Medium INTEGER +4 / -1 PYQ · JEE Mains 2025

Let $\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and $\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their ecentricities be $e_1=\sqrt{\frac{5}{2}}$ and $e_2$ respectively. If the product of the lengths of their transverse axes is $100 \sqrt{10}$, then $25 \mathrm{e}_2^2$ is equal to _________ .

Answer (integer) 55

Solution

<p>$$\begin{aligned} & \frac{2 b^2}{\mathrm{a}}=15 \sqrt{2} \\ & 1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{5}{2} \\ & \mathrm{a}=5 \sqrt{2} \\ & \mathrm{~b}=5 \sqrt{3} \\ & \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=12 \sqrt{5} \\ & 2 \mathrm{a} \cdot 2 \mathrm{~B}=100 \sqrt{10} \\ & 2.5 \sqrt{2} .2 \mathrm{~B}=100 \sqrt{10} \\ & \mathrm{~B}=5 \sqrt{5} \\ & \mathrm{~A}=5 \sqrt{6} \\ & \mathrm{e}_2^2=1+\frac{\mathrm{A}^2}{\mathrm{~B}^2} \\ & =1+\frac{150}{125} \\ & \mathrm{e}_2^2=1+\frac{30}{25} \\ & 25 \mathrm{e}_2^2=55 \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

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