Explanation Given information: Polar form of a conic is given as r = 6 4 − 3 cos θ . Concept used: Polar form of a conic having focus at the pole(origin), eccentricity e , and the distance between focus ( F ) and directrix ( l ) equal to p is given as below. 1. When polar form of conic is r = e p 1 ± e cos θ , the directrix is a vertical line and graph of conic is symmetric about x − axis. 2. When polar form of conic is r = e p 1 ± e sin θ , the directrix is a horizontal line and graph of conic is symmetric about y − axis. Also, 1. If e < 1 , the conic is an ellipse. 2. If e = 1 , the conic is a parabola and 3. If e > 1 , the conic is a hyperbola. Calculation: Now, polar form of a conic is given as r = 6 4 − 3 cos θ . So, associated standard form is r = e p 1 − e cos θ . Now, converting given polar form into standard form, r = 6 4 − 3 cos θ ⇒ r = 6 4 4 4 − 3 cos θ 4 { Dividing both numerator and denominator by 4 } ⇒ r = 3 2 1 − 3 <

2nd Edition

ISBN: 9780321825391

Author: Ratti

Publisher: PEARSON

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Chapter 8, Problem 18PTA

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The conic section and sketch the graph.

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Chapter 8 Solutions

Precalculus: A Unit Circle Approach

Ch. 8.2 - Prob. 1E Ch. 8.2 - Prob. 2E Ch. 8.2 - Prob. 3E Ch. 8.2 - Prob. 4E Ch. 8.2 - Prob. 5E Ch. 8.2 - Prob. 6E Ch. 8.2 - Prob. 7E Ch. 8.2 - Prob. 8E Ch. 8.2 - Prob. 9E Ch. 8.2 - Prob. 10E

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The conic section and sketch the graph.

Solution Summary: The author describes the polar form of a conic with focus at the pole, eccentricity, and distance between focus and directrix.

The conic section and sketch the graph.

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Chapter 8 Solutions