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 $y\ge 4$.