Let \(\mathrm{E}: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}\) and…
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)]
8
\(\text{ }\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\text{ foci are }(\pm a{e}_{1},0)\\ \frac{{x}^{2}}{{A}^{2}}-\frac{{y}^{2}}{{B}^{2}}=1\text{ foci are }\left(\pm A{e}_{2},0\right)\\ \Rightarrow 2{\mathrm{ae}}_{1}=2\sqrt{3}\Rightarrow {\mathrm{ae}}_{1}=\sqrt{3}\\ \text{ and }2A{e}_{2}=2\sqrt{3}\Rightarrow A{e}_{2}=\sqrt{3}\\ \text{and}2{\mathrm{ae}}_{1}=2A{e}_{2}\Rightarrow \frac{{e}_{1}}{{e}_{2}}=\frac{A}{a}=\frac{1}{3}\left(\text{Given}\right)\\ \Rightarrow a=3A..\text{.}\left(1\right)\\ \text{ Now }a-A=2\Rightarrow a-\frac{a}{3}=2\\ \Rightarrow a=3\text{ and }A=1\\ a{e}_{1}=\sqrt{3}\Rightarrow {\mathrm{e}}_{1}=\frac{1}{\sqrt{3}}\text{ and }{\mathrm{e}}_{2}=\sqrt{3}\\ {\mathrm{b}}^{2}={\mathrm{a}}^{2}\left(1-{{\mathrm{e}}_{1}}^{2}\right)\\ {\mathrm{b}}^{2}=6\\ \text{ and }{\mathrm{B}}^{2}={\mathrm{A}}^{2}\left({\left({\mathrm{e}}_{2}\right)}^{2}-1\right)=(2)\Rightarrow {\mathrm{B}}^{2}=2\\ \text{sum of }\mathrm{LR}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}+\frac{2{\mathrm{B}}^{2}}{\mathrm{A}}=8\\\)
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