JEE Main 2015 — Ellipse Question with Solution

Equation of the hyperbola
$\frac{x^2}{4}-\frac{y^2}{9}=1$



Focus of hyperbola $\left({\mathrm{a}}_1{\mathrm{e}}_1,\mathrm{ }0\right)$ and $(-{\mathrm{a}}_1{\mathrm{e}}_1,\mathrm{ }0)$  

${\mathrm{a}}_1=2,\mathrm{ }{\mathrm{e}}_1=\mathrm{ }\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$

$\therefore$ Foci would be $\left(+\sqrt{13}, 0\right)$ and $\left(-\sqrt{13}, 0\right)$ 

Product of eccentricity would be 

$\frac{\sqrt{13}}{2} \cdot e_2=\frac{1}{2}$

$\therefore e_2= \frac{1}{\sqrt{13}}$ .

As the major & minor axis of the ellipse coincide with axis of hyperbola then the value of $a_2$ for ellipse would be $\sqrt{13}$ ,

${\mathrm{e}}_2=\sqrt{1-\frac{{\mathrm{b}}_2^2}{{\mathrm{a}}_2^2}}$

$\frac{1}{\sqrt{13}}=\sqrt{1-\frac{{\mathrm{b}}_2^2}{13}}$

${\mathrm{b}}_2^2=12$

$\therefore$ Equation of the ellipse would be 

$\frac{x^2}{13}+\frac{y^2}{12}=1.$

Option (1)  $\frac{39}{4 \cdot \left(13\right)}+\frac{3}{12}=1$

Satisfies the equation, hence it lies on the ellipse. 

Option (2) $\frac{13}{4 \left(13\right)}+\frac{3}{4.12}=1$

does not lie on the ellipse.

Option (3)  $\frac{13}{2\left(13\right)}+\frac{6}{12}=1$ satisfy.

Option (4) $\frac{13}{13}+0=1$ satisfy.

So, option $\left(\frac{\sqrt{13}}{2}, \frac{\sqrt{3}}{2}\right)$ is the answer.