In the plane, let there be a parabola. Let its vertex be $O$ and an arbitrary point on it be $M$. We draw a perpendicular to the line $OM$ at point $O$, and draw a line parallel to the axis of the parabola through point $M$. Determine the geometric locus of the intersection point of the two lines as $M$ runs along the parabola.
I. Solution: Consider the parabola with its directrix $i$ and focus $F$, and let the projection of $F$ onto $i$ be $F'$. Then the vertex $O$ of the parabola is the midpoint of the segment $F F'$, and the axis $t$ is the line $F F'$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9a6513bca071469f9081g-1.jpg?height=548&width=705&top_left_y=230&top_left_x=710)

(The segment $T O$ should correctly be labeled as $F^{\mathbf{1}}$.)

Let's examine the geometric locus of the intersection point $P$ of the lines described in the problem in a Cartesian coordinate system where the origin is $O$, the $X$-axis is the line $O F$, the unit length is the segment $O F$, the coordinates of $F$ are $(1,0)$, the parameter of the parabola is $p=2$ units, and its equation is $y^{2}=4 x$.

Choose the direction tangent $m$ of the line $O M$ as the auxiliary variable. Then the coordinates of $M$, as the non-origin intersection point of the parabola and the line $y=m x$, are $M\left(4 / m^{2}, 4 / m\right)$. For every (real) $m \neq 0$, there is exactly one point $M$; and conversely, for every point $M$ different from $O$, there is exactly one $m$, the direction coefficient of the line $O M$. If $m=0$, the line $y=m x$ intersects the parabola only at $O$, which cannot be chosen as $M$ because the line $O M$ is not uniquely determined in this case.

The equation of the line perpendicular to $O M$ at $O$ is $y=-x / m$, and the equation of the line parallel to the axis of the parabola through $M$ is $y=4 / m$. From these, the coordinates of $P$ are $(-4,4 / m)$, meaning the abscissa of $P$ is independent of $m$, constant, and all points $P$ lie on the line $x=-4$, which is parallel to $i$. The geometric locus consists of all points on $e$ whose ordinate $y=4 / m$ is given by some allowed $m$.

Since for every $m \neq 0$, the function $y=4 / m$ takes all real values except 0, the sought geometric locus includes all points of $e$ except the point $(-4,0)$, which is the projection of $F$ onto $e$.

Returning to the pure geometric (coordinate-free) formulation of the problem, the sought geometric locus is obtained by taking the line parallel to $i$ whose distance from $i$ and $F$ is $3/2$ and $5/2$ times the parameter of the parabola, respectively, and removing the intersection point $T$ with the axis of the parabola.

Fanta Katalin (Szombathely, Kanizsai Dorottya High School, 4th grade)

II. Solution: To show that every point of the sought geometric locus is at the same distance from the directrix and the parallel tangent at the vertex, and on the opposite side of the parabola, we can use both coordinate geometry and "pure" geometric methods as follows. Let (in addition to the above notation) the intersection point of the line through $M$ parallel to the axis with the tangent at the vertex (the $Y$-axis) be $N$. In the right triangle $O M P$, $O N^{2}=P N \cdot N M$. Here, $N M$ and $O N$ are the coordinates of $M$ in the system used above, so $y^{2}=P N \cdot x$. Equating this to the equation of the parabola, we get $P N=4$, a constant, meaning every point of the geometric locus is 4 units away from the tangent at the vertex - and on the opposite side of the tangent from the parabola, because the altitude $O N$ separates the hypotenuse endpoints $P$ and $M$ - or on the line $e \| i$ found above. The point $T$ of $e$ does not belong to the geometric locus because the line through $T$ parallel to the axis intersects the parabola at $O$, making the direction of $O M$ indeterminate, and the construction of $P$ impossible. However, every point $P^{*}$ of $e$ different from $T$ belongs to the geometric locus; indeed, the line through $P^{*}$ parallel to $t$ intersects the parabola at a point $M$ different from $O$, and to this $M$, $P^{*}$ belongs.

Surguta László (Gyöngyös, Vak Bottyán High School, 3rd grade)

Note. Many solutions used the equation of the parabola in the form $y^{2}=2 p x$. This also leads to the same result. In the first solution, we saw that after posing a geometric question, we can freely choose not only the position of the axes but also the unit length when placing a coordinate system on the figure. Naturally, it is convenient to choose one of the segments occurring in the figure.