JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Hyperbola

The eccentricity of an ellipse E with centre at the origin

O

\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) $\frac{8}{7}$ ✓ Correct

Answer: (D)

Solution

For the ellipse the directrix gives

\frac{a}{e} = \frac{4 6}{3}

, so with

e = \frac{3}{2}

a = \frac{3}{2} \cdot \frac{4 6}{3} = 22

, and

b^{2} = a^{2} (1 - e^{2}) = 8 \cdot \frac{1}{4} = 2

. Thus the semi-major axis is

22

and the minor axis is

2 b = 22

. For the hyperbola the eccentricity is

e_{H} = 22

and the latus rectum

\frac{2 b _{H}^{2}}{a _{H}} = 22

. From

e_{H}^{2} = 1 + \frac{b _{H}^{2}}{a _{H}^{2}}

8 = 1 + \frac{b _{H}^{2}}{a _{H}^{2}}

gives

b_{H}^{2} = 7 a_{H}^{2}

, and substituting into

\frac{2 b _{H}^{2}}{a _{H}} = 22

gives

7 a_{H} = 2

, so

a_{H} = \frac{2}{7}

. The distance between the foci is

2 a_{H} e_{H} = 2 \cdot \frac{2}{7} \cdot 22 = \frac{8}{7}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Hyperbola

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