Let the foci of a hyperbola coincide with the foci of the ellipse \; x²/36 + \; y²/16 = 1. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is : [21 Jan 2026 Shift 1]

Correct Answer is : 965\; {96}{{5}}596

Correct Answer is : 965\; {96}{{5}}596

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Hyperbola

Section: Mathematics

Question : 13 of 98

Marks: +1, -0

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

Solution:

Concept:

For confocal ellipse and hyperbola, the distance from center to focus is same:

c^2 = a^2 + b^2

for hyperbola equals

A^2 - B^2

for ellipse.

Explanation:

From ellipse

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

, we have

A^2 = 36

B^2 = 16

Focal distance:

c^2 = A^2 - B^2 = 36 - 16 = 20

For hyperbola with same foci:

c^2 = a^2 + b^2 = 20

Given eccentricity of hyperbola

e_H = 5

. For hyperbola,

e_H = \frac{c}{a}

, so

c = 5a

Then

c^2 = 25a^2 = 20

, giving

a^2 = \frac{20}{25} = \frac{4}{5}

and

a = \frac{2}{\sqrt{5}}

Now

b^2 = c^2 - a^2 = 20 - \frac{4}{5} = \frac{96}{5}

Latus rectum length for hyperbola is

\frac{2b^2}{a} = \frac{2 \times \frac{96}{5}}{\frac{2}{\sqrt{5}}} = \frac{96}{\sqrt{5}}

Answer:

\frac{96}{\sqrt{5}}

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