5.2.1. Derive the equation of the tangent plane and the equation of the normal to the surface $4 z=x y-2 x-4 y+8$ at the point $M(-4, -2, 8)$.
Solution. Let's check if point $M$ belongs to the surface:

$$
4 \cdot 8=(-4) \cdot(-2)-2 \cdot(-4)-4(-2)+8,32=32
$$

thus, point $M$ belongs to the surface.

The equation of the tangent plane is:

$$
z-z_{M}=\frac{\partial z}{\partial x}_{M}\left(x-x_{M}\right)+\frac{\partial z}{\partial y}_{M}\left(y-y_{M}\right)
$$

Let's find the values of the partial derivatives at point $M$:

$$
\begin{gathered}
\frac{\partial z}{\partial x}=\frac{1}{4}(x y-2 x-4 y+8) \quad=\frac{1}{4}(y-2) ; \quad \frac{\partial z}{\partial x}{ }_{M}= \\
=\frac{1}{4}(-2-2)=-1 ; \\
\frac{\partial z}{\partial y}=\frac{1}{4}(x y-2 x-4 y+8)=\frac{1}{4}(x-4) ; \frac{\partial z}{\partial y}_{M}= \\
=\frac{1}{4}(-4-4)=-2,
\end{gathered}
$$

and substitute them into the equation of the tangent plane:

$$
z-8=(-1)(x-(-4))+(-2)(y-(-2)) \text { or } x+2 y+z=0 \text {. }
$$

The equation of the normal line is taken in the form:

$$
\begin{gathered}
\frac{x-x_{M}}{\frac{\partial z}{\partial x}_{M}}=\frac{y-y_{M}}{\frac{\partial z}{\partial y}_{M}}=\frac{z-z_{M}}{-1}, \text { or } \frac{x+4}{-1}=\frac{y+2}{-2}=\frac{z-8}{-1} \\
\text { or } x+4=\frac{y+2}{2}=z-8
\end{gathered}
$$