| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.8 This is a substantial Further Maths polar curves question requiring calculator investigation, geometric insight about limaçon behavior for different parameter values, and analytical work to find crossing angles and tangent equations. While the individual techniques (sketching, solving r=0, finding tangents) are standard FP2 material, the multi-part investigation of curve features across parameter ranges and connecting the analytical result back to the geometric observation requires sustained reasoning and synthesis across several concepts. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Symmetry in horizontal axis | G1 | |
| \((3, 0)\) to \((0, 0)\) | G1 | |
| \((0, 0)\) to \((0, 1)\) | G1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a > 0.5\) | B1 | |
| \(a < -0.5\) | B1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Circle: \(r\) is constant | B1 | Shape and reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The two loops get closer together | B1 | |
| The shape becomes more nearly circular | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Cusp | B1 | |
| \(a = -0.5\) | B1 | |
| Total (ii) overall | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 + 2a\cos\theta = 0 \Rightarrow \cos\theta = -\dfrac{1}{2a}\) | B1 | Equation |
| If \(a > 0.5\), \(-1 < -\dfrac{1}{2a} < 0\) and there are two values of \(\theta\) in \([0, 2\pi]\): \(\pi - \arccos\!\left(\dfrac{1}{2a}\right)\) and \(\pi + \arccos\!\left(\dfrac{1}{2a}\right)\) | M1 | |
| These differ by \(2\arccos\!\left(\dfrac{1}{2a}\right)\) | A1 (ag) | www |
| \(\arccos\!\left(\dfrac{1}{2a}\right) = \arctan\sqrt{4a^2-1}\) | M1 | Relating arccos to arctan by triangle or \(\tan^2\theta = \sec^2\theta - 1\) |
| Tangents are \(y = x\sqrt{4a^2-1}\) | A1 | |
| and \(y = -x\sqrt{4a^2-1}\) | A1ft | Negative of above |
| \(\sqrt{4a^2-1}\) is real for \(a > 0.5\) if \(a > 0\) | E1 | |
| Total | 8 |
# Question 5:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Symmetry in horizontal axis | G1 | |
| $(3, 0)$ to $(0, 0)$ | G1 | |
| $(0, 0)$ to $(0, 1)$ | G1 | |
| **Total** | **3** | |
## Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a > 0.5$ | B1 | |
| $a < -0.5$ | B1 | |
| **Total** | **2** | |
## Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| Circle: $r$ is constant | B1 | Shape and reason |
## Part (ii)(C)
| Answer | Mark | Guidance |
|--------|------|----------|
| The two loops get closer together | B1 | |
| The shape becomes more nearly circular | B1 | |
## Part (ii)(D)
| Answer | Mark | Guidance |
|--------|------|----------|
| Cusp | B1 | |
| $a = -0.5$ | B1 | |
| **Total (ii) overall** | **7** | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 + 2a\cos\theta = 0 \Rightarrow \cos\theta = -\dfrac{1}{2a}$ | B1 | Equation |
| If $a > 0.5$, $-1 < -\dfrac{1}{2a} < 0$ and there are two values of $\theta$ in $[0, 2\pi]$: $\pi - \arccos\!\left(\dfrac{1}{2a}\right)$ and $\pi + \arccos\!\left(\dfrac{1}{2a}\right)$ | M1 | |
| These differ by $2\arccos\!\left(\dfrac{1}{2a}\right)$ | A1 (ag) | www |
| $\arccos\!\left(\dfrac{1}{2a}\right) = \arctan\sqrt{4a^2-1}$ | M1 | Relating arccos to arctan by triangle or $\tan^2\theta = \sec^2\theta - 1$ |
| Tangents are $y = x\sqrt{4a^2-1}$ | A1 | |
| and $y = -x\sqrt{4a^2-1}$ | A1ft | Negative of above |
| $\sqrt{4a^2-1}$ is real for $a > 0.5$ if $a > 0$ | E1 | |
| **Total** | **8** | |5 The limaçon of Pascal has polar equation $r = 1 + 2 a \cos \theta$, where $a$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Use your calculator to sketch the curve when $a = 1$. (You need not distinguish between parts of the curve where $r$ is positive and negative.)
\item By using your calculator to investigate the shape of the curve for different values of $a$, positive and negative,\\
(A) state the set of values of $a$ for which the curve has a loop within a loop,\\
(B) state, with a reason, the shape of the curve when $a = 0$,\\
(C) state what happens to the shape of the curve as $a \rightarrow \pm \infty$,\\
(D) name the feature of the curve that is evident when $a = 0.5$, and find another value of $a$ for which the curve has this feature.
\item Given that $a > 0$ and that $a$ is such that the curve has a loop within a loop, write down an equation for the values of $\theta$ at which $r = 0$. Hence show that the angle at which the curve crosses itself is $2 \arccos \left( \frac { 1 } { 2 a } \right)$.
Obtain the cartesian equations of the tangents at the point where the curve crosses itself. Explain briefly how these equations relate to the answer to part (ii)(A).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2009 Q5 [18]}}