JEE Main 2021 (Online) 26th August Morning ShiftQuestion 1
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 $(5-e^{2})\cdot A$ is :
Ellipse