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I got that equation of path is conic section $u=\frac{1}{3c}(1+2\cos\theta)$ where $c$ is constant and one vertex of hyperbola is $(-c,0)$ and $u=r^{-1}$. So, $r=\frac{3c}{1+2\cos\theta}$. Since $e=2>1$ is eccentricity, the path is hyperbolic.
How can I find equations of asymptotes and latus rectum?
I know that $e=2$ and $a=c$.
Also, $e=\frac{\sqrt{a^2+b^2}}{a}$. From this I get $b=c\sqrt3$.
Asymptote that I need is $y=-\frac{b}{a}x=\sqrt3 x$ and latus rectum is $(-c,\frac{b^2}{a})=(-c,3c)$.
Is this good?

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up vote 1 down vote accepted

For the asymptotes, $u=0$, which means that $\cos{\theta}=-1/2$, or $\theta = 2 \pi/3$ or $4 \pi/3$, which means that the slope of the asymptotes is $\tan{\theta}=\pm \sqrt{3}$.

For the latus rectum, the length is $2 b^2/a = 6 c$.

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Why does it hold that $u=0$ for asymptotes? – gov Jul 23 '13 at 13:54
$u = 1/r$, so this would correspond to $r \to \infty$, which is by definition the behavior near the asymptotes. – Ron Gordon Jul 23 '13 at 13:54
I see, thanks :) What about latus rectum? – gov Jul 23 '13 at 13:56

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