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find an equation for the tangent plane at the point (1,-1,4) for the surface given by $$z^2-2x^2-2y^2-12=0$$

my answer/work: $$<-4x,-4y,2z>$$ plug in points to obtain $$<-4,4,8>$$ plug into tangent plane equation. $$z-8=-4(x-1)+4(y+1)$$ This was just a simple quiz and my professor marked -1 off this problem.. what did i do wrong? was i supposed to continue and distribute everything? because from my knowledge i thought the tangent equation was just $$z-z_o=f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o)$$

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One way

Note that

$$z^2-2x^2-2y^2-12=0\implies z=\pm \sqrt{2x^2+2y^2+12}.$$ Since the point $(1,-1,4)$ has $z>0$ we consider $$z=\sqrt{2x^2+2y^2+12}.$$ Now

$$\frac{\partial z}{\partial x}=\frac{2x}{\sqrt{2x^2+2y^2+12}}$$ and

$$\frac{\partial z}{\partial y}=\frac{2y}{\sqrt{2x^2+2y^2+12}}.$$ Evaluating at $(1,-1)$ we have $$\frac{\partial z}{\partial x}=\frac12,\quad \frac{\partial z}{\partial y}=-\frac12.$$ Thus, the equation of the tangent plane is $$z-4=\frac12(x-1)-\frac12(y+1).$$

Another way

If you call $f(x,y,z)=z^2-2x^2-2y^2-12$ and you get $$\nabla f=(f_x,f_y,f_z)$$ and evaluates it at the point $(1,-1,4)$ you get the normal vector of the plane at such a point. Thus you can write the equation of the plane as $$4(x-1)-4(y+1)+8(z-4)=0.$$

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