We throw objects from a point at equal speeds but at different angles. Determine the geometric place of the apexes of those parabolic trajectories, which are in the same vertical plane.
We place a coordinate system in a certain vertical plane, whose origin (0) is at the starting point of the motion, and whose abscissa axis is horizontal.

If we launch a body from 0 with an initial velocity $c$ at an angle $\alpha$, then as is known from physics, the coordinates of the vertex of the parabolic trajectory are:

$$
y=\frac{c^{2}}{2 g} \cdot \sin ^{2} \alpha
$$

and

$$
x=\frac{c^{2}}{2 g} \cdot \sin 2 \alpha
$$

To obtain the equation of the geometric locus of the vertices, we need to eliminate $\alpha$ from these two equations, which can be arbitrary. For this purpose, we transform equation (1) using the equation

$$
2 \sin ^{2} \alpha=1-\cos 2 \alpha
$$

Thus, from (1) we get

$$
y-\frac{c^{2}}{4 g}=-\frac{c^{2}}{4 g} \cdot \cos 2 \alpha
$$

or

$$
\cos ^{2} 2 \alpha=\frac{\left(y-\frac{c^{2}}{4 g}\right)^{2}}{\left(\frac{c^{2}}{4 g}\right)^{2}}
$$

and from (2)

$$
\sin ^{2} 2 \alpha=\frac{x^{2}}{\left(\frac{c^{2}}{2 g}\right)^{2}}
$$

Therefore, the sum of (3) and (4) yields

$$
\frac{x^{2}}{\left(\frac{c^{2}}{2 g}\right)^{2}}+\frac{\left(y-\frac{c^{2}}{4 g}\right)^{2}}{\left(\frac{c^{2}}{4 g}\right)^{2}}=1
$$

If we now transition to a coordinate system with the same ordinate axis and an abscissa axis that is $\frac{c^{2}}{4 g}$ higher than the previous one, using the equations

$$
\begin{gathered}
x=X \\
y-\frac{c^{2}}{4 g}=Y
\end{gathered}
$$

then the equation of the curve becomes:

$$
\frac{X^{2}}{\left(\frac{c^{2}}{2 g}\right)^{2}}+\frac{Y^{2}}{\left(\frac{c^{2}}{4 g}\right)^{2}}=1
$$

This is the equation of an ellipse with semi-axis lengths $\frac{c^{2}}{2 g}$ and $\frac{c^{2}}{4 g}$. Therefore, the sought geometric locus is an ellipse, whose minor axis is the path the body travels when launched vertically upward, and whose major axis is twice the length of the minor axis.