Let $H_1 : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ and $H_2 : -\dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1$ be two hyperbolas having lengths of latus rectums $15\sqrt{2}$ and $12\sqrt{5}$ respectively. Let their eccentricities be $e_1 = \sqrt{\dfrac{5}{2}}$ 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 _____.
Step-by-Step Solution
Key Concept: Use $\ell(LR)=2b^2/a$ and $e_1^2=1+b^2/a^2$ for $H_1$; use $\ell(LR)=2A^2/B$ and $e_2^2=1+A^2/B^2$ for $H_2$; the product of transverse axes is $(2a)(2B)=100\sqrt{10}$.
For $H_1$: $2b^2/a=15\sqrt{2}$ and $e_1^2=1+b^2/a^2=5/2$, so $b^2/a^2=3/2$, giving $b^2=(3/2)a^2$. Then $3a=15\sqrt{2}$, $a=5\sqrt{2}$, $b^2=75$. For $H_2$: $2A^2/B=12\sqrt{5}$, and transverse axes product $(2a)(2B)=100\sqrt{10} \Rightarrow B=\tfrac{100\sqrt{10}}{4\cdot5\sqrt{2}}=5\sqrt{5}$. Then $A^2=\tfrac{12\sqrt{5}\cdot B}{2}=\tfrac{12\sqrt{5}\cdot5\sqrt{5}}{2}=150$. $e_2^2=1+A^2/B^2=1+150/125=1+\tfrac{6}{5}=\tfrac{11}{5}$. $25e_2^2=55$.
Correct Answer: 55