed in a circle. If S is any point on the </span><span style="font-family: times new roman,times; font-size: medium;">arc QR of the circle,</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) prove that \({\rm{ST}} = {\rm{SQ}} + {\rm{SR}}\) ;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) show that triangle RST is similar to triangle PSQ where P is the intersection </span><span style="font-family: times new roman,times; font-size: medium;">of [TS] and [QR];</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iii) using your results from parts (i) and (ii) deduce that \(\frac{1}{{{\rm{SP}}}} = \frac{1}{{{\rm{SQ}}}} + \frac{1}{{{\rm{SR}}}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[10]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Perpendiculars are drawn from a point P on the circumcircle of triangle LMN to </span><span style="font-family: times new roman,times; font-size: medium;">the three sides. The perpendiculars meet the sides [LM], [MN] and [LN] at the </span><span style="font-family: times new roman,times; font-size: medium;">points E, F and G respectively.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Prove that \({\rm{PL}} \times {\rm{PF}} = {\rm{PM}} \times {\rm{PG}}\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><img src="images/football.png" alt></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">triangle ABC has interior angle bisectors AH, BG and exterior angle bisector CK </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">using the angle bisector theorem, <em><strong>M1</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{CH}}}}{{{\rm{HB}}}} = \frac{{{\rm{CA}}}}{{{\rm{AB}}}}\) , \(\frac{{{\rm{AG}}}}{{{\rm{GC}}}} = \frac{{{\rm{AB}}}}{{{\rm{CB}}}}\) , \(\frac{{{\rm{BK}}}}{{{\rm{AK}}}} = \frac{{{\rm{CB}}}}{{{\rm{CA}}}}\) <em><strong>A2</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">hence, \(\frac{{{\rm{CH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{AG}}}}{{{\rm{GC}}}} \times \frac{{{\rm{BK}}}}{{{\rm{AK}}}} = \frac{{{\rm{CA}}}}{{{\rm{AB}}}} \times \frac{{{\rm{AB}}}}{{{\rm{CB}}}} \times \frac{{{\rm{CB}}}}{{{\rm{CA}}}} = 1\) <strong><em>M1A1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">but \(\frac{{{\rm{BK}}}}{{{\rm{AK}}}} = - \frac{{{\rm{BK}}}}{{{\rm{KA}}}}\) <strong><em>R1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so, \(\frac{{{\rm{CH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{AG}}}}{{{\rm{GC}}}} \times \frac{{{\rm{BK}}}}{{{\rm{AK}}}} = - 1\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">hence, by converse Menelaus’ theorem, G, H and K are collinear <strong><em>R1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[8 marks] </span></em></strong></p>
<div class="question_part_label">A.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">(i)</span><br><img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAVQAAAFACAIAAABstxNvAAAgAElEQVR4nOy9d0BTyff/fVJIgUDoRXpHKQoWFEEUxYYKdkF27VjWhmIBCxbsFRXEXrAtiopl7YqCDVSwrIouKkVUQDokhJTnj/n+8vABpISb3ATv6y+9hJkTzTszc+YUEBEQEPyWAN4GEBAQ4AMhfgKC3xRC/AQEvymE+AkIflMI8RMQ/KYQ4icg+E0hxE+gqFRUVJSWllZUVOBtiKJCiJ9AITlw4ECHDh2sra1DQkLwtkVRIcRPoGBUVFTs2rUL/h8ODg7E4i8ZhPgJFIwjR44AAI1Gc3JyGjp0KJPJfPnyJd5GKSSE+AkUibNnz6qoqADApEmTpk+fPn36dACYMGHCzp07jxw5grd1CgYhfgLFoKamJj4+XllZGQC8vb2dnZ2hHoMGDdq2bdvnz5/xNlYxIMRPoBjk5+erq6uz2ewDBw4kJSXRaLSOHTv27Nnz6tWr0dHRVlZWHTt2JJFIAGBtbZ2eno63vQoAIX4CBUAgEMyePRsAVq5cKRKJ5s+fTyaTCwoKxC8oKCgoKyuLi4vr2rUrADg7O+NnrMJAiJ9AMfDw8ACAjRs3ikSiiIgIJpNZVlZW/2U8Hi8wMBAAYmNj+Xy+zM1UJAjxEygAz58/19HRUVJSio2NFYlEGRkZqqqqOTk5tV8jFArRH27fvq2hoeHm5iYQCHCwVXEgxE+gABw/flxFRcXGxqayslIkEnG53JMnT/L5/Orq6tLS0vqvHzt2bKdOnQjxNw4hfgIF4Pz582pqao6OjrW3+uXl5b96/ejRo52dnYltf+MQ4idQAE6cOKGkpNStWzcul4ueJCYmWllZbd26VSQS1Vnhv3792qVLF2tr6+rqahxsVRwI8RMoAFeuXCGTycbGxoWFhejJixcvNDU1AWDVqlW1X1leXu7m5gYAo0aNIlb+xiHET6AAnDx5kkwmm5iYVFVViR9u2LABAJSUlPbs2VNTU4MeXrhwAQA0NTUzMzNxMlZhIMRPoADcuXOHTqfTaLTbt2+LH+bn5/fs2RMA6HQ6Ov+npKTo6OgAwKZNm/AzVmEgxN/24fF4NTU11dXVQqGwsrKyrKyspqaGy+UWFRWtWbNm3Lhx/v7+mzZtWrly5dChQ4cOHRoSEhIUFHTz5s38/PwvX74UFBQUFRX9+PHjx48fOJ6iUTxvcHBw7YepqakAoK6ufvny5YCAAHV1dQAICwvDy0jFghB/myI7O/uff/65efPmlStX4uLioqKipkyZ4uDg4Orqam9vb2dnZ2pqamhoaG1tbW1tbWpqikLilZWVTUxM0J+1tLSMjY1NTEzs7OxsbW07duyora2tp6enqampqanp4eGxbNmy8PDwiIiIqKiomJiYAwcOyOat3b59G23yAwMDo6OjKyoqysrKkpOTmUwmi8Wi0+kAYGdnt3nzZtnY0wYgxK/wVFRUzJ49OzAw0N/f38zMrHaiC4VCQeGuCGdnZxcXF09PTy8vrz///HP16tWenp6LFy/OysrKy8ubMWNGfHx8fn7+z58/y8vLExMTY2Nj09LSfH19vb29BwwYMGrUqMGDB9va2tZJp/Hw8HBzc3Nxcenfv396enpGRsaHDx+k8U43bNiA1nYAMDY2NjQ01NLSQm+TyWQuX768tkeAoEkI8SsA4tg1kUj0+fPnc+fOnT17duPGjfb29vb29ubm5ii/ncFgsNnsdu3a6enp9evXb/78+Xfu3BGJRBcvXgwLC7t27RomxnC53Ojo6NOnTx87dmz37t2rVq2ytbW1srJq166dubm5vr4+mUymUCjjx49fvnx5cHAwJlsD8b9Aenr60KFD3d3de/Xq1bdvXzc3N3d3dxqNRqfTb9682fqJfisI8SsGCQkJwcHBM2bMsLKyEi+5NjY27u7ubm5uly5dyszM/Pz58+fPnzMzM//77z9p33LNnj374cOH3759y8nJKSsry8vLQ+6AEydOBAYGBgUF1d5x2Nvbjxw58sWLFxkZGZh7DcrKytCe/+zZs9iO3OYhxC+/ZGVlXbt2LTQ01M3NjUajAYCKigqdTre0tFy0aNGWLVvwKl8lEAiOHTvm5eXl6+vr5OTk6OhoYWExfvz42udtgUBw5cqVy5cvX7p0yc/Pr3379srKyjQarX///iEhIZcvX8bKmJqamiFDhgCAl5cXh8PBatjfAUL88khCQsKKFSvs7e3RyqmkpNS9e/cdO3ZkZ2e/e/cuPz8fbwNFov93iXDv3r3AwMCxY8cOGjQIALp06dK9e3d3d/cLFy5kZ2d///49KysrOztbJBIlJCT06NEDvSNtbW0PD4+dO3e+e/euuLi4lZY8fvxYXMzj4cOHt2/fvn///uPHj4nY/sYhxC8vlJSUvHv3bsuWLQ4ODqgoha6u7owZM6ZOnXr37l28rWsWt27d2rZtGwq809TU1NDQUFJSAgAGg7Ft27abN2+eOHFixowZSkpKFApFRUVFV1dXR0fHz88vMTGxNULlcrnh4eF13JBdu3Yl/H+NQ4hfLli9erW9vb2SkhKVSnVwcPjrr7/Onz///v17vO2ShJycnNevX//333/x8fEzZ86cMGHC1KlTxbsYdH4RQyaTxVrt1avX+fPn37x5I9m8x44dGzhwIDocRUZGFhUVYfu+2h6E+HHj48eP//777/Dhw+3s7JAALC0tjx07VudltV39isvXr1+PHDly8eLF69evh4WFmZqa6unp6erqUqnUOis2k8k8ceIEusbncrlcLre8vLyZh/mqqqpHjx69fv1a2m+nbUCIHwfu3Lnz559/KisrUygUKpXq4uISHh5+6dKl2nWpatM29F+bgoKCrKysz58/JycnL168ePXq1YsWLZowYQI679BotC5dujg6Orq6urq5udnb23fu3DkhIYHYxmMLIX6ZEh0d3a1bNxSaMnTo0JkzZyYlJdV5jVjqfD4f/bnNiF8oFDb+XlasWGFoaCg+C4j/gBg8ePC6deu2bdsmTuwlaA2E+GVBTU3N/fv3IyMj0YeYwWCMGTOmyd9qM5pvEXl5eVFRURoaGmLN6+joeHt7Dx06tGPHjuiJra2tl5fX5cuXU1NT8bZXgSHEL10KCgr27t0r/tSOGzfu6NGj79+/J3awjTN48GB0xwkAc+fORQ8FAsGTJ08uXrzo6OiIWncAwMyZMzds2PCrExNBIxDilxZFRUVXr14VR8Lb2NhgFWDb5ikrK3NzcyOTyahFB6rYWxsOh5OZmTlp0iQ9PT30z2tmZrZ27dpr167duXOHx+PhYrbCQYgfe7Kzszdv3mxpaYk+l6NGjUpISPjx4wfedikGDx48QPFCYlxdXb9+/drgi3Nzc5OTk0ePHm1hYSF+vZ+fH/E92xwI8WNJSUnJnj17DAwM0KfQ2tr6xIkTeBulSBw9ehSd9kkkUkBAwOrVq1VVVQHA3t6+8co8VVVVe/bsmTBhAirvDwBDhgw5efJkg7X9CRCE+DFj3rx54vUnKCho5cqVUmoa11YdgfHx8bUX/IsXL4pEovv376upqQGAh4dHcypzVVZWxsbGduzYEX2J9OzZc+PGjeIiXwS1IcSPAW/fvl25ciX6yA4fPjw5OVl6czV5W6agHDp0CLn3/P39UQXOc+fOoR/FxMSg+39nZ+eSkpLmjMbn8//777/Jkyej/5RevXqlpaU1WOH/d4YQf6vIyspatGgRhUJBR9P4+Pja3qY2qVJpcOTIERTwHxUVJRKJtm3bVlv8IpFoz5493bt3BwBfX99///23+SOnpaVt3LgR1Szq2LHjzp07sbdeYSHELznR0dEoicXf3//Bgwd4m6OonD59GgX879mzBz1Zt24dABw6dKj2y0pKSthsNrrk//TpU4umyM7ODg0NRb/eu3dvIv4XQYi/xZSUlJw4ceKPP/4AAC0trdOnTxMBZ5Lx6tUrHx8fNpttYWERFxcnfr5z504SiXT69OnaLxYKhYcPH0aXfy4uLhLsqtLT04OCggBAQ0NjxowZ4gyi37a8PyH+lvH06VN0IlVSUlq+fHmdXpEEzef169eoKpG5uXmdEJ1Dhw4BQB3xI3bv3o2O8WfOnJFs3ri4OBR8YWBgsGTJkt85KIAQf3P58eNHRESE+A4vIyMDb4sUmI8fPzo4OKALuTq192pqapYsWUImk+ts+8Xs3btXV1dXTU1t+/btks1eWFgYGBjIYrEAwNvbOz4+XrJxFB1C/M3i/v376Bqvc+fOW7duTUxMxNsiBWbVqlW6uroqKiqzZ8+u/9OSkhJLS0symXz06NFfjYB69QBAI69pkuTk5KlTp6JUi+PHj0s8juJCiL8JsrOzw8PDUcXo5cuXi3vFEZ58CcjNzfX09ASAdu3aPX/+vMHXlJaWoiv6Rjb2ubm56LuYzWajcACJCQ4OptPpOjo6x44da9E9QhuAEP8vKS0tXbRoEQrX69ixY/0Ic4IWUV5ejirtaGtrP3369Fcv4/P5fn5+ABAbG9vgC9DX7ps3b1CvLiUlpUuXLklsFZfLffv27dChQ1H+RVpaWmVlpcSjKRaE+BsmPT3d19cXXSylpKTUaQXfViNtpEdaWhqq5KWpqdnkoWnLli0A0GTB/5ycHHSBHxAQ0PiFi7gywq8oKCg4cuSIkZERhUKZNm1aM0OJFB1C/HUpLy8/efKklpYWhUKRdrjeb8Lly5e9vLxQiO6zZ88af7FQKESluPfv39/kyOnp6WjkkSNHNtIRoEnxI1JSUpAbMiAg4P37922++C8h/v+hpKRk2LBhyJkksTOZQExlZeX+/fsZDAYAREVFNUeBHA4HlT9A0X5NkpCQwGAwqFTq8uXLWx/D/+XLF1RSkclk1q+n2MYgxP9/cLncvXv3WltbA4CpqenWrVvxtkjhyc/PR2t4/XC9RqioqLCwsBg2bFgzk6BLSkqWLFmCZvH395fYWqFQiO78v379Gh0draSkRKfTPT0969dZazMQ4v8/5syZg77v/fz8vn37hrc5Cs/nz59RNH7v3r0bDNf5FVVVVSYmJuLqPc1BKBQuXLhQQ0PD2toaq9Dd9PT0UaNGAcCgQYPaai0GQvyilJQU5OzV1NR8+fIlji3o2ww/fvxwdHQEADs7uzq+0iaprKy0srKaO3dui1yqNTU1CxYsQE3EWxr5/yt4PN68efNQcEdKSkrbcwH81uL/9u3b1q1b0cd0/vz59+7dw9uitsD06dPRJby3t/etW7da+uv79+9XVlYeOXJkS+9TOBwOygiytbV99epVS+dtkJqamps3b6ICIRMmTEhJScFkWDnh9xV/eXm5u7s74dvDlq1bt6J/0vHjx0u2hwoJCVFVVZWsQ9m///6L8iw9PT0xXKh//vyJ9oa/yjhQUH5H8fN4vIcPH44bNw4A3Nzcfs/QTsy5fv06OuR36NDh3r17EifMhISEGBkZSZwomZycjCr/HDlyRLIRfsXjx4+NjIwAYNmyZdiOjBe/o/jnz5+PvsUHDRrU0hMpQYNcuXJFX18fANasWdNKd+n06dNpNBqqvSdZJNXZs2cBQEdHB3NH3d27d5lMJio90AZKg/1e4k9NTUX1tvT19UNDQwnfXuuprKzctm0bKrP1q4Dc5pOYmKihoTFx4sTW5Njz+fw///wTpQxinrH77NkzMzMzAFi8eDG2I8ueti9+8eqRmpqKPqPm5uYfP37E16q2wbdv31CAPQCsX7++9QPGxMQ0J7C3SRISEgBAV1dXGs08cnJyUFhhSEgI5oPLkjYufqT8mpqalJQUVEg/ICDgV/lkBC2ipKSkf//+ADB69OgXL15gMuaePXsAIDIyspXjVFRUGBoaIs8fVrbV5vz586gSyY4dOzAfXGa0cfGLRKLS0tKxY8eiVtCbN2/G2xyFh8/nczicx48f9+nTB3n1MUyDu3btWu1ifq3h1q1bqJ+PlpZWkwkFEpCVldWhQwcA2LBhA+aDy4Y2Lv6SkpJevXqhOI29e/fibU4b4cyZM2irP3z4cGz9JkeOHAGA1atXt3IctONLTExELf2srKxu376NhYH/w6dPn8zMzDQ0NH7VUEjOacvi37t3r7OzMwD06dNHXISDoJX8+PEDJedOnDgR23ajAoHAw8PD1NQUqxA9kUh069YtdD9namoqjRC9169foxBmRWwK0GbFj7z6ZmZmy5YtKy4uxtucNsLHjx/R92lUVBTmjYY5HI65ubm3t3crx6lzQfjs2bN27dqRyeTo6OhWjlwfgUAwYcIEAOjbt6/CVQFum+KPjY0FABUVFaKcPlaUlZUtWbKkXbt2AODj4yONFuMCgcDd3X3UqFGYj5yYmGhlZaWkpLRu3TrMBy8pKbGysiKTyQpX2bGtib+yshIleDo4ODRSK4qgpSxcuBDdkj548EBK8RECgaBHjx7Dhw+XxuDIm0ChUCIiIjDf/x89ehQA/vjjD2yHlTZtSvyvXr1CORj6+vrp6el4m9NGOHfunK+vr5KSkqGhobRvST08PHx9faUx8o8fP9D+nEqlvnz5EvPxT506xWKx5syZo0DJf21H/HFxcaidc3R09JcvX/A2p+3QqVMnADAxMWlOk9zWcOnSJXV1dVdXVykdnjkcDlobhg8fLo0qnYMGDQKAWbNmYT6ylGgj4n/79i0K6li4cGHt50SZzdZw6tSpbt26sVis4cOHS5Zm1yI2b94MAMOGDZOe5+zp06eoHHOLioU0kxcvXnTp0oVCoShKQ5e2IP4LFy5oamoqKSkdPHiQUDtWJCUloc6WzSmkiQmrVq1iMpnS3l9cvXqVTqerqKicPXsW88HLy8s9PT3NzMywKiggVRRe/E+ePEFrfjPrPRI0SWFh4YgRI1BLzJEjR8ps3rCwMB0dHWl3Pa2pqTE2NkaNQ6RRn2///v0oYRTzkTFHscX/6NEjExMTAJgzZw7etrQR8vPzUdaKkZHR6NGjHz9+LLOpV6xYwWKxZFBAMTU1tVu3bqhRF+YR31VVVUFBQVQqFfOCApijwOLPyMhAwdtr164VPyS2/a2hrKwM9SAeMWKE7KtWHjhwgEwmS3vbj8jLy0OlRxgMBuaRvzk5OUwmk8Fg5ObmYjsytiiq+B8+fGhubg4AS5cuVaDLFblFKBQeP34cVdo2NzfPzs6WsQGfP392cXHp0KGDNJJw64BWiKysrM6dOwOAmpoaJqlEtfnnn3/QtQK2w2KLQor/69evNjY2ALBq1Sq8bWkjbNy4EeXqBAYG4lKp+saNGwAgm3YJ4u3hly9fPDw8yGTy0KFDMb9icHJy0tHRyc7OltvdqOKJ/+PHj6iAxMyZM8XOIbn995V/bt++HRYWRqFQjIyM9u3bh1d1qps3b2KSz9dSHj16JM5QRMk5WH2W7t27J074lc/Pp4KJv7q6+q+//kJVIsVf1UTbTIm5desW+uibmJhgmEsnASiTPyIiQsbzCoXCpUuXon+EP//8E3X1w+TjJBAIpk+frqqqKrc9PxRM/CiImslkxsfH422LYsPj8a5cuWJoaEgmk5WVlf/++2987UlPT6dSqWFhYbKfWigUTp48GQDIZHJ4eLgIu4X6+vXrADBv3jxMRsMcRRL//fv3WSwWi8U6efIk3rYoNj9+/Bg4cCBa7nbu3Pn27Vt87amqqkL2zJw5ExcDuFxuQEAAcv69efMGq2F5PF6HDh3U1NTkc/FXGPELBIIxY8YoKysTPbNbSUVFBbrJ9/T0PHz4MN7miEQiUWlpqaqqqpOTE45fQ5WVlehE2bVr14cPH2I17JMnTwDAy8tLDq/9FEP8hYWFKFF32rRpIpFIKBQS13uSUVhY2Lt3b1TWuqKiAm9z/o/KykpDQ0M3Nzd8zSgqKtLQ0ECXnVh5QGpqapYtWwYAGG4osEIxxL9v3z4A6N69O7p/JsQvGUFBQaampgCwYMECDoeDtzn/P2VlZTo6Ol27dsW9E8bLly9HjBgBAD169MAqf/nGjRt0Or31JYkxRwHEn5GR4eLiYmlp+f37d7xtUVSuXLni6+sLAGw2e+rUqXibU5fi4mI9Pb0uXbrIQxuVt2/fooZ/+vr6jx49av2AQqHQ3d2dxWJ9+PCh9aNhiLyL/9OnT05OTm2pQZrsyc3NRYX3/P395bO6UVlZWfv27Tt37iztrJ5mEh8fjwqWDRgwAJMB0cl/+vTpmIyGFfIuftT1dcqUKYpYHVUeQO2l6HT6zp078bbll3A4HDs7u549e8pPDcwjR45oaGioqan5+/v/999/rRytpKTEw8NDU1NTNpkLzUR+xc/n848ePaqvr6+srCw/rikFgsvlnj9/XklJCQDkxKv/KyoqKkxNTeVn5UfExcVRKBTk/2v9+f/kyZMAcOvWLUxswwT5FX95eTnyTvn4+KCerQTNp7y8HGXpeHt7Y3JwlSpVVVUODg69evWShzN/bQ4ePGhtbQ0ATk5OrUx2+u+//9zc3Dw9PeXnC05+xY8aLeMS8qXoREZG2tnZoWLyRUVFeJvTNFwut1u3bo6OjtIorddKrl69itr+tD6LbMWKFRQKRX62sXIq/qKiou7du5ubm//8+RNvWxSMtWvXotC9/fv3K8qOqbq62srKqn379nJ1ASkmLi5OQ0NDXV09JiamNev2hQsXdHR0du/ejaFtrUFOxT9gwAAAeP36Nd6GKBJCoTAkJERFRcXc3Pzq1at4m9MCMjIyVFVVnZ2dcb/n/xVhYWHo/q81W9HKyko9Pb3BgwdjaFhrkEfxP336lEwmS6NzS9smODgYAAwNDaVRl16qHDlyhMlkuri4lJeX421Lw/D5fBQZSSaTJfaeot4+qqqq0ugaKgFyJ/7c3FzUWVGuLkXknJ8/f0ZFRQGAvb39+fPn8Tanxezbt09ZWdnV1VWeL3Rzc3N9fX1JJBKLxZKs6Aifz9+2bRsAXLx4EXPzJEC+xC8QCAIDA1EHWHlz/MotL1++tLW1BYB169ZJo4WeDECZ2j4+PnK77UfweLzBgwejyz/J/Haowsfp06cxt00C5Ev8paWlurq6DAaDOO03k+vXr6urq6NuJYpb0WTfvn00Gk0ahbQx58uXL506dWIwGKNHj/78+XNLf720tHTKlCkGBgYyKFXYJHIk/urq6mHDhqF+W3jbogAUFxfv3LkTAHx9fa9du4a3Oa0iOjqawWDIW+j7rzh48CCDwQCATZs2SfDrW7ZsAQB5SFSRI/HfuXMHANzd3eXW6yM/cDgcdCHi4eFRUlKCtzmtZf/+/SQSSYH6qaOexbq6uhL0Ndi9ezeJRIqLi8N9pyYv4q+pqUEJPEePHhXJa8FDOaGwsHDmzJkA4OzsLA8LSOs5deoUAChQaba0tDTU80+CnsLl5eUmJiaWlpa4h/rJi/gPHDgAAOrq6ugsRIj/VxQWFqJWsz4+Pm0mAurEiRM0Gu3Jkyd4G9IC/v33X2NjYxaLFR4e3iIZczgcU1NTZ2dn3LOY5EX8fn5+Kioq586dw9sQ+YXL5d6/f9/Ly4vNZgcEBChK9F5zOHXqFJVKxap4hsy4c+cO6go/bdq05t9TlJeXa2pqGhoa4n68lQvxf/z4EQBCQ0PxNkSuWb16NQBoaGi8f/8eb1uwJD8/v0+fPlQq9d69e3jb0jIEAgFyUaNPbzMv/wQCwd69ewEA9zq0+Iufw+E4OTk5Ojrm5+fjbYucUl1dvWrVKnQsSkhIwNscjHn79i3KO5aTuLcW8e3bNzMzM6T/O3fuNPO30tPTAWDv3r1Sta1J8Bf/5s2bAUD+W5riiLiqbEpKCt62YM+3b9/U1NQcHBw+fvyIty2S8OjRI7T5Hzx4cDNLdGdmZqqpqe3bt0/atjUO/uJ3d3dXU1NTlDteGfPu3btBgwYBwPz589uqEzQvL49Go23btg1vQyTn9evX7du3b1Haz6pVq7y9vfFN78VZ/CkpKcRp/1e8evVKX18fXenJTxI45hQVFdFoNNQqR3G5ePEii8VSV1ffvHlzcyLTd+3aBQAYNgiQAJzF36tXLxcXF+K0X5/ExEQTExNjY+Ply5e3YeWLRKJXr14pKSktXboUb0NaC4r8AYDm1PxJSEhQVlaeOHEijukMeIr/9u3bSkpKxLJfh+rq6qCgIBUVla5du8pnmyesQP0X/vjjDwBoA7e8nz9/tra2JpPJEydObDIEg8/n29jYaGtr41i8CDfxCwQCT09PAJAgQLIN8/37d3TI79KlS5uJ4WkEgUDQs2dPNpvdBoKURSIR6itFo9F8fHwaf0d8Pn/atGkmJiY4pq7jJv579+5RKBQXF5d3797hZYO8sW/fvnbt2ikrKw8ePPjr