Let H 1 : frac x 2 a 2 -frac y 2 b 2 =1 and H 2 :-frac x 2 A 2 +frac y 2 B 2 =1 be two hyperbolas having length of latus rectums 15 √(2) and 12 √(5) respectively. Let their eccentricities be e 1 =√(5/2) and e 2 respectively. If the product of the lengths of their transverse axes is 100 √(10) , then 25 e 2 2 is equal to \ \ \ \ \

Mathematics · Hyperbola

JEE Main 2025 — 24 January, Evening Shift — Question 22

Let $H_{1}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $H_{2}:-\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their eccentricities 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

Numerical answer — enter this value.

Step-by-step solution

$\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=15 \sqrt{2}$

$1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{5}{2}$

$\mathrm{a}=5 \sqrt{2}$ $\mathrm{b}=5 \sqrt{3}$

$\frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=12 \sqrt{5}$

$2 \mathrm{a} .2 \mathrm{~B}=100 \sqrt{10}$

$2.5 \sqrt{2} .2 \mathrm{~B}=100 \sqrt{10}$

$B=5 \sqrt{5}$ $A=5 \sqrt{6}$

$\mathrm{e}_{2}^{2}=1+\frac{\mathrm{A}^{2}}{\mathrm{~B}^{2}}$

$=1+\frac{150}{125}$

$\mathrm{e}_{2}^{2}=1+\frac{30}{25}$

$25 \mathrm{e}_{2}^{2}=55$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola