Determine the tangent plane to the curve 2x^2 + y^2 - z = 0, at the point (1,1,3)?

The tangent plane could be determined using partial
derivatives:

fx = 4x => f(1,1) =
4

fy = 2y => f(1,1) =
2

The equation of the tangent plane at the point (1,1,3)
is:

z - 3 = 4(x-1) + 2(y -
1)

We'll remove the brackets and we'll
get:

z - 3 = 4x - 4 + 2y -
2

We'll add 3 both sides:

z =
4x + 2y - 3

The equation of the tangent plane
to the given curve, at the point (1,1,3), is: z = 4x + 2y -
3.