| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Show polar curve has Cartesian form |
| Difficulty | Challenging +1.3 This is a multi-part FP2 question involving coordinate geometry, parametric equations, and curve sketching. Parts (i)-(ii) require careful algebraic manipulation but follow standard techniques (finding line-circle intersections). Part (iii) involves asymptotic behavior and sketching, requiring some insight but guided by calculator exploration. Part (iv) requires geometric reasoning about angles. While lengthy and requiring multiple techniques, each step is relatively standard for FP2 level, making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| If \(a = 1\), angle \(OCP = 45°\) so P is \((1 - \cos 45°, \sin 45°)\) | M1 | Completion www |
| \(\Rightarrow P\left(1 - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) | A1 (ag) | |
| OR Circle \((x-1)^2 + y^2 = 1\), line \(y = -x + 1\) | ||
| \((x-1)^2 + (-x+1)^2 = 1\) | M1 | Complete algebraic method to find x |
| \(\Rightarrow x = 1 \pm \frac{1}{\sqrt{2}}\) and hence P | A1 | |
| OR \(Q\left(1 + \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos OCP = \frac{a}{\sqrt{a^2+1}}\) | M1 | Attempt to find cos OCP and sin OCP in terms of a |
| \(\sin OCP = \frac{1}{\sqrt{a^2+1}}\) | A1 | Both correct |
| P is \((a - a\cos OCP, a\sin OCP)\) | ||
| \(\Rightarrow P\left(a - \frac{a^2}{\sqrt{a^2+1}}, \frac{a}{\sqrt{a^2+1}}\right)\) | A1 (ag) | Completion www |
| OR Circle \((x-a)^2 + y^2 = a^2\), line \(y = -\frac{1}{a}x + 1\) | ||
| \(\left(x - a\right)^2 + \left(-\frac{1}{a}x + 1\right)^2 = a^2\) | M1 | Complete algebraic method to find x |
| \(\Rightarrow x = \frac{2a^2 + \sqrt{\left(2a + \frac{2}{a}\right)^2 - 4\left(1 + \frac{1}{a^2}\right)}}{{2\left(1 + \frac{1}{a^2}\right)}}\) | A1 | Unsimplified |
| \(\Rightarrow x = a \pm \frac{a^2}{\sqrt{a^2+1}}\) and hence P | A1 | |
| OR \(Q\left(a + \frac{a^2}{\sqrt{a^2+1}}, -\frac{a}{\sqrt{a^2+1}}\right)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Locus | ||
| Locus of P (1st & 3rd quadrants) through (0, 0) | G1 | |
| Locus of P terminates at (0, 1) | G1 | |
| Locus of P: fully correct shape | G1 | |
| Locus of Q (2nd & 4th quadrants: dotted) reflection of locus of P in y-axis | G1ft | |
| Stated separately | B1 | |
| Stated | B1 | |
| As \(a \to \infty\), \(P \to (0, 1)\) | M1 | Attempt to consider y as \(a \to \infty\) |
| \(\frac{a}{\sqrt{a^2+1}} \to \frac{a}{-a} = -1\) as \(a \to \infty\) | A1 | Completion www |
| Answer | Marks | Guidance |
|---|---|---|
| \(POQ = 90°\) | B1 | o.e. |
| Angle in semicircle | B1 | |
| Loci cross at 90° | B1 | |
| 18 marks for Question 5 |
## Part (i)
If $a = 1$, angle $OCP = 45°$ so P is $(1 - \cos 45°, \sin 45°)$ | M1 | Completion www
$\Rightarrow P\left(1 - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ | A1 (ag) |
**OR** Circle $(x-1)^2 + y^2 = 1$, line $y = -x + 1$ | |
$(x-1)^2 + (-x+1)^2 = 1$ | M1 | Complete algebraic method to find x
$\Rightarrow x = 1 \pm \frac{1}{\sqrt{2}}$ and hence P | A1 |
**OR** $Q\left(1 + \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$ | B1 | | | 3 marks total
## Part (ii)
$\cos OCP = \frac{a}{\sqrt{a^2+1}}$ | M1 | Attempt to find cos OCP and sin OCP in terms of a
$\sin OCP = \frac{1}{\sqrt{a^2+1}}$ | A1 | Both correct
P is $(a - a\cos OCP, a\sin OCP)$ | |
$\Rightarrow P\left(a - \frac{a^2}{\sqrt{a^2+1}}, \frac{a}{\sqrt{a^2+1}}\right)$ | A1 (ag) | Completion www
**OR** Circle $(x-a)^2 + y^2 = a^2$, line $y = -\frac{1}{a}x + 1$ | |
$\left(x - a\right)^2 + \left(-\frac{1}{a}x + 1\right)^2 = a^2$ | M1 | Complete algebraic method to find x
$\Rightarrow x = \frac{2a^2 + \sqrt{\left(2a + \frac{2}{a}\right)^2 - 4\left(1 + \frac{1}{a^2}\right)}}{{2\left(1 + \frac{1}{a^2}\right)}}$ | A1 | Unsimplified
$\Rightarrow x = a \pm \frac{a^2}{\sqrt{a^2+1}}$ and hence P | A1 |
**OR** $Q\left(a + \frac{a^2}{\sqrt{a^2+1}}, -\frac{a}{\sqrt{a^2+1}}\right)$ | B1 | | | 4 marks total
## Part (iii)
**Locus** | |
Locus of P (1st & 3rd quadrants) through (0, 0) | G1 |
Locus of P terminates at (0, 1) | G1 |
Locus of P: fully correct shape | G1 |
Locus of Q (2nd & 4th quadrants: dotted) reflection of locus of P in y-axis | G1ft |
Stated separately | B1 |
Stated | B1 |
As $a \to \infty$, $P \to (0, 1)$ | M1 | Attempt to consider y as $a \to \infty$
$\frac{a}{\sqrt{a^2+1}} \to \frac{a}{-a} = -1$ as $a \to \infty$ | A1 | Completion www | | 8 marks total
## Part (iv)
$POQ = 90°$ | B1 | o.e.
Angle in semicircle | B1 |
Loci cross at 90° | B1 | | | 3 marks total
| | 18 marks for Question 5Fig. 5 shows a circle with centre C $(a, 0)$ and radius $a$. B is the point $(0, 1)$. The line BC intersects the circle at P and Q. P is above the $x$-axis and Q is below.
\includegraphics{figure_5}
\begin{enumerate}[label=(\roman*)]
\item Show that, in the case $a = 1$, P has coordinates $\left(1 - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. Write down the coordinates of Q. [3]
\item Show that, for all positive values of $a$, the coordinates of P are
$$x = a\left(1 - \frac{a}{\sqrt{a^2 + 1}}\right), \quad y = \frac{a}{\sqrt{a^2 + 1}} \quad (*)$$
Write down the coordinates of Q in a similar form. [4]
Now let the variable point P be defined by the parametric equations (*) for all values of the parameter $a$, positive, zero and negative. Let Q be defined for all $a$ by your answer in part (ii).
\item Using your calculator, sketch the locus of P as $a$ varies. State what happens to P as $a \to \infty$ and as $a \to -\infty$.
Show algebraically that this locus has an asymptote at $y = -1$.
On the same axes, sketch, as a dotted line, the locus of Q as $a$ varies. [8]
(The single curve made up of these two loci and including the point B is called a right strophoid.)
\item State, with a reason, the size of the angle POQ in Fig. 5. What does this indicate about the angle at which a right strophoid crosses itself? [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2009 Q5 [18]}}