Question

Sol. $\angle SQP = \angle S^{'}QP = \frac{\pi}{2}$

$\Rightarrow \ S^{'},Q,S$ are collinear

$T$ is radical centre of $C,C_{1}$ and $C_{2}$.

Equation of circle $C_{2}$

$${C_{2}:(x - acos\theta)(x + ae) + (y - bsin\theta)y = 0 }{C:x^{2} + y^{2} - a^{2} = 0 }$$Equation of tangent to $C_{2}$ and $C$ is $a(cos\theta - e)x + bsin\theta y + a^{2}(ecos\theta - 1) = 0$

Put $x = acos\theta$

$$a^{2}\left( \cos^{2}\theta - ecos\theta \right) + bsin\theta y + a^{2}(ecos\theta - 1) = 0$$

$$\begin{matrix} \Rightarrow \ bsin\theta y & \ = a^{2}\sin^{2}\theta \\ \Rightarrow \ y & \ = \frac{a^{2}}{b}sin\theta \\ y & \ = \left( acos\theta,\frac{a^{2}}{b}sin\theta \right) \end{matrix}$$

Equation of chord of contact of $T$ w.r.t.

$${x^{2} + y^{2} = a^{2} }{xcos\theta + y\frac{a}{b}sin\theta = a }$$$\Rightarrow \frac{xcos\theta}{a} + \frac{ysin\theta}{b} = 1$ which is same as equation of tangent to ellipse at $P$.

$\because\ SA$ and $S^{'}B$ both are ⟂ to $AB$

$$\Rightarrow \ S^{'}B\| SA$$