Verified Solution • 2025–26 Session moderate

QJEEM25Ma7fda: JEE Main 2025 Mathematics - Class 11 JEEM25Ma7fda Solution: Let and be two hyperbolas having length of latus rectums and…

Let H1:x2a2−y2 b2=1\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1 and H2:−x2 A2+y2 B2=1\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1 be two hyperbolas having length of latus rectums 15215 \sqrt{2} and 12512 \sqrt{5} respectively. Let their ecentricities be e1=52e_1=\sqrt{\frac{5}{2}} and e2e_2 respectively. If the product of the lengths of their transverse axes is 10010100 \sqrt{10} , then 25e2225 \mathrm{e}_2^2 is equal to _________ .

Step-by-Step Verified Solution

amp;2b2a=152amp;1+b2a2=52amp;a=52amp; b=53amp;2 A2 B=125amp;2a⋅2 B=10010amp;2.52.2 B=10010amp; B=55amp; A=56amp;e22=1+A2 B2amp;=1+150125amp;e22=1+3025amp;25e22=55\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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