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  4. On the ellipse (x2/8)+(y2/4)=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' 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) . A is :

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Q. 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 $S$' 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 :

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Solution:

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Equation of tangent : $y=2 x+6$ at $P$
$\therefore P (-8 / 3,2 / 3) $
$e =\frac{1}{\sqrt{2}}$
$ S \,\&\, S '=(-2,0) \&(2,0)$
Area of $\Delta SPS '=\frac{1}{2} \times 4 \times \frac{2}{3}$
$A=\frac{4}{3}$
$\therefore \left(5-e^{2}\right) A=\left(5-\frac{1}{2}\right) \frac{4}{3}=6$