JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 25)
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
55
Explanation
$$\begin{aligned}
& \frac{2 b^2}{\mathrm{a}}=15 \sqrt{2} \\
& 1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{5}{2} \\
& \mathrm{a}=5 \sqrt{2} \\
& \mathrm{~b}=5 \sqrt{3} \\
& \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=12 \sqrt{5} \\
& 2 \mathrm{a} \cdot 2 \mathrm{~B}=100 \sqrt{10} \\
& 2.5 \sqrt{2} .2 \mathrm{~B}=100 \sqrt{10} \\
& \mathrm{~B}=5 \sqrt{5} \\
& \mathrm{~A}=5 \sqrt{6} \\
& \mathrm{e}_2^2=1+\frac{\mathrm{A}^2}{\mathrm{~B}^2} \\
& =1+\frac{150}{125} \\
& \mathrm{e}_2^2=1+\frac{30}{25} \\
& 25 \mathrm{e}_2^2=55
\end{aligned}$$
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