2. The points $A(-c, 0)$ and $B(c, 0)$ are vertices of $\triangle A B C$. The angles $\angle A=\alpha$ and $\angle B=\beta$ satisfy the equation $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=-k^{2}$. Find the geometric locus of the centroids of $\triangle A B C$.
Solution. If $T(x, y)$ is the centroid of triangle $ABC$ and $C\left(x_{1}, y_{1}\right)$, then

$$
x=\frac{-c+c+x_{1}}{3}=\frac{x_{1}}{3} \quad \text { i.e. } \quad x_{1}=3 x
$$

and

$$
y=\frac{0+0+y_{1}}{3}=\frac{y_{1}}{3} \text { i.e. } y_{1}=3 y
$$

Thus, $C(3 x, 3 y)$.

![](https://cdn.mathpix.com/cropped/2024_06_05_df4cd6b4472bb43b0444g-10.jpg?height=327&width=584&top_left_y=1682&top_left_x=881)

The projection of point $C$ onto the $x$-axis will be denoted by $C^{\prime}(3 x, 0)$.

If point $C$ lies in the first or third quadrant, from the right triangle $B C^{\prime} C$ we have $\operatorname{tg} \beta=\frac{3 y}{c-3 x}$, and from the right triangle $A C^{\prime} C$ we have $\operatorname{tg} \alpha=\frac{3 y}{c+3 x}$. According to this, from the equation $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=-k^{2}$ we get

$$
\frac{3 y}{c+3 x} \cdot \frac{3 y}{c-3 x}=-k^{2} \text { i.e. } \frac{x^{2}}{\left(\frac{c}{3}\right)^{2}}-\frac{y^{2}}{\left(\frac{c k}{3}\right)^{2}}=1
$$

The case when point $C$ lies in the second or fourth quadrant is considered analogously.