JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 12)
Let $\mathrm{H}$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is :
$\frac{5}{2}$
3
2
$\frac{3}{2}$
Explanation
$ 2 \mathrm{ae}=|(1+\sqrt{2})-(1+\sqrt{2})|=2 \sqrt{2}$
$$ \Rightarrow $$ $\mathrm{ae}=\sqrt{2}$
$$ \Rightarrow $$ $\mathrm{a}=1$
$\Rightarrow \mathrm{b}=1 \because \mathrm{e}=\sqrt{2} \Rightarrow$ Hyperbola is rectangular
$\Rightarrow \mathrm{L} . \mathrm{R}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2$
$$ \Rightarrow $$ $\mathrm{ae}=\sqrt{2}$
$$ \Rightarrow $$ $\mathrm{a}=1$
$\Rightarrow \mathrm{b}=1 \because \mathrm{e}=\sqrt{2} \Rightarrow$ Hyperbola is rectangular
$\Rightarrow \mathrm{L} . \mathrm{R}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2$
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