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 $S^{\prime }$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $SP{S}^{\prime },$ then the value of $(5-e^2)\cdot A$ is

• A. $12$
• B. $14$
• C. $6$
• D. $24$

Answer 1

The correct option is C $6$


Slope of line $=-\frac{1}{2}$
Slope of tangent $=2$
Equation of tangent
$\frac{x}{2\sqrt{2}}\cos \theta +\frac{y}{2}\sin \theta =1$
$\text{Slope}=-\frac{\cos \theta }{\sqrt{2}\sin \theta }=2$
$\tan \theta =\frac{-1}{2\sqrt{2}}\Rightarrow \cos \theta =\frac{-2\sqrt{2}}{3},\sin \theta =\frac{1}{3}$
$P(2\sqrt{2}\cos \theta ,2\sin \theta )\equiv (\frac{-8}{3},\frac{2}{3})$
$PM=\frac{2}{3}$
$A=\frac{1}{2}\times S{S}^{\prime }\times PM$
$=\frac{1}{2}\times 2ae\times \frac{2}{3}$
$=\sqrt{a^2-b^2}\times \frac{2}{3}$
$=\sqrt{8-4}\times \frac{2}{3}$
$=\frac{4}{3}$
$e=\sqrt{1-\frac{4}{8}}=\frac{1}{\sqrt{2}}$
$\therefore (5-e^2)\cdot A=(5-{\left(\frac{1}{\sqrt{2}}\right)}^2)\times \frac{4}{3}=6$