Explanation Results used: Polar equations of conic sections : Let d > 0. The conic section with a focus at the origin, the directrix x = ± d and eccentricity e has the polar equation r = e d 1 + e cos θ or r = e d 1 − e cos θ . It represents the conic section with focus at the origin the directrix y = ± d and eccentricity e has the polar equation r = e d 1 + e sin θ or r = e d 1 − e sin θ . If 0 < e <1, the conic section is an ellipse; if e = 1, it is a parabola; and if e > 1 it is a hyperbola. The curves are defined over any interval in θ of length 2 π . Calculation: The given equation of the curve is r = 4 1 + cos θ . Compare with r = e d 1 + e cos θ and obtain that e d = 4 and e = 1 . Therefore, obtain that d = 4 . As the term cos θ is in the equation the directrix are vertical. Therefore, the directrix is x = 4 . As e = 1 the curve is a parabola. As the directrix is vertical the vertex lies on the line which is parallel to the x -axis. So to calculate the vertex, substitute θ = 0 in r = 4 1 + cos θ

MyLab Math with Pearson eText -- Standalone Access Card -- for Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134856926

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Question

Chapter 12.4, Problem 57E

To determine

To graph: The curve r=41+cosθ and mention the vertices, foci, directrix and asymptotes ( if exists ) on the graph; verify the answer by using graphing utility.

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Chapter 12.4, Problem 56E

Chapter 12.4, Problem 58E

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