Let $\mathrm{E}: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}\mathrm{b}$ and $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1$. Let the distance between the foci of E and the foci of H be $2 \sqrt{3}$. If $\mathrm{a}-\mathrm{A}=2$, 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 : [JEE Main 2025, 22 Jan (Shift 2)]

Question

Let $\mathrm{E}: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ and $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1$. Let the distance between the foci of E and the foci of H be $2 \sqrt{3}$. If $\mathrm{a}-\mathrm{A}=2$, 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 : [JEE Main 2025, 22 Jan (Shift 2)]

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

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

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