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Hyperbola

Section: Mathematics

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Question : 7 of 98

Marks: +1, -0

Let the ellipse

E: \;\frac{x^2}{144}+\;\frac{y^2}{169}=1

and the hyperbola

H: \;\frac{x^2}{16}-\;\frac{y^2}{\lambda^2}=-1

have the same foci. If

e

L

respectively denote the eccentricity and the length of the latus rectum of

H

, then the value of

24(e+L)

[28 Jan 2026 Shift 2]

296
126
67
148

Solution:

Concept:

Foci of ellipse and hyperbola coincide. Use relation between

a

,

b

,

c

for ellipse and hyperbola to find

\lambda

. Then compute eccentricity and latus rectum.

Explanation:

Ellipse:

\frac{x^2}{144}+\frac{y^2}{169}=1

is vertical (

b>a

).

a=12

,

b=13

.

c^2=b^2-a^2=169-144=25\Rightarrow c=5

.

Foci:

(0,\pm5)

.

Hyperbola:

\frac{x^2}{16}-\frac{y^2}{\lambda^2}=-1

is vertical:

\frac{y^2}{\lambda^2}-\frac{x^2}{16}=1

.

Here

B^2=\lambda^2

,

A^2=16

.

For hyperbola,

c^2=A^2+B^2

.

Same foci ⇒

c=5

.

25=16+\lambda^2\Rightarrow\lambda^2=9\Rightarrow\lambda=3

.

Eccentricity

e=\frac{c}{B}=\frac{5}{3}

.

Latus rectum

L=\frac{2A^2}{B}=\frac{2\times16}{3}=\frac{32}{3}

.

e+L=\frac{5}{3}+\frac{32}{3}=\frac{37}{3}

.

24(e+L)=24\times\frac{37}{3}=8\times37=296

.

Answer:

296 (Option A)

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