1. In the plane, a rectangular coordinate system with origin $O$ is given. Points $A$ and $B$ are given by their coordinates $A(a, 0)$ and $B(-a, 0)$, where $a>0$. On the normal to the $x$-axis at $A$, a point $C$ is given, and point $D$ is on the normal to the $x$-axis at $B$, such that the angle $C O D$ is a right angle. Determine and investigate the geometric locus of points defined by the intersections of the lines $B C$ and $A D$.
Solution. Let the equation of the line $OC$ be $y=kx$, where $k$ is an arbitrary parameter. Since the line $OD$ is perpendicular to the line $OC$, its equation is

$OD: \quad y=-\frac{1}{k} x$.

From this, the coordinates of points $C$ and $D$ can be easily obtained, i.e.,

![](https://cdn.mathpix.com/cropped/2024_06_05_b87d47f24deda06cfa28g-09.jpg?height=356&width=582&top_left_y=1865&top_left_x=920)

$$
C(a, ka) \quad \text{and} \quad D\left(-a, \frac{1}{k} a\right)
$$

The equations of the lines $AD$ and $BC$ are

$$
AD: \quad y=-\frac{1}{2k}(x-a) \quad \text{and} \quad BC: \quad y=\frac{k}{2}(x+a)
$$

The coordinates of the point $AD \cap BC$ are obtained by solving system (1), which means the mutual dependence of the coordinates of the intersection point is obtained by eliminating the parameter $k$ from system (1).

From the equation of $BC$, we get $k=\frac{2y}{x+a}$, and substituting this into the equation of $AD$, we get

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

Thus, the desired geometric locus of points is an ellipse whose major axis is $AB$, and the minor axis is half of the major axis.

From this ellipse, the points $A$ and $B$ must be excluded, because if the intersection of the lines $AD$ and $BC$ lies on the line $AB$, then the points $D$ and $C$ also lie on the line $AB$, and thus the angle $COD$ cannot be a right angle.