JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Ellipse

Let

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b

and

H : \frac{x ^{2}}{A ^{2}} - \frac{y ^{2}}{B ^{2}} = 1

.

Let the distance between the foci of E and the foci of H be

23

. If

a - A = 2

, and the ratio of the eccentricities of E and H is

\frac{1}{3}

, then the sum of the lengths of their latus rectums is equal to :

Answer: (C)

Solution

For ellipse E, foci are at

(\pm a e, 0)

, and for hyperbola H at

(\pm A e^{'}, 0)

. Equal foci-separation gives

a e = A e^{'} = 3

.

Given

\frac{e}{e ^{'}} = \frac{1}{3}

, so

e^{'} = 3 e

; combined with

a e = A e^{'}

, this gives

a = 3 A

.

With

a - A = 2

3 A - A = 2 \Rightarrow A = 1, a = 3

.

Then

a e = 3 \Rightarrow e = \frac{1}{3}

, and

e^{'} = 3 e = 3

b^{2} = a^{2} (1 - e^{2}) = 9 (1 - \frac{1}{3}) = 6, B^{2} = A^{2} (e^{'2} - 1) = 1 (3 - 1) = 2.

Latus rectum of E:

\frac{2 b ^{2}}{a} = \frac{12}{3} = 4

. Latus rectum of H:

\frac{2 B ^{2}}{A} = \frac{4}{1} = 4

.

Sum

= 4 + 4 = 8.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option