Question

How do you convert `9=(4x-2y)^2-y-6` into polar form?

Answer 1

`r^2 (4 cos theta - 2 sin theta)^2 - r sin theta = 15`

Explanation:

`9=(4x-2y)^2-y-6`

We just substitute `x=r cos theta` and `y=r sin theta`

`9 = (4r cos theta - 2 r sin theta)^2 - r sin theta -6`

`r^2 (4 cos theta - 2 sin theta)^2 - r sin theta = 15`

The original equation we're given wasn't simplified so it's probably OK to stop here.