equatoin of tangent plane to point on surface

Question

How would I go about finding the tangent plane to the surface:

$$x^2+y^2-z^2-4xy+4xz=-68$$

at the point $(3,4,-3)$? Partial derivatives have gotten me nowhere so far.

Answer 1

Let $f(x,y,z)=x^2+y^2-z^2-4xy+4xz+68.$

Thus, in the point $(3,4,-3)$ we obtain: $$\frac{\partial f}{\partial x}=2x-4y+4z=6-16-12=-22,$$ $$\frac{\partial f}{\partial y}=2y-4x=8-12=-4$$ and $$\frac{\partial f}{\partial z}=-2z+4x=6+12=18.$$ Thus, the equation of the tangent plane is $$-22(x-3)-4(y-4)+18(z+3)=0$$ or $$11x+2y-9z-68=0.$$