Explanation

(b)
Given , $a^{2} = 25$ and $b^{2} = 16$
$\therefore\ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{16}{25}} = \ \frac{3}{5}$
So, the coordinates of foci $S$ and $S^{'}$ are (3, 0) and $( - 3,\ 0)$ respectively. Let $P\left( 5\cos\theta,4\sin\theta \right)$ be a variable point on the ellipse.
Then, $\mathrm{\Delta} = area\ of\ \mathrm{\Delta}PSS^{'} = \left| \begin{matrix}
3 & 0 & 1 \\
- 3 & 0 & 1 \\
5\cos\theta & 4\sin\theta & 1
\end{matrix} \right| = 12\sin\theta\ \ \ \ \ \ \ \ \ \ \ \$
[Since, value of $\sin\theta\$lies between $- 1$ and 1]
So, maximum value of area of $\mathrm{\Delta}PSS^{'}$ is $12$