JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

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 $E_{2}$ be $ℓ_{1}$ 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 $E_{2}$ is

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

Solution

Ellipse $E_{1}$ (major axis along $x$ ): $2 a e = 8 \Rightarrow a e = 4$ . With $e = 4/5$ , $a = 5$ . Then $b^{2} = a^{2} (1 - e^{2}) = 25 \cdot \frac{9}{25} = 9$ .

Latus rectum: $ℓ_{1} = \frac{2 b ^{2}}{a} = \frac{18}{5}$ .

Use the relation $2 ℓ_{1}^{2} = 9 ℓ_{2}$ :

$2 \cdot \frac{324}{25} = 9 ℓ_{2} \Rightarrow ℓ_{2} = \frac{72}{25}$ .

Ellipse $E_{2}$ (major axis along $y$ since $A < B$ ): $A^{2} = B^{2} (1 - e^{2}) = \frac{9 B ^{2}}{25}$ .

Latus rectum: $ℓ_{2} = \frac{2 A ^{2}}{B} = \frac{2 \cdot 9 B ^{2} /25}{B} = \frac{18 B}{25} = \frac{72}{25}$ .

$\Rightarrow B = 4$ .

Distance between foci of $E_{2}$ : $2 B e = 2 \cdot 4 \cdot \frac{4}{5} = \frac{32}{5}$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →