FP3 June 2012 Q6
6. The ellipse \(E\) has equation \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
The line \(l_1\) is a tangent to \(E\) at the point \(P\ (a\cos\theta, b\sin\theta)\).
The circle \(C\) has equation \[x^2 + y^2 = a^2\]
The line \(l_2\) is a tangent to \(C\) at the point \(Q\ (a\cos\theta, a\sin\theta)\).
Given that \(l_1\) and \(l_2\) meet at the point \(R\),
| Scheme | Marks |
|---|---|
| \(\dfrac{2x}{a^2} + \dfrac{2y}{b^2}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{xb^2}{ya^2} = -\dfrac{b\cos\theta}{a\sin\theta}\) | M1 A1 |
| \(\therefore y - b\sin\theta = -\dfrac{b\cos\theta}{a\sin\theta}(x - a\cos\theta)\) | M1 |
| Uses \(\cos^2\theta + \sin^2\theta = 1\) to give \(\dfrac{x\cos\theta}{a} + \dfrac{y\sin\theta}{b} = 1\) * | A1cso |
| (4) |
Notes
a1M1: Finding gradient in terms of \(\theta\). Must use calculus.
a1A1: cao
a2M1: Finding equation of tangent
a2A1: cso (answer given). Need to get \(\cos^2\theta + \sin^2\theta\) on the same side.
| Scheme | Marks |
|---|---|
| Gradient of circle is \(-\dfrac{\cos\theta}{\sin\theta}\) and equation of tangent is \(y - a\sin\theta = -\dfrac{\cos\theta}{\sin\theta}(x - a\cos\theta)\) or sets \(a = b\) in previous answer | M1 |
| So \(y\sin\theta + x\cos\theta = a\) | A1 |
| (2) |
Notes
b1M1: Finding gradient and equation of tangent, or setting \(a = b\).
b1A1: cao need not be simplified.
| Scheme | Marks |
|---|---|
| Eliminate \(x\) or \(y\) to give \(y\sin\theta\left(\tfrac{a}{b} - 1\right) = 0\) or \(x\cos\theta\left(\tfrac{b}{a} - 1\right) = b - a\) | M1 |
| \(l_1\) and \(l_2\) meet at \(\left(\dfrac{a}{\cos\theta}, 0\right)\) | A1, B1 |
| (3) |
Notes
c1M1: As scheme
c1A1: \(x = \dfrac{a}{\cos\theta}\), need not be simplified.
c1B1: \(y = 0\), need not be simplified.
| Scheme | Marks |
|---|---|
| The locus of \(R\) is part of the line \(y = 0\), such that \(x \geqslant a\) and \(x \leqslant -a\) Or clearly labelled sketch. Accept “real axis” | B1, B1 |
| (2) | |
| (11 marks) |
Notes
d1B1: Identifying locus as \(y = 0\) or real/’\(x\)’ axis.
d2B1: Depends on previous B mark, identifies correct parts of \(y = 0\). Condone use of strict inequalities.