997 10.1 Parabolas The equation of the parabola is found as follows. 1y - k22 = 4p1x - h2 Parabola with horizontal axis of symmetry 1y - 322 = 41-221x - 12 Let p = -2, h = 1, and k = 3. 1y - 322 = -81x - 12 Multiply. The domain is 1-∞, 14, and the range is 1-∞, ∞2. S Now Try Exercise 43. Headlight Solar furnace Focus 32 ft 210 ft Figure 12 An Application of Parabolas Parabolas have a special reflecting property that makes them useful in the design of telescopes, radar equipment, auto headlights, and solar furnaces. When a ray of light or a sound wave traveling parallel to the axis of symmetry of a parabolic shape bounces off the parabola, it passes through the focus. For example, in a solar furnace, a parabolic mirror collects light at the focus and thereby generates intense heat at that point. If a light source is placed at the focus, then the reflected light rays will be directed straight ahead. x y (–105, 32) (105, 32) 50 25 –50 50 100 –100 0 Figure 13 EXAMPLE 6 Modeling the Reflective Property of Parabolas The Parkes radio telescope has a parabolic dish shape with diameter 210 ft and depth 32 ft. Because of this parabolic shape, distant rays hitting the dish will be reflected directly toward the focus. A cross section of the dish is shown in Figure 12. (Data from Mar, J., and H. Liebowitz, Structure Technology for Large Radio and Radar Telescope Systems, The MIT Press, Massachusetts Institute of Technology.) (a) Determine an equation that models this cross section by placing the vertex at the origin with the parabola opening up. (b) The receiver must be placed at the focus of the parabola. How far from the vertex of the parabolic dish should the receiver be located? SOLUTION (a) Locate the vertex at the origin as shown in Figure 13. The form of the parabola is x2 = 4py. The parabola must pass through the point A210 2 , 32B, or 1105, 322. Use this information to solve for p. x2 = 4py Parabola with vertical axis of symmetry 110522 = 4p1322 Let x = 105 and y = 32. 11,025 = 128p Multiply. p = 11,025 128 Solve for p. The cross section can be modeled by the following equation. x2 = 4py Parabola with vertical axis of symmetry x2 = 4 a 11,025 128 b y Substitute for p. x2 = 11,025 32 y Simplify. (b) The distance between the vertex and the focus is p. In part (a), we found p = 11,025 128 ≈86.1, so the receiver should be located at 10, 86.12, or 86.1 ft above the vertex. S Now Try Exercise 51.
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