Q:

Finding the tangent plane to a paraboloid which is parallel to a given one.

Find the point on the paraboloid $z = 9 − 4x^2 − y^2$ at which the tangent plane is parallel to the plane $z = 4y$.

My work: I've gotten gradient $f<x,y,z> = <8x,2y, 1>$ and $g<y,z> = <-4,1>$. Where do I go from here?

A:

The tangent plane at $(x_0,y_0,z_0)$, a point on the surface $f(x,y,z)=4x^2 +y^2 +z -9=0$, is
$$f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$$
that is
$$8x_0(x-x_0)+2y_0(y-y_0)+1(z-z_0)=0.$$
Therefore the tangent plane is parallel to the given plane $0x-4y+z=0$ iff 
$$\mbox{$(8x_0,2y_0,1)$ is parallel to $(0,-4,1)$.}$$
Can you take it from here?

