Let and be two hyperbolas having length of latus rectums and respectively. Let their eccentricities be and respectively. If the product of the lengths of their transverse axes is , the is equal to __________. [2025]

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Solution

Q.
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 $25 e_{2}^{2}$ is equal to __________. [2025]

Ans.
(55)

Given, Hyperbola : $H_{1} : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and $e_{1} = \sqrt{\frac{5}{2}}$ and $H_{2} : \frac{- x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1$

Using $H_{1}$ , length of latus rectum = $15 \sqrt{2}$

$\Rightarrow \frac{2 b^{2}}{a} = 15 \sqrt{2}$ ... (i)

Since, $e_{1} = \sqrt{\frac{5}{2}} \Rightarrow 1 + \frac{b^{2}}{a^{2}} = \frac{5}{2} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{3}{2}$ ... (ii)

Using (i) and (ii), we get $a = 5 \sqrt{2} and b = 5 \sqrt{3}$

Now, for $H_{2}, \frac{2 A^{2}}{B} = 12 \sqrt{5}$ ... (iii)

Since, product of transverse axes is $100 \sqrt{10}$ , then

$2 a \cdot 2 B = 100 \sqrt{10}$

$\Rightarrow 2 \times 5 \sqrt{2} \times 2 B = 100 \sqrt{10} \Rightarrow B = 5 \sqrt{5}$

$∴ A = 5 \sqrt{6}$ [Using (iii)]

Now, eccentricity of $H_{2}$ is given by

$e_{2}^{2} = 1 + \frac{A^{2}}{B^{2}} = 1 + \frac{150}{125} = 1 + \frac{30}{25} = \frac{55}{25}$

$\Rightarrow 25 e_{2}^{2} = 55$ .

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