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(1)
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将三个点代入方程，求解方程组
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$$
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\large \left\{\begin{matrix}
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0=a(-1)^2-b+c & \\
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0=9a+3b+c & \\
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c=-3
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\end{matrix}\right.
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$$
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整理:
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$$
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\large \left\{\begin{matrix}
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a-b-3=0 & \\
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3a+b-1=0 & \\
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c=-3
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\end{matrix}\right.
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$$
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$$ \Rightarrow
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\large \left\{\begin{matrix}
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a=1& \\
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b=-2& \\
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c=-3
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\end{matrix}\right.
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$$
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即$y=x^2-2x-3$
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顶点$D$的坐标,根据顶点坐标公式：
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$$
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\large \left\{\begin{matrix}
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x=-\frac{b}{2a} & \\
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y=\frac{4ac-b^2}{4a} &
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\end{matrix}\right.
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$$
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$$ \Rightarrow
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\large \left\{\begin{matrix}
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x=1 & \\
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y=-4 &
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\end{matrix}\right.
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$$
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---
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**重点是第二问**：
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因为是等腰三角形的情况共三种，需要分情况讨论：
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- $AD=AP$ 在以$A$为圆心，以$AD$长为半径的圆对称轴的交点上
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- $AD=DP$ 在以$D$为圆心，以$AD$长为半径的圆对称轴的交点上
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- $AP=DP$ 在$AD$垂直平分线与对称轴的交点上
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有两种方法，几何法和代数法，分别来计算一下：
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#### 几何法
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- $AD=AP$ 时，$P$点坐标就是$D$关于$X$轴的对称点，$P(1,4)$
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- $AD=DP$ 时，利用勾股定理，可以求解$AD$长度，也就是$DP$长度=$\sqrt{(-1-1)^2+(-4)^2}=2\sqrt{5}$,$P_2=({-1,2\sqrt{5}-4})$
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<font color='red' size=4><b>注意</b></font>
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这里非常容易丢失一组解！也可能是以$D$为圆心的圆与对称轴的下方交点! 即$P_3=(-4-2\sqrt{5})$
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垂直平分线的交点:
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此时$AP=PD$
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设 上面一小段为$m$
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$(4-m)^2=m^2+2^2 \Rightarrow $
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$16-8m+m^2=m^2+4$
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$m=\frac{3}{2}$
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$P_4(1,-\frac{3}{2})$
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#### 代数法
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- 表示点
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$A(-1,0),D(1,-4),P(1,t)$
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- 表示边
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利用两点间距离公式 $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
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分别计算$AD^2,AP^2,DP^2$,就可以免去开根号了
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- 列方程求解
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$AD^2=4+16=20$
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$AP^2=4+t^2$
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$DP^2=(t+4)^2$
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三个方程式分别联立成三个方程组：
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$$
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\large \left\{\begin{matrix}
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4+t^2=20 & ① \\
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4+t^2=t^2+8t+16 & ② \\
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(t+4)^2=20 & ③
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\end{matrix}\right.
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$$
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① $t=4$ (根据题意舍掉$-4$)
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② $t=-\frac{3}{2}$
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③ $t=\pm2\sqrt{5}-4$
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