8.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-28_735_892_264_529}
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\caption{Figure 1}
\end{figure}
The curve \(C\) shown in Figure 1 has polar equation
$$r = 1 + \sin \theta \quad - \frac { \pi } { 2 } < \theta \leqslant \frac { \pi } { 2 }$$
The point \(P\) lies on \(C\) such that the tangent to \(C\) at \(P\) is perpendicular to the initial line.
- Use calculus to determine the polar coordinates of \(P\).
The tangent to \(C\) at the point \(Q\) where \(\theta = \frac { \pi } { 2 }\) is parallel to the initial line.
The tangent to \(C\) at \(Q\) meets the tangent to \(C\) at \(P\) at the point \(S\), as shown in Figure 1.
The finite region \(R\), shown shaded in Figure 1, is bounded by the line segments \(Q S , S P\) and the curve \(C\). - Use algebraic integration to show that the area of \(R\) is
$$\frac { 1 } { 32 } ( a \sqrt { 3 } + b \pi )$$
where \(a\) and \(b\) are integers to be determined.
(6)