Question

Let circle $S$ be drawn considering the endpoints of the diameter as the endpoints of minor axis of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$. If the locus of the point of intersection of perpendicular tangents to this circle is passing through the foci of the ellipse, then the eccentricity of the ellipse is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

The correct option is D $\sqrt{\frac{2}{3}}$

Given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$
Endpoints of the diameter $=(0,\pm b)$
Circle whose endpoints of the diameter is $=(0,\pm b)$ is
$x^2+(y+b)(y-b)=0\Rightarrow x^2+y^2=b^2$

Locus of the point of intersection of perpendicular tangents is director circle,
$x^2+y^2=2b^2$

Foci of the ellipse $=(\pm ae,0)$, which lies on the director circle, so
$a^2{e}^2=2b^2\Rightarrow a^2{e}^2=2a^2(1-e^2)\Rightarrow e^2=2-2e^2(\because a^2\ne 0)\therefore e=\sqrt{\frac{2}{3}}$