Explanation Theorem used: A polar equation of the form r = e d 1 ± e sin θ or r = e d 1 ± e cos θ represents a conic section with eccentricity e . The conic is an ellipse if e < 1 , a parabola if e = 1 , or a hyperbola if e > 1 . Calculation: The given polar equation is r = 4 5 + 6 cos θ . Obtain the eccentricity of the given equation. The polar equation for the given equation will be either r = e d 1 ± e sin θ or r = e d 1 ± e cos θ . Since the given polar equation is r = 4 5 + 6 cos θ , divide the given equation by 5 as follows. r = 4 5 + 6 cos θ r = 4 5 5 5 + 6 5 cos θ r = 4 5 1 + 6 5 cos θ From the above mentioned theorem, it is clear that e = 6 5 . Here, the eccentricity is greater than one, so the conic is Hyperbola. Since the eccentricity e = 6 5 , the polar equation can be rewrite as r = 4 5 1 + 6 5 cos θ

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ISBN: 9781337613927

Author: Stewart

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Question

Chapter 10.6, Problem 24E

To determine

To graph: The conic r=45+6cosθ and its directrix and also graph the conic obtained by rotating this curve about the origin through the angle π3.

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Chapter 10.6, Problem 23E

Chapter 10.6, Problem 25E

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Chapter 10 Solutions

Calculus, Early Transcendentals

Ch. 10.1 - Prob. 1E Ch. 10.1 - For the given parametric equations, find the...Ch. 10.1 - Sketch the curve by using the parametric equations...Ch. 10.1 - Sketch the curve by using the parametric equations...Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E

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To graph : The conic r = 4 5 + 6 cos θ and its directrix and also graph the conic obtained by rotating this curve about the origin through the angle π 3 .

To graph: The conic r=45+6cosθ and its directrix and also graph the conic obtained by rotating this curve about the origin through the angle π3.

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Chapter 10 Solutions