Explanation Result used: For polar equation of the form r = e p 1 − e sin θ . e is the eccentricity of the conic If e = 1 , the conic is parabola with focus at the pole. If e > 1 , the conic is hyperbola with focus at the pole. If e < 1 , the conic is ellipse with focus at the pole. The directrix is parallel to the polar axis at a distance p units to the below to the pole. Calculation: Consider the given equation, r = 6 2 − sin θ . Simplify the equation as shown follows. r = 6 2 − sin θ = 6 2 ( 1 − 1 2 sin θ ) = 3 1 − 1 2 sin θ Compare the general form r = e p 1 − e sin θ with the given equation, e = 1 2 and e p = 3 . Substitute 1 2 for e in the equation e p = 3 to get p . 1 2 ( p ) = 3 p = 6 Here, e = 1 2 , the given conic is an ellipse with the focus at the pole. As the general form of the equation is r = e p 1 − e sin θ , the directrix is parallel to the polar axis at a distance p units that is 6 units to the below of the pole. Note that, the vertices of the parabola is found out by substituting π 2 , 3 π 2 for θ in the equation

11th Edition

ISBN: 9780135227688

Author: Sullivan

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Chapter 11, Problem 28RE

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To identify: The conic that polar equation r=62−sinθ, represents, and graph it.

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Chapter 11, Problem 27RE

Chapter 11, Problem 29RE

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Chapter 11 Solutions

EBK ALGEBRA+TRIGONOMETRY

Ch. 11.2 - Prob. 1AYU Ch. 11.2 - Prob. 2AYU Ch. 11.2 - Prob. 3AYU Ch. 11.2 - Prob. 4AYU Ch. 11.2 - Prob. 5AYU Ch. 11.2 - Prob. 6AYU Ch. 11.2 - Prob. 7AYU Ch. 11.2 - Prob. 8AYU Ch. 11.2 - Prob. 9AYU Ch. 11.2 - Prob. 10AYU

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The conic that polar equation r = 6 2 − sin θ , represents, and graph it.

To identify: The conic that polar equation r=62−sinθ, represents, and graph it.

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Chapter 11 Solutions