Q.Show that angle between the two asymptotes of a hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $2\tan^{-1}(b/a)$ or $2\sec^{-1}(e)$.

Hyperbola

The standard hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has centre at the origin,

transverse axis $2a$, conjugate axis $2b$, and eccentricity $e$ with

$b^2=a^2(e^2-1)$, so $e>1$. Its foci are $(\pm ae,0)$, directrices $x=\pm\dfrac ae$, and

each latus rectum has length $\dfrac{2b^2}{a}$; the asymptotes are $y=\pm\dfrac ba x$.

The conjugate hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=-1$, whose eccentricity

$e'$ satisfies $\dfrac1{e^2}+\dfrac1{e'^2}=1$.

The line $y=mx+c$ touches the hyperbola iff $c^2=a^2m^2-b^2$, giving the tangent

$y=mx\pm\sqrt{a^2m^2-b^2}$; the tangent at $(a\sec\theta,b\tan\theta)$ is

$\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$. A general hyperbola is put in this

form by completing squares (translation of centre) or by a rotation. These tools …