Question

Let $S$ and $S^{\prime }$ be the two foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ If the circle described on $S{S}^{\prime }$ as diameter touches the ellipse in real points, then $6e^2=$(where $e$ is eccentricity of ellipse)

Answer 1

$\because$ Circle with $S{S}^{\prime }$ as diameter touches the ellipse at real points
$($ i.e. at $(0,b)$ and $(0,-b))$
Equation of circle will be
$\Rightarrow x^2+y^2=a^2{e}^2=a^2\left(\frac{a^2-b^2}{a^2}\right)=a^2-b^2$
Circle passes through $(0,b)$ and $(0,-b)$
$\Rightarrow 0+b^2=a^2-b^2\Rightarrow 2b^2=a^2\Rightarrow 2a^2(1-e^2)=a^2\Rightarrow 1-e^2=\frac{1}{2}\Rightarrow e^2=\frac{1}{2}\therefore 6e^2=3$