Question

The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ is given by _________

Answer 1

Given: Equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$

Which represents hyperbola having foci on $y-$axis

$\because a^2=b^2(e^2-1)$

$\Rightarrow \frac{a^2}{b^2}=e^2-1$

$\Rightarrow 1+\frac{a^2}{b^2}=e^2$

$\Rightarrow e=\sqrt{1+\frac{a^2}{b^2}}$

Hence, the eccentricity of the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ is given by $e=\sqrt{1+\frac{a^2}{b^2}}$