Let E: (x^2)/(a^2)+(y^2)/(~b^2)=1, agtb 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 the lengths of their latus rectums is equal to:

Correct answer: (C) 8. (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^′ ⇒ fracee^′=A/a ⇒ 1/3=A/a ⇒ a=3 ~A Now a-A=2 ⇒ a-a/3-2 ⇒ a=3 and A=1 Ae=√3 ⇒ e=1/(√3) and e^′=√3 b^2=a^2 (1-e^2 ) b^2=6 and B^2=A^2 ( (e^′ )^2-1 )=(2) ⇒ B^2=2 sum of LR=(2 ~b^2)/a+(2 ~B^2)/(~A)=8

Q45 / 75 · JEE Main 2025 (22 Jan Shift 2) · Mathematics

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 $H$ be $2 \sqrt{3}$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$, then the sum of the lengths of their latus rectums is equal to:

A
10
B
9
Correct answer
C
8
D
7

Solution

$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { foci are }(\mathrm{ae}, 0) \text { and }(-\mathrm{ae}, 0) \\ & \frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1 \text { foci are }\left(\mathrm{Ae}^{\prime}, 0\right) \text { and }\left(-\mathrm{Ae}^{\prime}, 0\right) \\ & \Rightarrow 2 \mathrm{ae}=2 \sqrt{3} \Rightarrow \mathrm{ae}=\sqrt{3} \\ & \text { and } 2 \mathrm{Ae}^{\prime}=2 \sqrt{3} \Rightarrow \mathrm{Ae}^{\prime}=\sqrt{3} \\ & \Rightarrow \mathrm{ae}=\mathrm{Ae}^{\prime} \Rightarrow \frac{\mathrm{e}}{\mathrm{e}^{\prime}}=\frac{\mathrm{A}}{\mathrm{a}} \\ & \Rightarrow \frac{1}{3}=\frac{\mathrm{A}}{\mathrm{a}} \Rightarrow \mathrm{a}=3 \mathrm{~A} \end{aligned}$ 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 $\mathrm{B}^2=\mathrm{A}^2\left(\left(\mathrm{e}^{\prime}\right)^2-1\right)=(2) \Rightarrow \mathrm{B}^2=2$ $\text { sum of } \mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}+\frac{2 \mathrm{~B}^2}{\mathrm{~A}}=8$

Hence, the correct answer is option: 'C'.

Solution

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