Question:medium

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 lengths of latus rectums \( 15\sqrt{2} \) and \( 12\sqrt{5} \) respectively. Let their eccentricities be \( e_1 = \frac{5}{\sqrt{2}} \) and \( e_2 \) respectively. If the product of the lengths of their transverse axes is \( 100\sqrt{10} \), then \( 25e_2^2 \) is equal to:

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For problems involving hyperbolas, use the formulas for the lengths of the latus rectum and the relationship between the transverse axes and eccentricity to solve for the unknowns.
Updated On: Jan 14, 2026
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Solution and Explanation

The latus rectum length of hyperbola \( H_1 \) is \( \frac{2b^2}{a} \), which equals \( 15\sqrt{2} \). Thus, \( \frac{2b^2}{a} = 15\sqrt{2} \). For hyperbola \( H_2 \), the latus rectum length is \( \frac{2B^2}{A} \), given as \( 12\sqrt{5} \). Therefore, \( \frac{2B^2}{A} = 12\sqrt{5} \). The product of their transverse axis lengths is \( 100\sqrt{10} \), meaning \( 2a \times 2A = 100\sqrt{10} \). We use these equations to find \( e_2 \) and subsequently calculate \( 25e_2^2 \).
Final Answer: \( 25e_2^2 = 50 \).

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