| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.2 This is a structured polar curves question with multiple parts guiding students through exploration of a family of curves. Parts (i)-(iv) involve standard polar curve sketching with calculator assistance and pattern recognition, while part (v) requires algebraic manipulation to convert to Cartesian form—all routine techniques for Further Maths students. The question is methodical rather than requiring novel insight, making it slightly above average difficulty due to the Further Maths content and multi-part nature, but well within expected scope. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks |
|---|---|
| Undefined for \(\theta = \frac{\pi}{2}\) and \(\frac{3\pi}{2}\) | B1 B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| [Graph showing vertical line through \((1, 0)\) with scale indicated] | B1 M1 A1 [3] | Vertical line through \((1, 0)\) (indicated, e.g. by scale); Use of \(x = r\cos \theta\) |
| \(r = \sec \theta \Rightarrow r\cos \theta = 1 \Rightarrow x = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 1\): [Graph showing section through \((2, 0)\) with asymptote] | B1 B2 | Section through \((2, 0)\) (indicated); Section through \((0, 0)\) (give B1 for one error); If asymptote included max. 2/3 |
| \(a = -1\) gives same curve; \(a = 1, 0 < \theta < \pi\) corresponds to \(a = -1, \pi < \theta < 2\pi\); \(a = -1, 0 < \theta < \pi\) corresponds to \(a = 1, \pi < \theta < 2\pi\) | B1 B1 B1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Loop e.g. \(a = 2\) [Graph showing loop] | B1 [3] | Give B1 for one error |
| B2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \sec \theta + a \Rightarrow r = \frac{1}{x} + a\) | M1 | Use of \(x = r\cos \theta\) |
| \(\Rightarrow r\left(1 - \frac{1}{x}\right) = a\) | M1 | Use of \(r = \sqrt{x^2 + y^2}\) |
| \(\Rightarrow \sqrt{x^2 + y^2}\left(\frac{x-1}{x}\right) = a\) | M1 | Correct manipulation |
| \(\Rightarrow \sqrt{x^2 + y^2}(x - 1) = ax\) | A1(ag) [4] | |
| \(\Rightarrow (x^2 + y^2)(x-1)^2 = a^2 x^2\) |
### (i)
Undefined for $\theta = \frac{\pi}{2}$ and $\frac{3\pi}{2}$ | B1 B1 [2] |
### (ii)
[Graph showing vertical line through $(1, 0)$ with scale indicated] | B1 M1 A1 [3] | Vertical line through $(1, 0)$ (indicated, e.g. by scale); Use of $x = r\cos \theta$
$r = \sec \theta \Rightarrow r\cos \theta = 1 \Rightarrow x = 1$ | |
### (iii)
$a = 1$: [Graph showing section through $(2, 0)$ with asymptote] | B1 B2 | Section through $(2, 0)$ (indicated); Section through $(0, 0)$ (give B1 for one error); If asymptote included max. 2/3
$a = -1$ gives same curve; $a = 1, 0 < \theta < \pi$ corresponds to $a = -1, \pi < \theta < 2\pi$; $a = -1, 0 < \theta < \pi$ corresponds to $a = 1, \pi < \theta < 2\pi$ | B1 B1 B1 [6] |
### (iv)
Loop e.g. $a = 2$ [Graph showing loop] | B1 [3] | Give B1 for one error
| B2 |
### (v)
$r = \sec \theta + a \Rightarrow r = \frac{1}{x} + a$ | M1 | Use of $x = r\cos \theta$
$\Rightarrow r\left(1 - \frac{1}{x}\right) = a$ | M1 | Use of $r = \sqrt{x^2 + y^2}$
$\Rightarrow \sqrt{x^2 + y^2}\left(\frac{x-1}{x}\right) = a$ | M1 | Correct manipulation
$\Rightarrow \sqrt{x^2 + y^2}(x - 1) = ax$ | A1(ag) [4] |
$\Rightarrow (x^2 + y^2)(x-1)^2 = a^2 x^2$ | |
---5 This question concerns curves with polar equation $r = \sec \theta + a$, where $a$ is a constant.\\
(i) State the set of values of $\theta$ between 0 and $2 \pi$ for which $r$ is undefined.
For the rest of the question you should assume that $\theta$ takes all values between 0 and $2 \pi$ for which $r$ is defined.\\
(ii) Use your graphical calculator to obtain a sketch of the curve in the case $a = 0$. Confirm the shape of the curve by writing the equation in cartesian form.\\
(iii) Sketch the curve in the case $a = 1$.
Now consider the curve in the case $a = - 1$. What do you notice?\\
By considering both curves for $0 < \theta < \pi$ and $\pi < \theta < 2 \pi$ separately, describe the relationship between the cases $a = 1$ and $a = - 1$.\\
(iv) What feature does the curve exhibit for values of $a$ greater than 1 ?
Sketch a typical case.\\
(v) Show that a cartesian equation of the curve $r = \sec \theta + a$ is $\left( x ^ { 2 } + y ^ { 2 } \right) ( x - 1 ) ^ { 2 } = a ^ { 2 } x ^ { 2 }$.
\hfill \mbox{\textit{OCR MEI FP2 2012 Q5 [18]}}