1. ## Question

How to find the Asymptote of the Curve-

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

I guess I posted this in the right section?

2. Plz help me

3. Originally Posted by cnpranav
How to find the Asymptote of the Curve-

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

I guess I posted this in the right section?

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.