Let H_1: x^2 a^2 - y^2 ~b^2 =1 and H_2:- x^2 ~A^2 + y^2 ~B^2 =1 be two hyperbolas having length of latus rectums 15 2 and 12 5 respectively. Let their ecentricities be e_1= (5)/(2) and e_2 respectively. If the product

Correct answer: 0 (x^2)/(a^2)-(y^2)/(b^2)=1 (2 b^2)/(a)=15 2 ...(i) 1+(b^2)/(a^2) = (5)/(2) ....(ii) From (i) and (ii) a=5 2 and b^2=75 (x^2)/(A^2)-(y^2)/(B^2)=-1 (2 A^2)/(B)=12 5 ....(iii) Since, product of transverse axis is =100 10 (2 A) ·(2 B)=100 10 From (iii) and (iv) aligned & A^2=150 and B=5 5 & e_2= 1+(A^2)/(B^2) = (11)/(5) & 25 e_2^2=25((11)/(5))=55 aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Hyperbola

Let H_1: x^2 a^2 - y^2 ~b^2 =1 and H_2:- x^2 ~A^2 + y^2 ~B^2 =1 be...

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 ________.

Correct answer

Step-by-step solution

\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}

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