FP3 June 2014 (R) Q5
5. The ellipse \(E\) has equation \[x^2 + 9y^2 = 9\]
The point \(P(a\cos\theta, b\sin\theta)\) is a general point on the ellipse \(E\).
(a) Write down the value of \(a\) and the value of \(b\). (1)
The line \(L\) is a tangent to \(E\) at the point \(P\).
(b) Show that an equation of the line \(L\) is given by \[3y\sin\theta + x\cos\theta = 3\] (3)
The line \(L\) meets the \(x\)-axis at the point \(Q\) and meets the \(y\)-axis at the point \(R\).
(c) Show that the area of the triangle \(OQR\), where \(O\) is the origin, is given by \[k\,\mathrm{cosec}\,2\theta\] where \(k\) is a constant to be found. (3)
The point \(M\) is the midpoint of \(QR\).
(d) Find a cartesian equation of the locus of \(M\), giving your answer in the form \(y^2 = \mathrm{f}(x)\). (4)
| Scheme | Marks |
|---|---|
| \(a = 3,\ b = 1\) Both | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{\cos\theta}{3\sin\theta}\) Complete correct gradient method including use of coordinates | M1 |
| \(y - \sin\theta = -\dfrac{\cos\theta}{3\sin\theta}(x - 3\cos\theta)\) (I) Correct straight line method | M1 |
| \(3y\sin\theta - 3\sin^2\theta = -x\cos\theta + 3\cos^2\theta\) Allow both M’s if working in \(a\) and \(b\) so far | |
| \(3y\sin\theta + x\cos\theta = 3\cos^2\theta + 3\sin^2\theta = 3\ ^*\) Correct completion to printed answer with no errors seen. Some working is needed from (I) to *. | A1* |
| (3) |
| Scheme | Marks |
|---|---|
| \(x = 0 \Rightarrow y = \dfrac{1}{\sin\theta},\ y = 0 \Rightarrow x = \dfrac{3}{\cos\theta}\) Both | B1 |
| Area \(= \dfrac{1}{2} \times \text{"}\dfrac{1}{\sin\theta}\text{"} \times \text{"}\dfrac{3}{\cos\theta}\text{"}\) Correct method for area | M1 |
| \(= \dfrac{3}{2\sin\theta\cos\theta} = 3\,\mathrm{cosec}\,2\theta\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(x = \dfrac{3}{2\cos\theta},\ y = \dfrac{1}{2\sin\theta}\) Correct follow through mid-point | B1ft |
| \(\sin\theta = \dfrac{1}{2y},\ \cos\theta = \dfrac{3}{2x}\) \(\left(\dfrac{3}{2x}\right)^2 + \left(\dfrac{1}{2y}\right)^2 = 1\) Attempt sin and cos in terms of \(x\) and \(y\) and attempt Pythagoras. Allow if \(x\) and \(y\) are exchanged. | M1 |
| \(9y^2 + x^2 = 4x^2y^2\) | |
| \(y^2(4x^2 - 9) = x^2 \Rightarrow y^2 = \ldots\) Attempt to isolate \(y^2\) | M1 |
| \(y^2 = \dfrac{x^2}{4x^2 - 9}\) Correct equation (oe) | A1 |
| (4) | |
| (11 marks) |
Notes
(d) Way 2
| Scheme | Marks |
|---|---|
| \(x = \dfrac{3}{2\cos\theta},\ y = \dfrac{1}{2\sin\theta}\) Correct follow through mid-point | B1ft |
| \(y^2 = \dfrac{1}{4\sin^2\theta}\) Attempt \(y^2\) in terms of sin | M1 |
| \(y^2 = \dfrac{1}{4(1 - \cos^2\theta)}\) Correct use of Pythagoras | M1 |
| \(y^2 = \dfrac{1}{4\left(1 - \frac{9}{4x^2}\right)}\) Correct equation (oe) | A1 |