Find points which make the tangent plane horizontal?

Question

I'm wondering if someone could give me some hints as to how to approach this question, or some theory to understand what it's actually asking:

Thank you!

Answer 1

The tangent plane at $(x_0,y_0,z_0)$ to a surface given in an implicit equation $F(x,y,z)=0$ is given by $$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.$$ Can you take it from here? How do you recognize an horizontal plane?

Answer 2

$F(x,y,z) =0$ is the surface, let $P := (x_0,y_0z_0) $ be a point on the surface.

Normal vector $\vec n$ to the surface at $P$ is:

$\vec n = (F_x(P),F_y(P),F_z(P)).$

This normal vector specifies the tangent plane to the surface.

If the normal points in the $z$ direction,

the plane is horizontal (|| to x-y plane).

Hence :

$(2x+2,2y-4,2z) =a(0,0,1)$,

where $a \in \mathbb{R}.$

$\rightarrow:$

$x=-1, y=2, z =a/2.$

The point $(x,y,z)$ is a point on the surface.

$F(x=-1,y=2, z=a/2) =0$.

Left to do: Find $a$.

Then your tangent plane:

$\vec n \cdot ( \vec r - (x_0,y_0,z_0))= 0$

where $(x_0=-2 ,y_0 =2 , z_0= a/2)$

and $\vec n=a(0,0,1).$

Note: You may find two values for $a$.

What does it mean?