I have realized that my previous answer was wrong. Thank you
Sonnhard. Now, I do it again.
Using
0 <= sqrt ( x^2 + y^2 ) = r and ( x, y ) = r ( cos theta, sin theta )0≤√x2+y2=rand(x,y)=r(cosθ,sinθ),
2 = x^2y -x/y^2 + xy^22=x2y−xy2+xy2 converts to
r^4sin theta cos theta ( cos theta+sin theta )r4sinθcosθ(cosθ+sinθ)
- 2 r - cos theta csc^2theta= 0, r ne 0.−2r−cosθcsc2θ=0,r≠0.
Graph, from the given Cartesian equation:
graph{ x^2y -x/y^2 + xy^2 - 2 = 0}
As x and y to 0, parabolic x/y^2 to - 2xandy→0,parabolicxy2→−2.
In polars, as r to 0, cos theta/sin^2theta to 0 rArr cos theta to 0r→0,cosθsin2θ→0⇒cosθ→0
rArr abs theta to pi/2⇒|θ|→π2
