Let the foci of a hyperbola coincide with the foci of the ellipse x^ 2 36 + y^ 2 16 =1 . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Correct answer: D. 96 5 For ellipse (x^2)/(36) + (y^2)/(16) = 1 : a = 6 , b = 4 , c = 36-16 = 2 5 . Foci of ellipse: (± 2 5 , 0) . For hyperbola with eccentricity e = 5 and foci at (± 2 5 , 0) : e = (C)/(A) A = 2 5 5 . C^2 = A^2 + B^2 20 = (4)/(5) + B^2 B^2 = (96)/(5) . Latus rectum = (2B^2)/(A) = 2 · (96)/(5) 2 5 5 = 96 5 .

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Hyperbola

Let the foci of a hyperbola coincide with the foci of the ellipse...

Let the foci of a hyperbola coincide with the foci of the ellipse

\frac{x^{2}}{36}+\frac{y^{2}}{16}=1

. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Options

A16
B $24 \sqrt{5}$
C12
D $\frac{96}{\sqrt{5}}$

Correct answer

D. $\frac{96}{\sqrt{5}}$

Step-by-step solution

For ellipse

\frac{x^2}{36} + \frac{y^2}{16} = 1

a = 6

b = 4

c = \sqrt{36-16} = 2\sqrt{5}

.
Foci of ellipse:

(\pm 2\sqrt{5}, 0)

.
For hyperbola with eccentricity

e = 5

and foci at

(\pm 2\sqrt{5}, 0)

e = \frac{C}{A} \Rightarrow A = \frac{2\sqrt{5}}{5}

C^2 = A^2 + B^2 \Rightarrow 20 = \frac{4}{5} + B^2 \Rightarrow B^2 = \frac{96}{5}

.
Latus rectum

= \frac{2B^2}{A} = \frac{2 \cdot \frac{96}{5}}{\frac{2\sqrt{5}}{5}} = \frac{96}{\sqrt{5}}

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