China Mathematical Competition

Suppose that $(y_1^2 - 4, y_1)$ are the coordinates of point $B$ and $(y^2 - 4, y)$ of point $C$. Obviously, $y_1^2 - 4 \ne 0$, so $k_{AB} = \frac{y_1 - 2}{y_1^2 - 4} = \frac{1}{y_1 + 2}$.

Since $AB \perp BC$, so $k_{BC} = - (y_1 - 2)$. Thus
$$y - y_1 = -(y_1 + 2)[y^2 - 4 - (y_1^2 - 4)].$$
Noting $y \ne y_1$, we obtain
$$(2 + y_1)(y + y_1) + 1 = 0,$$
and that is
$$y_1^2 + (2 + y)y_1 + (2y + 1) = 0.$$

From $\Delta \ge 0$, we obtain $y \le 0$ or $y \ge 4$.
When $y = 0$, the coordinates of $B$ are $(-3, -1)$ and when $y = 4$, they are $(5, -3)$. They both satisfy the conditions given by the problem. So, the range of values for the y-coordinate of point $C$ is $y \le 0$ or $y \ge 4$.