Conversion of polar equation to rectangular equation

[Burger2010]

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


I'm lost!
Please Help!

 

[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

 

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

 

[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.