Question

How do you convert `8=(6x-4y)^2+2y-x` into polar form?

Answer 1

`r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0`

Explanation:

The relation between polar coordinates `(r,theta)` and Cartesian coordinates `(x,y)` is `x=rcostheta`, `y=rsintheta` and `x^2+y^2=r^2`.

Hence `8=(6x-4y)^2+2y-x` can be written as

`8=36x^2+16y^2-48xy+2y-x`

or `8=36r^2cos^2theta+16r^2sin^2theta-48r^2sinthetacostheta+2rsintheta-rcostheta`

or `8=20r^2cos^2theta+16r^2-24r^2sin2theta+2rsintheta-rcostheta`

or `r^2(20cos^2theta-24sin2theta+16)+r(2sintheta-costheta)-8=0`

or `r^2(10(cos2theta-1)-24sin2theta+16)+r(2sintheta-costheta)-8=0`

or `r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0`