3. (2026·泸州期中)如图,抛物线$y=-\dfrac{3}{5}x^2+bx+c$与x轴交于点A和点B(5,0),与y轴交于点C(0,-3),连接BC,点E是对称轴上的一个动点.点P在抛物线上.
(1)求抛物线的解析式.
(2)在抛物线上是否存在点P,使△BPE是以BE为斜边的等腰直角三角形?若存在,请求出点P的坐标;若不存在,请说明理由.

(1)$\because$ 抛物线$y=-\dfrac{3}{5}x^2+bx+c$经过$B(5,0)$,$C(0,-3)$,$\therefore \begin{cases} -15+5b+c=0, \\ c=-3, \end{cases}$解得$\begin{cases} b=\dfrac{18}{5}, \\ c=-3, \end{cases}$
∴该抛物线的解析式为$y=-\dfrac{3}{5}x^2+\dfrac{18}{5}x-3$.
(2)设$E(3,m)$,$P(n,-\dfrac{3}{5}n^2+\dfrac{18}{5}n-3)$,
①当点$P$在$x$轴上方时,$1＜n＜5$,如图,过点$P_1$作对称轴的垂线,垂足为$F$,过点$B$作$BG⊥ FP_1$的延长线于点$G$(点$P_1,P_2$同理),

$\because △ BP_1E_1$是以$BE_1$为斜边的等腰直角三角形,$\therefore ∠ BP_1E_1=90°$,$P_1B=P_1E_1$,$\therefore ∠ BP_1G+∠ E_1P_1F=90°$.$\because ∠ G=∠ P_1FE_1=90°$,$\therefore ∠ BP_1G+∠ P_1BG=90°$,$\therefore ∠ P_1BG=∠ E_1P_1F$,在$△ BP_1G$和$△ P_1E_1F$中,$\begin{cases} ∠ G=∠ P_1FE_1=90°, \\ ∠ P_1BG=∠ E_1P_1F, \\ P_1B=P_1E_1, \end{cases}$$\therefore △ BP_1G≌ △ P_1E_1F$(AAS),$\therefore BG=P_1F$,$\therefore -\dfrac{3}{5}n^2+\dfrac{18}{5}n-3=|n-3|$,解得$n_1=\dfrac{13}{3}$,$n_2=0$(舍),$n_3=\dfrac{5}{3}$,$n_4=6$(舍),$\therefore P_1(\dfrac{13}{3},\dfrac{4}{3})$,$P_2(\dfrac{5}{3},\dfrac{4}{3})$;
②当点$P$在$x$轴下方时,$n＞5$或$n＜1$,如图,过点$P_4$作$x$轴的垂线,垂足为$H$,过点$E_4$作$E_4K⊥ HP_4$的延长线于点$K$(点$P_3,P_4$同理),$\because △ BP_4E_4$是以$BE_4$为斜边的等腰直角三角形,$\therefore ∠ BP_4E_4=90°$,$P_4B=P_4E_4$,$\therefore ∠ BP_4H+∠ E_4P_4K=90°$.$\because ∠ K=∠ P_4HB=90°$,$\therefore ∠ BP_4H+∠ P_4BH=90°$,$\therefore ∠ P_4BH=∠ E_4P_4K$,在$△ BP_4H$和$△ P_4E_4K$中,$\begin{cases} ∠ K=∠ P_4HB=90°, \\ ∠ P_4BH=∠ E_4P_4K, \\ BP_4=P_4E_4, \end{cases}$$\therefore △ BP_4H≌ △ P_4E_4K$(AAS),$\therefore P_4H=E_4K$,$\therefore \dfrac{3}{5}n^2-\dfrac{18}{5}n+3=|n-3|$,解得$n_1=0$,$n_2=\dfrac{13}{3}$(舍),$n_3=\dfrac{5}{3}$(舍),$n_4=6$,$\therefore P_3(0,-3)$,$P_4(6,-3)$,
综上所述,点$P$的坐标为$(0,-3)$,$(\dfrac{13}{3},\dfrac{4}{3})$,$(6,-3)$,$(\dfrac{5}{3},\dfrac{4}{3})$.

