FP3 June 2010 Q8
8. The hyperbola \(H\) has equation \(\dfrac{x^2}{16} - \dfrac{y^2}{4} = 1\).
The line \(l_1\) is the tangent to \(H\) at the point \(P(4\sec t, 2\tan t)\).
(a) Use calculus to show that an equation of \(l_1\) is \[2y\sin t = x - 4\cos t\] (5)
The line \(l_2\) passes through the origin and is perpendicular to \(l_1\).
The lines \(l_1\) and \(l_2\) intersect at the point \(Q\).
(b) Show that, as \(t\) varies, an equation of the locus of \(Q\) is \[\left(x^2 + y^2\right)^2 = 16x^2 - 4y^2\] (8)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4\sec t\tan t \quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\sec^2 t\) | B1 (both) |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\sec^2 t}{4\sec t\tan t} \quad \left(= \dfrac{1}{2\sin t}\right)\) | M1 |
| \(y - 2\tan t = \dfrac{1}{2\sin t}\left(x - 4\sec t\right)\) | M1 A1 |
| \(2y\sin t - \dfrac{4\sin^2 t}{\cos t} = x - \dfrac{4}{\cos t}\) \(2y\sin t = x - \dfrac{4 - 4\sin^2 t}{\cos t} = x - 4\cos t\) * | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Gradient of \(l_2\) is \(-2\sin t\) | M1 |
| \(y = -2x\sin t \quad (2)\) | A1 |
| \(2\left(-2x\sin t\right)\sin t = x - 4\cos t \Rightarrow x = \dfrac{4\cos t}{1 + 4\sin^2 t} \quad (1)\) | M1 A1 |
| \(y = \dfrac{-8\sin t\cos t}{1 + 4\sin^2 t}\) | M1 A1 |
| \(\left(x^2 + y^2\right)^2 = \left(\dfrac{16\cos^2 t}{\left(1 + 4\sin^2 t\right)^2} + \dfrac{64\sin^2 t\cos^2 t}{\left(1 + 4\sin^2 t\right)^2}\right)^2\) \(= \dfrac{256\cos^4 t}{\left(1 + 4\sin^2 t\right)^4}\left(1 + 4\sin^2 t\right)^2 = \dfrac{256\cos^4 t}{\left(1 + 4\sin^2 t\right)^2}\) | M1 |
| \(16x^2 - 4y^2 = \dfrac{256\cos^2 t}{\left(1 + 4\sin^2 t\right)^2} - \dfrac{256\sin^2 t\cos^2 t}{\left(1 + 4\sin^2 t\right)^2} = \dfrac{256\cos^4 t}{\left(1 + 4\sin^2 t\right)^2}\) | A1 |
| (8) | |
| (13 marks) |