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