# Conversion of polar equation to rectangular equation

1. May 31, 2009

### Burger2010

r= (15)/(3-2cos(theta))

I'm lost!!

2. May 31, 2009

### vandanak

in polar coordinates r=(x^2+y^2)^1/2 and also x=rcosQ ,y=rsinQ so putting these in above equation and cosQ=x/r we get( x^2+y^2)^1/2=15/(3-2*x/( x^2+y^2)^1/2) now this can be solved

3. May 31, 2009

### protonchain

In case Vandanak's brilliant solution is hard to read, I've formatted his statements in TeX

Given:
$$r = \frac{15}{3-2cos(\theta)}$$

$$r = \sqrt{x^2+y^2}$$

$$x = rcos(\theta)$$

$$y = rsin(\theta)$$

substituting
$$\sqrt{x^2+y^2} = \frac{15}{3-2(\frac{x}{\sqrt{x^2+y^2}})}$$

4. Jun 1, 2009

### arildno

If I did this correctly, you should end up with:
$$y=\pm\frac{\sqrt{5(15-x)(x+3)}}{3}$$

Thus, the solution is a closed curve.