Aspire Faculty ID #18042 · Topic: JEE Main 2026 (24 January Morning Shift) · Just now
JEE Main 2026 (24 January Morning Shift)

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 $\ell_1,\ell_2$ be lengths of latus rectum of $E_1,E_2$ respectively such that $2\ell_1=\ell_2$. If the distance between the foci of $E_1$ is $8$, then the distance between foci of $E_2$ is:

Solution

$2ae=8 \Rightarrow a=\frac{5}{2}$

$e=\frac{4}{5} \Rightarrow b^2=a^2(1-e^2)=\frac{25}{4}\cdot\frac{9}{25}=\frac{9}{4}$

$\ell_1=\frac{2b^2}{a}=\frac{2\cdot\frac{9}{4}}{\frac{5}{2}}=\frac{18}{5}$

$\ell_2=2\ell_1=\frac{36}{5}$

For second ellipse

$\ell_2=\frac{2B^2}{A}=\frac{36}{5}$

Also

$e=\frac{4}{5}\Rightarrow B^2=A^2\frac{9}{25}$

Substitute

$\frac{2A^2\cdot\frac{9}{25}}{A}=\frac{36}{5}$

$\Rightarrow \frac{18A}{25}=\frac{36}{5}$

$\Rightarrow A=10$

Distance between foci

$=2Ae=2\cdot10\cdot\frac{4}{5}=\frac{32}{5}$
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