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Mathematics · 2025

JEE Main · 24 January 2025, Shift 2 · Q25

Let H_1: x^2/(a^2)-y^2/(b^2)=1 and H_2:-x^2/(A^2)+y^2/(B^2)=1 be two hyperbolas having length of latus rectums 15 √ 2 and 12 √ 5 respectively. Let…

Let $\displaystyle \mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and $\displaystyle \mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$ be two hyperbolas having length of latus rectums $\displaystyle 15 \sqrt{2}$ and $\displaystyle 12 \sqrt{5}$ respectively. Let their ecentricities be $\displaystyle e_1=\sqrt{\frac{5}{2}}$ and $\displaystyle e_2$ respectively. If the product of the lengths of their transverse axes is $\displaystyle 100 \sqrt{10}$, then $\displaystyle 25 \mathrm{e}_2^2$ is equal to $\displaystyle \_\_\_\_$.
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.