跟踪训练
4.（2024·苏州高新区二模）如图①，抛物线$y = (x - m)^{2} - 2m + 1$（$m$为常数）与$x$轴交于$A$，$B$两点（点$B$在点$A$的右侧），与$y$轴交于点$C$.
（1）给出下列说法：①抛物线的开口向上；②点$C$在$y$轴正半轴上；③$m ＞ \frac{1}{2}$；④抛物线顶点在直线$y = - 2x + 1$上. 其中，正确的是________（填序号）.
（2）如图②，若直线$y = - 2x + 1$与该抛物线交于$M$，$N$两点（点$M$在点$N$的下方），试说明线段$MN$的长是一个定值，并求出这个值.
（3）在（2）的条件下，设直线$y = - 2x + 1$与$y$轴交于点$D$，连接$BM$，$BN$，$BD$. 当$DN:MN = 1:2$时，求$m$的值，并判断$\triangle MBN$与$\triangle MDB$是否相似.

（1）由$y=(x - m)^{2}-2m + 1$，得顶点的坐标为$(m,-2m + 1)$，二次项系数为$1＞0$.$\therefore$抛物线的开口向上.故①正确.当$x = 0$时，$y = m^{2}-2m + 1=(m - 1)^{2}\geqslant0$，$\therefore$点$C$不一定在$y$轴正半轴上.故②错误.令$y = 0$，得$(x - m)^{2}-2m + 1 = 0$，即$x^{2}-2mx + m^{2}-2m + 1 = 0$.$\because$抛物线与$x$轴交于$A$，$B$两点，且点$B$在点$A$的右侧，$\therefore$该方程有两个不同的实数根.$\therefore b^{2}-4ac=(-2m)^{2}-4(m^{2}-2m + 1)＞0$，解得$m＞\frac{1}{2}$.故③正确.将顶点坐标代入直线$y = - 2x + 1$，得$-2m + 1 = - 2m + 1$.故④正确.综上所述，①③④正确.（2）联立$\begin{cases}y = - 2x + 1\\y=(x - m)^{2}-2m + 1\end{cases}$，消去$y$、整理，得$x^{2}+(2 - 2m)x + m^{2}-2m = 0$.$\because b^{2}-4ac=(2 - 2m)^{2}-4(m^{2}-2m)=4＞0$，$\therefore$设点$M$的坐标为$(x_{1},y_{1})$，点$N$的坐标为$(x_{2},y_{2})$，则$x_{1}+x_{2}=2m - 2$，$x_{1}\cdot x_{2}=m^{2}-2m$.$\therefore MN=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}=\sqrt{(x_{1}-x_{2})^{2}+(-2x_{1}+1 + 2x_{2}-1)^{2}}=\sqrt{5}\cdot\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}$.$\therefore MN=\sqrt{5}\cdot\sqrt{(2m - 2)^{2}-4(m^{2}-2m)}=2\sqrt{5}$.$\therefore$线段$MN$的长是定值，该定值为$2\sqrt{5}$.（3）$\because DN:MN = 1:2$，$\therefore\frac{DN}{2\sqrt{5}}=\frac{1}{2}$.$\therefore DN=\sqrt{5}$.在$y = - 2x + 1$中，令$x = 0$，得$y = 1$.$\therefore$点$D$的坐标为$(0,1)$.$\therefore OD = 1$.设点$N$的坐标为$(n,-2n + 1)$，则$DN=\sqrt{n^{2}+(-2n + 1 - 1)^{2}}=\sqrt{5}$，解得$n = - 1$或$n = 1$.$\therefore$点$N$的坐标为$(-1,3)$或$(1,-1)$.当点$N$的坐标为$(-1,3)$时，$3=(-1 - m)^{2}-2m + 1$，解得$m = 1$或$m = - 1$.当$m = 1$时，抛物线对应的函数表达式为$y = x^{2}-2x$，令$y = 0$，得$x = 0$或$x = 2$.$\therefore$点$A$的坐标为$(0,0)$，点$B$的坐标为$(2,0)$.联立$\begin{cases}y = x^{2}-2x\\y = - 2x + 1\end{cases}$，解得$\begin{cases}x = 1\\y = - 1\end{cases}$或$\begin{cases}x = - 1\\y = 3\end{cases}$.$\therefore$点$M$的坐标为$(1,-1)$，点$N$的坐标为$(-1,3)$，符合条件.$\therefore MB=\sqrt{(1 - 2)^{2}+(-1 - 0)^{2}}=\sqrt{2}$，$MD=\sqrt{(1 - 0)^{2}+(-1 - 1)^{2}}=\sqrt{5}$.$\because MN = 2\sqrt{5}$，$\therefore MB:MN\neq MD:MB$.$\therefore\triangle MBN$与$\triangle MDB$不相似.当$m = - 1$时，抛物线对应的函数表达式为$y = x^{2}+2x + 4$，令$y = 0$，得$x^{2}+2x + 4 = 0$，无解，不符合题意，舍去.当点$N$的坐标为$(1,-1)$时，$-1=(1 - m)^{2}-2m + 1$，解得$m = 1$或$m = 3$.当$m = 1$时，已讨论.当$m = 3$时，抛物线对应的函数表达式为$y = x^{2}-6x + 4$，当$y = 0$时，$x^{2}-6x + 4 = 0$，解得$x = 3-\sqrt{5}$或$x = 3+\sqrt{5}$.$\therefore$点$A$的坐标为$(3-\sqrt{5},0)$，点$B$的坐标为$(3+\sqrt{5},0)$.联立$\begin{cases}y = x^{2}-6x + 4\\y = - 2x + 1\end{cases}$，解得$\begin{cases}x = 1\\y = - 1\end{cases}$或$\begin{cases}x = 3\\y = - 5\end{cases}$.$\therefore$点$N$的坐标为$(1,-1)$，点$M$的坐标为$(3,-5)$.$\therefore$易得$BM=\sqrt{30}$，$DM = 3\sqrt{5}$.$\because MN = 2\sqrt{5}$，$\therefore MN:BM=2\sqrt{5}:\sqrt{30}=\sqrt{6}:3$，$BM:DM=\sqrt{30}:3\sqrt{5}=\sqrt{6}:3$.$\therefore MN:BM = BM:DM$.$\because\angle BMN=\angle DMB$，$\therefore\triangle BMN\backsim\triangle DMB$.综上所述，当$m = 1$时，$\triangle MBN$与$\triangle MDB$不相似；当$m = 3$时，$\triangle MBN$与$\triangle MDB$相似.