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Hyperbola

Section: Mathematics

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Question : 22 of 98

Marks: +1, -0

Let

H_1: \;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1

H_2:-\;\frac{x^2}{A^2}+\;\frac{y^2}{B^2}=1

be two hyperbolas having length of latus rectums

15 \sqrt{2}

12 \sqrt{5}

respectively. Let their eccentricities be

e_1=\sqrt{\;\frac{5}{2}}

e_2

respectively. If the product of the lengths of their transverse axes is

100 \sqrt{10}

25 e_2^2

\;\;\;\;

[24 Jan 2025 Shift 2]

Your Answer:

Solution:

\;\;\;\;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1

\;\;\;\;\frac{2 b^2}{a}=15 \sqrt{2} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)

\;\;\;\sqrt{1+\;\frac{b^2}{a^2}}=\sqrt{\;\frac{5}{2}} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)

From (i) and (ii)

a=5 \sqrt{2}

and

b^2=75

\;\frac{x^2}{A^2}-\;\frac{y^2}{B^2}=-1

\;\frac{2 A^2}{B}=12 \sqrt{5} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (iii)

Since, product of transverse axis is

=100 \sqrt{10}

(2 A) \cdot (2 B)=100 \sqrt{10} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (iv)

From (iii) and (iv)

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

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