# Find all Points on the Surface at which the Tangent is Parallel to the Plane

The problem: Find all points on the surface $z=x^3+xy^2$ at which the tangent plane is parallel to the plane $2x+2y+z=0$

So I established $f(x,y,z)=x^3+xy^2-z$ and the normal vector determined from the plane being $<2,2,1>$.

The normal vector for a tangent plane at the point $(x_0,y_0,z_0)$ on the surface is such:

$$\nabla f(x_0,y_0,z_0)=<3x_0^2+y_0^2, 2x_0y_0,-1>=k<2,2,1>$$

Anyway I try to solve this equation, I can't find any points. Is this a possible solution, or am I wrong?

• Yes it is, and you are correct. – Doug M Apr 25 '16 at 23:12

It seems to me that one might suspect this outcome, since the surface has odd symmetry about the $\ yz-$ plane ($\ x \ = \ 0 \$) and even symmetry about the $\ xy-$ plane ( $\ y \ = \ 0 \$ ) , but the given plane, $\ z \ = \ -2x \ - \ 2y \$ , has only diagonal symmetry (the normal vector is in the plane $\ y \ = \ x \$ ) . So there is no value of $\ c \$ for which a plane $\ 2x \ + \ 2y \ + \ z \ = \ c \$ will not simply "cut through" the surface. Here is a picture of the surface and the given plane: 