JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Ellipse

Let each of the two ellipses

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

and

E_{2} : \frac{x ^{2}}{A ^{2}} + \frac{y ^{2}}{B ^{2}} = 1, (A < 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}

Answer: (B)

Solution

For

E_{1}

2 a e = 8

with

e = \frac{4}{5}

gives

a = 5

, then

b^{2} = a^{2} (1 - e^{2}) = 9

and

ℓ_{1} = \frac{2 b ^{2}}{a} = \frac{18}{5}

. From

2 ℓ_{1}^{2} = 9 ℓ_{2}

ℓ_{2} = \frac{72}{25}

. For

E_{2}

, whose major axis is along

y

A^{2} = B^{2} (1 - e^{2}) = \frac{9}{25} B^{2}

and

ℓ_{2} = \frac{2 A ^{2}}{B} = \frac{18 B}{25} = \frac{72}{25}

, so

B = 4

. The distance between the foci of

E_{2}

2 B e = 2 \cdot 4 \cdot \frac{4}{5} = \frac{32}{5}

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option