I know partial derivatives should be used to solve this, but I'm wondering is there any to find the plane using the vector method of getting the Normal vector and a point on the plane. In this case, we already have a point, we just need to somehow find the Normal vector.
Also, is there any connection between Normal vectors and partial derivatives in this case of finding equations for tangent planes?
This question should be approached as follows.
The algebraic (implicit) equation of the given paraboloid can be expressed as
$ F(x,y,z) = 0$
where
$ F(x,y,z) = 2 x^2 + y^2 - z $
The normal to the plane is along the gradient vector of $F(x,y,z) $,
and this is given by
$ N = ( F_x, F_y, F_z) = ( 4 x, 2 y , -1 ) $
At $(x,y,z) = (1, 1, 3)$ we get
$ N = (4, 2, -1) $
So that the equation of the tangent plane is
$ 4 (x - 1) + 2 (y - 1) - (z -3) = 0 $
which gives
$ \boxed{ 4x + 2y - z - 3 = 0 } $