4A. Let $AB$ be a given segment of length $2a$ with midpoint $O$. Points $C$ and $D$ move along the normals to $AB$ drawn at points $A$ and $B$ such that the angle $\angle COD$ is a right angle. Determine the geometric locus of the intersection points of the lines $AD$ and $BC$.
Solution. Let us introduce a coordinate system with the center at $O$ and the $x$-axis as the line $A B$. Then $A(-a, 0)$ and $B(a, 0)$. Let $C(-a, r)$ and $D(a, s)$. Then the line $A D$ has the equation $y=\frac{s}{2 a}(x+a)$, and the line $B C$ has the equation $y=-\frac{r}{2 a}(x-a)$. Therefore,

$$
y^{2}=-\frac{r s}{4 a^{2}}\left(x^{2}-a^{2}\right)
$$

The lines $C O$ and $D O$ have equations $y=-\frac{r}{a} x$ and $y=\frac{s}{a} x$, respectively. Since these are perpendicular, we have $r s=a^{2}$. Substituting $r s=a^{2}$ into (1) gives

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

![](https://cdn.mathpix.com/cropped/2024_06_05_84a5520b68919fd51f4cg-11.jpg?height=428&width=733&top_left_y=1206&top_left_x=749)

Thus, the desired locus of points is the ellipse

$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{\left(\frac{a}{2}\right)^{2}}=1
$$