Suppose S and S′ are foci of the ellipse x^ 2 25 + y^ 2 16 = 1 . If...

Given , $a^{2} = 25$ and $b^{2} = 16$

$∴ e = \sqrt{1 - \dfrac{b^{2}}{a^{2}}} = \sqrt{1 - \dfrac{16}{25}} = \dfrac{3}{5}$

So, the coordinates of foci $S$ and $S^{′}$ are $( 3, 0 )$ and $( - 3, 0 )$ respectively.

Let, P \left(5cos\theta,$4sin$\theta\right) be a variable point on the ellipse.

Then, ∆ = areaof ∆ PSS^{′} = \dfrac{1}{2}\left|\begin{cases}3 & 0 & 1 \\ - 3 & 0 & 1 \\ 5cos \theta \text{ and } 4sin \theta \text{ and } 1\end{cases}\right| = 12sin\theta

Since the value of $sin$\theta lies between $- 1$ and $1$

So, the maximum value of area of $∆ PSS^{′}$ is $12$ sq. units

Also,& $a^{8} + b^{4} ≥ 2a^{4}b^{2} ( since, A.M ≥ G.M )$

$⇒ \dfrac{3\left(a^{8} + b^{4}\right)}{a^{4}b^{2}} \ge6$