Find a point through which every surface tangent to z=xe^(y/x) passes

Question

Find a point through which every plane tangent to the surface $$ z=xe^{\frac{y}{x}} $$ passes. It's not a homework. I know, that I need a normal vector and the point of tangency to find a tangent plane.

Answer 1

Equivalently; if $F=z-x\exp(y/x)=0$ then by taking three point as @Christian suggested: $$A(1,0,1),~~B(-1,0,-1),~~C(1,1,e)$$ and an assumed point, say $P(x_0,y_0,z_0)$, we should have a system of $$\nabla F|_A\cdot \vec{AP}=0,~~ \nabla F|_A\cdot \vec{BP}=0,~~\nabla F|_A\cdot \vec{CP}=0$$ And this system should be at least a solution, if presumably there is such that point. I added a graph about the function to think better what would be that point(s).

Answer 2

A hint:

If there is such a point you can find it by intersecting the three tangent planes belonging to the three points $\bigl(1,0,z(1,0)\bigr)$, $\bigl(-1,0,z(-1,0)\bigr)$, $\bigl(1,1,z(1,1)\bigr)\in S$. Then verify that the tangent plane belonging to an arbitrary point $\bigl(u,v,z(u,v)\bigr)\in S$ passes through this point as well.