# Finding points on a surface $\{ z= f(x, y)\}$ with horizontal tangent plane

Could someone please explain in detail how this is done? For example there is a surface

$$M = \{ (x, y, z) : z = x^4 - 4xy^3 + 6y^2 - 2\}$$

and the question is to find the points on $$M$$ where this surface has a horizontal tangent plane.

What is implied from having a horizontal tangent plane?

• The term 'horizontal' means parallel to the $z=0$ plane, I presume – Shubham Johri Apr 7 at 6:14

'Horizontal' presumably means parallel to the $$z=0$$ plane. If the tangent plane at a point on the surface is parallel to the $$xy$$ plane, the normal to the plane, and thus the normal to the surface at the point is along the $$z$$ direction.
Your surface is $$z(x,y)=x^4-4xy^3+6y^2-2$$, so the direction-ratios of the normal at any point are given by its gradient, i.e. $$(4(x^3-y^3),12y-12xy^2,-1)$$. Now, you have two equations:$$4(x^3-y^3)=0\\12y-12xy^2=0$$The common solutions $$(x,y)$$ are $$(0,0),(1,1),(-1,-1)$$.