Q34 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
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 $E_{2}$ 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
- Correct answerA$\frac{32}{5}$
- B$\frac{8}{5}$
- C$\frac{16}{5}$
- D$\frac{96}{5}$
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 $E_2 = 2Be = 2 \cdot 4 \cdot \frac{4}{5} = \frac{32}{5}$
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 $E_2 = 2Be = 2 \cdot 4 \cdot \frac{4}{5} = \frac{32}{5}$
Hence, the correct answer is option: 'A'.