Question

Sol. $\left( S_{1}S_{2} \right)^{2} = \left( S_{1}P \right)^{2} + \left( S_{2}P \right)^{2}$

$$(2ae)^{2} = a^{2}\left\lbrack (1 - ecos\theta)^{2} + (1 + ecos\theta)^{2} \right\rbrack $$

$$\begin{array}{r} 2e^{2}\ = 1 + e^{2}\cos^{2}\theta\#(1) \\ \ \therefore b^{2}\ = a^{2}\left( 1 - e^{2} \right)\#(1) \\ \Rightarrow 1\ = 2\left( 1 - e^{2} \right) \Rightarrow e^{2} = \frac{1}{2}\#(1) \end{array}$$

∴ From eqn. (1), we have

$$cos\theta = 0 \Rightarrow \theta = \frac{\pi}{2},\frac{3\pi}{2}$$

Area of

$$\begin{matrix} & S_{1}PS_{2}Q = \left( S_{1}P \right)\left( S_{2}P \right) = a^{2} = 2 \\ & \left\lbrack \because S_{1}P = S_{2}P = a \right\rbrack \end{matrix}$$