Question

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 $\left(5-e^2\right).A$ is :

• A. 6
• B. 12
• C. 14
• D. 24

Answer 1

Answer: (A)

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