How to find the Asymptote of the Curve-

$\displaystyle 2a/r= 1 + 2sin(\theta)

$

I guess I posted this in the right section?

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- Oct 5th 2009, 10:30 AM #1

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- Oct 25th 2009, 09:55 AM #2

- Oct 25th 2009, 12:41 PM #3

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You should have stated what $\displaystyle r$ is. That's probably why you didn't recieve any help right away. When I looked at the problem, I couldn't tell what the curve is. However, I'm going to assume that $\displaystyle r$ is a variable that depends on $\displaystyle \theta$, and then I can solve this equation for $\displaystyle r$ to obtain:

$\displaystyle r=\frac{2a}{1+2sin(\theta)}$

So you can think of $\displaystyle r$ as a function of $\displaystyle \theta$. In this situation, there is vertical asymptote to the curve. This occurs when:

$\displaystyle 1+2sin(\theta)=0$

$\displaystyle sin(\theta)=-\frac{1}{2}$

This occurs when $\displaystyle \theta=\frac{7\pi}{6}$ and $\displaystyle \theta=\frac{11\pi}{6}$ on the interval $\displaystyle [0,2\pi]$

There are really an infinite number of asymptotes to this curve since the sine function is periodic.

Two of the asymptotes are just the verticle lines $\displaystyle x=\frac{7\pi}{6}$ and $\displaystyle x=\frac{11\pi}{6}$, really I should say theta, but the x-axis is usually represented by x even when we are using angles. So you could just replace theta with x in all of the work above.