Q:

How to find the normal plane from a tangent plane?

$$f(x,y,z)=\frac{x^2}{4} +\frac{y^2}{9} +\frac{z^2}{25}=3 $$
I found the tangent plane from this surface at $P(2,3,5)$ by using the gradient vector, $\nabla F=\langle f_x, f_y, f_z\rangle$. I was wondering if I could find the normal plane at that same point using the information already given. 
The tangent plane is $(x-2)+2/3(y-3)+2/5(z-5)=0$
The missing key is to find the vector that is perpendicular to the normal.

A:

I think you may be a little bit confused -- in three dimensions, a surface has a tangent plane and a normal line. The normal line is spanned by the gradient.

