JEE Main 2026 Apr 6 Shift 2, Mathematics Q11: Introduction to Hyperbola

JEE Main2026Apr 6, Shift 2Mathematics

The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{3}{2}$ and its directrices are $x = \pm \frac{4 6}{3}$ . Let $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ 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:

A
$\frac{4 2}{7}$
B
$\frac{4 2}{7}$
C
$\frac{4}{7}$
D
$\frac{8}{7}$

Solution

Ellipse $E$ . $e_{E} = \frac{3}{2}$ , directrix $x = \frac{a _{E}}{e _{E}} = \frac{4 6}{3}$ $\Rightarrow$ $a_{E} = 22$ .

$b_{E}^{2} = a_{E}^{2} (1 - e_{E}^{2}) = 8 \cdot \frac{1}{4} = 2$ , so $b_{E} = 2$ . Minor axis $= 2 b_{E} = 22$ .

Hyperbola $H$ . Eccentricity $e_{H} = a_{E} = 22$ . Latus rectum $= \frac{2 b ^{2}}{a} = 22$ $\Rightarrow$ $b^{2} = a 2$ .

Also $b^{2} = a^{2} (e_{H}^{2} - 1) = 7 a^{2}$ . Equating: $a 2 = 7 a^{2}$ $\Rightarrow$ $a = \frac{2}{7}$ .

$2 a e_{H} = 2 \cdot \frac{2}{7} \cdot 22 = \frac{8}{7}$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q11: Introduction to Hyperbola

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