Let [math]P[/math] be a point on the hyperbola [math]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/math][br]Draw in the focal radii.[br]Let [math]Q[/math] be a point on [math]\overline{F_1P}[/math] such that [math]QP=PF_2[/math].[br] Notice[math]F_1Q=PF_1-QP[/math][br] [math]F_1Q=PF_1-PF_2[/math] [br] [math]F_1Q=2a[/math][br][br]Let [math]M[/math] be the midpoint of [math]\overline{QF_2}[/math]Draw line [math]\overline{PM}[/math]. What can you conclude about it? [br][list=1][*]It appears to be tangent[br][/*][*][math]\overline{PM}[/math] bisects[math]\angle QPF_2[/math] [/*][*][math]\overline{PM}[/math] is the perpendicular bisector of [math]\overline{QF_2}[/math][br][/*][/list][br]Can we prove it is tangent?[br]Let [math]P'[/math] be any point on [math]\overline{PM}[/math] where[math]P'\ne P[/math] and connect it to [math]F_1[/math], [math]F_2[/math] and [math]Q[/math].[br][math]P'Q=P'F_2[/math] because [math]P'[/math] is on the perpendicular bisector of [math]\overline{QF_2}[/math][br][math]F_1Q+P'Q>P'F_1[/math] by the triangle inequality[br][math]F_1Q+P'F_2>P'F_1[/math][br][math]F_1Q>P'F_1-P'F_2[/math][br][math]2a>P'F_1-P'F_2[/math][br]So [math]P'[/math] is not on the hyperbola! Which means [math]P[/math] is the only point on [math]\overline{PM}[/math], which is also on the hyperbola. In other words, [math]\overline{PM}[/math] is tangent to the hyperbola at [math]P[/math].[br][br]Now extend the focal radii. We can see that all four angles formed are congruent to one another.[br][br]Since the angles formed with the tangent are congruent, we can see that a ray following the path starting [b]from[/b] one focus will reflect off of the hyperbola directly [b]away[/b] from the other focus.[br][br]Similarly, a ray directed [b]towards[/b] one focus will reflect off of the hyperbola [b]towards[/b] the other focus.[br]