Problem 856

Let's place the plane of the parabola $g$ in front of us in a vertical position so that the axis $t$ of the parabola is vertical, and the vertex $C$ is at the bottom, meaning the parabola is above the line $i$. Let's denote the focus of the parabola by $F$ and its parameter by $p$.

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We can say that the point $Q$ arises from the point $P$ by a translation towards the axis. If $P_{1}$ is on the right branch of the parabola, the translation is to the left, and if $P_{2}$ is on the left branch, the translation is to the right. For the case where $P$ is on the axis, i.e., coincides with $C$, the problem does not specify the direction of the translation; we will take it that $C$ can be translated either to the left or to the right. Since the magnitude of the translation is always $p$, it is clear that the geometric locus of $Q$ consists of two branches of a parabola: the branch $g_{j}$ of $g$ to the right of $t$ is translated to the left by $p$, and the branch $g_{b}$ to the left of $t$ is translated to the right by $p$. $Q$ can only be on the resulting $g_{j}^{\prime}$ or $g_{b}^{\prime}$, and for any point $Z$ on these curves, there is a $P$ from which we exactly reach $Z$. (The two branches of the geometric locus of $Q$ intersect at $F$ because the length of the chord passing through the focus and perpendicular to the axis is equal to twice the parameter, and its endpoints are translated to $F$.)

The magnitude of the translation from $Q$ to $R$ varies with the movement of $P$, which we will follow by calculation. Let the equation of $g$ be $y = a x^2$, where $a > 0$. Comparing this with the standard form $y^2 = 2px$, $x = y^2 / 2p$, we get $a = 1 / 2p$, and the length of the parameter is $p = 1 / 2a$. Let the abscissa of a point $P_{1}$ on the branch $g_{j}$ be $x_{1}$, i.e., $x_{1} \geq 0$ (we allow the case $P_{1} = C$ as well), so the distance of $P_{1}$ from the axis is $x_{1}$, and its ordinate is $y_{1} = a x_{1}^2$. Thus, the coordinates of $Q_{1}$ arising from $P_{1}$ are $\left(x_{1} - \frac{1}{2a}, y_{1}\right)$, and the coordinates $(u, v)$ of the corresponding point $R_{1}$ are

$$u = x_{1} - \frac{1}{2a}, \quad v = y_{1} - x_{1}$$

Expressing the coordinates of $P_{1}$ in terms of the coordinates of $R_{1}$:

$$x_{1} = u + \frac{1}{2a}, \quad y_{1} = v + x_{1} = u + v + \frac{1}{2a}$$

Substituting these into the relationship between $x_{1}$ and $y_{1}$, the equation of $g$, we get the relationship between $u$ and $v$, the equation of the curve containing the geometric locus of $R_{1}$:

$$\begin{gathered} u + v + \frac{1}{2a} = a\left(u^2 + \frac{u}{a} + \frac{1}{4a^2}\right) = a u^2 + u + \frac{1}{4a} \\ v = a u^2 - \frac{1}{4a} \end{gathered}$$

Here, $-1 / 4a = -p / 2 < 0$, so $R_{1}$ is always on the parabola $\bar{g}$, which is obtained from $g$ by translating it parallel to the axis; towards the directrix by half the parameter; in other words, the focus of $\bar{g}$ is the point $C$, and its vertex tangent is the line $i$.

Conversely, if a point $R_{1}(u, v)$ on $\bar{g}$ has $u \geq -1 / 2a$, then by (1) $x_{1} \geq 0$ and

$$y_{1} = u + a u^2 - \frac{1}{4a} + \frac{1}{2a} = a\left(u^2 + \frac{u}{a} + \frac{1}{4a^2}\right) = a\left(u + \frac{1}{2a}\right)^2 = a x_{1}^2$$

so there is a point $P_{1}$ on $g_{j}$ from which we reach $R_{1}$ as prescribed, and $R_{1}$ belongs to the geometric locus. If $u < -1 / 2a$, however, there is no such point $P_{1}$. Therefore, the arc of $\bar{g}$ to the right of the line $u \leq -1 / 2a$ belongs to the geometric locus.

Similarly, if $P_{2}$ is a point on the branch $g_{b}$, i.e., $x_{2} \leq 0$, then the distance of $P_{2}$ from the axis is $|x_{2}| = -x_{2}$, its ordinate is $y_{2} = a x_{2}^2$, and the calculation proceeds as follows:

$$\begin{aligned} & Q_{2}\left(x_{2} + \frac{1}{2a}, y_{2}\right), \quad R_{2}\left(x_{2} + \frac{1}{2a} = u, \quad y_{2} - |x_{2}| = y_{2} + x_{2} = v\right) \\ & x_{2} = u - \frac{1}{2a}, \quad y_{2} = v - u + \frac{1}{2a} \\ & v - u + \frac{1}{2a} = a\left(u^2 - \frac{u}{a} + \frac{1}{4a^2}\right), \quad v = a u^2 - \frac{1}{4a} \end{aligned}$$

Accordingly, $R_{2}$ is also on $\bar{g}$, specifically on the arc of $\bar{g}$ where $u \leq 1 / 2a$, and similarly, it can be shown that this arc is also part of the geometric locus of $R$, and together with the previous arc, it gives all the points of the geometric locus, as we have constructed $R$ for every point of $g$ in the two ways described above.

The two arcs completely cover $\bar{g}$, and for the points of the arc $-1 / 2a \leq u \leq 1 / 2a$, we can reach them from two positions of $P$. Therefore, the geometric locus of $R$ is obtained from the original parabola by translation, parallel to the axis, towards the directrix by half the parameter.

Zsuzsa Fodor (Bp., Radnóti M. Gym. IV. class)