| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola focus and directrix properties |
| Difficulty | Standard +0.3 This is a straightforward application of standard parabola properties (focus, directrix, focal distance) with routine coordinate geometry. Parts (a)-(b) are direct recall of formulas for y²=4ax, while (c)-(e) apply the defining property that distance to focus equals distance to directrix, followed by basic area calculation. Slightly above average difficulty due to being Further Maths content, but mechanically straightforward with no novel problem-solving required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| \(C: y^2 = 36x \Rightarrow a = \frac{36}{4} = 9\); focus \(S(9, 0)\) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + 9 = 0\) or \(x = -9\) | B1\(\sqrt{}\) | \(x + 9 = 0\) or \(x = -9\); or ft using their \(a\) from part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| \(PS = 25 \Rightarrow QP = 25\) | B1 | Either 25 by itself or \(PQ = 25\). Do not award if just \(PS = 25\) is seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\)-coordinate of \(P \Rightarrow x = 25 - 9 = 16\) | B1\(\sqrt{}\) | \(x = 16\) |
| \(y^2 = 36(16)\) | M1 | Substitutes their \(x\)-coordinate into equation of \(C\) |
| \(y = \sqrt{576} = 24\) | A1 | \(y = 24\) |
| Therefore \(P(16, 24)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(OSPQ = \frac{1}{2}(9 + 25)(24)\) | M1 | \(\frac{1}{2}(\text{their } a + 25)(\text{their } y)\) or rectangle and 2 distinct triangles, correct for their values |
| \(= 408 \text{ (units)}^2\) | A1 | 408 |
## Question 6:
### Part (a):
| $C: y^2 = 36x \Rightarrow a = \frac{36}{4} = 9$; focus $S(9, 0)$ | B1 | (1 mark) |
|---|---|---|
### Part (b):
| $x + 9 = 0$ or $x = -9$ | B1$\sqrt{}$ | $x + 9 = 0$ or $x = -9$; or ft using their $a$ from part (a) |
|---|---|---|
(1 mark)
### Part (c):
| $PS = 25 \Rightarrow QP = 25$ | B1 | Either 25 by itself or $PQ = 25$. Do not award if just $PS = 25$ is seen. |
|---|---|---|
(1 mark)
### Part (d):
| $x$-coordinate of $P \Rightarrow x = 25 - 9 = 16$ | B1$\sqrt{}$ | $x = 16$ |
|---|---|---|
| $y^2 = 36(16)$ | M1 | Substitutes their $x$-coordinate into equation of $C$ |
| $y = \sqrt{576} = 24$ | A1 | $y = 24$ |
| Therefore $P(16, 24)$ | | |
(3 marks)
### Part (e):
| Area $OSPQ = \frac{1}{2}(9 + 25)(24)$ | M1 | $\frac{1}{2}(\text{their } a + 25)(\text{their } y)$ or rectangle and 2 distinct triangles, correct for their values |
|---|---|---|
| $= 408 \text{ (units)}^2$ | A1 | 408 |
(2 marks)
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d20fa710-2d91-4ac2-adbc-46ccdcb93380-07_789_791_228_566}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the parabola $C$ with equation $y ^ { 2 } = 36 x$. The point $S$ is the focus of $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $S$.
\item Write down the equation of the directrix of $C$.
Figure 1 shows the point $P$ which lies on $C$, where $y > 0$, and the point $Q$ which lies on the directrix of $C$. The line segment $Q P$ is parallel to the $x$-axis.
Given that the distance $P S$ is 25 ,
\item write down the distance $Q P$,
\item find the coordinates of $P$,
\item find the area of the trapezium $O S P Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q6 [8]}}