Let $ B = \{(x,y,z)| Ax^2 + By^2 + Cz^2 = D \}$ and a plane $ax + by + cz = k $. Find all the parallel planes to the given plane that are also tangent to $B$.
Clearly the normal vectors of those planes are proportional to $(a,b,c)$, thus $$ ax + by + cz + d=0. $$ And the tangent plane of $B$ is of the form $\nabla F(x_0, y_0, z_0) \cdot(x-x_0, x-y_0, z- z_0)= a(x-x_0)+b(y -y_0) + c(z-z_0)=0 $.
But I'm not sure what I'm missing in order to find how $d$ is related to $k$? Would appreciate any hint.
hint
The normal vector to the tangent plane at the point $M(x_0,y_0,z_0) $ is $$\nabla F(M)=(\frac {\partial F}{\partial x}(M),\frac {\partial F}{\partial y}(M),\frac {\partial F}{\partial z}(M)) $$
$$=(2Ax_0,2By_0,2Cz_0)$$