Explanation Notice that r = k 1 + e cos θ and x = r cos θ , y = r sin θ . r 2 = ( k 1 + e cos θ ) 2 r 2 ( 1 + e cos θ ) 2 = k 2 r 2 ( 1 + e 2 cos 2 θ + 2 e cos θ ) = k 2 ( r 2 + r 2 e 2 cos 2 θ + 2 r 2 e cos θ ) = k 2 Further simplification of above equation. [ r 2 + e 2 ( r cos θ ) 2 + 2 r e ( r cos θ ) ] = k 2 x 2 + y 2 + e 2 x 2 + 2 r e ( x ) = <

Thomas' Calculus - MyMathLab Integrated Review

14th Edition

ISBN: 9780134786223

Author: Hass

Publisher: PEARSON

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Chapter 11, Problem 87PE

To determine

To show: That (1−e2)x2+y2+2kex−k2=0, where x=rcosθ,y=rsinθ and r=k1+ecosθ.

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Chapter 11, Problem 86PE

Chapter 11, Problem 88PE

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Chapter 11 Solutions

Thomas' Calculus - MyMathLab Integrated Review

Ch. 11.1 - Finding Cartesian from Parametric...Ch. 11.1 - Finding Cartesian from Parametric...Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Finding Cartesian from Parametric...

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That ( 1 − e 2 ) x 2 + y 2 + 2 k e x − k 2 = 0 , where x = r cos θ , y = r sin θ and r = k 1 + e cos θ .

To show: That (1−e2)x2+y2+2kex−k2=0, where x=rcosθ,y=rsinθ and r=k1+ecosθ.

Want to see the full answer?

Chapter 11 Solutions