1. Question

How to find the Asymptote of the Curve-

$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-

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

I guess I posted this in the right section?

You should have stated what $r$ 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 $r$ is a variable that depends on $\theta$, and then I can solve this equation for $r$ to obtain:

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

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

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

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

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

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 $x=\frac{7\pi}{6}$ and $x=\frac{11\pi}{6}$, 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.