16. (12分)(2026·山东期末)在平面直角坐标系$xOy$中,已知抛物线$C_1:y=ax^2+bx-\frac{1}{2}$顶点$P$的坐标$(1,-1)$.
(1)求抛物线$C_1$的表达式.
(2)将抛物线$C_1$沿射线$PO$平移$\sqrt{2}$个单位长度,得到抛物线$C_2$,$A(-2,m)$为抛物线$C_2$上的点.
①直接写出抛物线$C_2$的表达式.
②若$M,N$为抛物线$C_2$上异于$A$的两点,且$AM⊥AN$.记点$M,N$到直线$x=2$的距离分别为$d_1,d_2$,$d_1d_2$是一个定值吗?若是,请求出该值;若不是,请说明理由.

(1)$\because$ 抛物线$C_1:y=ax^2+bx-\frac{1}{2}$顶点P的坐标为$(1,-1)$,
$\therefore \begin{cases}-\frac{b}{2a}=1,\\a+b-\frac{1}{2}=-1,\end{cases}$ 解得$\begin{cases}a=\frac{1}{2},\\b=-1,\end{cases}$ $\therefore$ 抛物线$C_1$的表达式为$y=\frac{1}{2}x^2-x-\frac{1}{2}$.
(2)①抛物线$C_2$的表达式为$y=\frac{1}{2}x^2$. 解析:由(1)知抛物线$C_1$的表达式为$y=\frac{1}{2}x^2-x-\frac{1}{2}=\frac{1}{2}(x-1)^2-1$,由勾股定理可得$PO=\sqrt{1^2+1^2}=\sqrt{2}$,$\because$ 将抛物线$C_1$沿射线PO平移$\sqrt{2}$个单位长度,得到抛物线$C_2$,又$\because PO=\sqrt{2}$,$\therefore$ 抛物线$C_2$的顶点为点$O(0,0)$,$\therefore$ 抛物线$C_2$的表达式为$y=\frac{1}{2}x^2$.
②$d_1d_2$是定值.如图,作直线$x=-2$,直线$x=2$,过点M作x轴的平行线,分别交直线$x=-2$和直线$x=2$于点B,C,过点N作x轴的平行线,分别交直线$x=-2$和直线$x=2$于点D,E,设点$M(x_1,\frac{1}{2}x_1^2)$,$N(x_2,\frac{1}{2}x_2^2)$,将$x=-2$代入$y=\frac{1}{2}x^2$,得$y=\frac{1}{2}×(-2)^2=2$,$\therefore$ 点A的坐标为$(-2,2)$,由题意可知,$MB=x_1+2$,$MC=x_1-2$,$ND=x_2+2$,$EN=2-x_2$,$AD=2-\frac{1}{2}x_2^2$,$AB=\frac{1}{2}x_1^2-2$. $\because MB⊥ BD$,$ED⊥ BD$,$\therefore ∠ MBA = ∠ NDA = 90°$,$\therefore ∠ MAB + ∠ AMB = 90°$. $\because AM⊥ AN$,$\therefore ∠ MAN = 90°$,$\therefore ∠ MAB + ∠ NAD = 90°$,$\therefore ∠ AMB = ∠ NAD$,$\therefore △ AMB ∽ △ NAD$,$\therefore \frac{AD}{MB}=\frac{ND}{AB}$,即$AD· AB = MB· ND$,$\therefore (2-\frac{1}{2}x_2^2)(\frac{1}{2}x_1^2-2)=(x_1+2)·(x_2+2)$,$\therefore \frac{1}{4}(4-x_2^2)(x_1^2-4)=(x_1+2)(x_2+2)$,$\therefore \frac{1}{4}(2-x_2)(2+x_2)(x_1-2)(x_1+2)=(x_1+2)(x_2+2)$.
$\because$ 点M,N不与点A重合,$\therefore x_1+2≠0$,$x_2+2≠0$,$\therefore \frac{1}{4}(2-x_2)(x_1-2)=1$,即$(2-x_2)(x_1-2)=4$.$\because d_1=MC=x_1-2$,$d_2=EN=2-x_2$,$\therefore d_1d_2=(2-x_2)(x_1-2)=4$,即$d_1d_2$为定值4.

17. (14分)(2026·成都期中)在平面直角坐标系$xOy$中,抛物线$y=ax^2+bx-3$与$x$轴交于$A(-1,0),B$两点,与$y$轴交于点$C$,对称轴为直线$x=1$,点$D$为抛物线第二象限上一点.
(1)求抛物线的函数表达式.
(2)如图①,$P$为抛物线第四象限上任意一点,过点$P$作$PG// AC$交$BC$于点$G$,求$PG$最大值及此时$P$点坐标.
(3)如图②,连接$BD$,$F$为$BD$上一点,射线$CF$交抛物线于$E$,若$△ CFB∽△ EFD$,点$F$横坐标是否为定值?若是,请求出$F$的横坐标;若不是,请说明理由.

