| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Show polar curve has Cartesian form |
| Difficulty | Standard +0.8 This is a multi-part Further Maths polar coordinates question requiring: verification by substitution (routine), solving a trigonometric equation for intersection points, calculating a triangle area using polar coordinates, and converting to Cartesian form. The Cartesian conversion requires algebraic manipulation with r² = x² + y² and x = r cos θ, involving squaring and rearranging to reach the specified form. While systematic, it demands fluency across multiple techniques and careful algebra, placing it moderately above average difficulty. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| When \(\theta = \pi\), \(r = \frac{2}{3 + 2\cos\pi} = \frac{2}{3+2(-1)} = 2\) | B1 | Total: 1; Correct verification |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{2}{3+2\cos\theta} = 1 \Rightarrow \cos\theta = -\frac{1}{2}\) | M1 | Equates \(r\)'s and attempts to solve |
| Points of intersection \(\left(1,\frac{2\pi}{3}\right),\ \left(1,\frac{4\pi}{3}\right)\) | A2,1 | Total: 3; Condone \(-2\pi/3\) for \(4\pi/3\); A1 if either one point correct or two correct solutions of \(\cos\theta = -0.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Area \(OMN = \frac{1}{2}\times 1\times 1\times \sin( | \theta_M - \theta_N | )\) |
| \(= \frac{1}{2}\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4}\) | A1 | Perp. from \(L\) to \(MN\) \(= 2 - 1\cos\frac{\pi}{3} = \frac{3}{2}\) M1A1 |
| Area \(OMLN = 2\times\frac{1}{2}\times 1\times 2\times\sin\frac{\pi}{3}\) | M1 | Area \(LMN = \frac{1}{2}\times\sqrt{3}\times\frac{3}{2} = \frac{3\sqrt{3}}{4}\) A1 |
| Area \(LMN = \sqrt{3} - \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3r + 2r\cos\theta = 2\) | M1 | |
| \(3r + 2x = 2\) | B1 | \(r\cos\theta = x\) stated or used |
| \(3r = 2 - 2x\) | A1 | \(3r = \pm(2-2x)\) |
| \(9(x^2 + y^2) = (2-2x)^2\) | M1 | \(r^2 = x^2 + y^2\) used |
| \(9y^2 = (2-2x)^2 - 9x^2\) | A1 | Total: 5; CSO; ACF for \(f(x)\) e.g. \(9y^2 = -5x^2 - 8x + 4\) |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|---------|
| When $\theta = \pi$, $r = \frac{2}{3 + 2\cos\pi} = \frac{2}{3+2(-1)} = 2$ | B1 | **Total: 1**; Correct verification |
### Part (b)(i):
| Working | Marks | Guidance |
|---------|-------|---------|
| $\frac{2}{3+2\cos\theta} = 1 \Rightarrow \cos\theta = -\frac{1}{2}$ | M1 | Equates $r$'s and attempts to solve |
| Points of intersection $\left(1,\frac{2\pi}{3}\right),\ \left(1,\frac{4\pi}{3}\right)$ | A2,1 | **Total: 3**; Condone $-2\pi/3$ for $4\pi/3$; A1 if either one point correct or two correct solutions of $\cos\theta = -0.5$ |
### Part (b)(ii):
| Working | Marks | Guidance |
|---------|-------|---------|
| Area $OMN = \frac{1}{2}\times 1\times 1\times \sin(|\theta_M - \theta_N|)$ | M1 | **ALT:** $MN = 2\times 1\times\sin\frac{\pi}{3}$ M1 |
| $= \frac{1}{2}\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4}$ | A1 | Perp. from $L$ to $MN$ $= 2 - 1\cos\frac{\pi}{3} = \frac{3}{2}$ M1A1 |
| Area $OMLN = 2\times\frac{1}{2}\times 1\times 2\times\sin\frac{\pi}{3}$ | M1 | Area $LMN = \frac{1}{2}\times\sqrt{3}\times\frac{3}{2} = \frac{3\sqrt{3}}{4}$ A1 |
| Area $LMN = \sqrt{3} - \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$ | A1 | **Total: 4** |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|---------|
| $3r + 2r\cos\theta = 2$ | M1 | |
| $3r + 2x = 2$ | B1 | $r\cos\theta = x$ stated or used |
| $3r = 2 - 2x$ | A1 | $3r = \pm(2-2x)$ |
| $9(x^2 + y^2) = (2-2x)^2$ | M1 | $r^2 = x^2 + y^2$ used |
| $9y^2 = (2-2x)^2 - 9x^2$ | A1 | **Total: 5**; CSO; ACF for $f(x)$ e.g. $9y^2 = -5x^2 - 8x + 4$ |
---5 The diagram shows a sketch of a curve $C$, the pole $O$ and the initial line.\\
\includegraphics[max width=\textwidth, alt={}, center]{f4fdffc7-5647-4462-a983-1564d4e76a4d-3_301_668_1644_689}
The curve $C$ has polar equation
$$r = \frac { 2 } { 3 + 2 \cos \theta } , \quad 0 \leqslant \theta \leqslant 2 \pi$$
\begin{enumerate}[label=(\alph*)]
\item Verify that the point $L$ with polar coordinates ( $2 , \pi$ ) lies on $C$.
\item The circle with polar equation $r = 1$ intersects $C$ at the points $M$ and $N$.
\begin{enumerate}[label=(\roman*)]
\item Find the polar coordinates of $M$ and $N$.
\item Find the area of triangle $L M N$.
\end{enumerate}\item Find a cartesian equation of $C$, giving your answer in the form $9 y ^ { 2 } = \mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q5 [13]}}