I have the surface: $z^2 = xy$
I want to find the tangent plane and normal line to this surface at the point $(x,y)=(1,4)$
I know how to do this but in this case I'm confused as to what I should use as my $z$ value? In order to find it I substituted the given $x$ and $y$ values into the surface equation, by doing this I obtained $z^2 = 4$.
Here I'm not sure whether I should use $2$ or $-2$ as my $z$ value since $\sqrt4 = +-(2)$ or does it not matter what $z$ value I use and either one is fine?
Any clarification would be much appreciated.
IF you have quoted the problem exactly as it was given to you then it is just a badly written problem! It ask you to find the tangent plane at the point where x= 1, y= 4 when, in fact, there are two such points and two such planes. $\nabla z^2- xy= -y\vec{i}- x\vec{j}+ 2z\vec{k}$.
The normal vector at (1,4,2) is $-4\vec{i}- \vec{j}+ 4\vec{k}$ and the tangent plane there is $-4(x-1)- (y-1)+ 4z= 0$.
The normal vector at (1,4,-2) is $-4\vec{i}- \vec{j}- 4\vec{k}$ and the tangent plane there is $-4(x-1)- (y-1)- 4z= 0$.