Tangent line to a hyperbola Can anyone explain and prove the following statement?
Any tangent line to a hyperbola touches the hyperbola halfway between the points of intersection of the tangent and the asymptotes.
Thank you very much for the help!

Derivating implicity the equation of the canonical hyperbola:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Longrightarrow \frac{2x}{a^2}dx-\frac{2y}{b^2}dy=0\Longrightarrow \frac{dy}{dx}=\frac{b^2x}{a^2y}$$
Thus, at any point $\,(x_1,y_1)\,$ on the hyperbola, the tangent line is
$$\text{I}\;\;\;y-y_1=\frac{b^2x_1}{a^2y_1}(x-x_1)\Longleftrightarrow y=\frac{b^2x_1}{a^2y_1}x-\frac{b^2}{y_1}$$
The hyperbola's asymptotes are the lines
$$\text{II}\;\;\;\;y=\pm\frac{b}{a}x$$
The intersection points between I and II are  
$$\frac{b^2x_1}{a^2y_1}x-\frac{b^2}{y_1}=\pm\frac{b}{a}x\Longrightarrow\left(\frac{b^2x_1}{a^2y_1}\mp\frac{b}{a}\right)x=\frac{b^2}{y_1}\Longrightarrow$$
$$x=\frac{a^2b}{bx_1\mp ay_1}\Longrightarrow y=\pm\frac{ab^2}{bx_1\mp ay_1}$$
so the points are
$$A=\left(\frac{a^2b}{bx_1-ay_1}\;,\;\frac{ab^2}{bx_1-ay_1}\right)\;\;,\;\;B=\left(\frac{a^2b}{bx_1+ay_1}\;,\;-\frac{ab^2}{bx_1+ay_1}\right)$$
The midpoint of $\,AB\,$ is
$$\frac{1}{2}\left(\left[\frac{a^2b}{bx_1-ay_1}+\frac{a^2b}{bx_1+ay_1}\right]\;,\;\left[\frac{ab^2}{bx_1-ay_1}-\frac{ab^2}{bx_1+ay_1}\right]\right)=$$
$$\frac{1}{2}(2x_1\,,\,2y_1)=(x_1,y_1)\;\;\;\;\;\;\;\;\;\;\;\;\square$$
Note: Be sure you can prove all the steps above, and remember: $\,(x_1,y_1)\,$ is a point on the hyperbola!
