i need help answering this question. find an equation of the tangent plane to the hyperboloid given by z^2 - 2x^2 - 2y^2 - 12 = 0 at the point (1,-1,4)

Question

 and i would like steps if possible to help explain how you got the answer 

Answer 1

z^2 - 2x^2 - 2y^2 - 12=0

z^2 - 2x^2 - 2y^2 = 12

U(x,y,z)= - 2x^2 - 2y^2+z^2

Gradient of U:

∇U=(-4x, -4y, 2z)

A normal to the tangent plane  at the point (1, -1, 4) is

n=∇U(1, -1, 4)=(-4, 4, 8)

From here it follows that the equation of the tangent plane is

-4(x-1)+4(y+1)+8(z-4)=0

-4x+4y-24=-8z

z=(1/2)x-(1/2)y+3.