A parabola's normal at point $M$ intersects the parabola again at point $N$. What is the geometric locus of the midpoint of chord $M N$? (Determine the essential properties of this geometric locus!)





Translating the text into English while preserving the original formatting and line breaks, the result is as above.
In problem 1405 (8th grade this year), we established that the equation of the normal to the parabola $y^{2}-2 p x=0$, if the direction coefficient of the normal is $m$,

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

can be written in this form. The parallel chords of the parabola with direction coefficient $m$ are bisected by the conjugate diameter to this direction. The equation of this diameter is:

$$
m y-p=0 \ldots
$$

The equations (1) and (2) determine, for a certain value of $m$, the midpoint of the chord with direction coefficient $m$, i.e., the coordinates of this midpoint as functions of $m$. If we eliminate $m$ from the two equations, we obtain a relationship between the coordinates of the midpoint of the normal chord; this will be the equation of the geometric locus of the midpoints. In the given case, from (2) we have $m=\frac{p}{y} ;$ substituting this into (1) and rearranging,

$$
\begin{gathered}
y^{4}-p(x-p) y^{2}+\frac{1}{2} p^{4}=0 \ldots \\
\text{or} \quad x-p=\frac{y^{2}}{p}+\frac{p^{4}}{2 y^{2}} \ldots
\end{gathered}
$$

From (3), it is clear that the sought geometric locus is a quartic curve, and it is symmetric with respect to the $X$-axis. Real values for $y^{2}$ are obtained from (3) if

$$
p^{2}(x-p)^{2}-2 p^{4} \geqq 0, \quad \text { i.e., if } \quad x \geqq p(1+\sqrt{2}) \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_5358a5a0626d59d63b24g-1.jpg?height=387&width=523&top_left_y=1243&top_left_x=779)

If $x$ satisfies condition (5), then from (3) we obtain two positive values for $y^{2}$, and thus four values for $y$, two positive and two negative, the latter being equal in absolute value to the former. (Symmetry with respect to the $X$-axis!) The curve $A$ has two branches, symmetric with respect to the $X$-axis!

If $x=p(1+\sqrt{2})$, then we obtain two equal values for $y^{2}$, i.e., the line $x=p(1+\sqrt{2})$ is tangent to both branches of the curve; the ordinates of the points of tangency are: $\pm \frac{p}{\sqrt[4]{2}}$.

From equation (4), it is clear that the curve (and each of its branches) asymptotically approaches the $X$-axis. (Indeed, if $y \rightarrow \pm 0$, then $x \rightarrow+\infty$)

If we consider the points of the curve whose ordinates increase indefinitely, we see that these points approach the

$$
x-p=\frac{y^{2}}{p} \quad \text { or } \quad y^{2}=P(x-p) \ldots
$$

parabola, remaining within it.

George Feher and Gyula Sebestyen (Mihaly Fazekas High School, 8th grade, Debrecen)