TITLE: Prove the slope of the asymptote of a hyperbola
QUESTION [1 upvotes]: Consider a hyperbola that opens horizontally (left to right):
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
How do we prove that the slopes of the asymptotes are $\pm \frac{b}{a}$?

REPLY [1 votes]: Not likely to be the proof you’re looking for, but a different approach: From the point of view of projective geometry, the asymptotes are the tangents to the curve at the points that it crosses the line at infinity. Since any given point at infinity is the intersection of a family of parallel lines, knowing these points for the hyperbola will tell you the slopes of its asymptotes.  
First, homogenize the equation of the hyperbola: $$ {(x-hw)^2 \over a^2}-{(y-kw)^2 \over b^2} = w^2 $$ and then, since points at infinity have $w=0$, set it to zero in the above equation: $${x^2 \over a^2}-{y^2 \over b^2} = \left(\frac xa+\frac yb\right) \left(\frac xa-\frac yb\right) = 0.\tag{*}$$ From this equation, we see that the hyperbola intersects the line at infinity at $[\pm a:b:0]$, which corresponds to lines with slopes of $\pm\frac ba$.  
Note that equation (*) is independent of $h$ and $k$, which we‘d expect since a translation shouldn’t affect the slopes of the asymptotes, but it’s also independent of the constant term in the equation of the hyperbola. This reflects the fact that the asymptotes are also the degenerate member of a family of hyperbolas with common asymptotes. Equation (*) is in fact the equation of the asymptotes for a family centered at the origin. This gives you a way to find an equation for the asymptotes of any hyperbola, not just one with axes parallel to the coordinate axes: translate to its center, set the constant term to zero, and translate back.
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