Question

How do you convert `r=6/(2-cos(theta))` into cartesian form?

Answer 1

The cartesian form is `3x^2-12x+4y^2-36=0`

Explanation:

The formulae used are `r^2=x^2+y^2`
and `x=rcostheta` and `y=rsintheta`

Rewring the polar form as `r(2-costheta)=6`
Replacing `costheta=x/r`
We get `r(2-x/r)=6` `=>` `2r-x=6`
`=>` `2r=x+6`
replacing `r`
`2sqrt(x^2+y^2)=x+6`
Squaring both sides
`4(x^2+y^2)=(x+6)^2`
`4x^2+4y^2=x^2+12x+36`
`3x^2-12x+4y^2-36=0`
graph{3x^2-12x+4y^2-36=0 [-7.9, 7.9, -3.95, 3.95]}