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

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:
  1. 10
  2. 9
  3. 8
  4. 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 $\mathrm{B}^2=\mathrm{A}^2\left(\left(\mathrm{e}^{\prime}\right)^2-1\right)=(2) \Rightarrow \mathrm{B}^2=2$
$\text { sum of } \mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}+\frac{2 \mathrm{~B}^2}{\mathrm{~A}}=8$

Asked in: JEE Main 2025 (22 Jan Shift 2)

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