## Explaining the Magical Property of Parabolic Reflectors – Part 2 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 = kx2 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 = kx2.

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, kx2 ). 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 = kx2 because translating, rotating, reflecting a graph over a line, and stretching the graph of a parabola preserves the property.

## Explaining the Magical Property of Parabolic Reflectors – Part 1 There is no coincidence that the reflective surface of car headlights, flashlights, spotlights, satellite dish antennas, solar dishes, solar cookers, microphones, reflecting telescopes, and radio telescopes have a unique curvature that can be described by the graph of the parabola y = kx2 where k = is a constant. The surface of a parabolic reflector is created by rotating the graph of a parabola about its axis of symmetry. 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. Because parabolic reflectors have this property, they are used to collect or project energy. The graph below shows the cross section of a parabolic reflector created by rotating the graph of the parabola y = 0.125x2 about its axis of symmetry. 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. So why do parabolic reflectors work the way they do? In this post, I will answer that question by showing how trig, pre-calculus, and calculus teachers can use modern computer graphing technology to help explain the properties of a parabolic reflector by verifying specific instances of the property. After completing the exercises themselves, students will be better prepared to follow or give a rational explanation of why this property is true for any point (x, y) on the parabola. My next post will explain why the angle of incidence equals the angle of reflection for any point (x, y) on the graph of the equation y = kx2. One of the cardinal principles of problem solving is to solve the problem for a simple or specific case of the problem before attempting to solve the general problem.

I would start my lesson by giving each student a handout that contains graphs similar to the graphs shown below. During the lesson, students would be required to use their calculators to make various calculations, verify specific facts during the lesson, and answer questions on the handout. Next, use a diagram similar to the diagram below to give a quick review of the definition of a parabola. Point (x, y) is on the parabola only if the distance from (x, y) to the focus point equals the distance from (x, y) to the directrix line. The class can check that points A, B, C, and D satisfy the definition of a parabola.  In the next part of the lesson, show students how they can use the slope of a line and the inverse tangent function on their calculators to find the positive angular direction of a ray. Adding + 3600 to the direction of a ray gives us an equivalent direction of a ray. Adding + 1800 to the direction of a ray gives us the direction of the opposite ray. The graph and the statements in the text box below illustrate how to determine the positive angular direction of a ray given the slope of the line that contains the ray.  The last part of the lesson is about demonstrating that the measure of angle ABC (angle of incidence) equals the measure of angle FBD (angle of reflection). 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. In the diagram below, the flat surface at point B = (6, 4.5) is the tangent line to the graph of the equation y = 0.125x2 at point B. From differential calculus, we learn that the slope of the tangent line to the graph of y = kx2 at any point (x, kx2) = 2kx. Using the results presented in the previous part of the lesson, the angular direction of a side of an angle can be calculated from the slope of the side of an angle. By knowing the direction of each side of an angle, we can find the degree measure of the angle by subtracting the directions of the sides of the angle. The text box below shows the verification that the measure of angle ABC equals the measure of angle FBD. For point M = (-4, 2), the slope of the tangent at point M = 0.125*2*(-4) = -1, the direction of ray ML = 1350, the direction of ray MD = 3150, and the measure of angle LMN = the measure of angle FMD = 450.   Teaching Points:

• Provide students with a handout that contains a graph similar to the last graph in this post.
• Have students show that the angle of incidence equals the angle of reflection for specific points on the parabola.
• There is no reason that students can’t be given the formula for the slope of a tangent line to a point (x, y) on a parabola. Tell them that they will learn how this formula comes about when they take calculus.
• Although it is not necessary to know the equation of the tangent line, have them find the equation of the tangent line. It’s a good review exercise and will better prepare them for calculus.
• A curious student will want to know why this magical property is true for any point (x, y) on any parabola. Have the student find the answer to this question, and let them present his/her solution to the class. This would make a wonderful Power Point presentation. Give the student the following hint: “Use the angle difference identity for the cotangent function to show Cot(900 – β) = Cot(2700 – β) = Cot(β) = 1/Tan(β).”
• Tell students that it’s sufficient to show that the magical property is true for parabolas of the form y = kx2 because translating, rotating, and stretching the graph of a parabola preserves the property.