From a point, we throw bodies at equal speeds, in the same vertical plane, but at different angles $\alpha$. Determine the geometric place of the vertices of the parabolic trajectories as $\alpha$ varies from $0^{\circ}$ to $90^{\circ}$.
I. Solution: It is known from physics that the coordinates of the vertex of a parabolic trajectory are

$$
\begin{aligned}
& x=\frac{c^{2} \sin 2 \alpha}{2 g} \\
& y=\frac{c^{2} \sin ^{2} \alpha}{2 g}
\end{aligned}
$$

The highest elevation occurs when $\alpha=90^{\circ}$, in which case $y_{\max }=\frac{c^{2}}{2 g}$. For simplicity, we can write $2 m$ instead of this value, and the equations (1) and (2) take the following form:

$$
\begin{aligned}
& x=2 m \sin 2 \alpha \\
& y=2 m \sin ^{2} \alpha
\end{aligned}
$$

We obtain the equation of the geometric locus by eliminating the parameter $\alpha$, which can take any value. Since $2 \sin ^{2} \alpha=1-\cos 2 \alpha$, the equation (4) becomes

$$
y=m(1-\cos 2 \alpha)=m-m \cos 2 \alpha
$$

from which

$$
\cos ^{2} 2 \alpha=\frac{(y-m)^{2}}{m^{2}}
$$

From (3),

$$
\sin ^{2} 2 \alpha=\frac{x^{2}}{(2 m)^{2}}
$$

Adding (5) and (6) and swapping the sides of the equation, we get

$$
\frac{x^{2}}{(2 m)^{2}}+\frac{(y-m)^{2}}{m^{2}}=1
$$

This is the equation of an ellipse whose center is $K(0, m)$, the major axis is parallel to the x-axis, and the semi-major axis is $2 m$, while the semi-minor axis on the y-axis is $m$. From these data, it is clear that the ellipse touches the x-axis at the origin. From this ellipse, only the half-ellipse to the right of the y-axis satisfies the condition $-0^{\circ} \leq \alpha \leq 90^{\circ}$. $m=\frac{c^{2}}{4 g}$, and if $g$ is considered a constant, then $m$ depends only on $c$. By varying $m$, the ellipses only differ in size, being similar to each other.

II. Solution: Let $P\left(4 m \sin \alpha \cos \alpha, 2 m \sin ^{2} \alpha\right)$ be a point on the geometric locus, where $2 m=\frac{c^{2}}{2 g}=y_{\max }$. Let's examine the geometric locus of points $P^{\prime}$ obtained by doubling the ordinates of the points $P$. (See the diagram, which also shows the construction.)

![](https://cdn.mathpix.com/cropped/2024_05_02_955f4fd51791c7965c7dg-1.jpg?height=692&width=415&top_left_y=1882&top_left_x=844)

Thus, the coordinates of $P^{\prime}$ are: $4 m \sin \alpha \cos \alpha, 4 m \sin ^{2} \alpha$.

The slope of $O P^{\prime}$ is $\frac{4 m \sin ^{2} \alpha}{4 m \sin \alpha \cos \alpha}=\frac{\sin \alpha}{\cos \alpha}=\operatorname{tg} \alpha$, meaning $O P^{\prime}$ forms the angle of projection with the x-axis.
Let the perpendicular from $P^{\prime}$ to the y-axis intersect the y-axis at point $B$. By the theorem of alternate interior angles, $P^{\prime} B O \varangle=\alpha$.

By the Pythagorean theorem,

$$
\begin{aligned}
O P^{\prime}=\sqrt{A P^{\prime 2}+A O^{2}} & =\sqrt{16 m^{2} \sin ^{4} \alpha+16 m^{2} \sin ^{2} \alpha \cos ^{2} \alpha}= \\
& =\sqrt{16 m^{2} \sin ^{2} \alpha\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)}=4 m \sin \alpha
\end{aligned}
$$

and thus $O B=\frac{O P^{\prime}}{\sin \alpha}=4 m$, independent of $\alpha$.

The geometric locus of points $P^{\prime}$ is therefore the half-circle constructed over $O B=4 m$ to the right of the y-axis.

The equation of this circle is

$$
x^{2}+(y-2 m)^{2}=(2 m)^{2}
$$

The geometric locus of points $P$ is the midpoints of the ordinates of these half-circle points.

Thus, the geometric locus of these midpoints is

$$
x^{2}+(2 y-2 m)^{2}=(2 m)^{2}
$$

or

$$
\frac{x^{2}}{(2 m)^{2}}+\frac{(y-m)^{2}}{m^{2}}=1, \quad \text { where } \quad x \geq 0
$$

This half-ellipse is formed by compressing the half-circle in the y-axis direction by a factor of $\lambda=\frac{1}{2}$. (See the 4th point of function transformations in the 4th-grade textbook of secondary schools, pages 43-44.)