2A. Given points $A(-2,0)$ and $B(2,0)$. Points $C$ and $D$ lie on the normals to the segment $AB$ at points $A$ and $B$ respectively, such that $\measuredangle COD$ is a right angle. Determine the geometric locus of the points of intersection of lines $AD$ and $BC$.
Solution. Let point $C$ have coordinates $C(-2, c)$. The equation of line $O C$ is $y=-\frac{c}{2} x$. Since $\angle C O D$ is a right angle, the equation of line $O D$ is $y=\frac{2}{c} x$ and point $D$ has coordinates $D\left(2, \frac{4}{c}\right)$. The equation of line $A D$ is

![](https://cdn.mathpix.com/cropped/2024_06_05_e2d2e4c7452aa39480e3g-08.jpg?height=252&width=363&top_left_y=1460&top_left_x=1126)
$y=\frac{1}{c}(x+2)$, while the equation of line $B C$ is $y=-\frac{c}{4}(x-2)$. The coordinates of the intersection point $P$, of lines $A D$ and $B C$, are the solution to the system

$$
\left\{\begin{array}{c}
y=\frac{1}{c}(x+2) \\
y=-\frac{c}{4}(x-2)
\end{array}\right.
$$

By eliminating the parameter $c$ from the equations, the equation $\frac{x^{2}}{4}+y^{2}=1$ is obtained.

Finally, from the arbitrariness of $\boldsymbol{c}$, it follows that the geometric locus is the ellipse

$$
\frac{x^{2}}{4}+y^{2}=1
$$