On the ellipse x^2/8+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+2 y=0.…

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