Suppose $S$ and $S\prime$ are foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1.$ If $P$ is a variable point on the ellipse and if $∆$ is the area (in sq. units) of the triangle $PSS\prime$, then the maximum value of $∆$ is double of

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 $S^{\prime }$ are $(3, 0)$ and $(-3, 0)$ respectively.

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

Then, $∆=\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f} ∆PS{S}^{\prime }=\frac{1}{2}\left|\begin{array}{ccc}3& 0& 1\\ -3& 0& 1\\ 5\cos \mathrm{\theta }& 4\sin \mathrm{\theta }& 1\end{array}\right|=12\sin \mathrm{\theta }$

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

So, the maximum value of area of $∆PS{S}^{\prime }$ is $12$ sq. units

Also, $a^8+b^4\ge 2a^4{b}^2 (\mathrm{since}, \mathrm{A.}M\ge \mathrm{G.}M)$

$\Rightarrow \frac{3\left(a^8+b^4\right)}{a^4{b}^2}\ge 6$