## Calculus (3rd Edition)

Using Theorem 2 and Theorem 4, we get $e = \sqrt {1 + {m^2}}$.
Consider the hyperbola in standard position. Let $a>0$ and $b>0$. By Theorem 2, we have $c = \sqrt {{a^2} + {b^2}}$. By Theorem 4, we have $e = \frac{c}{a}$. So, $e = \frac{{\sqrt {{a^2} + {b^2}} }}{a} = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}$ Since the hyperbola has two asymptotes given by $y = \pm \frac{b}{a}x$, we see that $\pm \frac{b}{a}$ are the slopes of the asymptotes. Setting $m = \pm \frac{b}{a}$, we get $e = \sqrt {1 + {m^2}}$.