Tangent plane to a surface parallel to another plane

Question

Find all the points $(x_0,y_0,z_0)$ on the surface with equation $$z=x^2+y^2$$

at which the tangent plane is parallel to the plane $$x+y=7$$ After finding the answer in this part, give a geometric interpretation.

I know the general equation of a tangent plane to a surface is $$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

But how do I make the coeffecient of $z$ be $0$ ? And I don't think there exists any point on the surface such that the tangent plane is parallel to $x+y=7$ since this plane is perpendicular to the $xy$-plane, and cuts the cone shape $z=x^2+y^2$. But how do I write it out? Thank you guys!

Answer 1

Given a surface in the form $f(x,y,z)=0$, the normal vector to this surface at the point $(x,y,z)$ is simply given by $\nabla f$.

So the normal to $z=x^2+y^2$, or equivalently $x^2+y^2-z=0$, is given by $(2x,2y,-1)$.

The normal to the plane $x+y=7$ is $(1,1,0)$.

Thus, these two normals can never be parallel, hence there does not exist a point where their tangent planes are parallel.