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 latusrectum 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)

Equation of ellipse
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$
Equation of parabola $x^{2}=4(y +b)$
Foci of ellipse $\left(\pm\,ae, 0\right)$
End of latusrectum of parabola
$=(\pm 2, 1-b)$

$ABCD$ is a square
$A B=C D \Rightarrow 2 a e=4$
$\Rightarrow a e=2$
$B C=C D \Rightarrow B C^{2}=C D^{2}$
$(2-a e)^{2}+(1-b)^{2}=4^{2}$
$\Rightarrow 0+(1-b)^{2}=4^{2}$
$\Rightarrow 1-b=\pm 4$
$b=5, b=-3 \Rightarrow a=\sqrt{29}$ or $\sqrt{13}$
$\therefore e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$\Rightarrow e=\sqrt{1-\frac{9}{13}}=\frac{2}{\sqrt{13}}$ or $2$