Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Review - Exercises - Page 691: 56


$\theta= \cos^{-1} (\pm\dfrac{1}{e})$

Work Step by Step

Given: $r=\dfrac{ed}{1-e \cos \theta}$ This can be re-written as: $1-e\cos \theta=\dfrac{ed}{r}$ or, $\cos \theta=\dfrac{1}{e}(1-\dfrac{ed}{r})$ This gives: $\theta=\cos^{-1} (\dfrac{1}{e}-\dfrac{d}{r})$ The asympototes for the hyperbola are: $\theta=\lim\limits_{r \to \infty}\cos^{-1} (\dfrac{1}{e}-\dfrac{d}{r})$ or, $\theta=\pm \cos^{-1} (\dfrac{1}{e})$ Hence, $\theta= \cos^{-1} (\pm\dfrac{1}{e})$
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