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 $mathrm{S}^{prime}$ be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS ' then, the value of $left(5-e^2right) cdot A$ is :

On the ellipse let P be a point in the second quadrant such that the tange

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 $\left ( 5-e^{2} \right )\cdot A \, \,is:$
Option: 1 12
Option: 2 6
Option: 3 14
Option: 4 24

Answers (1)

$a= 2\sqrt{2},b= 2$
Let point P be $\left ( 2\sqrt{2}\cos \theta ,2\sin \theta \right )$
$Slope \: of\: \: x+2y= 0\; \; is\; \frac{-1}{2}$
$\therefore$ Slope of tangent = 2

Equation of tangent at P:

T = 0

$\frac{x\cdot 2\sqrt{2}\cos \theta }{8}+\frac{y\cdot 2\sin \theta }{4}= 1$
$\Rightarrow \frac{x \cos \theta }{2\sqrt{2}}+\frac{y\sin \theta }{2}= 1$
$Its\,\, slope= \frac{-\cos \theta \cdot 2}{2\sqrt{2}\sin \theta }= 2$
$\Rightarrow -\cot \theta= 2\sqrt{2}$
$\Rightarrow \cot \theta= -2\sqrt{2}$

$\therefore e^{2}= 1-\frac{b^{2}}{a^{2}}= 1-\frac{4}{8}= \frac{1}{2}.$
$Area\, of\, \Delta SP{S}'= \frac{1}{2}\cdot \left ( S{S}' \right )\cdot 2\sin \theta$
                                   $= \frac{1}{2}\cdot \left ( 2ae\right )2\cdot \left ( \frac{1}{3} \right )$
$=2\sqrt{2}\cdot \frac{1}{\sqrt{2}}\cdot 2\cdot \frac{1}{3}$
                                  $= \frac{4}{3}$
$\therefore \left ( 5-e^{2} \right )\cdot A=\left ( 5-\frac{1}{2} \right )\cdot \frac{4}{3}= \frac{9}{2}\cdot \frac{4}{3}$
                              $= 6$