Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve

$$x = 2t^2, y = 4t, \quad -\sqrt{2} < t < \sqrt{2}.$$

P$(2t^2, 4t)$ is a point on the curve with parameter $t$. TS is the tangent to the curve at P, and PR is the line through P parallel to the $x$-axis. Q is the point (2, 0). The angles that PS and QP make with the positive $x$-direction are $\theta$ and $\phi$ respectively.

\includegraphics{figure_8}

\begin{enumerate}[label=(\roman*)]
\item By considering the gradient of the tangent TS, show that $\tan \theta = \frac{1}{t}$. [3]

\item Find the gradient of the line QP in terms of $t$. Hence show that $\phi = 2\theta$, and that angle TPQ is equal to $\theta$. [8]
\end{enumerate}

[The above result shows that if a lamp bulb is placed at Q, then the light from the bulb is reflected to produce a parallel beam of light.]

The inside surface of the headlight has the shape produced by rotating the curve about the $x$-axis.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that the curve has cartesian equation $y^2 = 8x$. Hence find the volume of revolution of the curve, giving your answer as a multiple of $\pi$. [7]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4 2012 Q8 [18]}}