730 CHAPTER 10 Analytic Geometry Analyzing and Graphing the Polar Equation of a Conic Analyze and graph the equation r 4 2 cos . θ = − Solution EXAMPLE 1 The equation is not quite in the form of equation (6), since the constant term in the denominator is 2 instead of 1. Divide the numerator and denominator by 2 to obtain r 2 1 1 2 cosθ = − r ep e 1 cosθ = − This equation is in the form of equation (6), with e ep 1 2 and 2 = = Then p p 1 2 2, so 4 = = Since e 1 2 1, = < the conic is an ellipse. One focus is at the pole, and the directrix is perpendicular to the polar axis, p 4 = units to the left of the pole. The major axis is along the polar axis. To find the vertices, let 0 θ = and . θ π = The vertices of the ellipse are 4, 0 ( ) and 4 3 , . π ( ) The center of the ellipse is the midpoint of the vertices, namely, 4 3 , 0 . ( ) [Do you see why? The vertices 4, 0 ( ) and 4 3 , π ( ) in polar coordinates are 4, 0 ( ) and 4 3 , 0 ( ) − in rectangular coordinates. The midpoint in rectangular coordinates is 4 3 , 0 , ( ) which is also 4 3 , 0 ( ) in polar coordinates.] Then a distance = from the center to a vertex 8 3 . = Using a 8 3 = and e 1 2 = in equation (2), e c a , = yields c ae 4 3 . = = Finally, using a 8 3 = and c 4 3 = in b a c 2 2 2 = − yields b a c b 64 9 16 9 48 9 4 3 3 2 2 2 = − = − = = Figure 58(a) shows the graph. Figure 58(b) shows the graph on a TI-84 Plus CE. Figure 58 r 4 2 cosθ = − Polar axis Directrix F (4, 0) 4 – 3( , 0) 4 – 3( , p) 4 3 3 (a) 3 23 24.8 4.8 (b) 4 2 2 cos u r 1 5 THEOREM Polar Equation of a Conic; Focus at the Pole; Directrix Perpendicular to the Polar Axis a Distance p to the Left of the Pole The polar equation of a conic with focus at the pole and directrix perpendicular to the polar axis at a distance p to the left of the pole is r ep e 1 cosθ = − (6) where e is the eccentricity of the conic. Now Work PROBLEM 11
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