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