Please check my work: Find an equation for the tangent plane to the surface $$z^2=x^2-y^2+3$$ at point (1,2,0)

So, I started by taking directional derivatives and got $$f_x=2x $$ and $$f_y=2y$$

Next I plugged them into the formula (which I hope is the correct formula for this) : $$z-c=f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

Question

Please check my work: Find an equation for the tangent plane to the surface $$z^2=x^2-y^2+3$$ at point (1,2,0)

So, I started by taking directional derivatives and got $$f_x=2x $$ and $$f_y=2y$$

Next I plugged them into the formula (which I hope is the correct formula for this) : $$z-c=f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

anonymous · Answer

so I get $$z^2=2x-8y+2$$

Did I do this correctly? And, have I finished the problem? I feel like I made a mistake somewhere because I dont think this is a plane...

anonymous · Answer

I think its because of the z^2. Does anyone know how to avoid getting the z^2?