If we draw perpendiculars from the focus of a parabola to its normals, what is the geometric locus of the feet of these perpendiculars?
The direction coefficient of the tangent drawn at the point $\left(x_{0} ; y_{0}\right)$ of the parabola $y^{2}=2 p x$ is $\frac{p}{y_{0}}$, so the normal is: $-\frac{y_{0}}{p}$. The equation of the normal corresponding to the point $M\left(x_{0}, y_{0}\right)$ is:

$$
y-y_{0}=-\frac{y_{0}}{p}\left(x-x_{0}\right) \text { or } y-y_{0}=-\frac{y_{0}}{p}\left(x-\frac{y_{0}^{2}}{2 p}\right)
$$

Let $-\frac{y_{0}}{p}=m$, so $y_{0}=-p m$, which means we introduce the direction coefficient of the normal as a parameter in the normal equation. Thus, the equation of the normal is

$$
y+p m=m\left(x-\frac{p m^{2}}{2}\right) \text { or } y=m(x-p)-\frac{p m^{3}}{2}
$$

Place the origin of the coordinate system at the focus of the parabola, but keep the direction of the axes the same. The transformation relations between the new and original coordinates are

$$
x^{\prime}=x-\frac{p}{2}, \quad y^{\prime}=y, \quad \text { or } \quad x=x^{\prime}+\frac{p}{2}, \quad y^{\prime}=y
$$

Thus, in the new coordinate system, the equation of the normal is

$$
y^{\prime}=m\left(x^{\prime}-\frac{p}{2}\right)-\frac{p m^{3}}{2}
$$

or, omitting the primes,

$$
y=m\left(x-\frac{p}{2}\right)-\frac{p m^{3}}{2} \cdots
$$

The equation of the line passing through the focus and perpendicular to the normal is:

$$
y=-\frac{1}{m} x \ldots
$$

Solving the system of equations (1) and (2), we obtain the coordinates of the intersection point of the two lines as functions of $m$. If we eliminate the parameter $m$ from the two equations, we get a relationship between the coordinates of the intersection point, which is the equation of the desired geometric locus. From (2), $m=-\frac{x}{y}$. Substituting this into (1):

$$
y=-\frac{x}{y}\left(x-\frac{p}{2}\right)+\frac{p}{2} \frac{x^{3}}{y^{3}} \quad \text { or } \quad y^{4}+x^{2} y^{2}-\frac{p}{2} x y^{2}-\frac{p}{2} x^{3}=0
$$

or

$$
y^{2}\left(y^{3}-\frac{p}{2} x\right)+x^{2}\left(y^{2}-\frac{p}{2} x\right) \equiv\left(x^{2}+y^{2}\right)\left(y^{2}-\frac{p}{2} x\right)=0 \ldots
$$

The geometric locus in question consists of two parts: one part satisfies the equation $x^{2}+y^{2}=0$ and is nothing but the focus itself (or the imaginary pair of lines passing through the focus). The other part is the parabola given by the equation $y^{2}-\frac{p}{2} x=0$, whose vertex is the focus of the original parabola and whose parameter is $\frac{p}{4}$, a quarter of the original parabola's parameter.

Jenő Krisztonosich (Szent László g. VIII. o. Bp. X.).

Second Solution. We illuminate the previous result using synthetic geometric methods. Let $I$ be the projection of the focus $F$ onto the normal $M N$. Since $F M=F N$, $I$ bisects $M N$. If $M P$ has a midpoint $H$, then $H I \| P N$ and $H I=\frac{1}{2}, \quad P N=\frac{1}{2} p$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bd2f13ad1eccaa0aa10bg-1.jpg?height=486&width=526&top_left_y=2229&top_left_x=797)

The midpoints of the ordinates $(M P)$ of the points on the parabola also form a parabola, whose vertex is the same as the given parabola $(A)$ and whose parameter is $\frac{p}{4}$. If we translate this parabola parallel to its axis by a distance $H I=\frac{p}{2}$, we obtain the geometric locus of the point $I$, which is a parabola with vertex $F$, parameter $\frac{p}{4}$, and axis coinciding with the axis of the given parabola.

Jenő Sauer (Bencés 2. VII. o. Győr.)