Finding the parametrization of a line perpendicular to a non-planar surface? There's a question from a Calc III practice problem set that has me confused. I understand parametrization pretty well, and I could answer this question if each variable were first order - but, the surface in question isn't a plane. I'm not sure how to proceed.
Find a parameterization of the line perpendicular to the surface $z=x^2+2y^2$ at the point $(2,1,6)$
With a curved surface, I imagine the process for finding a parametric is a bit more complicated. I have no idea how to go about finding this parametric.

We take the gradient of this. Let us have $0=x^2+2y^2-z=f(x,y,z)$
$$\nabla f(x,y,z)=\langle2x, 4y, -1\rangle $$ $$\nabla f(2,1,6)=\langle 2\cdot 2, 4\cdot 1,-1\rangle = \langle 4,4,-1\rangle $$  By definition, the normal line to a surface $S$ at point $P$ on $S$ is the line that has its direction vector as the normal vector to the surface, which was just derived from the gradient.  Thus, for the line the parametrization would be something along the lines of 
$$r(t)=\langle 2, 1, 6\rangle +t\langle 4, 4, -1\rangle$$  This is because as we mentioned the direction vector of this line will be the gradient at the point (2,1,6) and (2,1,6) is a point on the line. 
