FP3 June 2017 Q2
2. The ellipse \(E\) has equation \[\frac{x^2}{36} + \frac{y^2}{25} = 1\]
The line \(l\) is the normal to \(E\) at the point \(P\,(6\cos\theta, 5\sin\theta)\), where \(0 < \theta < \dfrac{\pi}{2}\)
The line \(l\) meets the \(x\)-axis at the point \(Q\).
The point \(R\) is the foot of the perpendicular from \(P\) to the \(x\)-axis.
| Scheme | Marks |
|---|---|
| \(\dfrac{2x}{36} + \dfrac{2y}{25}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{25x}{36y} = \dfrac{5\cos\theta}{-6\sin\theta}\) or \(x = 6\cos\theta,\ y = 5\sin\theta \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{5\cos\theta}{-6\sin\theta}\) or \(\dfrac{y^2}{25} = 1 - \dfrac{x^2}{36} \Rightarrow y = 5\sqrt{1 - \dfrac{x^2}{36}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{5x}{36}\left(1 - \dfrac{x^2}{36}\right)^{-\frac{1}{2}} = -\dfrac{5\cos\theta}{6\sin\theta}\) | M1 |
| \(= -\dfrac{5\cos\theta}{6\sin\theta}\) | A1 |
| \(m_N = \dfrac{6\sin\theta}{5\cos\theta}\) | M1 |
| \(y - 5\sin\theta = \textit{Their } m_N\,(x - 6\cos\theta)\) | M1 |
| \(6x\sin\theta - 5y\cos\theta = 11\sin\theta\cos\theta\) * | A1* |
| (5) |
Notes
M1: Correct attempt at \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) using implicit or parametric or explicit differentiation
\(\left(ax + by\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\dfrac{a\cos\theta}{b\sin\theta},\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = ax\left(1 - bx^2\right)^{-\frac{1}{2}}(oe) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\right)\)
A1: Correct tangent gradient in terms of \(\theta\). May be implied in their attempt the normal gradient.
M1: Correct perpendicular gradient rule. May be awarded if working in terms of \(x\) and \(y\).
M1: Correct straight line method for the normal using a “changed” \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\) which must have come from calculus. If using \(y = mx + c\), must reach as far as \(c = \ldots\)
A1*: Correct completion to printed answer with no errors.
Note that if the candidate uses e.g \(y - 5\sin\theta = -\dfrac{36y}{25x}(x - 6\cos\theta)\) before introducing \(\theta\), the final mark can be withheld.
| Scheme | Marks |
|---|---|
| \(b^2 = a^2\left(1 - e^2\right) \Rightarrow 25 = 36\left(1 - e^2\right) \Rightarrow e^2 = \dfrac{11}{36}\) or \(e = \sqrt{\dfrac{11}{36}}\) | M1 |
| \(y = 0 \Rightarrow x = \dfrac{11\cos\theta}{6}\) or \(\dfrac{11\sin\theta\cos\theta}{6\sin\theta}\) | B1 |
| \(\left(\dfrac{OQ}{OR} =\right)\dfrac{11\cos\theta}{6} \times \dfrac{1}{6\cos\theta}\) | M1 |
| \(= \dfrac{11}{36}\) | A1 |
| (4) | |
| (9 marks) |
Notes
M1: Uses the correct eccentricity formula to obtain a value for \(e\) or \(e^2\). Ignore \(\pm\) values for e.
B1: Correct \(x\) coordinate for \(Q\)
M1: Attempts \(\dfrac{\text{their } OQ}{\text{their } OR}\). May be implied by their ratio.
A1: Correct completion with no errors to obtain \(\dfrac{11}{36}\) both times.
Ignore any references to the foci or directrices but the final mark can be withheld if there are any incorrect statements such as e.g. using \(\cos\theta = 1\) in their ratio.