JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 82

The hyperbola $\frac{x^2}4-\frac{y^2}9=1$ has $e=\frac{\sqrt{13}}2$ and foci $(\pm\sqrt{13},0)$. So the ellipse has $a=\sqrt{13}$ and $e=\frac1{\sqrt{13}}$, giving $b^2=13-1=12$:
$$\frac{x^2}{13}+\frac{y^2}{12}=1.$$
Checking the points:

• $\left(\frac{\sqrt{39}}2,\sqrt3\right)$: $\frac34+\frac14=1$, on the ellipse.
• $\left(\sqrt{\frac{13}2},\sqrt6\right)$: $\frac12+\frac12=1$, on the ellipse.
• $\left(\frac{\sqrt{13}}2,\frac{\sqrt3}2\right)$: $\frac14+\frac1{16}\ne1$, not on it.