On the ellipse \(\frac{{x}^{2}}{8}+\frac{{y}^{2}}{4}=1\) let P be a point in the second quadrant such that the tangent a…

On the ellipse \(\frac{{x}^{2}}{8}+\frac{{y}^{2}}{4}=1\) let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line \(x+2y=0\). Let S and S ' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of \(\left(5-{e}^{2}\right)\) . A is:

[JEE Main 2021, 26 Aug (Shift 1)]