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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?

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  • $\begingroup$ Yes it is, and you are correct. $\endgroup$ – Doug M Apr 25 '16 at 23:12
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You are correct, there are no such points.

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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:

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