| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question combining parabola properties (focus, directrix) with coordinate geometry. Part (a) is routine algebra finding a line equation, but part (b) requires knowing the focus and directrix of y²=16x, finding a perpendicular line, and solving simultaneously—multiple conceptual steps beyond standard A-level. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = \frac{4k-8k}{k^2-4k^2}\left(= \frac{4}{3k}\right)\) | M1 | Valid attempt to find gradient in terms of \(k\) |
| \(y - 8k = \frac{4}{3k}(x - 4k^2)\) or \(y - 4k = \frac{4}{3k}(x-k^2)\) or \(y = \frac{4}{3k}x + \frac{8k}{3}\) | M1A1 | M1: correct straight line method with their gradient; A1: correct equation (if using \(y=mx+c\), award when \(c = \frac{8k}{3}\) obtained) |
| \(3ky - 24k^2 = 4x - 16k^2 \Rightarrow 3ky - 4x = 8k^2\) * | A1* | Correct completion to printed answer with at least one intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Focus \((4, 0)\) | B1 | Seen or implied as a number |
| Directrix \(x = -4\) | B1 | Seen or implied as a number |
| Gradient of \(l_2\) is \(-\frac{3k}{4}\) | M1 | Attempt negative reciprocal of gradient of \(l_1\) as a function of \(k\) |
| \(y - 0 = -\frac{3k}{4}(x-4)\) | M1, A1 | Use of their changed gradient and numerical Focus in either formula |
| \(x = -4 \Rightarrow y = -\frac{3k}{4}(-4-4)\) | M1 | Substitute numerical directrix into their line |
| \((y =) 6k\) | A1 | oe |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \frac{4k-8k}{k^2-4k^2}\left(= \frac{4}{3k}\right)$ | M1 | Valid attempt to find gradient in terms of $k$ |
| $y - 8k = \frac{4}{3k}(x - 4k^2)$ or $y - 4k = \frac{4}{3k}(x-k^2)$ or $y = \frac{4}{3k}x + \frac{8k}{3}$ | M1A1 | M1: correct straight line method with their gradient; A1: correct equation (if using $y=mx+c$, award when $c = \frac{8k}{3}$ obtained) |
| $3ky - 24k^2 = 4x - 16k^2 \Rightarrow 3ky - 4x = 8k^2$ * | A1* | Correct completion to printed answer with at least one intermediate step |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Focus $(4, 0)$ | B1 | Seen or implied as a number |
| Directrix $x = -4$ | B1 | Seen or implied as a number |
| Gradient of $l_2$ is $-\frac{3k}{4}$ | M1 | Attempt negative reciprocal of gradient of $l_1$ as a function of $k$ |
| $y - 0 = -\frac{3k}{4}(x-4)$ | M1, A1 | Use of their changed gradient and numerical Focus in either formula |
| $x = -4 \Rightarrow y = -\frac{3k}{4}(-4-4)$ | M1 | Substitute numerical directrix into their line |
| $(y =) 6k$ | A1 | oe |
---8. The points $P \left( 4 k ^ { 2 } , 8 k \right)$ and $Q \left( k ^ { 2 } , 4 k \right)$, where $k$ is a constant, lie on the parabola $C$ with equation $y ^ { 2 } = 16 x$.
The straight line $l _ { 1 }$ passes through the points $P$ and $Q$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the line $l _ { 1 }$ is given by
$$3 k y - 4 x = 8 k ^ { 2 }$$
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 $R$.
\item Find, in terms of $k$, the $y$ coordinate of the point $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q8 [11]}}