My apologies for the length of the solution. I just wish to provide a quick answer.
From 8=(2x+4y)^2-5y+x
Use the transformation equivalent x=r cos theta and y=r sin theta
Use them in the equation, so that 8=(2x+4y)^2-5y+x
becomes
8=4(r cos theta+2(r sin theta))^2-5(r sin theta)+r cos theta
simplify so that it becomes
(4 cos^2 theta+16 sin^2 theta +16 sin theta cos theta)*r^2+(cos theta-5 sin theta)*r-8=0
Using Quadratic Equation formula for r
r=(-b+-sqrt(b^2-4*a*c))/(2a)
Use a=4 cos^2 theta+16 sin^2 theta +16 sin theta cos theta
Use b=cos theta-5 sin theta
Use c=-8
So that
r=(-(cos theta-5 sin theta)+-sqrt((cos theta-5 sin theta)^2-4*(4 cos^2 theta+16 sin^2 theta +16 sin theta cos theta)*(-8)))/(2(4 cos^2 theta+16 sin^2 theta +16 sin theta cos theta))
r==(5 sin theta-cos theta+-sqrt(537 sin^2 theta+ 129 cos^2 theta+ 502*sin theta cos theta))/(8*cos^2 theta+32 sin^2 theta+ 32 sin theta cos theta)
have a nice day!
