We draw a tangent at the variable point $M$ of the parabola and erect a perpendicular line $e$ through the vertex of the parabola to this tangent. This line $e$ intersects the line $M F$ at point $P$. What is the geometric locus of point $P$? $F$ is the focus of the parabola.
I. Solution. Let the $X$-axis of our rectangular coordinate system be the axis of the parabola, and the $Y$-axis be the tangent at the vertex of the parabola; in this system, the equation of the parabola is:

$$
y^{2}=2 p x \ldots
$$

The direction coefficient of the tangent drawn at point $M(\xi, \eta)$ of the parabola is: $\frac{p}{\eta}$. The equation of the line passing through point $O(0,0)$ and perpendicular to the aforementioned tangent is:

$$
y=-\frac{\eta}{p} x \ldots
$$

The equation of the line passing through points $M$ and $F\left(\frac{p}{2}, 0\right)$ is:

$$
\frac{y}{x-\frac{p}{2}}=\frac{\eta}{\xi-\frac{p}{2}} \quad \text { or } \quad\left(\xi-\frac{p}{2}\right) y=\eta\left(x-\frac{p}{2}\right) \cdots
$$

The pair of values $(x, y)$ that satisfy the system of equations (2) and (3) provides the coordinates of point $P$ as functions of $\xi$ and $\eta$. However, between $\xi$ and $\eta$ there is the relationship:

$$
\eta^{2}=2 p \xi \ldots
$$

To obtain an equation between the $x, y$ coordinates of point $P$ that does not contain $\xi$ and $\eta$,

we eliminate $\xi$ and $\eta$ from equations (2), (3), and (4).

From (2), $\eta=-p \frac{y}{x} ; \quad$ from (3), $\xi=\frac{p(p-x)}{2 x}$.

Substituting these values of $\eta$ and $\xi$ into (4):

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

![](https://cdn.mathpix.com/cropped/2024_05_02_acbf5c41b0c33172bb6eg-1.jpg?height=393&width=382&top_left_y=1495&top_left_x=861)

Accordingly, the equation of the geometric locus of point $P$ is:

$$
x^{2}+y^{2}-p x=0
$$

This represents a circle whose center is the focus $F\left(\frac{p}{2}, 0\right)$ of the parabola and whose radius is $\frac{p}{2}$. The circle passes through the origin. Baán Sándor (Bencés g. VII. o. Kőszeg.)

Supplement. This circle has two points on the $Y$-axis: $O(0,0)$ and $P_{0}(p, 0)$. These, as limiting positions, belong to the geometric locus. As $M$ approaches $O$, $P$ approaches $P_{0}$; as $M$ approaches infinity, $P$ approaches $O$.

[Indeed, from (3), $x-\frac{p}{2}=\left(\xi-\frac{p}{2}\right) \frac{y}{\eta}$.

From (2), $\frac{y}{\eta}=-\frac{x}{p}$, so $\frac{2 x-p}{x}=-\frac{1}{p}(2 \xi-p)$.

If $\xi=0, \frac{2 x-p}{x}=1$ and hence $x=p$. This is the abscissa of $P_{0}$. [^0]

If $\xi \rightarrow \infty, \frac{2 x-p}{x}=\left(2-\frac{p}{x}\right) \rightarrow \infty$, so $x \rightarrow 0$. (This is the $\left.O\right)$.]

II. Solution. Let the directrix of the parabola be the line $d$, and the projection of $M$ on $d$ be $N$, so $M N=M F$. It is known that the tangent $t$ at point $M$ is perpendicular to $F N$. Since $O P \perp t, O P \| N F$; furthermore, $O F \| M N$. From this it follows that

$$
O F P \varangle=N M F \varangle \quad \text { and } \quad O P F \varangle=N F M \varangle,
$$

so

$$
O F P \triangle \sim M N F \triangle
$$

However, $\triangle N M F$ is isosceles, with $M N=M F$, so $\triangle O F P$ is also isosceles, with $F P=O F=$ constant. Therefore, the geometric locus of point $P$ is a circle whose center is $F$ and whose radius is $=O F$.

Káli L. Tibold (Bencés $g$. VIII. o. Pápa.)


[^0]:    ${ }^{1} \mathrm{Equation}(5)$ is obtained by multiplying both sides of the previous equation by $x^{2}$. $x^{2}=0$ is the equation of the $Y$-axis; however, the $Y$-axis cannot belong to the geometric locus, because the points $P$ are points of lines passing through the point $O$.