9.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-30_714_1079_251_495}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{figure}
The curve shown in Figure 5 has equation
$$x = 4 \sin ^ { 2 } y - 1 \quad 0 \leqslant y \leqslant \frac { \pi } { 2 }$$
The point \(P \left( k , \frac { \pi } { 3 } \right)\) lies on the curve.
- Verify that \(k = 2\)
- Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\)
- Hence show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 \sqrt { x + 1 } \sqrt { 3 - x } }\)
The normal to the curve at \(P\) cuts the \(x\)-axis at the point \(N\).
- Find the exact area of triangle \(O P N\), where \(O\) is the origin.
Give your answer in the form \(a \pi + b \pi ^ { 2 }\) where \(a\) and \(b\) are constants.