1123 APPENDIX A Polar Form of Conic Sections EXAMPLE 2 Finding a Polar Equation Find the polar equation of a parabola with focus at the pole and vertical directrix 3 units to the left of the pole. SOLUTION The eccentricity e must be 1, p must equal 3, and the equation must be of the following form. r = e p 1 - e cos u r = 1 # 3 1 - 1 cos u Substitute for e and p. r = 3 1 - cos u Multiply. The calculator graph in Figure 4 supports our result. When u = 180°, r = 1.5. The distance from F10, 0°2 to the directrix is 2r = 211.52 = 3 units, as required. S Now Try Exercise 11. Conversion from Polar to Rectangular Form EXAMPLE 3 Identifying and Converting from Polar Form to Rectangular Form Identify the type of conic represented by r = 8 2 - cos u . Then convert the equation to rectangular form. SOLUTION To identify the type of conic, we divide both the numerator and the denominator on the right side of the equation by 2. r = 4 1 - 1 2 cos u From the table, we see that this is a conic that has a vertical directrix, with e = 1 2 , making it an ellipse. To convert to rectangular form, we start with the given equation. r = 8 2 - cos u Given equation r 12 - cos u2 = 8 Multiply by 2 - cos u. 2r - r cos u = 8 Distributive property 2r = r cos u + 8 Add r cos u to each side. 12r22 = 1r cos u + 822 Square each side. 12r22 = 1x + 822 r cos u = x 4r2 = x2 + 16x + 64 Multiply. 41x2 + y22 = x2 + 16x + 64 r2 = x2 + y2 4x2 + 4y2 = x2 + 16x + 64 Distributive property 3x2 + 4y2 - 16x - 64 = 0 Standard form The coefficients of x2 and y2 are both positive and are not equal, further supporting our assertion that the graph is an ellipse. S Now Try Exercise 19. −4.1 −6.6 4.1 6.6 Figure 4 Degree mode
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