We have an ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$, we have a general point $P(\sqrt{8}\cos \theta ,2\sin \theta )$

Equation of tangent at point $p$

$\frac{x}{\sqrt{8}}\cos \theta +\frac{y}{2}\sin \theta -1=0$

Slope $=\frac{-1}{\sqrt{2}}\cot \theta$

Given that, tangent at $P$ is perpendicular to the line $x+2y=0$.

So, product of their slopes are perpendicular $\Rightarrow \frac{-1}{\sqrt{2}}\cot \theta \times \frac{-1}{2}=-1$

$\Rightarrow \frac{-1}{\sqrt{2}}\cot \theta =2$

$\Rightarrow \cot \theta =-2\sqrt{2}$

$\Rightarrow \cos \theta =\frac{-2\sqrt{2}}{3},\sin \theta =\frac{1}{3}$

So, point $P\left(\frac{-8}{3},\frac{2}{3}\right)$

$\mathrm{A}=\frac{1}{2}\times 2\mathrm{ae}\times \frac{2}{3}$

Where, $\mathrm{e}=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{8-4}{8}}=\frac{1}{\sqrt{2}}$

$\therefore A=\frac{1}{2}\times 2\times 2\sqrt{2}\times \frac{1}{\sqrt{2}}\times \frac{2}{3}=\frac{4}{3}$So, $\left(5-{\mathrm{e}}^2\right)\mathrm{A}=\left(5-\frac{1}{2}\right)\times \frac{4}{3}=6$