4.22. Form the equation of the geometric locus of points equidistant from the point $A(-2,0,0)$ and the plane $x=2$.
Solution. Let's equate the distances from the point to the plane and between two points:

$$
\begin{aligned}
& |1 \cdot x+0 \cdot y+0 \cdot z-2| / \sqrt{1^{2}+0^{2}+0^{2}}= \\
& \quad=\sqrt{(x+2)^{2}+(y-0)^{2}+(z-0)^{2}}
\end{aligned}
$$

Square both sides of the equation:

$$
(x-2)^{2}=(x+2)^{2}+y^{2}+z^{2}, x=-\left(y^{2}+z^{2}\right) / 8
$$

If we replace \( x=-x^{\prime}, y=y^{\prime}, z=z^{\prime} \), then in the new coordinates, the equation will take the form of the canonical equation of an elliptic paraboloid:

$$
2 x^{\prime}=\left(\left(y^{\prime}\right)^{2}+\left(z^{\prime}\right)^{2}\right) / 4
$$