## Calculus: Early Transcendentals 8th Edition

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