Let H be the hyperbola, whose foci are (1 √2,0) and eccentricity is √2.

Question

Let $H$ be the hyperbola, whose foci are $(1\pm \sqrt{2},0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is :

Accepted Answer

$ 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

$\Rightarrow \mathrm{L} . \mathrm{R}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2$

Solutions · 1