<s>[INST] 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!
 [/INST] 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!
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