Let each of the two ellipses E_ 1 : x^ 2 a^ 2 + y^ 2 b^ 2 =1,(a>b) and E_ 2 : x^ 2 ~A^ 2 + y^ 2 ~B^ 2 =1,(~A<B) have eccentricity (4)/(5) . Let the lengths of the latus recta of E_ 1 and E_ 2 be l_ 1 and l_ 2

Correct answer: A. (32)/(5) Both ellipses have e = 4/5 . For E_1 ( a > b ): b^2 = a^2(1-e^2) = 9a^2/25 . Latus rectum l_1 = 2b^2/a = 18a/25 . Distance between foci of E_1 : 2ae = 8a/5 = 8 a = 5 , so l_1 = 18/5 . For E_2 ( A A^2 = B^2(1-e^2) = 9B^2/25 . Latus rectum l_2 = 2A^2/B = 18B/25 . Given 2l_1^2 = 9l_2 : 2 · (324)/(25) = 9 · (18B)/(25) 648 = 162B B = 4 . Distance between foci of E_2 = 2Be = 2 · 4 · (4)/(5) = (32)/(5)

JEE Main 202624 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Ellipse

Let each of the two ellipses E_ 1 : x^ 2 a^ 2 + y^ 2 b^ 2 =1,(a>b)...

Let each of the two ellipses

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

and

\mathrm{E}_{2}: \frac{x^{2}}{\mathrm{~A}^{2}}+\frac{y^{2}}{\mathrm{~B}^{2}}=1,(\mathrm{~A}<\mathrm{B})

have eccentricity

\frac{4}{5}

. Let the lengths of the latus recta of

E_{1}

and

E_{2}

l_{1}

and

l_{2}

, respectively, such that

2 l_{1}^{2}=9 l_{2}

. If the distance between the foci of

E_{1}

is 8, then the distance between the foci of

E_{2}

Options

A $\frac{32}{5}$
B $\frac{8}{5}$
C $\frac{16}{5}$
D $\frac{96}{5}$

Correct answer

A. $\frac{32}{5}$

Step-by-step solution

Both ellipses have

e = 4/5

.
For

E_1

(

a > b

b^2 = a^2(1-e^2) = 9a^2/25

.
Latus rectum

l_1 = 2b^2/a = 18a/25

.
Distance between foci of

E_1

2ae = 8a/5 = 8 \Rightarrow a = 5

, so

l_1 = 18/5

.
For

E_2

(

A < B

, major axis along

y

A^2 = B^2(1-e^2) = 9B^2/25

.
Latus rectum

l_2 = 2A^2/B = 18B/25

.
Given

2l_1^2 = 9l_2

2 \cdot \frac{324}{25} = 9 \cdot \frac{18B}{25} \Rightarrow 648 = 162B \Rightarrow B = 4

.
Distance between foci of

E_2 = 2Be = 2 \cdot 4 \cdot \frac{4}{5} = \frac{32}{5}

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