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(1) 将三个点代入二次函数，得到一个二元一次方程组，求解即可求出$b,c$
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$$
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\large \left\{\begin{matrix}
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0=-1-b+c & \\
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3=-4+2b+c & \\
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\end{matrix}\right.
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$$
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$\therefore a=-1,b=2,c=3,方程：y=-x^2+2x+3$
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交$y$轴于点$C$,则$x=0,y=3$ $\therefore C坐标(0,3)$
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$D$点坐标可求:
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$x_d=-\frac{b}{2a}=\frac{2}{2}=1,y_d=\frac{4ac-b^2}{4a}=4$
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$AB$的直线方程也很好求：
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$y=kx+b$
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$$
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\large \left\{\begin{matrix}
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0=-k+b& \\
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3=2k+b&
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\end{matrix}\right.
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$$
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$\therefore k=1,b=1,y=x+1$
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---
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**重点是第二问**
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由于没有说明$A,C,M$哪个顶点是直角顶点，需要分类讨论：
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- $A$是直角顶点
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此时，过$A$点引$AC$的垂线
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- $C$是直角顶点
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此时，过$C$点引$AC$的垂线
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- $M$是直角顶点
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此时，以$AC$为直径画圆，此圆与$y$轴有两个交点
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#### 几何法
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1. $AM_1$需要做相似三角形，通过比例关系来求解
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$\triangle AM_1F \sim \triangle CEA$
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$\frac{CE}{AF}=\frac{AE}{M_1F}$
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$1/m=3/2$ $\Rightarrow m=\frac{2}{3}$
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$\therefore M1(1,-\frac{2}{3})$
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2. 
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#### 代数法
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- 当$A$为直角顶点时：
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$AC^2+AM^2=CM^2$
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$10+t^2+4=t^2-6t+10$
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$t=-\frac{2}{3}$
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$\therefore M(1,-\frac{2}{3})$
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- 当$C$为直角顶点时：
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$AC^2+CM^2=AM^2$
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$10+t^2-6t+10=t^2+4$
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$t=\frac{8}{3}$
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$\therefore M(1,\frac{8}{3})$
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- 当$M$为直角顶点时：
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$AC^2=AM^2+CM^2$
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$10=t^2+4+t^2-6t+10$
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$t^2-3t+2=0$
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$t=2,t=1$
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$\therefore M(1,2)$ 或 $M(1,1)$
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