Explanation Given information: Polar form of a conic is given as r = 4 2 − 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 = 4 2 − 3 cos θ . So, associated standard form is r = e p 1 − e cos θ . Now, converting given polar form into standard form, r = 4 2 − 3 cos θ ⇒ r = 4 2 2 2 − 3 cos θ 2 { Dividing both numerator and denominator by 2 } ⇒ r = 2 1 − 3 2 cos θ Comparing given polar form with standard polar form, we get - e = 3 2 and e p = 2 So, p = 4 3

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Chapter 8, Problem 63RE

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The polar graph of given conic.

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Chapter 8, Problem 62RE

Chapter 8, Problem 64RE

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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 polar graph of given conic.

Solution Summary: The author describes the polar graph of a given conic with focus at the pole, eccentricity, and distance between focus and directrix equal to p.

The polar graph of given conic.

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Chapter 8 Solutions