JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q11

← JEE Main 2026 · 6 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 11 of 75

The eccentricity of an ellipse E with centre at the origin O is $\dfrac{\sqrt3}2$ and its directrices are $x=\pm\dfrac{4\sqrt6}3$ . Let $H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :

\dfrac{4\sqrt2}{\sqrt7}

\dfrac{4\sqrt2}7

\dfrac4{\sqrt7}

\dfrac87

Answer and worked solution

Official answer: D

Ellipse E. $\dfrac ae=\dfrac{4\sqrt6}3$ with $e=\dfrac{\sqrt3}2$ , so $a=2\sqrt2$ and $b^2=8\cdot\tfrac14=2$ . The minor axis is $2\sqrt2$ .

Hyperbola H. $e_H=2\sqrt2$ and $\dfrac{2b^2}a=2\sqrt2$ , so $b^2=\sqrt2\,a$ . Also $b^2=a^2(e_H^2-1)=7a^2$ . So $a=\dfrac{\sqrt2}7$ .

Foci distance. $2ae_H=2\cdot\dfrac{\sqrt2}7\cdot2\sqrt2=\dfrac87$ .

(D)

Topics:Hyperbola Ellipse

← Previous question Next question →

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 11