# Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 731: 56

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

#### Work Step by Step

Given: $r=\dfrac{ed}{1-e \cos \theta}$ Need to arrange the above equation as: $1-e\cos \theta=\dfrac{ed}{r}$ This implies that $\cos \theta=\dfrac{1}{e}(1-\dfrac{ed}{r})$ $\theta=\cos^{-1} (\dfrac{1}{e}-\dfrac{d}{r})$ (Simplify) Thus, the asympototes for the hyperbola can be found as : $\theta=\lim\limits_{r \to \infty}\cos^{-1} (\dfrac{1}{e}-\dfrac{d}{r})$ $\implies \theta=\pm \cos^{-1} (\dfrac{1}{e})$ Hence, $\theta= \cos^{-1} (\pm\dfrac{1}{e})$

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