# Question 11:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\sin t = 3 \Rightarrow \sin t = 0.5 \Rightarrow t = \frac{\pi}{6}$ | B1 | Accept awrt 0.5236 (4dp); answers in degrees: $30°$ is B0 |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6\cos t}{-20\sin 2t}$ | M1, A1 | Uses $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ and differentiates to obtain gradient function of form $\frac{A\cos t}{B\sin 2t}$; $\frac{dy}{dx} = \frac{6\cos t}{-20\sin 2t} = \left(-\frac{3\cos t}{10\sin 2t}\right)$ |
| Sub $t=\frac{\pi}{6}$: $\frac{dy}{dx} = \frac{6\cos(\pi/6)}{-20\sin(\pi/3)} = -\frac{3}{10}$ | M1, A1 | Substitutes their $t$ from part (a); achieves correct numerical answer $-\frac{3}{10}$ (must be attributed to correct derivative) |
| Uses normal gradient with $(5,3)$: $\frac{y-3}{x-5} = \frac{10}{3}$ | M1 | Must use $(5,3)$ and their numerical value of $-\frac{dx}{dy}$ |
| $\Rightarrow 3y = 10x - 41$ | A1* | cso; must see intermediate line between $\frac{y-3}{x-5}=\frac{10}{3}$ and $3y=10x-41$ |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $x=10\cos 2t$, $y=6\sin t$ into $3y=10x-41$: $\Rightarrow 18\sin t = 100\cos 2t - 41$ | M1 | Attempt to substitute both parametric equations into $3y=10x-41$ forming trig equation in just $t$ |
| $\Rightarrow 18\sin t = 100(1-2\sin^2 t)-41$ using identity $\cos 2t = 1-2\sin^2 t$ | M1 | Uses identity $\cos 2t = 1-2\sin^2 t$; if other forms used, must see further step using $\cos^2 t = 1-\sin^2 t$ |
| $\Rightarrow 200\sin^2 t + 18\sin t - 59 = 0 \Rightarrow (2\sin t-1)(100\sin t+59)=0$ | M1, A1 | Correct 3TQ$=0$ in $\sin t$: $200\sin^2 t + 18\sin t - 59 = 0$ |
| $\Rightarrow \sin t = -\frac{59}{100}$ $(\Rightarrow t = -0.63106\ldots)$ | M1, A1 | Correct method solving 3TQ; award for either $\sin t = -\frac{59}{100}$ or $t=-0.63\ldots$ |
| Using either their $t$ or $\sin t$ to find either coordinate of $B$; $B$ has coordinates $(3.038, -3.54)$ | M1 | Accept as evidence sight of $10\cos 2\times t'$ or $6\sin t'$ or one correct answer (awrt 2dp) |
| Both coordinates correct and exact: $\left(\frac{1519}{500}, -\frac{177}{50}\right)$ | A1, A1 | One coordinate both correct and exact (A1); both coordinates correct and exact, cso and cao (A1) |

## Question 11 (Alternative solution using Cartesian equation):

### Part 11(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 10(1-2\sin^2 t) \Rightarrow x = 10 - \frac{5}{9}y^2$ | M1A1 | Uses double angle formula $\cos 2t = 1-2\sin^2 t$ to get equation of C in form $x = f(y^2)$. A1: correct equation $x = 10-\frac{5}{9}y^2$ or equivalent $x=10\left(1-2\left(\frac{y}{6}\right)^2\right)$ |
| $\frac{dx}{dy}\bigg_{y=3} = -\frac{10y}{9} = -\frac{10\times3}{9} = -\frac{10}{3}$ | M1A1 | M1: differentiates wrt $y$ and subs $y=3$ for numerical value. A1: $\frac{dx}{dy}=-\frac{10}{3}$ |
| $\frac{y-3}{x-5} = \frac{10}{3} \Rightarrow 3y = 10x - 41$ | M1A1 | M1: correct method for equation of normal using $(5,3)$ and $-\frac{dx}{dy}$. A1*: cso, proof must be convincing |

### Part 11(c) Alt 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $x = 10-\frac{5}{9}y^2$ into $3y=10x-41 \Rightarrow 3y = 10\left(10-\frac{5}{9}y^2\right)-41$ | M1 | Substitutes their $x=10-\frac{5}{9}y^2$ into $3y=10x-41$ to produce equation in $y$ |
| $\Rightarrow 50y^2 + 27y - 531 = 0$ | M1A1 | M1: forms quadratic in $y$. A1: correct quadratic |
| $\Rightarrow (y-3)(50y+177) = 0$ | M1A1 | M1: correct attempt at solving. A1: correct factors |
| $\Rightarrow y = \ldots$ then substitutes into $x = 10-\frac{5}{9}y^2$ | M1 | Substitutes $y$ value into $x=f(y^2)$ |
| $\Rightarrow x = 3.038,\ y = -3.54$ | A1A1 | A1: one correct value. A1: both correct. Accept $x=\frac{1519}{500},\ y=-\frac{177}{50}$ |

### Part 11(c) Alt 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $y = 6\sqrt{\frac{10-x}{20}}$ into $3y=10x-41 \Rightarrow 36\sqrt{\frac{10-x}{20}} = 10x-41$ | M1 | Substitutes their $y=6\sqrt{\frac{10-x}{20}}$ into $3y=10x-41$ to produce equation in $x$ |
| $\Rightarrow 36^2\left(\frac{10-x}{20}\right) = (10x-41)^2$ | | |
| $\Rightarrow 500x^2 - 4019x + 7595 = 0$ | M1A1 | M1: forms quadratic in $x$. A1: correct quadratic |
| $\Rightarrow (500x-1519)(x-5) = 0$ | M1A1 | M1: correct attempt at solving. A1: correct factors |
| Substitutes $x=\ldots$ into $y = 6\sqrt{\frac{10-x}{20}}$ | M1 | |
| $\Rightarrow x = 3.038,\ y = -3.54$ | A1A1 | A1: one correct. A1: both correct |

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