| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.2 This is a structured multi-part polar curves question with significant scaffolding. Part (i) involves routine algebraic manipulation to convert between Cartesian and polar forms. Parts (ii)-(iv) require sketching polar curves for various parameter values, which is a standard FP2 skill. While the question involves multiple parts and requires understanding of polar coordinate behavior, each step is guided and uses familiar techniques. The conceptual demand is moderate—understanding when r² can be negative and how curves behave as parameters change—but this is typical FP2 content rather than requiring novel insight. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(((x+1)^2+y^2)((x-1)^2+y^2)=1\) | ||
| \(\Rightarrow (r^2+2r\cos\theta+1)(r^2-2r\cos\theta+1)=1\) | M1 | Using \(r^2 = x^2+y^2\) and \(x = r\cos\theta\) |
| \(\Rightarrow (r^2+1)^2 - 4r^2\cos^2\theta = 1 \Rightarrow r^4 + 2r^2 = 4r^2\cos^2\theta\) | E1 | Correctly obtained |
| \(\Rightarrow r^4 = 2r^2(2\cos^2\theta - 1)\) | M1 | Using \(\cos 2\theta = 2\cos^2\theta - 1\) |
| \(\Rightarrow r^2 = 2\cos 2\theta\) | E1 | Correctly obtained |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r^2 \geq 0 \Rightarrow \cos 2\theta \geq 0\) | M1 | Considering \(\cos 2\theta \geq 0\) |
| \(\Rightarrow 0 \leq \theta \leq \dfrac{\pi}{4},\quad \dfrac{3\pi}{4} \leq \theta \leq \dfrac{5\pi}{4},\quad \dfrac{7\pi}{4} \leq \theta \leq 2\pi\) | A1 | Or \(-\dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{4}\), \(\dfrac{3\pi}{4}\leq\theta\leq\dfrac{5\pi}{4}\). Inequalities must be non-strict |
| Correct complete curve (lemniscate figure-of-eight shape) | G2 | Curve must be complete. Award G1 for a curve with one error |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k=1\): correct complete lemniscate curve | G1 | Curve must be complete |
| \(k=2\): correct complete curve (larger, rounder loops) | G2 | Curve must be complete. Award G1 for one error. For G2 curve must appear sufficiently different from \(k=1\) |
| For \(k=1\), gradients at the pole are finite; for \(k=2\), they appear infinite | B1 | |
| [7 total for (iii)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k=4\): correct complete curve | G2 | Curve must be complete. Award G1 for one error |
| Tends to circle as \(k\) tends to infinity | B1 | |
| [included in 7 above] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k=-1\): correct complete figure-of-eight curve | G2 | Curve must be complete. Award G1 for one error |
| As \(k \to -2\), curve retains its figure-of-eight shape but contracts towards the origin | B1 | |
| [3] |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $((x+1)^2+y^2)((x-1)^2+y^2)=1$ | | |
| $\Rightarrow (r^2+2r\cos\theta+1)(r^2-2r\cos\theta+1)=1$ | M1 | Using $r^2 = x^2+y^2$ and $x = r\cos\theta$ |
| $\Rightarrow (r^2+1)^2 - 4r^2\cos^2\theta = 1 \Rightarrow r^4 + 2r^2 = 4r^2\cos^2\theta$ | E1 | Correctly obtained |
| $\Rightarrow r^4 = 2r^2(2\cos^2\theta - 1)$ | M1 | Using $\cos 2\theta = 2\cos^2\theta - 1$ |
| $\Rightarrow r^2 = 2\cos 2\theta$ | E1 | Correctly obtained |
| **[4]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r^2 \geq 0 \Rightarrow \cos 2\theta \geq 0$ | M1 | Considering $\cos 2\theta \geq 0$ |
| $\Rightarrow 0 \leq \theta \leq \dfrac{\pi}{4},\quad \dfrac{3\pi}{4} \leq \theta \leq \dfrac{5\pi}{4},\quad \dfrac{7\pi}{4} \leq \theta \leq 2\pi$ | A1 | Or $-\dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{4}$, $\dfrac{3\pi}{4}\leq\theta\leq\dfrac{5\pi}{4}$. Inequalities must be non-strict |
| Correct complete curve (lemniscate figure-of-eight shape) | G2 | Curve must be complete. Award G1 for a curve with one error |
| **[4]** | | |
### Part (iii)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k=1$: correct complete lemniscate curve | G1 | Curve must be complete |
| $k=2$: correct complete curve (larger, rounder loops) | G2 | Curve must be complete. Award G1 for one error. For G2 curve must appear sufficiently different from $k=1$ |
| For $k=1$, gradients at the pole are finite; for $k=2$, they appear infinite | B1 | |
| **[7 total for (iii)]** | | |
### Part (iii)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k=4$: correct complete curve | G2 | Curve must be complete. Award G1 for one error |
| Tends to circle as $k$ tends to infinity | B1 | |
| **[included in 7 above]** | | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k=-1$: correct complete figure-of-eight curve | G2 | Curve must be complete. Award G1 for one error |
| As $k \to -2$, curve retains its figure-of-eight shape but contracts towards the origin | B1 | |
| **[3]** | | |5 The points $\mathrm { A } ( - 1,0 ) , \mathrm { B } ( 1,0 )$ and $\mathrm { P } ( x , y )$ are such that the product of the distances PA and PB is 1 . You are given that the cartesian equation of the locus of P is
$$\left( ( x + 1 ) ^ { 2 } + y ^ { 2 } \right) \left( ( x - 1 ) ^ { 2 } + y ^ { 2 } \right) = 1 .$$
\begin{enumerate}[label=(\roman*)]
\item Show that this equation may be written in polar form as
$$r ^ { 4 } + 2 r ^ { 2 } = 4 r ^ { 2 } \cos ^ { 2 } \theta$$
Show that the polar equation simplifies to
$$r ^ { 2 } = 2 \cos 2 \theta$$
\item Give a sketch of the curve, stating the values of $\theta$ for which the curve is defined.
\item The equation in part (i) is now to be generalised to
$$r ^ { 2 } = 2 \cos 2 \theta + k$$
where $k$ is a constant.\\
(A) Give sketches of the curve in the cases $k = 1 , k = 2$. Describe how these two curves differ at the pole.\\
(B) Give a sketch of the curve in the case $k = 4$. What happens to the shape of the curve as $k$ tends to infinity?
\item Sketch the curve for the case $k = - 1$.
What happens to the curve as $k \rightarrow - 2$ ?
\section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE.}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2012 Q5 [18]}}