4. Let $A B C$ be an isosceles triangle with base $\overline{A B}=2 a$ and let $P$ be the intersection point of line $B C$ with line $p$ passing through vertex $A$ and perpendicular to side $A C$. Find the geometric locus of points $P$ when points $A$ and $B$ are fixed, and point $C$ moves such that triangle $A B C$ is always isosceles.
Solution. From the conditions of the problem, it follows that point $C$ moves along the perpendicular bisector of segment $AB$. Therefore, let's choose a coordinate system such that the line $AB$ is the $x$-axis, and the perpendicular bisector of segment $AB$ is the $y$-axis (see diagram). Then we have $A(-a, 0), B(a, 0)$, and $C(0, c)$, where $c \neq 0$ varies. The equations of the lines $BC$ and $p$ are:

![](https://cdn.mathpix.com/cropped/2024_06_05_faf4f26986d5dd6dd724g-9.jpg?height=507&width=624&top_left_y=1431&top_left_x=856)

$$
\frac{x}{a}+\frac{y}{c}=1 \text { and } \quad y=-\frac{a}{c}(x+a)
$$

respectively. By eliminating the parameter $c$, we obtain the equation $x^{2}-y^{2}=a^{2}$, which represents the relationship between the coordinates $x$ and $y$ of point $P$. Therefore, the geometric locus of points $P$ is a hyperbola.