Question ID: #423
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 $25e_2^2$ is equal to:
Solution:
For $H_1$, we use the eccentricity relation $b^2 = a^2(e_1^2 – 1)$:
$$ b^2 = a^2\left(\frac{5}{2} – 1\right) = \frac{3}{2}a^2 $$
Using the latus rectum formula $\frac{2b^2}{a} = 15\sqrt{2}$:
$$ \frac{2(3a^2/2)}{a} = 15\sqrt{2} \Rightarrow 3a = 15\sqrt{2} \Rightarrow a = 5\sqrt{2} $$
The transverse axis length of $H_1$ is $2a = 10\sqrt{2}$.
For $H_2$, the transverse axis is $2B$. Given the product of axes is $100\sqrt{10}$:
$$ (10\sqrt{2})(2B) = 100\sqrt{10} \Rightarrow 20\sqrt{2}B = 100\sqrt{10} \Rightarrow B = 5\sqrt{5} $$
Using the latus rectum of $H_2$, $\frac{2A^2}{B} = 12\sqrt{5}$:
$$ 2A^2 = 12\sqrt{5}(5\sqrt{5}) = 300 \Rightarrow A^2 = 150 $$
Now, finding $e_2^2$:
$$ e_2^2 = 1 + \frac{A^2}{B^2} = 1 + \frac{150}{125} = 1 + \frac{6}{5} = \frac{11}{5} $$
$$ 25e_2^2 = 25\left(\frac{11}{5}\right) = 55 $$
Ans. (55)
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