FP1 June 2014 Q8
8. The points \(P(4k^2, 8k)\) and \(Q(k^2, 4k)\), where \(k\) is a constant, lie on the parabola \(C\) with equation \(y^2 = 16x\).
The straight line \(l_1\) passes through the points \(P\) and \(Q\).
(a) Show that an equation of the line \(l_1\) is given by \[3ky - 4x = 8k^2\] (4)
The line \(l_2\) is perpendicular to the line \(l_1\) and passes through the focus of the parabola \(C\). The line \(l_2\) meets the directrix of \(C\) at the point \(R\).
(b) Find, in terms of \(k\), the \(y\) coordinate of the point \(R\). (7)
| Scheme | Marks |
|---|---|
| \(m = \dfrac{4k - 8k}{k^2 - 4k^2}\ \left(= \dfrac{4}{3k}\right)\) Valid attempt to find gradient in terms of \(k\) | M1 |
| \(y - 8k = \dfrac{4}{3k}(x - 4k^2)\) or \(y - 4k = \dfrac{4}{3k}(x - k^2)\) or \(y = \dfrac{4}{3k}x + \dfrac{8k}{3}\) M1: Correct straight line method with their gradient in terms of \(k\). If using \(y = mx + c\) then award M provided they attempt to find \(c\) A1: Correct equation. If using \(y = mx + c\), awardwhen they obtain \(c = \dfrac{8k}{3}\) oe | M1A1 |
| \(3ky - 24k^2 = 4x - 16k^2 \Rightarrow 3ky - 4x = 8k^2\) * or \(3ky - 12k^2 = 4x - 4k^2 \Rightarrow 3ky - 4x = 8k^2\) * Correct completion to printed answer with at least one intermediate step. | A1* |
| (4) |
| Scheme | Marks |
|---|---|
| (Focus) \((4, 0)\) Seen or implied as a number | B1 |
| (Directrix) \(x = -4\) Seen or implied as a number | B1 |
| Gradient of \(l_2\) is \(-\dfrac{3k}{4}\) Attempt negative reciprocal of grad \(l_1\) as a function of \(k\) | M1 |
| \(y - 0 = \dfrac{-3k}{4}(x - 4)\) Use of their changed gradient and numerical Focus in either formula, as printed oe | M1, A1 |
| \(x = -4 \Rightarrow y = \dfrac{-3k}{4}(-4 - 4)\) Substitute numerical directrix into their line | M1 |
| \((y =)\, 6k\) oe | A1 |
| (7) | |
| (11 marks) |