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Question : 33 of 117

Marks: +1, -0

Let

E:\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1, a>b

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

. Let the distance between the foci of

E

and the foci of

H

2\sqrt{3}

a-A=2

, and the ratio of the eccentricities of

E

H

\;\frac{1}{3}

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

[22 Jan 2025 Shift 2]

8
7
10
9

Solution:

\;\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1 \;\text{ foci are }\;(a e, 0) \;\text{ and }\;(-a e, 0)

\;\;\frac{x^2}{A^2}+\;\frac{y^2}{B^2}=1 \;\text{ foci are }\;(A e', 0) \;\text{ and }\;(-A e', 0)

\;\Rightarrow 2 a e=2\sqrt{3} \Rightarrow a e=\sqrt{3}

\;\Rightarrow 2 A e'=2\sqrt{3} \Rightarrow A e'=\sqrt{3}

\;\Rightarrow a e=A e' \Rightarrow \;\frac{e}{e'}=\;\frac{A}{a}

\;\Rightarrow \;\frac{1}{3}=\;\frac{A}{a}, a=3A

\;\;\text{ Now, }\; a-A=2 \Rightarrow a-\;\frac{a}{3}-2=2

\;\Rightarrow a=3 \;\text{ and }\; A=1

\;A e=\sqrt{3} \Rightarrow e=\;\frac{1}{\sqrt{3}}, e'=\sqrt{3}

\;b^2=a^2 (1-e^2)

\;b^2=6

\;\;\text{ and }\; B^2=A^2 ((e')^2-1) =2

\;\Rightarrow B^2=2

\;\;\text{ sum of }\; L . R=\;\frac{2 b^2}{a}+\;\frac{2 B^2}{A}=8

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