1694myMLhEKhp6cnm80uLi7G2xZs2L9/P+r5NXbs2MbPa/n5+TQaDXWgwwXcxO/h4aGurq5wQV3NhMfj8Xi8Jl9W24GP6vDo6uo+ePBAPkvZSQlvb28rKysFrUTQIKGhoWQyGQD8/PzQrr7Bm5ovX74AwMiRI2Vu4P+Bj/iTkpIMDQ1ZLNbly5dxMUCuyM7OjomJAQAGg/Eb9iAeMWKEvr5+k07fnJycz58/K8TpQCAQzJ8/n06nA8CkSZMKCwsbfFl1dfW4cePYbParV69kbCECB/FXVFQEBQUBgBw2jZM99+/fR/d5ixcvVrjae5jg7+/PYDAaKYzx5cuXuXPnGhoa6ujoDB06VK76XjTCkiVLmEwm6vn1q0SMuLg4AMCr8hoO4k9ISFBVVbWyskpPT5f97PLD9+/fJ06ciMpCLFy4EG9z8KGwsLBz584A0GDCQm5ubnh4uImJCdSCQqEcPHhQ/o8J1dXVy5Yt09HRQW2RG2ygEB8fT6PR8CpkgIP4Z8+eTSaT161bJ/up5Yf8/HxXV1fUd/zChQttNXqvSa5fv46++xrMivXx8UGC37hx48mTJ48cOSJ+cuzYMdlbKwEREREkEgkA+vXrV/9Wj8fjTZkyZdSoUbh8l8la/NevX9fU1GSz2RcuXJDx1HJCVlbWli1bnJycjIyMtvdcOPQAACAASURBVG/fjnteJ77cvn2bTqc32GUgOTkZlcqfPXu2+OH379/RKUmyElq4gNIxaTSal5dX/dJ9u3fvBgAJAjpaXw5A1uLfsGHD73zaj4+PRxVgHB0d5b+Fngy4c+cOk8k8c+ZMnefZ2dnm5uZKSkrr16+vEwM3adIkAFCsXIDg4GANDQ0SiTRs2LA6Ol+9erW6unoza5YLBAL0h8rKyoKCgurq6rKysqqqKsl2jjIVf1FRkYeHR5spPtUiqqqqbt68iZYyf3//NpmfJwEvX75UV1ePi4ur8/z58+dUKtXExKTOR4XD4VhYWChcPohQKFy0aBHq+du7d+/am/yVK1fSaLQpU6Y0c6jMzMwlS5bY29s7OTnZ2dmZmpq2b99+1qxZEkQKylT8J06cUKwNG1YUFhb6+fmhy7zp06c3JwTgN+HZs2caGhqnTp2q/ZDL5aIwxxUrVtR5vnTpUgCwsrJKTEyUraUYgDp2AMCoUaPE/v///vtPVVW1W7du4lX9VyQlJfn7+4vdnyQSqXv37m5ubuive/bsaak9MhX/2LFj1dTU6n/Nt20iIyONjY0BwM3NjehHUBs+nz979mxlZeU6DqDKykojIyMqlXr27Fnxw3/++adHjx4A4OLi0spu2TiyePFipFVxUZaKigoVFRU7O7sm67Ih74CGhkZISEhoaKj4G3Pr1q0MBkNTU7OlkfKyE/8///yjqqo6YMAAmc2IO1++fFm2bBkAsFisixcvEsmLdaisrLSxsWEymTt37qz9nM/nOzk5WVlZob/m5ubu3r1bSUmJTCZPnTpVQWt+iFm5ciWVSjU2Nkabl6qqKi8vLxKJ1GTzhePHjwOAkZFRnefV1dV0Ol1PT09OxZ+dnd2zZ08ymawQXVkw4eLFi9ra2gDQrl07RSxQJQPKysratWunoqJSZ+X/8eOHkZGRgYHBsmXLFixYgA75tra2it6bBJGVlWVnZwcAQUFB6MnLly9pNNrff//d+C9u3ryZTCZbWFhkZGTUfs7hcNatW5eWltZSS2QkftRVyt/fXzbT4UtBQcGxY8f09PSQXyorKwtvi+QUVMd2zJgxtR+mpKTMnTu3dlRP+/btg4ODFSKwt5mkpaUZGhpSKJRhw4aFhYXNmDGDQqE0WdXr3r17NjY2nTp1mjJlCiaBITIS/5o1a9hsdtsrPlmfxMRE9L3etWvXNpCjLlWqq6vt7OyGDRuG/lpRUREaGopC4hkMhoqKiqOj46lTp3DMeMec79+/X7lypWfPnkwmU01NDVUuRdtDKyurJlMb79y5065dO1VVVXd39y9fvoha1+FC6uKvrq7++fOnvb29i4uLtOfCnRUrVoivc4gFv0k4HI6RkdHEiRPRX/Pz811cXACASqVSKBQWi5WTk4OvhViRnp5+48aNyMhI1H4evUctLS0ajUYmk+l0OpVKVVFR+fTpU5ND7d+/H33GXF1dUUaQxPqXxcq/ZcsWGo327NkzGcyFF5cuXRo3bhz6fx04cKA895mXHyorKzU0NObMmSN+UlxcvH37drQeDhw4EPeGVhJTUVHx/v37Z8+eDRw4sHPnzjQaTXyKcXJyCgoKunv37uHDh9E79fX1nT59OpvN/lX+Xx2uXbuG3Em2trbfvn2T2Eipi//Dhw9mZmYmJiZt9XK7uLh406ZN6H/RyMjoypUruLdhUhSqq6vNzMzGjh1b+2FeXp6Ghoaamlrzy+DLD5mZmVevXr18+bK3t7d4S48YMmTImTNnTp06JVb42rVrSSQShUIpLCw8duwYABw9erTJ237EmTNnUMqgq6urxAFjUhf/zZs3UcWCNimJHz9+9O/fH/3venp6EnF7LaKmpqZ9+/Z1el1++PBBU1MTXY7iZVgzqaio+PjxY3Z29o8fP44fP+7h4YEK+CMcHR0nTJiwbt26ly9fpqSk1PFcoFgGAJg0aRKfz9+zZw8AWFtbN7+Oy507d/T09Egkko6OjmQ9TqUu/pCQEDabfePGDWlPJGOEQuE///zTt29fANDX1z98+DDeFikefD7fwcFh+vTptR+WlJS0a9dOSUlJbu9Hy8vLL168ePfu3alTp7JYLFVVVZS3CwB2dnb79u07ePBgfHx8gzm8Yk6fPg0AdDod1fD7+vWrgYGBpaVli4o4XblyxdzcHAD69+8vwRuRrvi/fftmb29vZ2cn1VlwISwsTLzgo44jzdywEYjhcrm2traBgYG1H96/f19bW5tEIsmD+Kurq/Pz84uKilJTU9evXx8WFhYWFtarVy/x8s5kMl1cXGxsbPz8/J4/f14/aU/UkEPuwoULNBpNVVX1wIED4te4uLjo6+s3UvaPw+G8e/euTvLvkSNHSCSSiorKrl27WvrupCt+1Ipr1qxZbSlf/dmzZ2vXrkUL/ooVK9rSRZSM4fF4tra2AQEB4id8Pt/T01NZWRkA8AoDf/HiRWJi4sOHD69fv96lSxc9PT1DQ0N0AQkANjY2u3fvPn36dExMzK5du1DxJS6X26Kvfl9fXwCoHdrE5/O7detmaWnZSMOPBw8esFiskJAQkUhUVlZ24MCBAwcOiDuFBgcHt/SdSlf8O3bsAIlyleWK2t9cT548UVFRAQBDQ0O8Sq+1GVAYb+0KlkKh0NvbG7nHWlm5vDnrDZ/PLy4urqqq+vz5c2Rk5IEDB2rHF6G7d3Nz82XLlj18+PDBgwfJycmt8a4jli9fjjwCtW/1uVyuo6Nj/WDH2qC8OAAYPXp0p06dansTvb29JWjZLkXx5+fn9+zZ097evm00nEEuWQMDAzqdHhQU9O+//+JtkcLD4/EcHBxqx33W1NTY29sDQEREhJQmzczMTExMfPLkSVJS0qBBg4yNja2trVGxbQCwt7c/evTo+vXrY2JiCgoKuFwuttVWMjIyUFp3nXg+Dofj7u5OJpMbqVNw48YNMzMzdXV1ZWVlIyMjdInWr1+/CxcuSGakFMX/4MEDAFi2bJn0ppAZXC4X+fbQfQze5rQReDyeo6PjuHHjxE+EQuHdu3fPnTuXm5uLyRTV1dWVlZWXL1/esmXLrl27Nm7ciG7IETo6Onp6elpaWsuXL3/8+PGDBw+kmnyVkZHRvXt3Fos1bdq0+qfFlJQUGo0WFRXVyAgFBQUfP37MysoqKioqKCjIy8trzcoqLfELBIK//vrLyMjo5MmTUppCZrx+/bpPnz4A4OXlJfbQELQeoVDYvXv32uJvJTU1NWlpaampqU+fPn369GlISIitra2VlVXt7fGaNWvWrl27ZMmSTZs25ebmfv/+vc5OXnr+qZ49e6IbgQa3jTk5OWpqalu2bJHS7PWRlvg5HI6GhkanTp2kNL7M2LRpEyq85eXlpbgBZ/KJQCDo0aOHn59f64dKTk5++/ZtYGBg7ZOwpqamurq6mpra3Llzk5KSkpKSmtM/V0rcuHGDRqOxWKzDhw8LBIL6luTm5qqpqe3YsUNmJklL/NeuXVNSUrKyslJEwYi/+0NDQ8UuFsKrjzkCgaB79+4jRoxo6S9WV1cXFha+efNmz549f/75p5eXF41GQ2dpBoPRo0eP7t27+/j4vH//Pjc3Vx4qfxQVFSFHZmRkpEgkalD8379/p9PpmzdvlplV0hL/pk2bAGDFihUKevvN5/NRxShdXd2jR49K4EolaBIej9ejR4/hw4c38/WJiYkHDhyIi4ubP38+Ko6EoFKpenp6rq6uly5dks8rmMmTJwNAx44dG0nd+fnzp5KS0oYNG2RmlbTEv3HjRgqFoqCFOhMTE9FVypgxYySokUDQfPr06TNkyJAGf1RYWJiVlZWWlhYaGjpx4kRxxX6x4D08PHr06DFs2LDXr19//fpVbhP+37x5o6+vb2ho2Hj0d0lJiZmZWZ2iRlJFWuKPjIykUCh5eXlSGl9KPHnyZMaMGSiiY+DAgTgeEX8H8vLyunTpUjvIRyQSffjw4fbt26dOnXJwcNDV1RXnwzEYDA0NjQULFly4cOHq1asKVA1x0qRJFAqlSU9eUVGRpqamLMvbSkX85eXl3t7eFAoF1RtQFKqqqrp37w4APXv2JDqISonavvRNmzaRSKSRI0dmZWV9//49Li4uJCSk9lUcAOjp6XXt2vXgwYPv379XxNiKkydPampqGhoaNrmQlJaWamlpKby3/82bNxoaGkOGDFGg8J4dO3agmO2ZM2fKfx84haP+/VlmZuaECRMoFAqDwdDS0qqteVVV1f379586derYsWO5ubmKGxt+5coVNputpqa2f//+Jl9cXFzMZrNl2YxEKuJPTU2l0+m16y7LMzweTxzU+ccff6CHivuBk2e+fPmya9euTZs2rVu3Dnns2Gw2AFAoFGtraxsbm5iYmBcvXrx79w5vSzGAz+ejZg1Dhw5tzutLS0s1NDQWLVpUp0OR9JCK+J89e0Ymkw8dOiSNwbHl1atX3bp1AwBjY+MDBw40noZJ0CJ4PN6TJ08ePXq0YcOGGTNmDB06FOWfiqHT6SQSydnZOT4+nsPhtLG2hTdu3EBpIEuXLm3O68vLy83NzSkUyj///CNt2xDYi7+qqqpnz540Gk3O+6gLhcLY2Fg1NTUAGD58uCJ2gJFbDh48uHfv3jFjxtSWura2to6Ojp+fX0pKSmpqampq6ty5c6lU6vXr1/G2VyqgiKNevXo1s4YVn88fNWoUANTvXCglsBd/ZWWlmZlZp06d5Ll0T0FBwdixY9GHMjw8HG9zFBsOh/P69evY2NiRI0cOGDDA09NTLHhra+thw4b1799/woQJ379///79e22/1/bt2wFAsio0ck5ycjKdTnd2dm5+SiuHw7GysvLx8WlmJb/WI5WV38rKSrLSIrIhKirK0tISHcZSU1MVMQZRHuBwOLGxsbGxsQEBAaieLEJFRaVDhw4JCQmJiYkNfvTF/pQ1a9YAQJ3+3DweT9EdLk+ePDExMSGRSC2q71RSUsJisVatWiU9w+qAvfiLi4u7d+/u7Owsnz5zcbPEgQMHNlkmnaAOOTk5b9682bVr15w5c1C8KsLIyMjb23vEiBF379798uVLM/9h161bB/9b06JtgGrYNL/xLqKqqsrAwADV6pAN2Iv/0qVLbDbbxMRE3kJi79y5M3LkSACws7O7c+cO4dtrPpmZmTdu3Dh8+LClpSWZTEaCp1Ao7du3v3Dhwt27dyUL5dyyZUvbEz+Hw+nUqZOBgcHdu3db9IuVlZVaWlp1GhNLFezFf/z4cTKZHBISIj9n/urq6vPnz6OCDWw2+8WLF3hbpBh8+PBh/vz5w4cP19LSEi/yNjY2bm5uf/7559u3b1t5Oo2OjiaTyW0pnqqqqmrGjBkAUKceeXMoLy/X0dFRbPFHR0dTKBT56bVSWlr6119/oQ8u+sjibZFck5SUdO3atZCQEGdnZxaLhf7dmExm//79FyxYEBYWhlUIvUAgWLp0KZVKlZlzWwagRrpubm4SxLZWVFRoaWnNnz9fGoY1CPbiR/dnqampmI/cTIRCodhj9Pjx444dOwKAq6vr6dOn8TJJznn06FFERMScOXPENyDIUe/o6Ni1a9cLFy68f/8e88iTiooKc3NzKpWqKMFgTVJWVobaNEqWk8/lci0sLBT7zH/27FkKhYJ7MpxAINizZw86oPbp00dB8wsxB3UQfPnyZVJS0u7du93d3V1dXcW9ZahU6rBhwyZNmnTkyJHq6mqBQFA7Ixvbc1x5ebm1tTUAxMbGYjgsjpw6dYpEIvXo0UOy0qPV1dXt27eXZdk7jMVfXFzct29fZWVlfMX/7t07VCNVS0tr3bp1cpvsKUvevn07ceJES0vLFStWoPbhCAsLCycnp44dOx4/fvzjx48ys6esrKxjx45KSkqovoWiU11djXo3SbzBrKysNDEx2b59O7aGNQLG4s/KyjI3N1dVVU1PT8d25OaTkZGBfHuampqPHj3Cywx5oLy8/MyZM2PHjh02bFjt6hcA0L59+7Fjx+7cubO6ulooFMo+2KGqqsrCwqJjx47v37+X8dSYw+PxfHx8yGTy4MGDJR6Ew+FYWFjMmjULQ8MaB3vxq6mpTZ48GZfImdLS0itXrpiZmQHAnDlzfk+v/ps3b3bv3r19+/bw8PABAwaI1a6np2dnZ+ft7Z2UlHT//v0Ge8vIkqqqKj09vXnz5uFrBiZ8+PBBQ0PDxMTk/v37Eg9SVlamp6e3cOFCDA1rHIzFX1JSoqamhstGTigUos+6iopK29hJNgcej1dcXPzz58/nz5+PGjVq4MCB+vr6tVd4X1/f8ePHT548OSsrS64yZyoqKoyMjGTp35IeKFopICCgNa2oORyOiYnJtGnTMDSscTAWf15eHpvNjo6OxnbYJjl06BAq8+Th4YG7r1E2JCYm7tixY8iQIba2tgYGBgwGAwB0dHSsra0nTJiwd+/emJgYmeWHSUBVVZW2tvbMmTPxNqS1pKamWlpaqqurt7IzlUAg6NSp05gxY7AyrEkwFn9RUZGhoaEEPQNbw5kzZ9AqFxQU1FYD9blcLofDKSoqCg4O7t27d79+/Wov7+7u7j4+PmFhYYWFhfIZVV0fDodjbm4uy4VOSixYsAAAWl97r7Ky0sbGZvbs2ZhY1RwwFv/379/V1NSk6rGsnfWRnp7+559/AoC+vn6bzAx99epVTEyMv7+/o6OjlZUV6iDAYrFMTEyCgoKOHj26du3axnu8yC08Hs/ExGTBggV4G9IqsrOzyWSynZ1d64tWcbncrl27zpkzBxPDmgPG4udwOHQ6XXqN1kS1xH/z5k3k1TczM3v69Kn0ZpQxlZWVERERgYGBgYGBtV305ubmXl5es2fP/vLli/SaCAiFQtlUWy8sLDQ2Ng4KCpLBXFKipKQEHTYxcVsKBAJPT89Ro0a1fqhmgqX4hULh0aNHVVRUTp06heGw9eFyucuWLSORSACwfv16qfZXkwFlZWUvXrxITU198uRJQECAqampWPDGxsbbt2/fsmVLTEyMbKIVZCZ+1Obcx8dHBnNJA4FAsGrVKgAYMGAAVkctW1tbWX4bYiz+/v37s1gsqd7cZmdnoxOvp6dnVFSUOOxM4ZLAq6qqkpOT58+fj0oGi+nRo0e/fv3Onz//8OFDFJAnRigUyqzAm7SZNWuWhobGkydP8DZEQgQCga6uLovFevnyJSYDXr16VVVVdcaMGZiM1hywFL9AIPDx8aFSqQcPHsRw2NqcOHEC7YRdXV3rpJQpivgzMjLOnTvn5uZmbm4uLkpvaWkZERERGhraeCPQtiT+2bNn6+npKejbKSkpmT59OgD8quOIBCDH4V9//YXVgE2C8Znfz89PR0dHSkUyNm7cCAB0On3s2LGShU/jRVJS0qRJk8aNG+fr6ytOlXNycvLw8AgNDb1x40adFf53YM6cOSYmJgraAfHSpUsoMhrD3iFhYWEgUS6wxGAs/oEDB5qZmUnjf3Tnzp0AoK6ufvv2bcwHlxKZmZmrV6/u0qULKhOKYDKZERERhw4dUpStipSYMWOGlpaWXMUdNR8vLy8AOH/+PIZjzp8/HwAw6VncTDAWv7+/v56eHrY1fN69e4dSJszMzOS8IrBIJPrw4cOiRYsmT54cGBiIAm90dXU7d+7cv3//uLi4a9euPXv2DG8b5YLQ0FATExNFjMuIj4+XRlAJch8q8LZ//Pjx4q9zTFa2J0+eGBoaotC95OTk1g+IORUVFe/fv3/27Nno0aN79uyJruIBgEwmz5s3b+3atb/hlr45rFixwtraWlFCksTU1NR06dIFAK5cuYLtyA2WM5UqGIt/7NixqqqqmJTHKysrmz17Nmq6LsvIhxaRkJDg6uqK2s4gwauqqs6aNevGjRu/SZSxZLx+/drExERdXb20tBRvW1pGWloanU63tLTE/Gy7bds2KpUqy+sPjMU/ZswYXV3dVgY5i0Si6Oho8QVYaGionCTkV1RUZGZmHjx40MvLq0ePHqjVjziyeMWKFa9evfr8+bNs7skVmri4OABwcXFRLIcfl8sdOnQoAOzbtw/zwZcvX06lUm/cuCGS1dUVZuLncDg8Hm/s2LGampqtWfk5HA7y6gNA9+7d//77b6wsbA3Z2dk3b9708fERX86RyWQymRwVFXXt2jX590TIGw8ePKBSqbt378bbkJYxc+ZMFIghjb4aKOopISEB85F/BWbir6ysrKqqGj9+PACcOHFCskHu3LmDFnwTE5NTp07he5/H4/F+/vy5fv363r17ozhibW3tqVOnBgUFnT179v37922jnyQu3Lt3DwBWrlyJtyEtAMUjA4CUmlAuWbIEAGTZNg4D8ZeUlIhDNUaMGEEikSRINampqblw4QK6EtPU1MSl/qe4k9T3798fPHjg7++PfI3oTv748eMZGRn1f4vP5//ml3YSkJiYSKPRZFmvrvVERESg8x22xzqhUFhdXV1TU+Pr60smkxVA/A8ePNi1a1dsbOz06dNtbGz69u2LUmsmTZpEJpM/fPjQotGKi4tRi0IAWLp0aYMakwa1Rfvz58+vX79u27atd+/eJiYmyBh9ff2VK1c+f/5cnD5A6BwT9u7dCwAKlMz/7ds3NpttaGiIuR8Xib+srMzIyAgAZFnLuFni5/P5yCubkJAwevToyZMno97DtRk9erRIJBo3bhyFQmmRejMzM3v16oUidvfv3y/Z25CYqqqqlJSU9evXW1paistaOjg4REVFHT58+NOnT7/6RcKr1xoGDhzIZDKlnQCGISiYd/ny5VIav6ysjM1mDxgwoPXO8ubTmPhLSko+fPiQkZEREBDQpUsXb2/v2lVflZSUUF4dWiFTUlJEIpGfn5+SklJubm4zp4+Li9PR0QGALl26yKy9F4/Hq6ysfPr06Z49e1BKJsLCwmLq1KmpqanNcecQWwCJEQqFvXr16tGjB96GNJcnT57o6uqqqqpKb09eUlJiYGCwYcMGKY3fII2Jf82aNerq6jo6Omw2m8lkotBUGo1Gp9O1tbWR3zsgIODevXviFRL1JD927FiTE1dWVu7Zs4dKpTKZzEWLFmVnZ2Pzhn7Nhw8fXrx4ER8fjwpjoJL+TCZzzZo1GzZsuHTpksLdOSsuw4YN8/T0xNuK5hIcHAwAM2bMkF48YkVFhYODg4ydIP8j/pqaGrEALl26RKfTUSIN9f9BIpFIJBKFQtHR0VFRUSGRSJcuXao9wuTJkwFg+vTpjc/6+vXrTp06AYCGhob0vk15PB6q53vkyJGdO3fq6uqiFZ5Coairq7u5uT18+JDw2OOCr69v3759FWL3FB0draSkpKurK9VEdbTyy7ic6S9X/sePH48YMcLBwYFGoxkbG9PpdAqFQqPRaDSauE9r/UV+3LhxTTpy7t69q62tjXIYMFd+QUHBu3fvpk2b1qNHD0dHR0tLy9qOiZkzZ27cuPHNmze5ubmKFV7SxhgyZIirq6v8i7+ysnLAgAFkMrk1NbmbRCAQoHu+qVOnSm+W+jQgfj6fn5WVVVJSEhERYWpqSiKRUIIKgkajoaO+t7d3SEhIHcf+kCFDAGDRokW/mg818AaAsLAwDN/Gly9fzp07d/r0aScnJ7EnAgBYLFZgYOCtW7f++eefx48fYzgjgcTEx8dramp6e3vLv/gPHz4MAH/88Ye0J/Lx8WGz2efOnZP2RLVpQPxz5861t7e3sbGpvcJ36NBh+PDhffr0QcFt7dq1a7Dd7cuXL5WVlZcuXVr/R2VlZYMGDUJDRUdHt76KQ2FhYWZm5tixYwcNGiTe0ltaWs6ePXvhwoUhISFpaWmfPn2Sn07hBIjIyEgKheLm5ibnNyZ8Pt/V1RUA4uPjpTpRQUGBm5ubs7OzVGepz/+I/+HDhz179qy9cgLAtGnTVq5ciYqTigNvf9VFXCAQmJqaBgcH13l+5swZBwcHABgwYECLvCb1F4fy8vLbt28nJCRYW1ujtB/ExIkTL168+O3bt+YPToALO3bsAIBu3brJ+ffyo0ePAMDd3b2oqEiq31Pbt28HAGtraxkfRf9H/Lt27RJrSVNT083N7ebNm7VfsH79enQN/ivnPGozXKcY+4MHD9BlgaOjI+qWKxAIWrrlKygoSEtL27Vrl7e3N7JQWVm5a9eunTp1Wr9+fXp6uiJmhv+ebNmyhUqlSvUU3Xqqq6v79OnDYDBQ8Ripij8yMpJMJq9Zs0bGW6H/Ef+rV6+sra179+4dExPz6dOn+rl0aOVvpKkIj8czMzOr7e2/dOmSsrIyAIwYMUKc8NN85T98+PD27duhoaH6+vroclFfX3/v3r2nT59+9eqVSCRS0CJwvzO7d+/W1NSUUq03rFi9ejWFQunZs6e0J+Lz+cHBwSQS6eTJk9Keqw51z/w5OTkcDkdsVp2frl69GgAGDRr0q+G4XG6HDh3E6fc3btyg0+l0Ot3NzS0vL6+ZNgmFwuzs7JiYmICAgNoROLa2trNnz27+OATyyfr165WUlOS5CmNOTo6WlhaDwZBB3mFJSQkKJ4+NjZX2XHVoWWz/ihUrAEBdXb1+kdmqqqqRI0d26dKFzWbPnDnz6dOnPj4+dDqdwWDcunWryfW5tLT00aNHL168CA4Onjhxorg8ho6OztatW69fv15RUSEOAZR/LzHBr8jNze3SpYuKiooMwrokZvHixQAwZcoUTMrSNE5eXp6mpuagQYNkX7SiAfHXkVbtv27YsIFCoQDAkiVL6vxWaWmpkZERiURSVVVlMBioRq2Ghkb9Ioffv38vKyurqKi4c+fO9u3bY2JioqKihg8fXtvLaGpq6uPjk5aW9vnzZ2zeKIF8kJubq62t7eTkhLw/csj58+e1tLR0dXVlE2+OivbOnz9fBnPVoWUrP/LTAkB4eHidH3E4HCcnJxqNpqysTCaTUVBQTExM7ddkZ2enp6fb2tra2dlZW1uj75Ha1wpHjx6NiYk5e/asnB8ICSRG/jP50Yd81apVsplulF3xwwAAIABJREFU0qRJBgYGOTk5spmuNhKKf82aNXV+VFlZaWxsTCKRUPA/k8k0MDDYtm3buXPnYmNjDx06FBUVZWJigpx/AEChUExNTdeuXXv//v2HDx+mpKSIfQ0EbZgHDx4ARs3tpMGnT5/c3NzU1NRkU0uvqKiof//+Tk5OuJxkWyb+7du3o8if+iu/OIYHCRut/FAPMzOzkSNHDh8+PDExUQYHKgJ548WLF2pqaqGhoXgb0jDz5s2DRkNUsSUyMhIA2rdvj8vK11zxo28m8cq/fv36+q/Jzc21srLS19enUqlkMllXV1dLS4vNZl+/fv3mzZsXL1588OCB3J70CGRDVlYWk8msHwYmD9TU1Hh7e2tra0ujRF+DHDx4kEwmjxgxApcb65at/Pv27QMAEom0du3a+j8VCATfvn17/vy5uro6AGhra+vr6+vp6b158wYjawkUntzcXCqViot/q3H4fP66desA4MyZMzKbceHChQBQxzUmM1q87Ufib6SvSGFhoaqqalhY2Lx581D4LZPJXL16dZ0UIOK67jdEKBQeP36cRCINHDhQ3j4A//33H51OZ7FYMruDzMvLQ0UrN2/eLJsZ69Ay8R88eBAAaDRaRETEr15TWlqqrq4+aNCgnJycz58/Dxw4EJ0U2rdvj1UzYwIFhcPh2NjY0On0yMhIeRP/pk2bQLZJtXl5eWw2u3PnzngVlWiZ+Pfs2aOkpEShUPz8/H51es/Ly0OX/A8ePBCJREVFRRs2bOjYsSMAsFisBQsWSFzYm0DRqaiocHJycnR0xNuQurx48aJdu3ZqamqydEJnZ2czmcwRI0bIbMY6tHjbL07v37VrV4OvKS0t7dChA5VKffjwofghh8NZunSptbU12gUcOnRIXCcbIW/rAIE0qKys7NKli729vVxVUqmqqnJwcFBSUpJxCb3w8HAAmDVrliwnrU3LxB8bG0smk0kkkpKSUiOlV5ctWwYA9Wvvl5eXR0VFOTs7A4CTk9PGjRu/fv0qidUECoubm5uWlpbMirU2h+PHjwPA0KFDZTxvaGgonU7HsRVdy8S/c+dOlO3v4+PTSDL2hg0bxNv++mRlZfn5+aEtQPfu3eU5xpsAW4qKijw9Pa2srORn5S8qKurduzcAnD59ukW/2Mq96pcvX7y8vMhkMo7RLi0T/4EDB5BoG6zVI2bBggUAMGnSpPo/Ev+THT9+HNXw1NHRCQ8Pl/OiLgSYcObMGVQqQn4KJe/fvx8ABg8eLLMZkQTOnz8PAP7+/jweT2ZT16Fl4q+pqdm8ebOfn1/jX1enT59WUVHp169f45KuqalZvHgxcgT06dMnKSlJ/COhUEh4Adoe6BMfGBhYx+ODF6WlpY6OjlQqVfb+9qioKACoUyxHxmDcoluMm5ubi4tLc+KWsrOzUS8QTU3Nbdu2ZWVloeeE+Nse0dHRANBg9UdciI2NBQBXV9eqqipZzltcXKypqWloaPjx40dZzluHFohfKBQ2s+iaUCj09PT09vZu5AW1/5qRkYEiCFBc4KRJk+S50gOBxBw4cEBFRUV+yigj35PsS+i8efOGRqP17dtXxvPWQSorv1AoHD16NIvFalG73kePHk2aNAl9BfTu3fv58+fSsI0AR44fP06hUFBVPHzh8XiTJ08mkUiNhKtJj4iICBKJtGbNGny3t9IS/4ABAyQoUcDn85OTk8ePH29gYIDcMHfu3BGPKQVLCWQKKoOPi/gFAkFVVZX4U5ScnAwAurq6BQUFMraksLDQ19eXRqPhnsMurTM/Ok2tXr1asl9/+vQpygvQ0NA4deoUUdujDVBTU4PKwF2/fh0vG5D4eTyeq6srhUKRcVQPIiEhAQD69esnY0dDfaQlfh6P165dO2dn59evX0s2wqdPn44ePYoighwdHXfs2CE/90MEEvDz508Gg2FsbPzvv//ia8mtW7fodLqOjo7s116hUIgK5F+5ckXGU9dHWuIvKyszNTUFgMOHD7dmnKqqKtT/DwA6dOggP74igpby9etXKpW6ZcsWvA35v/qcO3bskP1ZsqKiQldXV068ntISv0gkCg0NBYC4uLhWjlNRUXH8+HFUz1dPT2/v3r0/f/4kgoIUjpKSEjU1NVwcbLV59+4dlUp1cnLC5Sz548cPGo3Wp08f2U9dHymKf/PmzQDg7OyMSaX9t2/fhoeHo+BiCwuLcePG4e4vIWgRpaWl2tra27Ztw9EGoVDo5eUFAC26h8IKHo83fvx4AKid84YjUhT//v37GQwGg8F48eIFVmOmp6f7+/ujU4CPj8+tW7fkvN8bgZiSkhJU1gVHG+7fv0+n02k0WmZmpuxnP3ToEAAEBQXJfuoGkaL4+Xz+2LFjqVSqxD6/BikvL4+Jidm1axcqFubl5ZWUlETUApV/SkpKaDQatq3ZWwrKOqlfflYG/Pjxw8jISIIMIukhRfGLRKJp06ZBQx0+MOHjx48zZ85Eu4CZM2eWl5fLScQ4QYOUlZWxWKzGU8Kkh1AoRJnmEydOxMWAjIwMHR0dc3NzvOr21Ee64o+Pj1dWVha37pMGZ8+e7dy5MwDY2Nh4eXkRCcJyS1lZWbt27fBy+D1+/FhZWVlVVRXDQ2iLmDJlCi5VAxpBuuLn8/mGhoajR4+Waqz+t2/fBg8ejLYAzs7Ohw8fJiIC5JCSkhJdXd3FixfjMvv69etZLJaLiwsu28OCggILCwsajSY/y75I2uLncrmBgYFsNvvIkSNSnUggEERFRaFKgQAwevToly9fEj0C5IqKigpDQ0NcHH7JycmoVdT9+/dlP7vo/zW8WLFiBS6z/wrpil8kEr158wZ1aJHBzRyPx0tKSho2bBhqFmRra/v+/XtpT0rQTIqKigwNDWVcsb+6uprL5aJaPYsWLcLlevjr16+2traYxLxgi9TFn5OTw2KxDAwMZHYaFwqFJ06c8Pb2BgBDQ0NfX99WRhkSYEJxcbGxsbHsPe27d++mUqlsNhsdBmUf1RcSEgIAAwYMyM/Pl/HUjSN18QsEguHDh2tpad29e1fac9UhPT195MiR6CAwb968T58+ydgAgtrMmDEDADZu3CjLSXNzczU1NZlM5t9//41XYuiIESMAYN++fbjM3ghSF79IJNq5c6eysnJAQAAuvpaEhAR9fX0AaNeuXUhISHJyMnEjiAvr1q1jMpkLFiyQ2Yw/fvwIDAwEgIULF8ps0toIhcKLFy+qqKiw2WzZ5w43iSzEX1NT07dvX2Vl5aNHj8pguvp8/fr18OHDFhYWaBcwbNiwp0+fin8qFAqJTAHZYGFhIct7/vXr1wOAm5ubzBpv1oHH49nZ2cl+v9NMZCR+dBWPV0NCRGlp6fTp083NzVHJ4EmTJuGeW/pbUVZWZmxs3GCLVymBwvhxbAe+fft2ZWVlS0tL+fykyUL8AoFg9erVNBpt3rx5XC5XBjPWofZh79mzZ6tXr+7duzeJRNLT01u8eDFxCpAN69evp1Kpc+fOlc10qGbGuHHjuFwuXju7UaNG4ZVE1BxkIX5E7969mUxmWlqazGZsBC6Xu2zZMgcHBxQXdOjQIRwbp/wm/PHHHwAwefJkGczF4/HMzMyYTObFixdlMF2DFBUV6evr0+l0OTztI2Qn/pUrV5JIpODgYJnN2CQFBQVDhw5FhcPNzc23bNkib5cxbYm5c+cCwIwZM371AoFAgMkurKamZsGCBWQyedy4ca0fTWJCQkIYDMahQ4dwtKFxZCf+U6dOmZmZ+fr6ymzGZlJQULB48WInJycAcHBwOHbsGFEyUBrMmTOHRCI1UsGGz+djEoRTWlqqpaUFAPHx8a0fTTJQKXr5yd5tENmJXyQSubm5MZlMeajcXJ/S0tIePXqg6wA7O7vLly/jXl+xjTF16lQ2m/3y5UtpT3T16lV9fX13d/dv375Je64GKSkpsbOz09HRqd2ESg6Rqfhv3ryprKzs6emJSW0fzMnKykpISPDx8UFfAS4uLitWrGgbBwEOh4P7debUqVN1dXWl7fRJTExUVlZmsViyb8Uh5u7du2Qy2cPDozkdq3BEpuKvrq62sbEhkUgJCQmynLelXL9+XdxH2NLSMjExEW+LWktxcTHuH8S//vqLTCZjW9mlPqh4HI6Zs7dv3zY1NWWz2eKWE3KLTMUvEonCwsJIJNKUKVPkvPxWeXn5tWvXUB9hOp0+evToV69eKWjjEKwcaa0kNDRUQ0MjIyNDelPk5eV1797dwMAAx3ZPY8aMkaBdDS7IWvxbt25VUVFxdnaW8bySkZubu2LFCiaTCQBsNjswMPDJkyd4GyUJ8vC1FRoaqqysLNVjFFr2ly1bJr0pGmfRokVMJtPKyurNmzd42dB8ZC3+T58+OTo6mpqa3rhxQ8ZTS8yrV68mTpyITgFkMjkiIkKxioXgGOVSm+DgYFVVValWzvTy8qLT6SdOnJDeFI3w7ds3NTU1ANi7dy8uBrQUWYtfJBINGTIEryKKreHWrVvDhg0T+wJv3bolD3vpJhEIBKWlpfIg/jVr1mBezVVMTU0N2lR6enpKY/wmyc/PR/F8vXv3VpQqMjiI/8WLF+rq6r6+vorSh1soFKJtM4fDOX78OCrMAAC+vr5nz57FK2mkmXC5XHnY84tEolWrVlEolA8fPkhj8GvXrlGpVBKJhFcjwMjISLQxvHDhAi4GSAAO4q+urra2tpbDwia/QigU1nZPVlRUJCYmBgQEoK8AGxub2jmC8oZAIJAT8c+bNw8A0tPTpTF4XFwcWnWlMXiTZGVloSAxHN0NEoCD+AUCAfocLFu2rLKyUvYGtAaxkKqrq48fP66rqwsABgYGERER79+/x/06rUGqq6vlQf9Lly5lsVhSuucfOHAgjUYTl8SX8ftFF8P9+/dXiJOgGBzELxKJ0tLS7OzsGAyGnOT5SExOTs6qVatUVVUBQEVFZdq0aSkpKXgbJaeEh4ezWKzU1FTMR46JiaFSqV27dsV85Oawf/9+ANDW1pbGW5Mq+IhfJBLNmjULANavX4+XARiSmZm5YMECDQ0NAKBQKEuWLHn37p3CbWqkzapVq6RxJk9KSqLT6QCAS9j4pk2bGAyGnp6eAt1eicFN/Dk5ORYWFtra2vfu3cPLBmzJyspavnw5i8UCAGVl5fHjxyv6vgZb1qxZAwCYh0s+fPiQwWD069cP22Ebh8vl8vn8bdu20Wg0KpWKGm/Kw9mqReAmfpFI5O7ujm+hFWnw4cOHJUuWoF0Ag8HYvHnzx48f8TZKLti4cSMAYFvHNScnp2/fvgAQGRmJ4bDN4dy5c6hCvL29vTzcpEoAnuJ/+vSpkZERnU5/+/Ytjma0nvrxs0lJSX/88QdKLNXR0Vm6dKkMstnkHCR+bDd6aEw7OzsZXxtzuVwUrmJvby/tbAXpgaf4RSJReHg4qquNrxmtpI74xdu/z58/BwcHo6gvCoUye/bsV69e4WQj/qxduxYAMDwbV1RUBAYG0un0K1euYDXmr6i9pefz+agDrYWFhZTCFmQDzuL//v27sbGxqamp4n59NsmTJ0+mTJmCLgXV1dVnz559//59xboTwgR05sewrlZwcDAATJ06FasBG0Esfj6fP3/+fFT3RdG7wuIsfpFIdP78eXRTgmMmlgzIyclZvny5qakpCg0aM2ZMG/6+a5ANGzZQKJSrV69iMlpxcXHnzp0pFMqzZ88wGbA5VFVVzZkzB635Uk1PlA34i5/P55uYmOBe2Fs2/PjxY/r06aiJiKqq6sKFC1NTU+U8uxkr9u7dq6SkhNVV3/bt2wFABi1Aampq0LLP5/NRcFr79u2/fPki7XllAP7iF4lE9+7dY7PZrq6uihLt30qys7PDw8MNDQ3RLmDy5Mm/Q2hQZGQkmUxu/flcKBTeunVLX1+fyWTKIL1S7MlfsGABiURq//+1d+7hUOdfHD+Mu5hGSlmXjWQJGxW6b1o90SKKTbJWbZt6drtHbZsu0iaXKCoVSekRXVRWriVt417apDyiRSJrkdyLz++Pz/P7Pp5SuczMdy6f138z872cmfm+v5/P95zzOUdPTwjGfAxfiB8htHLlSgA4fPgw6rOQRrh59erVunXrcLEAFovl4uLS0tJCt1FcJDg4GADi4+OHeZyOjg7cecXNzY1nkyZvb298p+bzsnyDgl/E/++//0pJSamrq/N/8SPOkpaW5uvrKy4uDgCLFi0KDw+npa8JDwgMDORIRd0XL14oKCjIy8vzbD0VrhEiKyt77ty57u5u3pyUB/CL+Ht7e7dt2wYAKioqIhgPi4yMxHFjAJg1a1ZiYqIwXWSYgIAAALhw4cIwj3P48GEAmDt3Lm+GfX9/f+p/4cHpeAm/iB8h1N3djTuZOzo6isK0/0NiYmJmz56NfYHm5uZBQUHCVD6cI+JPTk7W0NAQFxfnTZ0sbDOLxTp9+rRglW8aCHwkfoyRkZG4uHhAQADdhtDGkSNH4P84ODg0NjbSbRFnwH0shlNjq62tDQeGPD09uZ1R29PTg7OSVFVVceq+8MF34i8pKWEymUwms7q6mm5baCMzM3PdunX4QtfT06Ox8wwHwQ6/4Yz8CQkJYmJiEhISXKoIQlFWVoYr86irq+OlGe/evRO+2SjfiR/9v6nb0qVLhdv7/VlevnyJ1z4BgK2traCHA/38/ADgzJkzQ9u9paXF2toaAJKSkjhq1/tkZWWpqqri2b7QLDntF34Uf0NDw9y5cwFgzpw5NTU1dJtDJ01NTcHBwXJycgCgqKi4a9eu+/fv023UEAkLCwOAqKiooe2ekJAAALq6ulz1g+Tn5+MULElJyZs3b3LvRPwAP4ofIdTa2jpx4kQA8PPzo9sW+ikuLt68efOECRPwLODAgQMClw3V3t6Os3EHVbiRmmk/fvzY0NAQALi6huf69eu4Loiuri5/dpTkLHwqfoRQYGAgg8HQ09PLysqi2xa+oKGhISAgQFJSEgDGjx/v6OjIkZ62vKGlpUVdXV1OTi4tLW3ge2Hx9/b24jU8mpqa3Bv2Hzx4gJu1+/v7C42T9dPwr/gRQjj7RUNDQxRuwwMkJSXFxsYGTwFMTU2tra0HJSe6aG1t1dTUlJSUHEKbxpycnJEjR8rKynJp2O/q6oqLixs9ejQAiFSYia/FjxAKCgoSExNjMpkDfNbt7e0V0LIqgyIlJYXqIDJmzJi4uDg+H6za29v19fVhkN1s3r5929PTgx3Av//+O5dsc3R0xL+kr68vl07Bn/C7+BFCBw4cAIDt27dXVVV9Vtjv1dgXYt68eRMSEoIVBQDGxsY+Pj6ZmZn8mYvS0dFhbGwMAOHh4YPasbS0lMViMZnMFy9ecNyqzs5OZ2dn/ANu2rSJ48fncwRA/LW1tfr6+tLS0kZGRljYwhdxHTJtbW3Xr1+3s7Oj8oLMzc35sLxMV1fX9OnTZWVlY2NjB75Xd3c3bpTs6OjIcZMiIyPnzJmDp07+/v4cPz7/IwDiRwg9f/4cP5KFhYXRbQs/8vr16ytXrsyYMQPnBRkYGAQFBSUkJPDPLKCzs1NLS2vp0qWDmpdFR0fjpHqOZ3zs379fQkICAKysrJ49e8bZgwsKgiF+hFB2draenh4ArF69WgRrYL3Hx55uGhoaPDw8qFmAiYnJpUuXmpqaeG/he3R0dBgZGQ1qap2QkKCoqKikpJSamspZY/A9BQB0dHRevXrF2YMLEAIjfoRQZWWlmpoaGf/RJ10bnZ2dubm5uJkEvsRtbGwSExPpjQviZ/41a9YMfJfvv/+e47V6SkpK3N3dGQwGk8lcu3atoJeNHiaCJH6EUGFhoZGRkYSEhJeXF9228Dt3794NCwtbtGgR5QtIS0ujKyjQ3d09adKkJUuWDHD7xsZG/ECOy361trYOP4hz//79r776CgDk5OQuXrw4zKMJAQImfoRQXV3duHHjAGDdunV02yIANDc329raYl8AAMybN4+WrNXu7m59fX0bG5uBbNzb22tpaQkAU6dO5VQu4/bt27HbaNq0aXS18eY3BE/8CKG8vDzcDyMiIoJuWwSD+vr63377TVdXFwAYDEZYWFhhYSEvfQGdnZ3Tp0/X1NQcSKWWoqIinNXDkeSu+Ph4KpLv4uLy5s2b4R9TOBBI8SOECgoKTExMAGDt2rUk8jdA6uvrbW1tcckwAFi0aBGbzebNqbu7u2fPni0nJzeQNXk49u7j4zP884aFheGEaFVV1Z07dw7/gMKEoIofIVRRUYE7YYhmkHbg9L059vT0REdHz58/X0pKCs8CXF1do6KiuL1SqKura+rUqWJiYgcPHvz0lrW1tRoaGuPGjRume+LRo0d2dnZMJlNOTm7hwoV8mPtAOwIsfoTQ33//vWDBAgBYvny5sNa95BKFhYXOzs441g0AM2bMCA4OzsnJoTbg7Hzq7du3ZmZmYmJiwcHBn9iso6PD3d0dht273dfXFzuGmEzmuXPnhnMoIUawxY8Qqq+vxwXw9+/fT7ctgkdMTMx3332Hs+hwyQBnZ+ebN29yPKmmra0Ne9qPHj36ic127NiB0xOG3AkrLi7OyckJf51ly5ZxsDWg8CHw4kcIFRcXT5s2DQDWr19Pty0CSWNjY1pamrm5Oa5gg90B+fn5HDxFR0eHmZmZjIzMJwo0sNlsDQ0NKSmpIfRuwvOU+Ph4KsGJBIM/izCIHyFUU1ODF2MfOHCAN3VdhYO+pek6Ojrq6+s3bNigoKAAAPLy8k5OTsXFxRxZKNXT0+Pt7f1p8Xt6euLzDqF8Y3t7e1RUlLy8PAAoKyuTNLCBICTiRwhlZWXh/L8RI0aQQO5wyMnJ8fLywpWUVFVVp06dOqjVOP3S3d2N6xEeOnSo3w2am5t1dHQA4Ndffx2U+6a1tXXDhg344AwGw9HRUcTz9gaO8IgfIfT06VP8/D/AkBLhE7x582bNmjXYbQYADg4Owykf2Nvbu2bNGgD4mLc/ICCAwWBoaGgMqnlxZWXlnj17qKn+cOqCiyBCJX6EUEVFRWhoqKSkJJPJvHHjhgAVuuJPKioqPD091dXV8SzA09NzyD9pSkoKg8EIDQ398KPi4uIJEyYoKCgMXPlNTU3YpS8mJqahobFy5Uqe5SwIDcIg/g+DUpcuXcJDwcyZM8vLy2mxSpioq6v74Ycf8E9qbW09tJXCycnJEhISu3bt+vCjvXv3Yuf8x/bt7u7u+y8/f/4cu3gBYNSoURUVFUOwhyAM4u+XkydPamlpAYCenl5ycjLJAhw+QUFB06ZNExcX19XV3bdvX3Z29qB2T0lJgf5KZWVkZKirq7NYrL5ZBh8jKSnJzc1NWVkZAAwNDY8ePfrkyRNE6rsMCaEVP0Koqanpl19+weODu7v7oB4mCf3S2dmZmJhIJQVs3rx54AsEMjMzP3T49fT0LF68GADc3Nz6vv+hmJ8/f25vb48jEePGjfPy8qIeQHp6eoj4h4Awix8h1Nvbm5mZqampCQAyMjJXr16l2yJhIDExcfny5fgWoKend+zYsZKSks+2Fb516xYA7N27t++b165dk5aWVlZWxmkFlIb7irmqqurUqVNUDsLWrVuHnAJE6IuQix+Tn5+PA1dMJnP9+vX//PMP3RYJPD09Pbdv36a6icnJyVlaWnp7e39iBMbinzdvHlV7v729HWdne3t797tLS0uLi4sLzuDAK3yJP5+DiIT4EUINDQ0nTpzAxW20tLRILQeO0NLS4uPjs2TJEirYNm3atNTU1CdPnnx4F7h8+TIAsFgsatz29fUFAGNj45cvX3548NzcXAsLC3xYaWnpkydPvn37lutfSZQQFfFjzp8/b2RkhK8nDw+P/Px8anU3eWgcDmw2OyIiglosKCUlNW/evPdSg1pbW1etWqWkpNTQ0IAQys3NHTNmjLS09OnTpxFCfZ8aoqOj582bh/8mfX39sLAw3Ja3p6dHFJoy8AzREj9CqKurKz8/38rKCl9be/bsodsioeL69etr166dMmUK/nktLCyys7OpO2xAQADOF0II7du3DwBUVFSo8by1tZXNZi9btgzvq6GhERERwdW2nCKOIIm/t7f3s16lAVJZWWlvb89isRgMhpOT09mzZ0W5iivHqa2tPXLkiIGBATV64zWXx48fB4Cvv/7a39+fxWIpKSnhpr05OTkODg7m5ubU9sHBwc+fP6f5awg7giR+jlNaWmpmZoYvODMzs4yMDLotEio6OjoOHjxISdrKyio6OlpbW1tBQQE7X7Zv3/7mzZu8vLypU6fibdTU1Hx8fEheJm8QafEjhNra2oKCgnAsEABcXV1v3779ie2Ja2CwvHz5Mjg4GNcOk5WVxU4BAFBSUtq5cydVSoDFYrm5uZHRnpeIuvgxtbW1vr6+eFXZqFGj3NzcLl682G+lRyL+QUH551JTU9euXTtixAhJSUlxcXFxcXEGg4Flr6ysvHv37qdPn/JDcxGRQtTF31fMbDb7m2++oaJWGzZsaG1tpdE2IaOurs7S0pLBYFC1w8aMGbNu3bqysjJqG3Jv5SWiLv4PSUhIsLe3Hzt2LABMnDhx9+7dJClomBQVFdna2uKEfMyECRMyMjKoBTlv374lsuc9RPz9U1FRQeWuTJkyZdWqVZcuXerriCIX62epr6+/ffv2li1bcEI+ACgqKlpZWSUlJVVWVtJtHYGI/5Okp6fPnz8fX7hiYmKTJk3y8vIihWIQQr29vf0O1zU1NeXl5VevXt2yZQvlzJOVlTUxMdmzZ09paelgn6TITZZ7EPF/hsbGxuTkZBcXFypkJSsrO3/+/BUrVjx58oQEpTDt7e13794NDQ0dN26crKwsdbt0d3e/cuVKUVHRZzPz+m7w9OlT6g5LxM89iPgHQWxsrJeXF3VxjxkzxsjI6OLFiyK4WLirq6upqam6ujokJMTb25vKxgUAAwMDIyOjmTNn5uTkDLb4Jy7LNX78eBUVlePHj3PJeAKGiH/QFBcXp6en//hhugebAAAFlElEQVTjj0wmk5oLzJkz56effurruKYQmrGrrq4uOzubzWZnZWXZ29tra2vjjolUDs/p06fj4+PfvXv3XuGdARIZGYlbMAGApKTkxIkTiWuAqxDxD52ioqJDhw7Z2dlRAlBUVHR1dV22bNm9e/cePnxIt4HD5d27d9XV1aGhoWfOnDly5AjOgxATE6O+r5qamqmpaVJSUn5+/nC6a3V2dsbFxeFbiZKSkqur64wZM3AKIAe/DuE9iPiHS1tb219//cVmszdt2mRjY4NdA2JiYtLS0ra2tk5OTtu2bWtqampsbGxsbMQL2viZhoaGvLy8goICPz+/yZMnU729+/Lzzz+HhoZeuHChsbHxvYf5oU1zbt26JS0tDQCzZs0qLS1FCNnY2MDH1/kTOAIRP+eJiIhQUlLqqxYNDQ11dXUdHR1zc/OsrKyHDx8WFhYOpFk1RVdXFzceH7q6ui5fvhwXFxcVFRUdHR0ZGTlnzpz3pM5isSZMmLBixQo2m11QUJCXl8fxdbVeXl4A4OzsTPULxfVCV69ezdkTEfpCxM8VHj16FBMTc+fOHT8/Py0tLTU1NaoKFQBIS0uLi4uzWCwfH5/jx48HBQUlJyc3NzfX19c3NTU1NzcPyk/W1dX1sY/a29vLysrKy8tLS0srKirKysoeP35cVVV1/vx5T09PDw+PvhmNFOLi4ubm5hYWFhYWFtbW1oWFhQ0NDRzp29MvsbGxUlJSampqN27cwO+UlZUZGBjIy8tfuHCBSyclICJ+HkBN+I8fP37gwIGNGzdSae0UEhISX3755dixYzU1NbW1tV1cXO7du5eTk3Pv3r3s7Oz8/PykpCT88u7du7h5Rmlp6Z07d+7evZuens5ms9PS0pKTkzMyMhITE2/cuBEfH5+RkeHs7Dxy5MhRo0ZJS0vLyMjgqTWVcoPPq6Sk9MUXX5w9ezY9PT09PT0zM7OwsJBnP86LFy9MTU0B4MGDB9SbCxcuBAAbGxuemSGaEPFzns/Oih8+fJibm5ubm5uTk+Pv76+urq6srDxq1Ch1dXVFRcUPh+L3kJKS8vT0pCrbfRYGg2FoaGhkZGRgYGBmZmZoaDhjxoy5c+d6eHgUFxc/e/ZsCL3xOIWbm5uUlNSSJUva29tbW1tTU1Pd3NxGjhxpbm4+2NLghMFCxE8/dXV11dXVNTU1jY2NlZWVV65cCQsLCwkJCQwMPHHixKlTp8LDw48dO6avr//hfMHMzAxHHCUkJOTk5Hbs2HHq1KmYmJiYmJjo6OhLly79+eefubm5uCEnrpmD703cm8YPnEePHuno6FhaWuKXoaGh+Ht9++23/S6pJHAWIn6BoaqqinK5lZSUFBQUFBUV/ffff8XFxbm5uUVFRY8fP6bbxkHw+vVrV1dXANDU1HR1dV28ePHo0aP19PQcHByePn1Kt3UiARE/gR5iY2MBoO9SPyCBfd5CxE+ghydPnkyePHns2LHjx4+3t7fX1NRkMBgmJiakYifPIOIn0EZISIiMjIyTkxNCKDw8HLstXFxcBtUIVGiyp3kPET+BNg4ePAgACxYswC+3bt2KJ/8bN26k1zARgYifQBs+Pj4AMHfuXBx6aGtrwy16Ro8ejVv3EbgKET+BNv744w8AsLCwoDIj7ty5w2QyVVRUTE1NcSV10qKLexDxE2jDz88PAObPn983LeratWtaWlojR440Njam0TZRgIifQBtY/Kampu8N7ykpKUwmU05O7tixY/yQjCSsEPETaANP+7W1tT+shubo6IgTmXfs2EGqJ3MJIn4CbTQ0NGzdurXfLmnl5eWzZ8/Gzn87Ozve2yYKEPET+JTy8vLAwEBvb+/k5GS6bRFOiPgJBBGFiJ9AEFGI+AkEEYWIn0AQUYj4CQQRhYifQBBRiPgJBBHlfyJwXRxqGpv5AAAAAElFTkSuQmCC" alt></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><span>\({\rm{ST}} \bullet {\rm{QR}} = {\rm{SQ}} \bullet {\rm{RT}} + {\rm{SR}} \bullet {\rm{QT}}\) </span><strong><span><span><em>M1A1</em> </span></span></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><span>but \({\rm{QT = QR = RT}}\) </span><em><span><span><strong>R1</strong> </span></span></em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><span>hence \({\rm{ST = SQ + SR}}\) </span><strong><span><span><em>AG</em> </span></span></strong></span></p>
<p align="JUSTIFY"><span style="font-size: medium;"><span style="font-family: times new roman,times;"><span> </span></span></span></p>
<p align="JUSTIFY"><span style="font-size: medium;"><span style="font-family: times new roman,times;"><span>(ii) \(\angle {\rm{STR = }}\angle {\rm{SQR}}\) <em><strong>R1</strong></em></span></span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\angle {\rm{QST = }}\angle {\rm{QRT = }}{60^ \circ }\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\angle {\rm{RST = }}\angle {\rm{RQT = }}{60^ \circ }\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence \(\angle {\rm{QST = }}\angle {\rm{RST}}\) and \(\Delta {\rm{RST}} \sim \Delta {\rm{PSQ}}\) <strong><em>R1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) \(\frac{{{\rm{ST}}}}{{{\rm{SQ}}}} = \frac{{{\rm{SR}}}}{{{\rm{SP}}}} \to {\rm{ST}} \times {\rm{SP}} = {\rm{SR}} \times {\rm{SQ}}\) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">but \({\rm{ST = (SQ + SR)}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{(SQ}} + {\rm{SR) \times SP}} = {\rm{SR}} \times {\rm{SQ}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{SQ}} \times {\rm{SP}} + {\rm{SR}} \times {\rm{SP}} = {\rm{SR}} \times {\rm{SQ}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence \(\frac{1}{{{\rm{SP}}}} = \frac{1}{{{\rm{SQ}}}} + \frac{1}{{{\rm{SR}}}}\) <strong><em>AG </em></strong></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em> </em></span></strong></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[10 marks]</em> </span></strong></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"><img src="images/annie.png" alt></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">sine \(m\angle {\rm{PEM}} \cong m\angle {\rm{PFM}} \cong {90^ \circ }\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">then quadrilateral PFEM is cyclic <strong><em>M1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so \(m\angle {\rm{PME}} \cong m\angle {\rm{PFG}}\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">since \(m\angle {\rm{PGL}} \cong m\angle {\rm{PEL}} \cong {90^ \circ }\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">then quadrilateral PGLE is cyclic </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so \(m\angle {\rm{PGE}} \cong m\angle {\rm{PLE}}\) <strong><em>R1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">now E, F and G are collinear since they are on the Simson line of \(\Delta {\rm{LMN}}\) <strong><em>R1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">hence \(\Delta {\rm{PFG}} \sim \Delta {\rm{PML}}\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so \(\frac{{{\rm{PL}}}}{{{\rm{PM}}}} = \frac{{{\rm{PG}}}}{{{\rm{PF}}}} \Rightarrow {\rm{PL}} \times {\rm{PF}} = {\rm{PM}} \times {\rm{PG}}\) <strong><em>R1AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[8 marks] </span></em></strong></p>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">As in paper 1 there is a sad lack of knowledge of geometry although some good solutions were seen and at least one school is using techniques very successfully that are not mentioned in the program. </span></p>
<div class="question_part_label">A.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part (a) was well done. </span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Few clear solutions to part (b) were seen. </span></p>
<div class="question_part_label">B.b.</div>
</div>
<br><hr><br><div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">ABCD is a quadrilateral. (AD) and (BC) intersect at F and (AB) and (CD) </span><span style="font-family: times new roman,times; font-size: medium;">intersect at H. (DB) and (CA) intersect (FH) at G and E respectively. This is </span><span style="font-family: times new roman,times; font-size: medium;">shown in the diagram below.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/run.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Prove that \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} = - \frac{{{\rm{HE}}}}{{{\rm{EF}}}}\) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">in \(\Delta {\rm{HFD}}\) , [HA], [FC] and [DG] are concurrent at B <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so, \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} \times \frac{{{\rm{FA}}}}{{{\rm{AD}}}} \times \frac{{{\rm{DC}}}}{{{\rm{CH}}}} = 1\) </span><span style="font-family: times new roman,times; font-size: medium;">by Ceva’s theorem <em><strong>A1R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">in \(\Delta {\rm{HFD}}\) , with CAE as transversal, <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{HE}}}}{{{\rm{EF}}}} \times \frac{{{\rm{FA}}}}{{{\rm{AD}}}} \times \frac{{{\rm{DC}}}}{{{\rm{CH}}}} = - 1\) </span><span style="font-family: times new roman,times; font-size: medium;">by Menelaus’ theorem <em><strong>A1R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">therefore, \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} = - \frac{{{\rm{HE}}}}{{{\rm{EF}}}}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">This was not a difficult question but again too many candidates often left gaps in their solutions perhaps thinking that what they were doing was obvious and needed no written support </span></p>
</div>
<br><hr><br><div class="specification">
<p><img src="images/7.png" alt></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">The diagram above shows the points \({\text{P}}(x,{\text{ }}y)\) and \({\rm{P'}}(x',{\text{ }}y')\) which are equidistant from the origin \({\text{O}}\). The line \(({\text{OP}})\) is inclined at an angle \(\alpha \) to the <em>x</em>-axis and \({\rm{P\hat OP'}} = \theta \).</span></p>
</div>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) (i) By first noting that \({\text{OP}} = x\sec \alpha \), show that \(x' = x\cos \theta - y\sin \theta \) and find a similar expression for \(y'\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) Hence write down the \(2 \times 2\) matrix which represents the anticlockwise rotation about \({\text{O}}\) which takes \({\text{P}}\) to \({\text{P'}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) The ellipse \(E\) has equation \(5{x^2} + 5{y^2} - 6xy = 8\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Show that if \(E\) is rotated <strong>clockwise </strong>about the origin through \(45^\circ\), its equation becomes \(\frac{{{x^2}}}{4} + {y^2} = 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence determine the coordinates of the foci of \(E\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) (i) \(x' = x\sec \alpha \cos (\theta + \alpha )\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = x\sec \alpha (\cos \theta \cos \alpha - \sin \theta \sin \alpha )\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = x\cos \theta - x\tan \alpha \sin \theta \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = x\cos \theta - y\sin \theta \) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y' = x\sec \alpha \sin (\theta + \alpha )\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = x\sec \alpha (\sin \theta \cos \alpha + \cos \theta \sin \alpha )\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = x\sin \theta + x\tan \alpha \cos \theta \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = x\sin \theta + y\cos \theta \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the matrix \(\left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right]\) represents the rotation <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) the above relationship can be written in the form</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left[ \begin{array}{l}x\\y\end{array} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right]\left[ \begin{array}{l}{x'}\\{y'}\end{array} \right]\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(\theta = - \frac{\pi }{4}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \frac{{x'}}{{\sqrt 2 }} - \frac{{y'}}{{\sqrt 2 }}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \frac{{x'}}{{\sqrt 2 }} + \frac{{y'}}{{\sqrt 2 }}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting in the equation of the ellipse,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(5{\left( {\frac{{x'}}{{\sqrt 2 }} - \frac{{y'}}{{\sqrt 2 }}} \right)^2} + 5{\left( {\frac{{x'}}{{\sqrt 2 }} + \frac{{y'}}{{\sqrt 2 }}} \right)^2} - 6\left( {\frac{{x'}}{{\sqrt 2 }} - \frac{{y'}}{{\sqrt 2 }}} \right)\left( {\frac{{x'}}{{\sqrt 2 }} + \frac{{y'}}{{\sqrt 2 }}} \right) = 8\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(5\left( {\frac{{{{x'}^2}}}{2} + \frac{{{{y'}^2}}}{2} - x'y'} \right) + 5\left( {\frac{{{{x'}^2}}}{2} + \frac{{{{y'}^2}}}{2} + x'y'} \right) - 6\left( {\frac{{{{x'}^2}}}{2} - \frac{{{{y'}^2}}}{2}} \right) = 8\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">leading to \(\frac{{{{x'}^2}}}{4} + {y'^2} = 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>M1A0M1A0 </em></strong>for using \(\theta = \frac{\pi }{4}\) leading to \(\frac{{{{y'}^2}}}{4} + {x'^2} = 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) in the usual notation, \(a = 2\), \(b = 1\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the coordinates of the foci of the rotated ellipse are \(\left( {\sqrt {3,} {\text{ 0}}} \right)\) and \(\left( { - \sqrt {3,} {\text{ 0}}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the coordinates of the foci of \({\text{E}}\) are therefore \(\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }},{\text{ }}\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)\) and \(\left( {\frac{{ - \sqrt 3 }}{{\sqrt 2 }},{\text{ }}\frac{{ - \sqrt 3 }}{{\sqrt 2 }}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p>A new triangle DEF is positioned within a circle radius <em>R</em> such that DF is a diameter as shown in the following diagram.</p>
<p style="text-align: center;"><img src="data:image/png;base64,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"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>In a triangle ABC, prove \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Prove that the area of the triangle ABC is \(\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that <em>R</em> denotes the radius of the circumscribed circle prove that \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}} = 2R\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence show that the area of the triangle ABC is \(\frac{{abc}}{{4R}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.iv.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find in terms of <em>R</em>, the two values of (DE)<sup>2</sup> such that the area of the shaded region is twice the area of the triangle DEF.</p>
<div class="marks">[9]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Using two diagrams, explain why there are two values of (DE)<sup>2</sup>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>A parallelogram is positioned inside a circle such that all four vertices lie on the circle. Prove that it is a rectangle.</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,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"></p>
<p>\({\text{sin}}\,{\text{B}} = \frac{h}{c}\) and \({\text{sin}}\,{\text{C}} = \frac{h}{b}\) <em><strong>M1A1</strong></em></p>
<p>hence \(h = c\,{\text{sin}}\,{\text{B}} = b\,{\text{sin}}\,{\text{C}}\) <em><strong>A1</strong></em></p>
<p>by dropping a perpendicular from B, in exactly the same way we find \(c\,{\text{sin}}\,{\text{A}} = a\,{\text{sin}}\,{\text{C}}\) <em><strong>R1</strong></em></p>
<p>hence \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}}\)</p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>area = \(\frac{1}{2}ah\) <em><strong>M1A1</strong></em></p>
<p>= \(\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,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"></p>
<p>since the angle at the centre of circle is twice the angle at the circumference \({\text{sin}}\,A = \frac{a}{{2R}}\) <em><strong>M1A1</strong></em></p>
<p>hence \(\frac{a}{{{\text{sin}}\,A}} = 2R\) and therefore \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}} = 2R\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>area of the triangle is \(\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}\) <em><strong>M1</strong></em></p>
<p>since \({\text{sin}}\,{\text{C}} = \frac{c}{{2R}}\) <em><strong>A1</strong></em></p>
<p>area of the triangle is \(\frac{1}{2}ab\,\frac{c}{{2R}} = \frac{{abc}}{{4R}}\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>area of the triangle is \(\frac{{\pi {R^2}}}{6}\) <em><strong>(M1)A1</strong></em></p>
<p>(DE)<sup>2 </sup>+ (EF)<sup>2</sup> = 4<em>R</em><sup>2 </sup> <em><strong>M1</strong></em></p>
<p>(DE)<sup>2 </sup>= 4<em>R</em><sup>2</sup> − (EF)<sup>2</sup> </p>
<p>\(\frac{1}{2}\left( {{\text{DE}}} \right)\left( {{\text{EF}}} \right) = \frac{{\pi {R^2}}}{6} \Rightarrow \left( {{\text{EF}}} \right) = \frac{{\pi {R^2}}}{{3\left( {{\text{DE}}} \right)}}\) <em><strong>M1A1</strong></em></p>
<p>\({\left( {{\text{DE}}} \right)^2} = 4{R^2} - \frac{{{\pi ^2}{R^4}}}{{9{{\left( {{\text{DE}}} \right)}^2}}}\) <em><strong>A1</strong></em></p>
<p>\(9{\left( {{\text{DE}}} \right)^4} - 36{\left( {{\text{DE}}} \right)^2}{R^2} + {\pi ^2}{R^4} = 0\) <em><strong>A1 </strong></em></p>
<p>\({\left( {{\text{DE}}} \right)^2} = \frac{{36{R^2} \pm \sqrt {1296{R^4} - 36{\pi ^2}{R^4}} }}{{18}}\) <em><strong>M1</strong></em></p>
<p>\({\left( {{\text{DE}}} \right)^2} = \frac{{36{R^2} \pm 6{R^2}\sqrt {36 - {\pi ^2}} }}{{18}}\left( { = \frac{{6{R^2} \pm {R^2}\sqrt {36 - {\pi ^2}} }}{3}} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[9 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,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"> <em><strong> A1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,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"></p>
<p>\(\mathop {\text{A}}\limits^ \wedge + \mathop {\text{C}}\limits^ \wedge = 180^\circ \) (cyclic quadrilateral) <em><strong>R1</strong></em></p>
<p>however \(\mathop {\text{A}}\limits^ \wedge = \mathop {\text{C}}\limits^ \wedge \) (ABCD is a parallelogram) <em><strong>R1</strong></em></p>
<p>\(\mathop {\text{A}}\limits^ \wedge = \mathop {\text{C}}\limits^ \wedge = 90^\circ \) <em><strong>A1</strong></em></p>
<p>\(\mathop {\text{B}}\limits^ \wedge = \mathop {\text{D}}\limits^ \wedge = 90^\circ \)</p>
<p>hence ABCD is a rectangle <em><strong>AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br>