1122 APPENDIX A Polar Form of Conic Sections The distance between the pole and P is PF = r , so the ratio of PF to PP′ is PF PP′ = r r e = r , r e = r # e r = e. Simplify the complex fraction. Thus, by the definition, the graph has eccentricity e and must be a conic. In the preceding discussion, we assumed a vertical directrix to the right of the pole. There are three other possible situations. Location of the Directrix of a Conic Section If the equation is: then the directrix is: r = ep 1 + e cos u vertical, p units to the right of the pole. r = ep 1 - e cos u vertical, p units to the left of the pole. r = ep 1 + e sinu horizontal, p units above the pole. r = ep 1 - e sinu horizontal, p units below the pole. EXAMPLE 1 Graphing a Conic in Polar Form Graph r = 8 4 + 4 sin u . ALGEBRAIC SOLUTION Divide both numerator and denominator by 4 to obtain r = 2 1 + sin u . Based on the preceding table, this is the equation of a conic with ep = 2 and e = 1. Thus, p = 2. Because e = 1, the graph is a parabola. The focus is at the pole, and the directrix is horizontal, 2 units above the pole. The vertex must have polar coordinates 11, 90°2. Letting u = 0° and u = 180° gives the additional points 12, 0°2 and 12, 180°2. See Figure 2. 0° 180° 90° 270° 2 (1, 90°) (2, 0°) (2, 180°) r = 8 4 + 4 sin u Figure 2 As expected, the graph is a parabola, and it opens downward because the directrix is above the pole. GRAPHING CALCULATOR SOLUTION Enter r1 = 8 4 + 4 sin u , where the calculator is in polar and degree modes with polar coordinate displays. Figure 3(a) shows the window settings, and Figure 3(b) shows the graph. (a) −4.1 −6.6 4.1 6.6 (b) Figure 3 Notice that the point 11, 90°2 is indicated at the bottom. S Now Try Exercise 1.
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