Let H_ 1 : x^ 2 a^ 2 - y^ 2 ~b^ 2 =1 and H_ 2 :- x^ 2 ~A^ 2 + y^ … - Vidyadip

Question

Let $\mathrm{H}_{1}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$ and $\mathrm{H}_{2}:-\frac{\mathrm{x}^{2}}{\mathrm{~A}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~B}^{2}}=1$ 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 $25 \mathrm{e}_{2}^{2}$ is equal to ____________ .

✓

Answer

(55)
Sol. $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=15 \sqrt{2}$
$1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{5}{2}$
$\mathrm{a}=5 \sqrt{2}$
$b=5 \sqrt{3}$
$\frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=12 \sqrt{5}$
$2 \mathrm{a} .2 \mathrm{~B}=100 \sqrt{10}$
$2.5 \sqrt{2} .2 \mathrm{~B}=100 \sqrt{10}$
$B=5 \sqrt{5}$
$A=5 \sqrt{6}$
$\mathrm{e}_{2}^{2}=1+\frac{\mathrm{A}^{2}}{\mathrm{~B}^{2}}$
$=1+\frac{150}{125}$
$\mathrm{e}_{2}^{2}=1+\frac{30}{25}$
$25 \mathrm{e}_{2}^{2}=55$

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