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

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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.**