典例3（2023·扬州）在平面直角坐标系中，点$A$在$y$轴正半轴上.
（1）已知四个点$(0,0)$，$(0,2)$，$(1,1)$，$(-1,1)$中恰有三个点在二次函数$y = ax^{2}$（$a$为常数，且$a\neq0$）的图像上.
①$a = $_______.
②如图①，菱形$ABCD$的顶点$B$，$C$，$D$在该二次函数的图像上，且$AD\perp y$轴，求菱形的边长.
③如图②，正方形$ABCD$的顶点$B$，$D$在该二次函数的图像上，点$B$，$D$在$y$轴的同侧，且点$B$在点$D$的左侧，设点$B$，$D$的横坐标分别为$m$，$n$，试探究$n - m$是否为定值. 如果是，求出这个值；如果不是，请说明理由.
（2）已知正方形$ABCD$的顶点$B$，$D$在二次函数$y = ax^{2}$（$a$为常数，且$a＞0$）的图像上，点$B$在点$D$的左侧，设点$B$，$D$的横坐标分别为$m$，$n$，直接写出$m$，$n$满足的等量关系式.
　　　

典例3(1)①1.
②设$BC$交$y$轴于点$E$.由①知，$y = x^{2}$.设菱形$ABCD$的边长为$2t$，则$AD// BC$，$AB = BC = CD = AD = 2t$.$\because AD\perp y$轴，$\therefore BC\perp y$轴.$\therefore$易得点$B$，$C$关于$y$轴对称.$\therefore BE = CE = t$.$\therefore B(-t,t^{2})$.$\therefore OE = t^{2}$.$\therefore AE=\sqrt{AB^{2}-BE^{2}}=\sqrt{3}t$，$\therefore OA = OE + AE = t^{2}+\sqrt{3}t$.$\therefore D(2t,t^{2}+\sqrt{3}t)$.把$D(2t,t^{2}+\sqrt{3}t)$代入$y = x^{2}$，得$t^{2}+\sqrt{3}t = 4t^{2}$，解得$t=\frac{\sqrt{3}}{3}$或$t = 0$（不合题意，舍去）.$\therefore$菱形的边长为$\frac{2\sqrt{3}}{3}$.
③$n - m$为定值.如图①，过点$B$作$BF\perp y$轴于点$F$，过点$D$作$DE\perp y$轴于点$E$，则$\angle AFB=\angle DEA = 90^{\circ}$.$\because$点$B$，$D$的横坐标分别为$m$，$n$，$\therefore B(m,m^{2})$，$D(n,n^{2})$.$\therefore BF = m$，$OF = m^{2}$，$DE = n$，$OE = n^{2}$.$\because$四边形$ABCD$是正方形，$\therefore\angle DAB = 90^{\circ}$，$AB = DA$.$\therefore$易得$\angle BAF=\angle ADE$.又$\because\angle AFB=\angle DEA = 90^{\circ}$，$\therefore\triangle ABF\cong\triangle DAE$.$\therefore BF = AE = m$，$AF = DE = n$.$\therefore AE + AF = OE - OF$，即$m + n = n^{2}-m^{2}=(n - m)(n + m)$.$\because$点$B$，$D$在$y$轴的同侧，$\therefore m + n\neq0$.$\therefore n - m = 1$.
(2)过点$B$作$BF\perp y$轴于点$F$，过点$D$作$DE\perp y$轴于点$E$.$\because$点$B$，$D$的横坐标分别为$m$，$n$，$\therefore B(m,am^{2})$，$D(n,an^{2})$.
①当点$B$，$D$在$y$轴的左侧时，如图②，$\therefore BF=-m$，$OF = am^{2}$，$DE=-n$，$OE = an^{2}$.同理(1)③，可得$\triangle ABF\cong\triangle DAE$，$\therefore BF = AE=-m$，$AF = DE=-n$.$\therefore AE + AF = OF - OE$，即$-m - n = am^{2}-an^{2}$.$\therefore m + n = a(n - m)(n + m)$.$\because m + n\neq0$，$\therefore n - m=\frac{1}{a}$.
②当点$B$在$y$轴的左侧，点$D$在$y$轴的右侧时，如图③，$\therefore BF=-m$，$OF = am^{2}$，$DE = n$，$OE = an^{2}$.同理(1)③，可得$\triangle ABF\cong\triangle DAE$，$\therefore BF = AE=-m$，$AF = DE = n$.$\therefore AE - AF = OF - OE$，即$-m - n = am^{2}-an^{2}$.$\therefore m + n = a(n + m)(n - m)$.$\therefore m + n = 0$或$n - m=\frac{1}{a}$.
③当点$B$，$D$在$y$轴的右侧时，如图④，$\therefore BF = m$，$OF = am^{2}$，$DE = n$，$OE = an^{2}$.同理(1)③，可得$\triangle ABF\cong\triangle DAE$，$\therefore BF = AE = m$，$AF = DE = n$.$\therefore AE + AF = OE - OF$，即$m + n = an^{2}-am^{2}=a(n + m)(n - m)$.$\because m + n\neq0$，$\therefore n - m=\frac{1}{a}$.
综上所述，$m$，$n$满足的等量关系式为$m + n = 0$或$n - m=\frac{1}{a}$.

