Determine a point $N$ on the directrix $FM$ of a parabola or its extension, such that the distances from $N$ to the focus and the directrix differ by a given length, $l$.
Draw perpendiculars from points $M$ and $N$ to the guide line, and let the feet of these perpendiculars be $G$ and $K$; transfer the distance $N K$ to $N F$ and let the endpoint of this be $S$. According to the problem, then $F S=1$.

But the isosceles triangles $K N S$ and $G M F$ are similar, so $S K \parallel F G$. Based on this now recognized property, $K$ and with it $N$ can be constructed. Since there are two points $S$ on the line $S M$ that are $l$ units away from $F$, the problem has two solutions.

(Meitner Elemér, För. VIII. o. t. Budapest.)