JEE Main 2026 — Ellipse Question with Solution

For the ellipse $E$, the eccentricity is $e_E=\frac{\sqrt{3}}{2}$ and the directrices are $x=\pm \frac{a_E}{e_E}=\pm \frac{4\sqrt{6}}{3}$.

The semi-major axis $a_E$ is given by:

$a_E=e_E\times \frac{4\sqrt{6}}{3}=\frac{\sqrt{3}}{2}\times \frac{4\sqrt{6}}{3}=\frac{12\sqrt{2}}{6}=2\sqrt{2}$

The semi-minor axis $b_E$ is given by:

$b_E^2=a_E^2(1-e_E^2)=(2\sqrt{2}{)}^2\left(1-\frac{3}{4}\right)=8\times \frac{1}{4}=2\Rightarrow b_E=\sqrt{2}$

For the hyperbola $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, its eccentricity $e_H$ is equal to the semi-major axis of $E$:

$e_H=a_E=2\sqrt{2}$

The length of the latus rectum of $H$ is equal to the length of the minor axis of $E$ ($2b_E$):

$\frac{2b^2}{a}=2b_E=2\sqrt{2}\Rightarrow b^2=\sqrt{2}a$

Using the standard relation for a hyperbola $b^2=a^2(e_H^2-1)$, we substitute $b^2$ and $e_H$:

$\sqrt{2}a=a^2((2\sqrt{2}{)}^2-1)$

$\sqrt{2}a=a^2(8-1)=7a^2$

Since $a>0$, dividing by $a$ gives:

$a=\frac{\sqrt{2}}{7}$

The distance between the foci of the hyperbola $H$ is $2a{e}_H$:

$2a{e}_H=2\times \frac{\sqrt{2}}{7}\times 2\sqrt{2}=\frac{8}{7}$

Answer: $\frac{8}{7}$