# 3d function and partial derivatvies

In order to get a tangent plane to a function $x^2+\frac{y^2}{4}-\frac{z^2}{9}=1$

I converted this as a form of $z=+-\sqrt{\cdots}$ and got $z=z_x(x-x_0)+z_y(y-y_0)$.

It was quite cumbersome. so I instead set $x^2+\frac{y^2}{4}-\frac{z^2}{9}-1= F(x,y,z)$ $($and $=0$ I guess$)$ and found $F_x,F_y,F_z$. and got the same result $F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0.$

I'm a bit confused about this process. what does $F_z$ mean? especially when $F(x,y,z)$ equals $0$?

Given a regular surface

$$S(x,y,z) = 0$$

Any displacement over the surface near $x_0,y_0,z_0 \in S$ can be represented as

$$S(x_0+dx,y_0+dy,z_0+dz) = S(x_0,y_0,z_0) + S_x dx+S_y dy + S_z dz +O(dx^2+dy^2+dz^2)= S_x dx+S_y dy + S_z dz = dS = 0$$

Now $(dx,dy,dz)$ are at the tangent plane so $\vec n = (S_x,S_y,S_z)$ the normal to the tangent plane at $(x,y,z)$

If $\vec n = 0$ at $(x_0,y_0,z_0)$ then the surface as such, degenerates at this point.

So an easy way to build a tangent plane $\Pi$ at $(x_0,y_0,z_0)$ pertaining to a given regular surface $S(x,y,z)=0$ is

$$\Pi\to (x-x_0)S_x^0+(y-y_0)S_y ^0+(z-z_0)S_z^0 = 0$$

where

$$(S_x^0,S_y^0,S_z^0) = \nabla S \;\;\mbox{at}\;\;(x_0,y_0,z_0)$$