Explanation All polar equations of conics are of the form: r = e d 1 ± e cos θ Or r = e d 1 ± e sin θ . If e < 1 , then the conic is an ellipse If e > 1 , then the conic is a hyperbola If e = 1 , then the conic is a parabola Given: r = d e 1 + e cos θ , Let e > 1 Formula Used: The polar equation of a conic with eccentricity e and directrix x = d is: r = e d 1 + e cos θ Calculation: The polar equation of the standard hyperbola with focus at origin and directrix x = d is given and the vertices will lie on the x -axis. The x -coordinates of the vertices can be obtained at θ = 0 and θ = π in the given polar equation...

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 12.5, Problem 62E

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We need to show that the vertices of the hyperbola $r = \frac{d e}{1 + e \cos θ}$ have x-coordinates $\frac{e d}{e + 1}$ and $\frac{e d}{e - 1}$ .

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Chapter 12.5, Problem 61E

Chapter 12.5, Problem 63E

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Chapter 12 Solutions

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Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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We need to show that the vertices of the hyperbola r = d e 1 + e cos θ have x -coordinates e d e + 1 and e d e − 1 .

Solution Summary: The polar equation of a conic with eccentricity e and directrix x = d is given and the vertices will lie on X -axis.

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We need to show that the vertices of the hyperbola $r = \frac{d e}{1 + e \cos θ}$ have x-coordinates $\frac{e d}{e + 1}$ and $\frac{e d}{e - 1}$ .

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Chapter 12 Solutions

We need to show that the vertices of the hyperbola r = d e 1 + e cos θ have x -coordinates e d e + 1 and e d e − 1 .

Solution Summary: The polar equation of a conic with eccentricity e and directrix x = d is given and the vertices will lie on X -axis.

Videos

We need to show that the vertices of the hyperbola r=de1+ecosθ have x-coordinates ede+1 and ede−1.

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Chapter 12 Solutions

We need to show that the vertices of the hyperbola $r = \frac{d e}{1 + e \cos θ}$ have x-coordinates $\frac{e d}{e + 1}$ and $\frac{e d}{e - 1}$ .