Question

How do you convert `-y+2=(3x+4)^2+(y-1)^2` into polar form?

Answer 1

Given that `-y+2=(3x+4)^2+(y-1)^2`

Substituting `x=r\cos\theta` & `y=r\sin\theta`

`-r\sin\theta+2=(3r\cos\theta+4)^2+(r\sin\theta-1)^2`

`-r\sin\theta+2=9r^2\cos^2\theta+16+24r\cos\theta+r^2\sin^2\theta+1-2r\sin\theta`

`(8\cos^2\theta+1)r^2+(24\cos\theta-\sin\theta)r+15=0`