China Mathematical Competition

Let $A(x_1,y_1)$, $B(x_2,y_2)$, $C(t^2,2t)$. From $$\{\begin{array}{l}x-2y-1=0,\\ y^2=4x,\end{array}$$ we get $y^2-8y-4=0$, which means $y_1+y_2=8$, $y_1\cdot y_2=-4$. Since $x_1=2y_1+1$, $x_2=2y_2+1$, we have $$\begin{array}{rl}{x}_1+x_2& =2(y_1+y_2)+2=18,\\ x_1\cdot x_2& =4y_1\cdot y_2+2(y_1+y_2)+1=1.\end{array}$$ Furthermore, by $\angle ACB=90^{\circ }$, we have $\overrightarrow{CA}\cdot \overrightarrow{CB}=0$, which means $$(t^2-x_1)(t^2-x_2)+(2t-y_1)(2t-y_2)=0,$$ that is $$t^4-(x_1+x_2)t^2+x_1\cdot x_2+4t^2-2(y_1+y_2)t+y_1\cdot y_2=0.$$ Then $$t^4-14t^2-16t-3=0,$$ or $$(t^2+4t+3)(t^2-4t-1)=0.$$ Obviously, $t^2-4t-1\ne 0$; otherwise, we have $t^2-2\cdot 2t-1=0$, which means $C$ is on $x-2y-1=0$, i.e., $C$ coincides with either $A$ or $B$. So $t^2+4t+3=0$. Then $t_1=-1$, $t_2=-3$. Therefore, the coordinate of $C$ is $(1,-2)$ or $(9,-6)$.