#### Answer

Using Theorem 2 and Theorem 4, we get $e = \sqrt {1 + {m^2}} $.

#### Work Step by Step

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}} $.