FP3 June 2016 Q5
5. The hyperbola \(H\) has equation \[\frac{x^2}{16} - \frac{y^2}{9} = 1\]
The point \(P\,(4\sec\theta, 3\tan\theta)\), \(0 < \theta < \dfrac{\pi}{2}\), lies on \(H\).
The line \(l\) is the directrix of \(H\) for which \(x > 0\)
The normal to \(H\) at \(P\) crosses the line \(l\) at the point \(Q\). Given that \(\theta = \dfrac{\pi}{4}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1\quad P(4\sec\theta, 3\tan\theta)\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3\sec^2\theta}{4\sec\theta\tan\theta}\left(= \dfrac{3}{4\sin\theta}\right)\) or \(\dfrac{2x}{16} - \dfrac{2y}{9}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8\sec\theta}{16} \times \dfrac{9}{6\tan\theta}\) or \(y = 3\left(\dfrac{x^2}{16} - 1\right)^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{2}\left(\dfrac{x^2}{16} - 1\right)^{-\frac{1}{2}}\dfrac{x}{8}\) \(= \dfrac{3}{2}\left(\dfrac{(4\sec\theta)^2}{16} - 1\right)^{-\frac{1}{2}}\dfrac{4\sec\theta}{8}\) | M1 A1 |
| Normal gradient \(-\dfrac{4\sin\theta}{3}\) | M1 |
| \(y - 3\tan\theta = -\dfrac{4\sin\theta}{3}(x - 4\sec\theta)\) | M1 |
| \(3y + 4x\sin\theta = 25\tan\theta\) * | A1* |
| (5) |
Notes
M1: Correct gradient method. Finds \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = p\sec^2\theta\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = q\sec\theta\tan\theta\) and uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}\theta} \times \dfrac{\mathrm{d}\theta}{\mathrm{d}x}\) or differentiates implicitly to give \(px + qy\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and substitutes for \(y\) and \(x\) to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or differentiates explicitly to give \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = px\left(qx^2 - 1\right)^{-\frac{1}{2}}\) and substitutes for \(x\)
A1: Correct derivative in terms of trig. functions, e.g. \(\dfrac{3\sec^2\theta}{4\sec\theta\tan\theta},\ \dfrac{8\sec\theta}{16} \times \dfrac{9}{6\tan\theta}\) Does not need to be simplified.
M1: Correct perpendicular gradient rule. Does not need to be simplified.
M1: Correct straight line method using a gradient (does not need to be simplified) in terms of \(\theta\) that has come from calculus and is not the tangent gradient. If they use \(y = mx + c\) then they must reach as far as finding \(c\).
A1*: Correct proof with no errors and one intermediate step from the previous line. Allow \(25\tan\theta = 3y + 4x\sin\theta\)
| Scheme | Marks |
|---|---|
| \(b^2 = a^2\left(e^2 - 1\right) \Rightarrow 9 = 16\left(e^2 - 1\right) \Rightarrow e = \dfrac{5}{4}\) | M1A1 |
| \(x = \dfrac{a}{e} \Rightarrow x = \dfrac{16}{5}\) or \(\dfrac{4}{5/4}\) etc. | A1 |
| \(\theta = \dfrac{\pi}{4},\ x = \dfrac{16}{5} \Rightarrow 3y + 2\sqrt{2} \times \dfrac{16}{5} = 25\) | M1 |
| \(y = \dfrac{25}{3} - \dfrac{32}{15}\sqrt{2}\) | B1B1 (A marks on EPEN) |
| (6) | |
| (11 marks) |
Notes
M1: Use of the correct eccentricity formula to obtain a value for \(e\)
A1: Correct value for \(e\). Ignore \(\pm\)
A1: Correct value for \(\dfrac{a}{e}\) Ignore \(\pm\)
M1: Substitutes \(\theta = \dfrac{\pi}{4}\) into the given normal equation and uses their positive directrix equation to obtain an equation in \(y\) or in \(y\) and e only.
B1: \(a = \dfrac{25}{3}\) oe or \(b = -\dfrac{32}{15}\) oe
B1: \(a = \dfrac{25}{3}\) oe and \(b = -\dfrac{32}{15}\) oe
Special Case: If the correct form of the answer is never seen but it appears correctly as a single fraction, allow B1B0 e.g. \(y = \dfrac{125 - 32\sqrt{2}}{15}\)