# Question 13:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2\cos t}{-12\sin 2t}$ | M1 | |
| $= \dfrac{2\cos t}{-24\sin t\cos t}$ | dM1 | Use of double angle $\sin 2t = 2\sin t\cos t$ |
| $= \dfrac{2\cos t}{-24\sin t\cos t} = -\dfrac{1}{12}\operatorname{cosec} t$ | M1 A1 | Cancel $\cos t$ to reach $-\frac{1}{12}\operatorname{cosec} t$ |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t=\dfrac{\pi}{3}$: $\dfrac{dy}{dx} = -\dfrac{1}{12\times\frac{\sqrt{3}}{2}} = -\dfrac{\sqrt{3}}{18}$ | M1 A1 | |
| Normal gradient $= -\dfrac{1}{m} = 6\sqrt{3}$ | M1 | |
| When $t=\dfrac{\pi}{3}$: $x=-3$ and $y=\sqrt{3}$ | B1 | |
| Equation of normal: $y - \sqrt{3} = 6\sqrt{3}(x+3)$, so $y = 6\sqrt{3}x + 19\sqrt{3}$ | M1 A1 | |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 6(1-2\sin^2 t) \Rightarrow x = f(y)$ | M1 | Use $\sin^2 t + \cos^2 t = 1$ or double angle |
| $x = 6 - 3y^2$ or $f(y) = 6 - 3y^2$ | dM1 A1 | |

## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2 < y < 2$ or $k = 2$ | B1 | |

# Mark Scheme

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates both $x$ and $y$ wrt $t$ and establishes $\frac{dy/dt}{dx/dt} = \frac{\pm A\cos t}{\pm B\sin 2t}$ | M1 | They may use any double angle formula for cos first. Condone sign slips |
| e.g. $\cos 2t = \pm 2\cos^2 t \pm 1$ to get $\frac{dy/dt}{dx/dt} = \frac{\pm A\cos t}{\pm B\sin t\cos t}$ | | |
| Correct double angle formula used: $\sin 2t = 2\sin t\cos t$ | dM1 | In alternative method the correct double angle formula must have been used |
| Cancels $\cos t$ and replaces $1/\sin t$ by $\cosec t$ correctly achieving a form $\frac{dy}{dx} = \lambda\cosec t$ | M1 | |
| $\frac{dy}{dx} = -\frac{1}{12}\cosec t$ | A1 | cao |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $t = \frac{\pi}{3}$ into their $\frac{dy}{dx} = \lambda\cosec t$ | M1 | |
| $\frac{dy}{dx} = -\frac{1}{12\times\sqrt{3}/2}$ or exact equivalent | A1 | May be implied by normal gradient of $6\sqrt{3}$. Accept decimals: $\frac{dy}{dx} = -0.096$ or implied by normal gradient of $10.4$ |
| Use of negative reciprocal in finding the gradient of the normal | M1 | |
| For $x = -3$, $y = \sqrt{3}$ | B1 | |
| Correct method for line equation using their **normal** gradient and $(-3, \sqrt{3})$ allowing a sign slip on one coordinate | M1 | Look for $y - y_1 = -\frac{dx}{dy}\bigg\|_{t=\pi/3}(x-x_1)$ or $x - x_1 = -\frac{dy}{dx}\bigg\|_{t=\pi/3}(y-y_1)$. If $y=mx+c$ used must proceed to $c=$ |
| $y = 6\sqrt{3}x + 19\sqrt{3}$ | A1 | cao |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to use double angle formula $\cos 2t = \pm 1 \pm 2\sin^2 t$ leading to equation linking $x$ and $y$ | M1 | If $\cos 2t = \cos^2 t - \sin^2 t$ used initially, must attempt to replace $\cos^2 t$ by $1 - \sin^2 t$ |
| Uses correct $\cos 2t = 1 - 2\sin^2 t$ and attempts to replace $\sin t$ by $\frac{y}{2}$ and $\cos 2t$ by $\frac{x}{6}$ | dM1 | Condone poor bracketing e.g. $\cos 2t = 1-2\sin^2 t \Rightarrow \frac{x}{6} = 1 - 2\frac{y^2}{2}$ |
| $x = 6 - 3y^2$ or $f(y) = 6 - 3y^2$ | A1 | Correct equation |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| States $k = 2$ or writes the range of $y$ as $-2 < y < 2$ | B1 | |