Find the set of points equidistant from F(0,0,tfrac a2) and…

Find the set of points equidistant from $F\left(0,0,\tfrac a2\right)$ and the plane $z=-\tfrac a2$. What is it?

Explanation of Method

Points equidistant from a point and a plane form a paraboloid of revolution.

Step 1: Equidistance

\[ \sqrt{x^2+y^2+\left(z-\tfrac a2\right)^2}=\left|z+\tfrac a2\right| . \]

Both sides are nonnegative, so squaring is an equivalence: $x^2+y^2+z^2-az+\tfrac{a^2}4=z^2+az+\tfrac{a^2}4$.

\[x^2+y^2=2az\]

Step 2: What it is

This is a paraboloid of revolution with vertex $O$, axis $Oz$, focus $F$ and directrix plane $z=-\tfrac a2$. Its meridian sections are parabolas $x^2=2az$ with parameter $p=a$. For $a\gt 0$ it opens upward, and for $a\lt 0$ downward.

Example ($a=1$): the point $M(0,2,2)$ lies on the paraboloid. $|MF|=\sqrt{4+\tfrac94}=\tfrac52$, and the distance to the directrix plane is $2+\tfrac12=\tfrac52$. $\checkmark$

$z$ $F\left(0,0,\tfrac a2\right)$ $M$ $O$ $z=-\tfrac a2$

The paraboloid $x^2+y^2=2az$ (drawn for $a=1$): every point $M$ is as far from the focus $F$ as from the directrix plane $z=-\tfrac a2$ (gray). Here $M(0,2,2)$: $|MF|=|MH|=\tfrac52$.

Result

\[\boxed{\,x^2+y^2=2az\,}\]