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</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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<br><hr><br>