(1)$\because A(-1,0)$,对称轴为直线$x=1$,$\therefore$ 点B的坐标为$(3,0)$.把$A(-1,0)$和$B(3,0)$代入抛物线$y=ax^2+bx-3$中得$\begin{cases}a-b-3=0,\\9a+3b-3=0,\end{cases}$ 解得$\begin{cases}a=1,\\b=-2,\end{cases}$ $\therefore$ 抛物线的函数表达式$y=x^2-2x-3$.
(2)如图①,作$AM// BC$交y轴于M,$DP// OC$交BC于D.
$\therefore ∠ AMC = ∠ BCM = ∠ GDP$,$∠ MAC + ∠ ACB = 180°$,$∠ ACG = ∠ CGP$. $\because ∠ CGP + ∠ DGP = 180°$,$\therefore ∠ MAC = ∠ DGP$,$\therefore △ AMC ∽ △ GDP$,$\therefore \frac{AC}{GP}=\frac{MC}{DP}$.当$x=0$时,$y=-3$,$\therefore C(0,-3)$. $\because AO=1$,$CO=3$,$∠ AOC=90°$,$\therefore AC=\sqrt{AO^2+CO^2}=\sqrt{10}$. $\because B(3,0)$,$C(0,-3)$,$\therefore OB=OC=3$. $\because ∠ BOC=90°$,$\therefore △ BOC$是等腰直角三角形,$\therefore ∠ OBC = ∠ OCB = 45°$. $\because AM// BC$,$\therefore ∠ AMO = ∠ BCO = 45°$,$\therefore AO=MO=1$,$\therefore MC=MO+CO=4$,$\therefore \frac{\sqrt{10}}{GP}=\frac{4}{DP}$,$\therefore GP=\frac{\sqrt{10}}{4}DP$,$\therefore DP$最大时,GP就最大.$\because B(3,0)$,$C(0,-3)$,设直线BC的表达式为$y=kx+n$,$\therefore \begin{cases}3k+n=0,\\n=-3,\end{cases}$ 解得$\begin{cases}k=1,\\n=-3,\end{cases}$ 则直线BC的表达式为$y=x-3$.设$P(m,m^2-2m-3)(0＜m＜3)$,$\therefore D$的坐标为$(m,m-3)$,$\therefore PD=(m-3)-(m^2-2m-3)=-m^2+3m=-(m-\frac{3}{2})^2+\frac{9}{4}$,$\therefore$ 当$m=\frac{3}{2}$时,PD有最大值为$\frac{9}{4}$,纵坐标为$y=(\frac{3}{2})^2-2×\frac{3}{2}-3=-\frac{15}{4}$,$\therefore PG$此时为最大值,$\therefore GP=\frac{\sqrt{10}}{4}×\frac{9}{4}=\frac{9\sqrt{10}}{16}$,$\therefore PG$最大值为$\frac{9\sqrt{10}}{16}$,此时P点坐标为$(\frac{3}{2},-\frac{15}{4})$.

(3)点F的横坐标是定值,这个定值是$\frac{3}{2}$.理由:由(2)可知,直线BC的表达式为$y=x-3$.如图②,设点E的坐标为$(d,d^2-2d-3)$,设直线DE交x轴于点K,过点E作$EG⊥ x$轴于G,过点D作$DH⊥ x$轴于H,则$∠ EGK = ∠ DHK = 90°$,设点D的坐标为$(t,t^2-2t-3)$.
$\because B(3,0)$,$C(0,-3)$,$\therefore OB=OC=3$. $\because ∠ BOC=90°$,$\therefore △ BOC$是等腰直角三角形,$\therefore ∠ OBC=45°$.用t表示出BD的表达式为$y=(t+1)x+(-3t-3)$. $\because △ CFB ∽ △ EFD$,$\therefore ∠ BCF = ∠ DEF$,$\therefore DE// BC$,$\therefore ∠ OBC = ∠ AKE = 45°$,$\therefore △ EGK$是等腰直角三角形,$\therefore EG=KG$.同理得$△ DHK$是等腰直角三角形,$\therefore DH=KH=t^2-2t-3$,$\therefore d^2-2d-3=d-t+t^2-2t-3$,$\therefore d^2-3d=t^2-3t$,$\therefore d^2-t^2=3d-3t$,$\therefore (d+t)(d-t)=3(d-t)$,$\therefore (d+t-3)(d-t)=0$. $\because d-t≠0$,$\therefore d+t-3=0$,$\therefore d=3-t$,$\therefore$ 点E的坐标为$(3-t,t^2-4t)$.同理得EC的表达式为$y=(1-t)x-3$,$\therefore (1-t)x-3=(t+1)x+(-3t-3)$,$\therefore 2tx=3t$.
$\because t≠0$,$\therefore x=\frac{3}{2}$,$\therefore$ 点F横坐标是定值,这个定值是$\frac{3}{2}$.