典例4（2024·镇江一模）已知点$M(m - 1,3 - m^{2})$在平面直角坐标系中，小明给了一些$m$的取值，$m - 1$，$3 - m^{2}$的部分对应值如下表：

他在平面直角坐标系中描出这些点后，观察出点$M$在以$A(-1,3)$为顶点的抛物线上.
（1）求该抛物线对应的函数表达式，并说明无论$m$取何实数值，点$M$都在此抛物线上.
（2）设（1）中的抛物线与$x$轴交于点$B$，$C$（点$B$在点$C$的左侧），点$D$在该抛物线的对称轴上，$\triangle PQM$是$\triangle ABC$以点$D$为位似中心的位似图形（点$A$，$B$，$C$的对应点分别是$P$，$Q$，$M$）. 若$\triangle PQM$与$\triangle ABC$的相似比是$1:\sqrt{3}$，求$m$的值.

典例4(1)$\because$点$M$在以$A(-1,3)$为顶点的抛物线上，$\therefore$设该抛物线对应的函数表达式为$y = a(x + 1)^{2}+3$.$\because$点$(-3,-1)$在该抛物线上，$\therefore-1 = a(-3 + 1)^{2}+3$，解得$a=-1$.$\therefore$该抛物线对应的函数表达式为$y=-(x + 1)^{2}+3$.当$x = m - 1$时，$y = 3 - m^{2}$，$\therefore$无论$m$取何实数，点$M$都在此抛物线上.
(2)在$y=-(x + 1)^{2}+3$中，令$y = 0$，得$-(x + 1)^{2}+3 = 0$，解得$x=-1\pm\sqrt{3}$.$\therefore$点$B$的坐标为$(-1-\sqrt{3},0)$，点$C$的坐标为$(-1+\sqrt{3},0)$.$\therefore BC = 2\sqrt{3}$，抛物线的对称轴垂直平分$BC$.又$\because$点$A$的坐标为$(-1,3)$，$\therefore$易得$AB = AC = 2\sqrt{3}$.$\therefore AB = AC = BC$.$\therefore\triangle ABC$是等边三角形.$\because$点$D$在抛物线的对称轴上，$\triangle PQM$是$\triangle ABC$以点$D$为位似中心的位似图形，$\therefore$点$P$在抛物线的对称轴上.$\because\triangle PQM$与$\triangle ABC$的相似比是$1:\sqrt{3}$，$\therefore BC// QM$.$\therefore\frac{QM}{BC}=\frac{1}{\sqrt{3}}$.$\therefore QM = 2$，抛物线的对称轴垂直平分$QM$.$\therefore$点$Q$与点$M$关于对称轴对称.当点$M$在对称轴左侧时，$-1-(m - 1)=1$，解得$m=-1$.当点$M$在对称轴右侧时，$(m - 1)-(-1)=1$，解得$m = 1$.综上所述，$m$的值为$1$或$-1$.