## Question 8:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2=a^2(1-e^2)\Rightarrow 4=9(1-e^2)\Rightarrow e=\ldots$ or $e=\sqrt{1-\frac{b^2}{a^2}}\Rightarrow e=\ldots$ | M1 | Uses a correct formula with $a$ and $b$ correctly placed to find a value for $e$ |
| $e=\frac{\sqrt{5}}{3}$ | A1 | Correct value (or equivalent). $e=\pm\frac{\sqrt{5}}{3}$ scores A0 |

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### Part (b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\pm ae,0)=(\pm\sqrt{5},0)$ or $\left(\pm3\cdot\frac{\sqrt{5}}{3},0\right)$ | B1ft | Correct foci. Must be coordinates but allow unsimplified and isw if necessary. Follow through their $e$ so allow for $(\pm3\times\text{their }e,0)$ |

### Part (b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\pm\frac{a}{e}=\pm\frac{9}{\sqrt{5}}$ or $x=\pm\frac{3}{\frac{\sqrt{5}}{3}}$ | B1ft | Correct directrices. Must be equations but allow unsimplified and isw if necessary. Follow through their $e$ so allow for $x=\pm3/\text{their }e$ |

**Special case:** Use of $a^2$ for $a$ and $b^2$ for $b$ consistently scores M0A0 in (a) and B1ft B1ft in (b). This gives $e=\frac{\sqrt{65}}{9}$, $(\pm\sqrt{65},0)$, $x=\pm\frac{81}{\sqrt{65}}$

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### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\mathrm{d}x}{\mathrm{d}\theta}=-3\sin\theta$, $\frac{\mathrm{d}y}{\mathrm{d}\theta}=2\cos\theta$ or $\frac{2x}{9}+\frac{2y}{4}\frac{\mathrm{d}y}{\mathrm{d}x}=0$ or $y=\left(4-\frac{4x^2}{9}\right)^{\frac{1}{2}}\Rightarrow\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{4x}{9}\left(4-\frac{4x^2}{9}\right)^{-\frac{1}{2}}\Rightarrow\frac{\mathrm{d}y}{\mathrm{d}x}=\ldots\left(=\frac{2\cos\theta}{-3\sin\theta}\right)$ | M1 | Correct strategy for the gradient of $l$ in terms of $\theta$. Allow $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\cos\theta}{-3\sin\theta}$ to be stated |
| $y-2\sin\theta=\frac{2\cos\theta}{-3\sin\theta}(x-3\cos\theta)$ | M1 | Correct straight line method (any complete method). Finding the equation of the normal is M0 |
| $-3y\sin\theta+6\sin^2\theta=2x\cos\theta-6\cos^2\theta \Rightarrow 2x\cos\theta+3y\sin\theta=6^*$ | A1* | Cso with at least one intermediate line of working |

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### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $l_2: y=\frac{3\sin\theta}{2\cos\theta}x$ | B1 | Correct equation for $l_2$ |
| $2x\cos\theta+3y\sin\theta=6$, $y=\frac{3\sin\theta}{2\cos\theta}x \Rightarrow x=\ldots, y=\ldots$ | M1 | Complete method for $Q$ |
| $Q:\left(\frac{12\cos\theta}{4\cos^2\theta+9\sin^2\theta},\frac{18\sin\theta}{4\cos^2\theta+9\sin^2\theta}\right)$ Correct coordinates. Allow as $x=\ldots, y=\ldots$ and allow equivalent correct expressions as long as they are single fractions e.g. $x=\frac{12\cos\theta}{4+5\sin^2\theta}$, $y=\frac{18\sin\theta}{4+5\sin^2\theta}$, $x=\frac{12\cos\theta}{9-5\cos^2\theta}$, $y=\frac{18\sin\theta}{9-5\cos^2\theta}$ | A1 | |

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### Part (e):

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $Q$, $\frac{y}{x}=\frac{3}{2}\tan\theta$ | M1 | Uses their coordinates of $Q$ to attempt an equation connecting $x$, $y$ and $\theta$ or states or uses the equation found in (d) |
| $x=\frac{12\cos\theta}{4\cos^2\theta+9\sin^2\theta}=\frac{12\sec\theta}{4+9\tan^2\theta}\Rightarrow x^2=\frac{144\sec^2\theta}{(4+9\tan^2\theta)^2}=\frac{144\left(1+\frac{4y^2}{9x^2}\right)}{\left(4+9\times\frac{4y^2}{9x^2}\right)^2}$ or $y=\frac{18\sin\theta}{4\cos^2\theta+9\sin^2\theta}=\frac{12\sec\theta\tan\theta}{4+9\tan^2\theta}\Rightarrow y^2=\frac{324\sec^2\theta\tan^2\theta}{(4+9\tan^2\theta)^2}=\frac{324\left(1+\frac{4y^2}{9x^2}\right)\frac{4y^2}{9x^2}}{\left(4+9\times\frac{4y^2}{9x^2}\right)^2}$ | dM1 | Eliminates $\theta$. Depends on the first mark |
| $\Rightarrow x^2=\frac{x^2(9x^2+4y^2)}{(x^2+y^2)^2}\Rightarrow(x^2+y^2)^2=9x^2+4y^2$ or $\Rightarrow 9\times16x^2y^2\left(1+\frac{y^2}{x^2}\right)^2=4\times18^2\left(1+\frac{4y^2}{9x^2}\right)\Rightarrow(x^2+y^2)^2=9x^2+4y^2$ | A1 | Correct equation or correct values for $\alpha$ and $\beta$ |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{12\cos\theta}{4+5\sin^2\theta}$, $y=\frac{18\sin\theta}{4+5\sin^2\theta}\Rightarrow(x^2+y^2)^2=\left(\frac{144\cos^2\theta+324\sin^2\theta}{(4+5\sin^2\theta)^2}\right)^2$ | M1 | Uses their $Q$ to obtain an expression for $(x^2+y^2)^2$ in terms of $\theta$ |
| $\left(\frac{144\cos^2\theta+324\sin^2\theta}{(4+5\sin^2\theta)^2}\right)^2=\left(\frac{144+180\sin^2\theta}{(4+5\sin^2\theta)^2}\right)^2=\left(\frac{36(4+5\sin^2\theta)}{(4+5\sin^2\theta)^2}\right)^2=\frac{1296}{(4+5\sin^2\theta)^2}$; $\frac{1296}{(4+5\sin^2\theta)^2}=\alpha x^2+\beta y^2=\alpha\frac{144\cos^2\theta}{(4+5\sin^2\theta)^2}+\beta\frac{324\sin^2\theta}{(4+5\sin^2\theta)^2}\Rightarrow\alpha=\ldots,\beta=\ldots$ | dM1 | Substitutes into the given answer and solves for $\alpha$ and $\beta$. Depends on the first mark |
| $(x^2+y^2)^2=9x^2+4y^2$ | A1 | Correct expression or correct values for $\alpha$ and $\beta$ |