Question

How do you change the polar equation `r(2+costheta)=1` into rectangular form?

Answer 1

`(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0`

Explanation:

Giving the pass equations

`{(x=rcostheta),(y=rsintheta):}`

`r(2+costheta)=1 = r(2+x/r)=2r+x=1` then

`r = (1-x)/2->x^2+y^2=(1-2x+x^2)/4` and finally

`3x^2+4y^2+2x-1=0` or

`(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0` which is an ellipse.