Question

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:

Hint:

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.

Answer 1

For the hyperbola \( H_1 \), the length of the latus rectum is given by \( \frac{2b^2}{a} \). The length of the latus rectum for \( H_1 \) is \( 15\sqrt{2} \), so we have: \[ \frac{2b^2}{a} = 15\sqrt{2}. \] Similarly, for \( H_2 \), the length of the latus rectum is \( \frac{2B^2}{A} \), and the given length is \( 12\sqrt{5} \), so: \[ \frac{2B^2}{A} = 12\sqrt{5}. \] Now, we are given that the product of the lengths of their transverse axes is \( 100\sqrt{10} \), so: \[ 2a \times 2A = 100\sqrt{10}. \] From these equations, we solve for \( e_2 \) and then compute \( 25e_2^2 \). 
Final Answer: \( 25e_2^2 = 50 \).