The figure below shows a cross-section $O B C E$ of a car headlight whose reflective surface is modelled in suitable units by the curve with parametric equations
$x=a(\theta-\sin \theta), \quad y=a(1-\cos \theta)$
for $0 \leq \theta \leq 2 \pi$, where $a$ is a positive constant.

(i) Find in terms of $a$
(a) the length of $O E$,
[2]
(b) the maximum height of the curve $O B C E$.
[1]
(ii) Show that $\frac{\mathrm{d} y}{\mathrm{~d} x}=\cot \frac{\theta}{2}$.
[3]
Point $B$ lies on the curve and has parameter $\beta . T S$ is tangential to the curve at $B$ and $B C$ is parallel to the $x$-axis. Given that $\angle T B C=\frac{\pi}{6}$,
(iii) show that $\beta=\frac{2 \pi}{3}$.
[2]
(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$+2 \pi a$
$3 y=-(\sqrt{3})^2 x+2 \pi a$