Mathematics · 2026
JEE Main · 24 January 2026, Shift 1 · Q10
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…
Let each of the two ellipses $\displaystyle \mathrm{E}_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ and $\displaystyle \mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1,(\mathrm{~A}<\mathrm{B})$ have eccentricity $\displaystyle \frac{4}{5}$. Let the lengths of the latus recta of $\displaystyle \mathrm{E}_1$ and $\displaystyle \mathrm{E}_2$ be $\displaystyle l_1$ and $\displaystyle l_2$, respectively, such that $\displaystyle 2 l_1^2=9 l_2$. If the distance between the foci of $\displaystyle E_1$ is $\displaystyle 8$ , then the distance between the foci of $\displaystyle E_2$ is
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{32}{5}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.