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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 $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :
$12$
$6$
$14$
$24$
Solution
Equation of tangent : $\mathrm{y}=2 \mathrm{x}+6$
at $\mathrm{P}$
$\therefore \mathrm{P}(-8 / 3,2 / 3)$
$\mathrm{e}=\frac{1}{\sqrt{2}}$
$\mathrm{~S} \& \mathrm{~S}^{\prime}=(-2,0) \&(2,0)$
Area of $\Delta$ SPS' $=\frac{1}{2} \times 4 \times \frac{2}{3}$
$A=\frac{4}{3}$
$\therefore\left(5-\mathrm{e}^{2}\right) \mathrm{A}=\left(5-\frac{1}{2}\right) \frac{4}{3}=6$