4. (2025·东营中考节选)如图,抛物线$y=-x^2+bx+c$交x轴于A,B两点,交y轴于点C,其中$A(-1,0),C(0,5)$.
(1)求抛物线的解析式；
(2)点P为对称轴上一点,当$△ ACP$的周长最小时,求点P的坐标；
(3)点M为对称轴上一点,点N为抛物线上一点,若以A,C,M,N为顶点的四边形是平行四边形,请直接写出点N的坐标.

(1)把$A(-1,0)$,$C(0,5)$代入$y=-x^2+bx+c$中得,$\begin{cases} 0=-1-b+c, \\ 5=c, \end{cases}$解得$\begin{cases} b=4, \\ c=5, \end{cases}$$\therefore y=-x^2+4x+5$.
(2)$\because A(-1,0)$,$C(0,5)$,$\therefore AC=\sqrt{26}$,$\therefore$ 当$AP+CP$的值最小时,则$△ ACP$的周长最小.作点$A$关于对称轴的对称点,即为点$B$,由(1)可知抛物线的解析式为$y=-x^2+4x+5$,$\therefore$ 对称轴为直线$x=-\dfrac{4}{2×(-1)}=2$,且$A(-1,0)$,$\therefore B(5,0)$.如图,连接$BC$,与对称轴的交点即点$P$.

设直线$BC$的解析式为$y=kx+b(k≠0)$,把$B(5,0)$,$C(0,5)$代入$y=kx+b(k≠0)$中得,$\begin{cases} 5=b, \\ 0=5k+b, \end{cases}$解得$\begin{cases} b=5, \\ k=-1, \end{cases}$$\therefore$ 直线$BC$的解析式为$y=-x+5$.
$\because$ 点$P$的横坐标为$x=2$,$\therefore$ 把$x=2$代入$y=-x+5$得$y=3$,$\therefore P(2,3)$.
(3)点$N$的坐标为$(-3,-16)$或$(1,8)$或$(3,8)$ 解析:设$M(2,m)$,$N(t,-t^2+4t+5)$,①当$AC$为对角线时,设$AC$中点为$E$,根据平行四边形的性质,点$E$也为$MN$的中点,$\because A(-1,0)$,$C(0,5)$,$\therefore E(-\dfrac{1}{2},\dfrac{5}{2})$,$\therefore \begin{cases} \dfrac{2+t}{2}=-\dfrac{1}{2}, \\ \dfrac{m+(-t^2+4t+5)}{2}=\dfrac{5}{2}, \end{cases}$解得$\begin{cases} t=-3, \\ m=21, \end{cases}$把$t=-3$代入$-t^2+4t+5$,得$-(-3)^2+4×(-3)+5=-16$,$\therefore N(-3,-16)$;
②当$AM$为对角线时,设$AM$中点为$F$,根据平行四边形的性质,点$F$也为$CN$的中点,$\because A(-1,0)$,$M(2,m)$,$\therefore F(\dfrac{1}{2},\dfrac{m}{2})$,$\therefore \begin{cases} \dfrac{0+t}{2}=\dfrac{1}{2}, \\ \dfrac{5+(-t^2+4t+5)}{2}=\dfrac{m}{2}, \end{cases}$解得$\begin{cases} t=1, \\ m=13, \end{cases}$把$t=1$代入$-t^2+4t+5$,得$-1^2+4×1+5=8$,$\therefore N(1,8)$;
③当$AN$为对角线时,设$AN$中点为$G$,根据平行四边形的性质,点$G$也为$CM$的中点,$\because A(-1,0)$,$N(t,-t^2+4t+5)$,$\therefore G(\dfrac{-1+t}{2},\dfrac{-t^2+4t+5}{2})$,$\therefore \begin{cases} \dfrac{-1+t}{2}=\dfrac{2+0}{2}, \\ \dfrac{-t^2+4t+5}{2}=\dfrac{m+5}{2}, \end{cases}$解得$\begin{cases} t=3, \\ m=3, \end{cases}$把$t=3$代入$-t^2+4t+5$,得$-3^2+4×3+5=8$,$\therefore N(3,8)$.
综上所述,若以$A,C,M,N$为顶点的四边形是平行四边形,此时点$N$的坐标为$(-3,-16)$或$(1,8)$或$(3,8)$.