## Calculus (3rd Edition)

Using Theorem 1 and Theorem 4, we get $\frac{b}{a} = \sqrt {1 - {e^2}}$.
Consider the ellipse in standard position. Let $a>b>0$. By Theorem 1 we have $c = \sqrt {{a^2} - {b^2}}$. By Theorem 4, we have $e = \frac{c}{a}$. So, $e = \frac{{\sqrt {{a^2} - {b^2}} }}{a}$, ${\ \ }$ ${e^2} = \frac{{{a^2} - {b^2}}}{{{a^2}}}$ ${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}$, ${\ \ }$ $\frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}$ Since $a>b>0$, so $\frac{b}{a} = \sqrt {1 - {e^2}}$.