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!
