# David's question at Yahoo! Answers (horizontal tangente plane).

#### Fernando Revilla

##### Well-known member
MHB Math Helper
David's question at Yahoo! Answers (horizontal tangent plane).

Here is the question:

Find the point(s) on the surface at which the tangent plane is horizontal.? z = 3 − x^2 − y^2 + 8y
(x, y, z) = ( )
Here is a link to the question:

Find the point(s) on the surface at which the tangent plane is horizontal.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

P.S.
Of course I meant in the title tangent instead of tangente (It is hard to forget our mother tongue).

Last edited:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello David,

The equation of the tangent plane to a surface $\phi :z=f(x,y)$ at the point $P_0(x_0,y_0,z_0)$ of $\phi$ is

$\pi: \phi_x(P_0)(x-x_0)+\phi_y(P_0)(y-y_0)-1(z-z_0)=0$

The plane $\pi$ is horizontal if and only if $\phi_x(P_0)=\phi_y(P_0)=0$. In our case if and only if $-2x_0=0$ and $-2y_0+8=0$. We get $x_0=0,y_0=4$.

As $P_0$ belongs to the surface, $z_0=3-0^2-4^2+8\cdot 4=19$. The solution is $(x_0,y_0,z_0)=(0,4,19)$.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Another way to do this: $$z= 3 − x^2 − y^2 + 8y$$ can be thought of as "level surface": $$f(x, y, z)= z+ x^2+ y^2- 8y= 3$$. The gradient, $$\nabla f= 2x\vec{i}+ (2y- 8)\vec{j}+ \vec{k}$$, is perpendicular to the surface and so the tangent plane (which is, of course, also perpendicular to the normal curve) is parallel to the xy-plane if and only if that gradient is vertical- that is, that 2x= 0 and 2y- 8= 0.