2024 NJC Promo Q12

The diagram above shows Curve \(D\) which has equation \(\left(y-3\sqrt{6}\right)^2-x^2=1\). Points \(R\) and \(S\) are stationary points on \(D\). Another curve \(E\) has equation \(y^2-x^2=1\).

1. By considering a single transformation that transforms the graph of \(E\) onto \(D\) or otherwise, find the equations of the asymptotes of \(D\) and the coordinates of \(S\).[2]
2. Verify that point \(P\left(\tan p, \sec p + 3\sqrt{6}\right)\) lies on \(D\), where \(-\pi < p \leq \pi\), \(p \neq \pm\frac{\pi}{2}\).[1]
3. Show that the equation of the normal to \(D\) at \(P\) is \((\sin p)y + x = 2\tan p + 3\sqrt{6}\sin p\).[4]

The story of the design of Gardens by the Bay was told in the book "Supernature". There are many cross-section diagrams in the book and one of them consists of the region bounded by the lower piece of \(D\) and the line \(g\) with equation \(y = -\frac{1}{\sqrt{3}}x + 2\sqrt{6}\). In this cross-section diagram, the lower piece of \(D\) traces out the Flower Dome while \(g\) traces out the ground, as shown below.

This cross-section diagram is used to illustrate the condition where light rays are perpendicular to the ground. One of these light rays is normal to the lower piece of \(D\) and is represented by line \(r\).

4. Find the coordinates of the point where \(r\) intersects the lower piece of \(D\).[5]