2023 CJC P1 Q8

Use the substitution \(x=\sqrt{15}\sin \theta \) to show that

\(\displaystyle\displaystyle \displaystyle\int{\sqrt{15-{{x}^{2}}}\mathrm{d}x}=\frac{1}{2}x\sqrt{15-{{x}^{2}}}+\frac{15}{2}{{\sin }^{-1}}\left( \frac{x}{\sqrt{15}} \right)+C\).[5]

The diagram below shows a sketch of the curves \({{C}_{1}}\) and \({{C}_{2}}\).

The curve \({{C}_{1}}\) has parametric equations

\(x=6\cos \theta \), \( \qquad y=2\sqrt{2}\sin \theta \)

for \(0\le \theta \le 2\pi \).

The curve \({{C}_{2}}\) has equation

\({{x}^{2}}+{{y}^{2}}=15\).

Given that \(P\) is a point of intersection between \({{C}_{1}}\) and \({{C}_{2}}\),

Determine the exact coordinates of \(P\).[4]