(i)(a)$O E=2 a \pi$

(b) $2a$

(iv)$k=\sqrt{3}$

(i)(a)

At $E, y=a(1-\cos \theta)=0$, Hence cos $\theta=1$
$\therefore \theta=2 \pi$

Hence $O E=2 a \pi$

(b) When $y$ is a maximum, $\cos \theta=-1$ OR $\frac{\mathrm{d} y}{\mathrm{~d} \theta}=a(\sin \theta)=0$ $\therefore \theta=\pi$ and $y=2 a$

(ii)

$\frac{\mathrm{d} y}{\mathrm{~d} \theta}=a(\sin \theta)=2 a \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and
$\frac{\mathrm{d} x}{\mathrm{~d} \theta}=a(1-\cos \theta)=a\left(1-1+2 \sin ^2 \theta\right)=2 a \sin ^2 \frac{\theta}{2}$

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} \theta}}{\frac{\mathrm{d} x}{\mathrm{~d} \theta}}=\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=\cot \frac{\theta}{2}$

 

(iii)

At $B, \frac{\mathrm{d} y}{\mathrm{~d} x}=\cot \frac{\beta}{2}=\frac{1}{\tan \frac{\beta}{2}}=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$ Hence $\tan \frac{\beta}{2}=\sqrt{3}$.
$\frac{\beta}{2}=\frac{\pi}{3} $
$\beta=\frac{2 \pi}{3}(\text { shown })$

(iv) Since $\frac{\mathrm{d} y}{\mathrm{~d} x}=\cot \frac{\theta}{2}$

Gradient of normal at point $B$ is $-\tan \frac{\pi}{3}-\sqrt{3}$

Equation of normal :
$y-\frac{3}{2} a=-\sqrt{3}\left(x-\left[a\left(\frac{2 \pi}{3}-\sin \frac{2 \pi}{3}\right)\right]\right)$

$y-\frac{3}{2} a=-\sqrt{3}\left(x-\left[a\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)\right]\right)$

$y=-\sqrt{3}\left(x-\frac{2 \pi a}{3}+\frac{a \sqrt{3}}{2}\right)+\frac{3}{2} a$

$y=-\sqrt{3} x+\frac{2 \pi a}{\sqrt{3}}$

$\sqrt{3} y=-3$x$+2 \pi a$

$3 y=-(\sqrt{3})^2 x+2 \pi a$