## Task A-4.1.

Given is the parabola $y^{2}=2 p x, p>0$. On the parabola, points $A, B$, and $C$ are given (with $A$ having the largest, and $C$ the smallest ordinate) such that the angle bisector of $\varangle A B C$ is parallel to the $x$-axis. If the length of the projection of segment $\overline{A C}$ on the $y$-axis is equal to $4 p$, determine the ordinate of the midpoint of segment $\overline{B C}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_77c5fa69d8ea891e3e07g-23.jpg?height=571&width=711&top_left_y=1031&top_left_x=684)

Let the coordinates of points $A, B, C$ be $\left(\frac{y_{A}^{2}}{2 p}, y_{A}\right),\left(\frac{y_{B}^{2}}{2 p}, y_{B}\right),\left(\frac{y_{C}^{2}}{2 p}, y_{C}\right)$, respectively.

(1 point)

Since the lines $A B$ and $B C$ form congruent angles with the $x$-axis, it follows that the slopes of these lines are opposite.

(2 points)

By writing the condition $k_{A B}=-k_{B C}$, we get: $\frac{y_{A}-y_{B}}{\frac{y_{A}^{2}}{2 p}-\frac{y_{B}^{2}}{2 p}}=-\frac{y_{B}-y_{C}}{\frac{y_{B}^{2}}{2 p}-\frac{y_{C}^{2}}{2 p}}$,

or $\frac{y_{A}-y_{B}}{y_{A}^{2}-y_{B}^{2}}=-\frac{y_{B}-y_{C}}{y_{B}^{2}-y_{C}^{2}}$.

Using the formula for the difference of squares, we get $y_{A}+y_{B}=-\left(y_{B}+y_{C}\right)$, so

$$
y_{A}+2 y_{B}+y_{C}=0
$$

The condition "the length of the projection of segment $\overline{A C}$ on the $y$-axis is $4 p$ ", given $y_{A}>y_{C}$, can be written as

$$
y_{A}-y_{C}=4 p . \quad(* *)
$$

Subtracting $(* *)$ from $(*)$ gives $2 y_{B}+2 y_{C}=-4 p$.

Thus, the $y$-coordinate of the midpoint $P$ of segment $\overline{B C}$ is $y_{P}=\frac{y_{B}+y_{C}}{2}=-p$.