Date | November 2018 | Marks available | 4 | Reference code | 18N.1.AHL.TZ0.H_7 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | H_7 | Adapted from | N/A |
Question
Consider the curves C1 and C2 defined as follows
C1:xy=4, x>0
C2:y2−x2=2, x>0
Using implicit differentiation, or otherwise, find dydx for each curve in terms of x and y.
Let P(a, b) be the unique point where the curves C1 and C2 intersect.
Show that the tangent to C1 at P is perpendicular to the tangent to C2 at P.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
C1:y+xdydx=0 (M1)
Note: M1 is for use of both product rule and implicit differentiation.
⇒dydx=−yx A1
Note: Accept −4x2
C2:2ydydx−2x=0 (M1)
⇒dydx=xy A1
Note: Accept ±x√2+x2
[4 marks]
substituting a and b for x and y M1
product of gradients at P is (−ba)(ab)=−1 or equivalent reasoning R1
Note: The R1 is dependent on the previous M1.
so tangents are perpendicular AG
[2 marks]