Example 5 (2002 National High School Competition Question) Given point $A(0,2)$ and two points $B, C$ on the parabola $y^{2}=x+4$ such that $A B \perp B C$, find the range of the y-coordinate of point $C$.

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Solve As shown in Figure 6-2, let the coordinates of point $B$ be $\left(y_{1}^{2}-4, y_{1}\right)$, and the coordinates of point $C$ be $\left(y^{2}-4, y\right)$. Clearly, $y_{1}^{2}-4 \neq 0$, so $K_{A B}=\frac{y_{1}-2}{y_{1}^{2}-4}=\frac{1}{y_{1}+2}$. Since $A B \perp B C$, we have $K_{B C}=-\left(y_{1}+2\right)$. The equation of line $B C$ is $y-y_{1}=-\left(y_{1}+2\right)\left[x-\left(y_{1}^{2}-4\right)\right]$. The coordinates of point $C$ satisfy $\left\{\begin{array}{l}y-y_{1}=-\left(y_{1}+2\right)\left[x-\left(y_{1}^{2}-4\right)\right] \\ y^{2}=x+4\end{array}\right.$, eliminating $x$, and noting that $y \neq y_{1}$, we get: $\left(y_{1}+2\right)\left(y+y_{1}\right)+1=0 \Leftrightarrow y_{1}^{2}+(2+y) y_{1}+(2 y+1)=0$, thus $\Delta=(2+y)^{2}-4(2 y+1) = y(y-4) \geqslant 0$, solving this gives $y \leqslant 0$ or $y \geqslant 4$, and when $y=0$, the coordinates of point $B$ are $(-3,-1)$, and when $y=4$, the coordinates of point $B$ are $(5,-3)$, both of which satisfy the conditions. Therefore, the range of the y-coordinate of point $C$ is $y \leqslant 0$ or $y \geqslant 4$.