My apologies for the length of the solution. I just wish to provide a quick answer.
From 8=(2x+4y)^2-5y+x8=(2x+4y)2−5y+x
Use the transformation equivalent x=r cos thetax=rcosθ and y=r sin thetay=rsinθ
Use them in the equation, so that 8=(2x+4y)^2-5y+x8=(2x+4y)2−5y+x
becomes
8=4(r cos theta+2(r sin theta))^2-5(r sin theta)+r cos theta8=4(rcosθ+2(rsinθ))2−5(rsinθ)+rcosθ
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(4cos2θ+16sin2θ+16sinθcosθ)⋅r2+(cosθ−5sinθ)⋅r−8=0
Using Quadratic Equation formula for rr
r=(-b+-sqrt(b^2-4*a*c))/(2a)r=−b±√b2−4⋅a⋅c2a
Use a=4 cos^2 theta+16 sin^2 theta +16 sin theta cos thetaa=4cos2θ+16sin2θ+16sinθcosθ
Use b=cos theta-5 sin thetab=cosθ−5sinθ
Use c=-8c=−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=−(cosθ−5sinθ)±√(cosθ−5sinθ)2−4⋅(4cos2θ+16sin2θ+16sinθcosθ)⋅(−8)2(4cos2θ+16sin2θ+16sinθcosθ)
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)r==5sinθ−cosθ±√537sin2θ+129cos2θ+502⋅sinθcosθ8⋅cos2θ+32sin2θ+32sinθcosθ
have a nice day!
