SECTION 10.6 Polar Equations of Conics 731 Polar Equations of Conics (Focus at the Pole, Eccentricity e ) Equation Description r ep e 1 cosθ = − Directrix is perpendicular to the polar axis at a distance p units to the left of the pole. r ep e 1 cosθ = + Directrix is perpendicular to the polar axis at a distance p units to the right of the pole. r ep e 1 sinθ = + Directrix is parallel to the polar axis at a distance p units above the pole. r ep e 1 sinθ = − Directrix is parallel to the polar axis at a distance p units below the pole. Eccentricity • If e 1, = the conic is a parabola; the axis of symmetry is perpendicular to the directrix. • If e 1, < the conic is an ellipse; the major axis is perpendicular to the directrix. • If e 1, > the conic is a hyperbola; the transverse axis is perpendicular to the directrix. Table 5 Exploration Graph r 4 2 cos 1 θ = + and compare the result with Figure 58. What do you conclude? Clear the screen and graph r 4 2 sin 1 θ = − and then r 4 2 sin . 1 θ = + Compare each of these graphs with Figure 58. What do you conclude? Equation (6) assumes that the directrix is perpendicular to the polar axis at a distance p units to the left of the pole. If the directrix is perpendicular to the polar axis at a distance p units to the right of the pole, then r ep e 1 cosθ = + See Problem 43. In Problems 44 and 45, you are asked to derive the polar equations of conics with focus at the pole and directrix parallel to the polar axis. Table 5 summarizes the polar equations of conics. Now Work PROBLEM 13 Analyzing and Graphing the Polar Equation of a Conic Analyze and graph the equation r 6 3 3sin . θ = + Solution EXAMPLE 2 To put the equation in proper form, divide the numerator and denominator by 3. r 2 1 sinθ = + See Table 5. This conic has its directrix parallel to the polar axis, a distance p units above the pole. e ep p 1 and 2 2 = = = e 1 = Since e 1, = the conic is a parabola with focus at the pole.The directrix is parallel to the polar axis 2 units above the pole; the axis of symmetry is perpendicular to the polar axis.The vertex of the parabola is at 1, 2 . π ( ) (Do you see why?) See Figure 59(a) for the graph. Notice that two additional points, 2, 0 ( ) and 2, , π ( ) are plotted to assist in graphing. Figure 59(b) shows the graph using Desmos. Figure 59 r 6 3 3sinθ = + Polar axis Directrix F (2, 0) (2, p) 2 p (1, ) (a) (b)
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