Draw a perpendicular to the tangent of a parabola at the point of tangency, and cut this perpendicular with a line passing through the focus and parallel to the tangent.

What is the geometric locus of the intersection points if the point of tangency runs along the parabola?
I. solution. Let's choose the axis of the parabola to be the $x$-axis, and the tangent at the vertex to be the $y$-axis. Then the equation of the parabola is

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

Let the point of tangency on the parabola be $(x_{1}, y_{1})$, then the equation of the tangent is

$$
y y_{1}=p\left(x+x_{1}\right)
$$

where

$$
y_{1}^{2}=2 p x_{1}
$$

that is,

$$
x_{1}=\frac{y_{1}^{2}}{2 p}
$$

Thus, the equation of the perpendicular at the point of tangency, and the line drawn parallel to the tangent from the focus $F\left(0, \frac{p}{2}\right)$ is

$$
y-y_{1}=-\frac{y_{1}}{p}\left(x-\frac{y_{1}^{2}}{2 p}\right)
$$

that is,

$$
2 p y_{1} x+2 p^{2} y=y_{1}\left(y_{1}^{2}+2 p^{2}\right)
$$

and

$$
y=\frac{p}{y_{1}}\left(x-\frac{p}{2}\right)
$$

that is,

$$
2 p x-2 y_{1} y=p^{2}
$$

Eliminating $y$, or $x$ from these,

$$
2 p\left(y_{1}^{2}+p^{2}\right) x=y_{1}^{4}+2 p^{2} y_{1}^{2}+p^{4}=\left(y_{1}^{2}+p^{2}\right)^{2} ; \quad 2\left(p^{2}+y_{1}^{2}\right) y=y_{1}\left(y_{1}^{2}+p^{2}\right)
$$

We can divide by $p^{2}+y_{1}^{2}$, since this cannot be 0, and eliminating $y_{1}$, we get $2 y=y_{1} ; 2 p x=y_{1}^{2}+p^{2}=4 y^{2}+p^{2}$, that is, $y^{2}=\frac{p}{2}\left(x-\frac{p}{2}\right)$; this is the equation of a parabola whose vertex is $\left(0_{1} \frac{p}{2}\right)$, the focus of the original parabola, and whose parameter is a quarter of the original parabola's parameter.

The problem can be solved without using coordinates and (which essentially means the same thing) the Pythagorean theorem. In this case, we will need the following relationships: let the projection of an arbitrary point $P$ on the parabola onto the directrix be $Q$, then the tangent at point $P$ is the angle bisector of $\angle F P Q$, or what is equivalent, the geometric locus of the feet of the perpendiculars dropped from the focus of the parabola to the tangents is the tangent at the vertex of the parabola.

Indeed, by drawing the angle bisector of $\angle F P Q$ (which intersects $Q F$ perpendicularly at the midpoint $R$ of the vertex tangent), we have $F P=P Q$, but for any other point $P^{\prime}$ on the line, $F P^{\prime}=P^{\prime} Q$, and the latter distance is greater than the distance from $P^{\prime}$ to the directrix (Figure 6).

![](https://cdn.mathpix.com/cropped/2024_05_02_6ff1719ca97be5a1d95ag-1.jpg?height=675&width=482&top_left_y=1953&top_left_x=819)

Figure 6

Thus, every point on the line other than $P$ is closer to the directrix than to the focus, so it lies on the same side of the parabola. It is also easy to see that every other line through $P$ has points on both sides of the par