TITLE: Eccentricity of a hyperbola given the angle between the x-axis and its asymptote
QUESTION [0 upvotes]: I need to find the eccentricity of a hyperbola whose asymptote makes an angle $\alpha$ with the $x$-axis.
So, I take the case where the transverse axis will be horizontal, $i.e.$
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
I note that $\tan(\alpha) = \frac{b}{a}$.
I need $\epsilon = \frac{a^2}{c}$, so I can get $a$ in terms of $\alpha$ and $b$, $i.e.$
$a^2 = \frac{b^2}{({tan\alpha})^2}$
by using the fact that the asymptote has equation $y=\frac{b}{a}x$  and ${tan\alpha} = \frac{b}{a}$
After this I get stuck.  I'm thinking that my solution must be in terms of $\alpha$ only, since I'm not given any information about $a$ or $b$.  And I know that $c^2 - a^2 = b^2$, but I'm not seeing the algebra necessary to properly substitute in so that the the eccentricity is represented only in terms of $\alpha$.
Am I on the right track, and, if so, where do I go from here?
I figure, once I have this case, I can easily make an analogous argument for a hyperbola with a vertical transverse axis.

REPLY [1 votes]: So $\tan\alpha = \frac{b}{a}$ and $c^2 - a^2 = b^2$
So we have: 
$c^2 = a^2 + b^2$
$c^2 = a^2 + (a^2)tan^2\alpha$
$c = \sqrt{a^2 + (a^2)tan^2\alpha}$
$c = a\sqrt{1 + tan^2\alpha}$
$c = a\sqrt{sec^2\alpha}$
$c = (a)(sec\alpha)$
$\frac{c}{a} = sec\alpha$
$\epsilon = sec\alpha$
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