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Q.
If hyperbola $\frac{x^2}{b^2}-\frac{y^2}{a^2}=1$ passes through the focus of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then eccentricity of hyperbola is.
Conic Sections
Solution:
$\frac{x^2}{b^2}-\frac{y^2}{a^2}=1$ passes through $(\pm a e, 0)$
$\therefore \frac{a^2 e^2}{b^2}=1$
$\therefore e^2=\frac{b^2}{a^2}$
$1-\frac{b^2}{a^2}=\frac{b^2}{a^2}$
$\therefore \frac{ b ^2}{ a ^2}=\frac{1}{2}$
$\therefore e_H=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3}$