Pre-U Pre-U 9794/2 2017 June — Question 7 10 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2017
SessionJune
Marks10
TopicParametric curves and Cartesian conversion
TypeProperties of specific curves
DifficultyStandard +0.3 This is a standard parametric differentiation question with routine algebraic manipulation. Part (i) uses the chain rule formula dy/dx = (dy/dθ)/(dx/dθ), part (ii) requires finding tangent intercepts and their midpoint (straightforward but multi-step), and part (iii) is simple substitution to verify the locus. All techniques are textbook exercises with no novel insight required, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
7 A curve, \(C\), is given parametrically by \(x = 2 \cos \theta , y = 3 \sin \theta , 0 < \theta < \frac { 1 } { 2 } \pi\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 } { 2 } \cot \theta\). A tangent to \(C\) intersects the \(x\)-axis and \(y\)-axis at \(P\) and \(Q\) respectively.
  2. Show that the midpoint of \(P Q\) has coordinates \(\left( \sec \theta , \frac { 3 } { 2 } \operatorname { cosec } \theta \right)\).
  3. Hence show that the midpoint of \(P Q\) lies on the curve \(\frac { 4 } { x ^ { 2 } } + \frac { 9 } { y ^ { 2 } } = 4\).
Question 7: Parametric curve \(x = 2\cos\theta\), \(y = 3\sin\theta\)
(i)
- \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta\), \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 3\cos\theta\) B1 Both derivatives correct
- \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{3\cos\theta}{-2\sin\theta}\) M1 Attempt at parametric differentiation soi
- \(= -\frac{3}{2}\cot\theta\) A1 Obtain correct unsimplified derivative and then simplify to given answer
(ii)
- \(y - 3\sin\theta = -\frac{3}{2}\cot\theta(x - 2\cos\theta)\) M1\* Attempt equation of tangent, in terms of \(\theta\)
- A1 Obtain correct equation aef – could be implied by correct \(c\) if using \(y = mx + c\)
- \(0 - 3\sin\theta = -\frac{3}{2}\cot\theta(x - 2\cos\theta)\) M1d\* Attempt \(x\)-intercept – substitute \(y = 0\) to get a value for \(x\)
- \(x = 2\sec\theta\) A1 Obtain \(x = 2\sec\theta\), with sufficient detail seen
- \(y - 3\sin\theta = -\frac{3}{2}\cot\theta(0 - 2\cos\theta)\) M1d\* Attempt \(y\)-intercept – substitute \(x = 0\) to get a value for \(y\)
- \(y = 3\cosec\theta\) A1 Obtain \(y = 3\cosec\theta\), with sufficient detail seen
- Midpoint is \(\left(\frac{1}{2}\times 2\sec\theta,\ \frac{1}{2}\times 3\cosec\theta\right) = \left(\sec\theta,\ \frac{3}{2}\cosec\theta\right)\) A1 Show given answer for midpoint – must show some working so A0 if straight from intercepts to given answer
(iii)
- \(\frac{4}{\sec^2\theta} + \frac{9}{\left(\frac{3}{2}\cosec\theta\right)^2} = 4\cos^2\theta + 4\sin^2\theta = 4\) M1 Substitute coords from (ii)
- A1 Convincingly show that midpoint is on curve
Total: 10 marks
**Question 7: Parametric curve $x = 2\cos\theta$, $y = 3\sin\theta$**

**(i)**
- $\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta$, $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 3\cos\theta$ **B1** Both derivatives correct
- $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{3\cos\theta}{-2\sin\theta}$ **M1** Attempt at parametric differentiation soi
- $= -\frac{3}{2}\cot\theta$ **A1** Obtain correct unsimplified derivative and then simplify to given answer

**(ii)**
- $y - 3\sin\theta = -\frac{3}{2}\cot\theta(x - 2\cos\theta)$ **M1\*** Attempt equation of tangent, in terms of $\theta$
- **A1** Obtain correct equation aef – could be implied by correct $c$ if using $y = mx + c$
- $0 - 3\sin\theta = -\frac{3}{2}\cot\theta(x - 2\cos\theta)$ **M1d\*** Attempt $x$-intercept – substitute $y = 0$ to get a value for $x$
- $x = 2\sec\theta$ **A1** Obtain $x = 2\sec\theta$, with sufficient detail seen
- $y - 3\sin\theta = -\frac{3}{2}\cot\theta(0 - 2\cos\theta)$ **M1d\*** Attempt $y$-intercept – substitute $x = 0$ to get a value for $y$
- $y = 3\cosec\theta$ **A1** Obtain $y = 3\cosec\theta$, with sufficient detail seen
- Midpoint is $\left(\frac{1}{2}\times 2\sec\theta,\ \frac{1}{2}\times 3\cosec\theta\right) = \left(\sec\theta,\ \frac{3}{2}\cosec\theta\right)$ **A1** Show given answer for midpoint – must show some working so A0 if straight from intercepts to given answer

**(iii)**
- $\frac{4}{\sec^2\theta} + \frac{9}{\left(\frac{3}{2}\cosec\theta\right)^2} = 4\cos^2\theta + 4\sin^2\theta = 4$ **M1** Substitute coords from (ii)
- **A1** Convincingly show that midpoint is on curve

**Total: 10 marks**
7 A curve, $C$, is given parametrically by $x = 2 \cos \theta , y = 3 \sin \theta , 0 < \theta < \frac { 1 } { 2 } \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 } { 2 } \cot \theta$.

A tangent to $C$ intersects the $x$-axis and $y$-axis at $P$ and $Q$ respectively.\\
(ii) Show that the midpoint of $P Q$ has coordinates $\left( \sec \theta , \frac { 3 } { 2 } \operatorname { cosec } \theta \right)$.\\
(iii) Hence show that the midpoint of $P Q$ lies on the curve $\frac { 4 } { x ^ { 2 } } + \frac { 9 } { y ^ { 2 } } = 4$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2017 Q7 [10]}}
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