Question

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latus rectum of parabola are the vertices of a square. The eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{13}}$
• B. $\frac{2}{\sqrt{13}}$
• C. $\frac{1}{\sqrt{11}}$
• D. $\frac{2}{\sqrt{11}}$

Answer 1

Answer: (B)

$\frac{2}{\sqrt{13}}$

Points of the foci of the given ellipse are $=(ae,0)$ and $(-ae,0)$

And the ends of the latus rectum of parabola $x^2=4(y+b)$, are

$(2,1-b)$ and $(-2,1-b)$

Given, these four points are vertices of a square,

$\therefore ae=2$ and $2ae=1-b$, where $e$ is the eccentricity of ellipse.

$\Rightarrow b=-3$

Now, we know

$a^2{e}^2=a^2-b^2\Rightarrow a=\sqrt{13}$

$\Rightarrow ae=2$

$\Rightarrow e=\frac{2}{\sqrt{13}}$