Would appreciate if someone can help me with this one.
The angle between the asymptotes of a hyperbola is $\dfrac{\pi}{3}$. Determine the eccentricity of the hyperbola.
Since I'm trying to self teach myself here, the only thing I could find was that the tangent of the angle between the asymptotes is $\dfrac{2ab}{a^2-b^2}$. I also know the formula for the eccentricity but I can't figure out how to find it out of that angle alone.
Best Answer
(Note: this answer shows how to work it out using the supplied formula for the tangent, so I'll keep it up, but there's a much simpler graphical method here.)
First off, there's an typo in your formula for eccentricity. It should be $\sqrt{(a^2 + b^2)/a^2}$.
With that out of the way, the key point to figuring this out is to realize that both the eccentricity and the angle really only depend on the ratio $b/a$, as they both measure how "squashed" the hyperbola is. So our job is now figuring out how to massage the formulas so they only depend on $b/a$, and not on $a$ or $b$ individually. For the tangent of the angle
$$\dfrac{2ab}{a^2-b^2} = \dfrac{1/a^2}{1/a^2}\cdot \dfrac{2ab}{a^2-b^2} = \dfrac{2ab/a^2}{a^2/a^2-b^2/a^2} = \dfrac{2(b/a)}{1-(b/a)^2}. $$
You can do something similar for the eccentricity formula.
Can you take it from there? Me, once I've rewritten the formulas in terms of $b/a$, I might introduce a new variable $k= b/a$ just to make the subsequent manipulations easier.