| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Circle sector area and angle |
| Difficulty | Challenging +1.2 This is a multi-part question requiring reflection of curves, finding intersections, and using tangency conditions without calculus. While it involves several steps and the constraint of avoiding differentiation adds mild complexity, the techniques are standard for Further Pure 1: reflecting y=x swaps x and y, solving simultaneous equations, and using discriminant conditions. The geometric insight about gradient -1 is straightforward given the symmetry. Slightly above average difficulty due to the multiple parts and algebraic manipulation required, but well within typical FP1 scope. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Elimination to give \(x = \frac{1}{8}x^2\) | M1 | OE |
| \(A\) is \((8, 8)\) | A1 | NMS 2/2 |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation of \(Q\) is \(x = \frac{1}{8}y^2\) | B1 | OE; condone \(y = \sqrt{8x}\) |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Points of contact are images in \(y = x\) | E1 | |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Eliminating \(y\) to give \(-x + c = \frac{1}{8}x^2\) | M1 | |
| (ie \(x^2 + 8x - 8c = 0\)) | ||
| Distinct roots if \(\Delta > 0\) | E1 | stated or implied |
| \(\Delta = 64 + 32c\), so \(c > -2\) | A1 | convincingly shown (AG) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For tangent \(c = -2\), so \(x^2 + 8x + 16 = 0\) | M1 | OE |
| ... and \(x = -4\), \(y = 2\) | A1 | |
| Reflection in \(y = x\) | M1 | or other complete method |
| \(x = 2\), \(y = -4\) | A1F | ft wrong answer for first point; allow NMS 2/2 |
| Total | 4 |
## Question 9:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Elimination to give $x = \frac{1}{8}x^2$ | M1 | OE |
| $A$ is $(8, 8)$ | A1 | NMS 2/2 |
| **Total** | **2** | |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation of $Q$ is $x = \frac{1}{8}y^2$ | B1 | OE; condone $y = \sqrt{8x}$ |
| **Total** | **1** | |
### Part (a)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Points of contact are images in $y = x$ | E1 | |
| **Total** | **1** | |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Eliminating $y$ to give $-x + c = \frac{1}{8}x^2$ | M1 | |
| (ie $x^2 + 8x - 8c = 0$) | | |
| Distinct roots if $\Delta > 0$ | E1 | stated or implied |
| $\Delta = 64 + 32c$, so $c > -2$ | A1 | convincingly shown (AG) |
| **Total** | **3** | |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For tangent $c = -2$, so $x^2 + 8x + 16 = 0$ | M1 | OE |
| ... and $x = -4$, $y = 2$ | A1 | |
| Reflection in $y = x$ | M1 | or other complete method |
| $x = 2$, $y = -4$ | A1F | ft wrong answer for first point; allow NMS 2/2 |
| **Total** | **4** | |9 The diagram shows a parabola $P$ which has equation $y = \frac { 1 } { 8 } x ^ { 2 }$, and another parabola $Q$ which is the image of $P$ under a reflection in the line $y = x$.
The parabolas $P$ and $Q$ intersect at the origin and again at a point $A$.\\
The line $L$ is a tangent to both $P$ and $Q$.\\
\includegraphics[max width=\textwidth, alt={}, center]{7441c4e6-5448-483b-b100-f8076e7e6cd8-5_1015_1089_623_479}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the point $A$.
\item Write down an equation for $Q$.
\item Give a reason why the gradient of $L$ must be - 1 .
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that the line $y = - x + c$ intersects the parabola $P$ at two distinct points, show that
$$c > - 2$$
\item Find the coordinates of the points at which the line $L$ touches the parabolas $P$ and $Q$.\\
(No credit will be given for solutions based on differentiation.)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2011 Q9 [11]}}