Question

If the pair of lines $b^2{x}^2-a^2{y}^2=0$ are inclined at an angle $\theta$, then find the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Answer 1

Consider $b^2{x}^2-a^2{y}^2=0$

$\Rightarrow b^2{x}^2=a^2{y}^2$

$\Rightarrow y^2=\frac{b^2{x}^2}{a^2}$

$\Rightarrow y=\pm \frac{b}{a}x$

Thus $b^2{x}^2-a^2{y}^2=0$ is a pair of intersecting lines $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

The angle between them is $2\theta =2ta{n}^{-1}\left(\frac{b}{a}\right)$.

Also $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$ represents the asymptotes of the hyperbola with eccentricity $e=sec\theta$ or $e=\frac{\sqrt{a^2+b^2}}{a}$