Let $\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and $\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their ecentricities be $e_1=\sqrt{\frac{5}{2}}$ and…

Question

Let $\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and $\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their ecentricities 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 ________.

Aicharya · Accepted Answer

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{2 b^2}{a}=15 \sqrt{2}$ ...(i)
$\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{5}{2}}$....(ii)
From (i) and (ii)
$a=5 \sqrt{2} \text { and } b^2=75$
$\frac{x^2}{A^2}-\frac{y^2}{B^2}=-1$
$\frac{2 A^2}{B}=12 \sqrt{5}$ ....(iii)
Since, product of transverse axis is $=100 \sqrt{10}$
$(2 A) \cdot(2 B)=100 \sqrt{10}$
From (iii) and (iv)
$\begin{aligned}
& A^2=150 \text { and } B=5 \sqrt{5} \\ & e_2=\sqrt{1+\frac{A^2}{B^2}}=\sqrt{\frac{11}{5}} \\ & \therefore 25 e_2^2=25\left(\frac{11}{5}\right)=55
\end{aligned}$

Let $\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ and…

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