Let E: x^2 a^2 + y^2 ~b^2 =1, a and H: x^2 ~A^2 - y^2 ~B^2 =1 . Let the distance between the foci of E and the foci of H be 2 3 . If a-A=2 , and the ratio of the eccentricities of E and H is (1)/(3) , then the sum of

Correct answer: C. 8 aligned & x^2 a^2 + y^2 b^2 =1 foci are (ae, 0) and (-ae, 0) & x^2 A^2 + y^2 B^2 =1 foci are (Ae^ , 0) and (-Ae^ , 0) & 2 ae=2 3 ae= 3 & and 2 Ae^ =2 3 Ae^ = 3 & ae=Ae^ e e^ = A a & (1)/(3)= A a a=3 A aligned Now a-A=2 a- a 3 -2 a=3 and A=1 Ae= 3 e= 1 3 and e^ = 3 aligned & b^2=a^2(1-e^2) & b^2=6 aligned and B^2=A^2((e^ )^2-1)=(2) B^2=2 sum of LR= 2 b^2 a + 2 B^2 A =8

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Hyperbola

Let E: x^2 a^2 + y^2 ~b^2 =1, a and H: x^2 ~A^2 - y^2 ~B^2 =1 . Let...

Let

\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}\gt\mathrm{b}

and

\mathrm{H}: \frac{x^2}{\mathrm{~A}^2}-\frac{y^2}{\mathrm{~B}^2}=1

. Let the distance between the foci of E and the foci of

H

2 \sqrt{3}

. If

a-A=2

, and the ratio of the eccentricities of

E

and

H

\frac{1}{3}

, then the sum of the lengths of their latus rectums is equal to:

Options

Correct answer

C. 8

Step-by-step solution

\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{ b}^2}=1 \text{ foci are }(\mathrm{ae}, 0) \text{\text{ and } }(-\mathrm{ae}, 0) \\ & \frac{\mathrm{x}^2}{\mathrm{ A}^2}+\frac{\mathrm{y}^2}{\mathrm{ B}^2}=1 \text{ foci are }\left(\mathrm{Ae}^{\prime}, 0\right) \text{ and }\left(-\mathrm{Ae}^{\prime}, 0\right) \\ & \Rightarrow 2 \mathrm{ae}=2 \sqrt{3} \Rightarrow \mathrm{ae}=\sqrt{3} \\ & \text{ and } 2 \mathrm{Ae}^{\prime}=2 \sqrt{3} \Rightarrow \mathrm{Ae}^{\prime}=\sqrt{3} \\ & \Rightarrow \mathrm{ae}=\mathrm{Ae}^{\prime} \Rightarrow \frac{\mathrm{e}}{\mathrm{e}^{\prime}}=\frac{\mathrm{A}}{\mathrm{a}} \\ & \Rightarrow \frac{1}{3}=\frac{\mathrm{A}}{\mathrm{a}} \Rightarrow \mathrm{a}=3 \mathrm{ A} \end{aligned}

Now

\mathrm{a}-\mathrm{A}=2 \Rightarrow \mathrm{a}-\frac{\mathrm{a}}{3}-2 \Rightarrow \mathrm{a}=3

and

\mathrm{A}=1 \mathrm{Ae}=\sqrt{3} \Rightarrow \mathrm{e}=\frac{1}{\sqrt{3}} \text{ and } \mathrm{e}^{\prime}=\sqrt{3}

\begin{aligned} & \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & \mathrm{b}^2=6 \end{aligned}

and \mathrm{B}^2=\mathrm{A}^2\left(\left(\mathrm{e}^{\prime}\right)^2-1\right)=(2) \Rightarrow \mathrm{B}^2=2 \text{ sum of } \mathrm{LR}=\frac{2 \mathrm{ b}^2}{\mathrm{a}}+\frac{2 \mathrm{ B}^2}{\mathrm{ A}}=8

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