Solve this


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 $x+2 y=0 .$ Let $S$ and $S^{\prime}$ be the foci of the ellipse and 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 :

  1. 6

  2. 12

  3. 14

  4. 24

Correct Option: 1,


Equation of tangent : $\mathrm{y}=2 \mathrm{x}+6$

at $\mathrm{P}$

$\therefore \mathrm{P}(-8 / 3,2 / 3)$


$S \& S^{\prime}=(-2,0) \&(2,0)$

Area of $\Delta S P S^{\prime}=\frac{1}{2} \times 4 \times \frac{2}{3}$


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

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