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 $(5 - e^{2}) A is_____$

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answer is 6.

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Detailed Solution

$e = \sqrt{\frac{8 - 4}{8}} = \sqrt{\frac{4}{8}} = \sqrt{\frac{1}{2}} slope of Tangent m = \frac{- 1}{\frac{- 1}{2}} = 2 point P = (\frac{- a^{2} m}{c}, \frac{b^{2}}{c}) = (\frac{- 8 \times 2}{\sqrt{32 + 4}}, \frac{4}{\sqrt{32 + 4}}) (∵ c^{2} = a^{2} m^{2} + b^{2}) = (\frac{- 16}{6}, \frac{4}{6}) area of ∆ {SPS}^{1} is = \frac{1}{2} \times 2 ae \times \frac{4}{6} = \sqrt{8 - 4} \times \frac{4}{6} = \frac{8}{6} = \frac{4}{3} = A (5 - e^{2}) A = (5 - \frac{1}{2}) . \frac{4}{3} = \frac{36}{6} = 6$