In the diagram below, you can move [math]P[/math] to be any point on the parabola.[br][list=1][br][*]First, consider a ray that is parallel to the axis of symmetry of the parabola and coming in to point [math]P[/math]. [i][Show the ray][/i] How will it be reflected off of the surface of the parabola? We can find the angles of incidence and reflection by using the tangent line to the parabola at [math]P[/math]. [i][Show the tangent][/i][br][/*][*]Now we can also look at the two segments [math]\overline{PF}[/math] and [math]\overline{PD}[/math]. By the definition of a parabola they are congruent. Also, notice that [math]\overline{PD}=y_0+p[/math]. [i][Show segments from [math]P[/math]][/i][br][/*][*]Consider [math]\overline{FA}[/math] where [math]A[/math] is the y-intercept of the tangent line. Notice that [math]FA=y_0+p[/math][br][/*][*]So we have [math]\overline{PD}\cong\overline{PF}\cong\overline{FA}[/math][br][/*][*]This means [math]\alpha=\beta[/math]. Why?[br][/*][*]Also we can conclude that [math]\gamma=\alpha[/math]. Why?[br][/*][*]In conclusion, because [math]\gamma=\beta[/math], we know that the ray to [math]P[/math] will be reflected along [math]\overline{PF}[/math] to [math]F[/math]. [i][Show the arrows][/i][br][/*][*]You can drag [math]P[/math] around to see that this will always be true at any point on the parabola.[br][/*][/list][br][b]Theorem[/b][br]Any ray parallel to the axis of symmetry will reflect off of the surface to the parabola to the focus.[br]Any ray emitted from the focus of the parabola will reflect off the surface such that its image is parallel to the axis of symmetry.[br][i][Show the arrows and reverse them to see this property][/i]