An ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$ and the parabola $\mathrm{x}^{2}=4(\mathrm{y}+\mathrm{b})$ 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

  $\frac{1}{\sqrt{13}}$
• Correct answer
  B

  $\frac{2}{\sqrt{13}}$
• C

  $\frac{1}{\sqrt{11}}$
• D

  $\frac{2}{\sqrt{11}}$

Solution

As it is square $$ \begin{array}{l} \mathrm{ae}=2 \\ 4=\sqrt{(\mathrm{ae}-2)^{2}+(1-\mathrm{b})^{2}} \\ \mathrm{~b}=-3 \quad \quad \mathrm{~b}=5 \\ \mathrm{a}^{2}=13 \quad \mathrm{a}^{2}=29 \\ \quad\left(\text { from realation } \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)\right. \\ \mathrm{ae}=2 \\ \mathrm{e}=\frac{2}{\sqrt{13}} \end{array} $$

Hence, the correct answer is option: 'B'.