SOLUTION —
We have, $a^2=25$ and $b^2=16$
$\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
So, the coordinates of foci $S$ and $S^{\prime}$ are $(3,0)$ and $(-3,0)$ respectively. Let $P(5 \cos \theta, 4 \sin \theta)$ be a variable point on the ellipse.
Then, $\Delta=$ area of $\triangle P S S^{\prime}=\frac{1}{2} \times 6 \times 4 \sin \theta=12 \sin \theta$
So, maximum value of area of $\triangle P S S^{\prime}$ is 12 , since value of $\sin \theta$ lies between -1 and 1
So, The correct option of this question will be (B).