My last post explains how parabolic reflectors have a magical property in that all incoming energy (light, sound, radio waves, etc.) traveling parallel to the axis of symmetry is reflected from the surface of the reflector to the focus point of the parabolic reflector. Conversely, all energy emanating from the focus point of a parabolic reflector is reflected to the surface of the reflector, and then in a direction that is parallel to its axis of symmetry. That post explained why parabolic reflectors work the way they do by verifying specific instances of the property.

This post will explain why parabolic reflectors have this property **for any point on the reflecting surface of a parabolic reflector** that is generated by rotating the graph of the equation y = kx^{2 }about its axis of symmetry. A principle of physics states that all energy in the form of particles or waves reflects off a flat surface in such a manner that the angle of incidence equals the angle of reflection. This post also demonstrates that the angle of incidence equals the angle of reflection where the flat surface is the tangent line at the point (x, y) on the graph of the equation y = kx^{2}.

All graphs in this post were created with my program Basic Trig Functions. Go to www.mathteachersresource.com to view multiple screen shots of the program’s modules. Click the ‘learn more’ button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

I will begin my demonstration by presenting some observations about the relationship between the slope of a line and the tangent of an angle that has its terminal side contained in the line. Because of this relationship, the trigonometric identity tan(α – β) will be used to calculate the tangent of an angle from the slopes of the sides of the angle. Diagram (1) below illustrates these relationships.

The next part of my demonstration presents some observations about the relationship between the slope of a line and the tangent function value of an angle that has its terminal side contained in the line. Because one side of the angle is contained in a vertical line, the trigonometric identity cot(α – β) will be used to calculate the cotangent of an angle from the reciprocals of the slopes of the sides of the angle. Diagram (2) below illustrates these observations.

The last part of my demonstration shows that Tan(angle ABC) = Tan(angle FBE) for any point B = (x, kx^{2 }). This in turn justifies the claim that m(angle ABC) = m(angle FBE) since the Tan(θ) function is one-to-one for values of θ that range from 0 to π/2 radians. Diagram (3) below illustrates these relationships.

The key steps in the demonstration are given in the text boxes below. Most of the steps depend on the concepts illustrated in diagrams (1) and (2). The points, lines and angles in the demonstration refer to the points, lines, and angles in diagram (3) above.

It’s sufficient to show that the magical property is true for parabolas of the form y = kx^{2} because translating, rotating, reflecting a graph over a line, and stretching the graph of a parabola preserves the property.