Find the set of points equidistant from the line…

Find the set of points equidistant from the line $\tfrac{x-1}0=\tfrac y0=\tfrac z1$ and the plane $x=0$. What is it?

Explanation of Method

Points equidistant from a line and a plane parallel to it form a parabolic cylinder.

Step 1: Distances

The line is $x=1,\ y=0$, parallel to $Oz$. Then

\[ d_1=\sqrt{(x-1)^2+y^2},\qquad d_2=|x| . \]

Step 2: Equation

\[ (x-1)^2+y^2=x^2\iff y^2=2x-1 . \]

\[y^2=2\left(x-\tfrac12\right)\]

The equation does not contain $z$, so the set is a parabolic cylinder with generators parallel to $Oz$. Each horizontal section is a parabola with focus at the trace $(1,0)$ of the line and directrix $x=0$. The vertex is at $x=\tfrac12$ and $p=1$.

$F(1,0)$ $M$ $\tfrac12$ $y^2=2x-1$

Every horizontal section $z=\text{const}$: the line becomes the point $F(1,0)$, the plane $x=0$ becomes the $y$-axis. Equidistant points form the parabola $y^2=2x-1$, e.g. $M\left(\tfrac52,2\right)$ with $MF=\tfrac52$.

Result

\[\boxed{\,y^2=2\left(x-\tfrac12\right)\,}\]