Question

What the is the polar form of `x^2-y^2 = 6x-x^2y-y`?

Answer 1

`r^2cos^2thetasintheta+rcos2theta+sintheta-6costheta=0`

Explanation:

The relation between polar coordinates `(r,theta)` and Cartesian or rectangular coordinates `(x,y)` is given by

`x=rcostheta` and `y=rsintheta` i.e. `r^2=x^2+y^2`

Hence we can write `x^2-y^2=6x-x^2y-y` as

`r^2cos^2theta-r^2sin^2theta=6rcostheta-r^3cos^2thetasintheta-rsintheta`

or `r^3cos^2thetasintheta+r^2cos2theta+r(sintheta-6costheta)=0`

or `r^2cos^2thetasintheta+rcos2theta+sintheta-6costheta=0`