Question

How do you convert `2y=3y^2-4x^2 -2x` into a polar equation?

Answer 1

`r=(2(sintheta+costheta))/(3sin^2theta-4cos^2theta)`

Explanation:

The relation between polar coordinates `(r,theta)` and Cartesian coordinates `(x,y)` is given by `x=rcostheta` and `y=rsintheta`

Hence, `2y=3y^2-4x^2-2x`

`hArr2rsintheta=3r^2sin^2theta-4r^2cos^2theta-2rcostheta`

i.e. `r^2(3sin^2theta-4cos^2theta)=2r(sintheta+costheta)`

or `r=(2(sintheta+costheta))/(3sin^2theta-4cos^2theta)`