tangent plane to a surface

Question

I have the following equation

x^2+y^2=2 in R^3

I also have the following point (1,1,t) where t is any number.

I have graphically checked that the tangent plane should equal to

x+y=2

But I don't understand how to get the variable t into the equation of the tangent plane?

Thanks in advance Tom

Answer 1

The equation of the surface is $$x^2+y^2-2=0.$$ The normal vector to the tangent plane to this surface is $$\nabla (x^2+y^2-2)=[2x,2y,0]$$ $$=[2,2,0]\text { at the point }(1,1,t).$$ The equation of the tangent plane to the surface at the point $(1,1,t)$ is $$[2,2,0 \bullet [x-1,y-1,z-t]=0,$$ which is the same as $$x+y=2,$$ so the tangent plane is the same whatever the value of $t.$