《2025年启东中学作业本九年级数学上册人教版》

6. 如图,在平面直角坐标系$xOy$中,抛物线$L_{1}:y = x^{2}-2x - 3的顶点为P$。直线$l过点M(0,m)(m\geqslant - 3)$,且平行于$x$轴,与抛物线$L_{1}交于A$,$B$两点(点$B在点A$的右侧)。将抛物线$L_{1}沿直线l翻折得到抛物线L_{2}$,抛物线$L_{2}交y轴于点C$,顶点为$D$。连接$BC$,$CD$,$DB$,若$\triangle BCD$为直角三角形,求此时$L_{2}$所对应的函数解析式。

答案：
6.解:由题意知抛物线$L_{1}$的顶点P(1,-4)与$L_{2}$的顶点D关于直线$y=m$对称,$\therefore$D(1,$2m+4$),抛物线$L_{2}:y=-(x-1)^{2}+(2m+4)=-x^{2}+2x+2m+3$,$\therefore$C(0,$2m+3$).①当$\angle BCD=90^{\circ}$时,如答图①,过点D作DN⊥y轴于点N.

$\because$D(1,$2m+4$),$\therefore$N(0,$2m+4$),$\because$C(0,$2m+3$),$\therefore DN=NC=1$,$\therefore \angle DCN=45^{\circ}$,$\because \angle BCD=90^{\circ}$,$\therefore \angle BCE=45^{\circ}$.$\because$当m=-3时,点C,E重合,此种情况不满足,$\therefore m\neq -3$.$\because$直线$l// x$轴,$\therefore \angle BEC=90^{\circ}$,$\therefore \angle CBE=\angle BCE=45^{\circ}$,$\therefore BE=CE=(2m+3)-m=m+3$,$\therefore$B($m+3$,$m$).$\because$点B在$y=x^{2}-2x-3$的图象上,$\therefore m=(m+3)^{2}-2(m+3)-3$,$\therefore m=0$或m=-3(舍去),将m=0代入$y=-x^{2}+2x+2m+3$,得$L_{2}:y=-x^{2}+2x+3$.②当$\angle BDC=90^{\circ}$时,如答图②,过点D作DN⊥y轴于点N,过点B作BT⊥ND交ND的延长线于点T.

同理,BT=DT,$\because$D(1,$2m+4$),$\therefore DT=BT=(2m+4)-m=m+4$,$\because DN=1$,$\therefore NT=DN+DT=1+(m+4)=m+5$,$\therefore$B($m+5$,$m$).$\because$点B在$y=x^{2}-2x-3$的图象上,$\therefore m=(m+5)^{2}-2(m+5)-3$,解得m=-3或m=-4,$\because m\geq -3$,$\therefore m=-3$,此时,B(2,-3),C(0,-3),符合题意,将m=-3代入$y=-x^{2}+2x+2m+3$得$L_{2}:y=-x^{2}+2x-3$.③易知,当$\angle DBC=90^{\circ}$时,此种情况不存在.综上所述,$L_{2}$所对应的函数解析式为$y=-x^{2}+2x+3$或$y=-x^{2}+2x-3$.

7. 抛物线$y = - x^{2}-2x + 3与x轴交于A(1,0)$,$B$两点,点$P$在抛物线的对称轴上,若线段$PA绕点P逆时针旋转90^{\circ}$后,点$A的对应点A'$恰好落在此抛物线上,则点$P$的坐标为(-1,1)或(-1,-2)。

答案： (-1,1)或(-1,-2)

8. (2024·眉山)如图,抛物线$y = - x^{2}+bx + c与x轴交于点A(-3,0)和点B$,与$y轴交于点C(0,3)$,点$D$在抛物线上。
(1)求该抛物线的函数解析式。
(2)当点$D$在第二象限内,且$\triangle ACD的面积为3$时,求点$D$的坐标。
(3)在直线$BC上是否存在点P$,使$\triangle OPD是以PD$为斜边的等腰直角三角形？若存在,请直接写出点$P$的坐标；若不存在,请说明理由。

答案：
8.解:
(1)把A(-3,0),C(0,3)代入$y=-x^{2}+bx+c$,得$\left\{\begin{array}{l} -9-3b+c=0,\\ c=3,\end{array}\right. $解得$\left\{\begin{array}{l} b=-2,\\ c=3,\end{array}\right. $$\therefore$抛物线的函数解析式为$y=-x^{2}-2x+3$.
(2)过点D作DK$// y$轴交AC于点K,如答图①.

由A(-3,0),C(0,3)得直线AC的函数解析式为$y=x+3$,设D(t,$-t^{2}-2t+3$),则K(t,$t+3$),$\therefore DK=-t^{2}-2t+3-(t+3)=-t^{2}-3t$.$\because \triangle ACD$的面积为3,$\therefore \frac{1}{2}DK\cdot |x_{A}-x_{C}|=3$,即$\frac{1}{2}(-t^{2}-3t)× 3=3$,解得t=-1或t=-2,$\therefore$点D的坐标为(-1,4)或(-2,3).
(3)在直线BC上存在点P,使$\triangle OPD$是以PD为斜边的等腰直角三角形,点P的坐标为(0,3)或$(\frac{25-\sqrt{193}}{18},\frac{-7+\sqrt{193}}{6})$或$(\frac{25+\sqrt{193}}{18},\frac{-7-\sqrt{193}}{6})$或$(\frac{11}{9},-\frac{2}{3})$.点拨:在$y=-x^{2}-2x+3$中,令y=0,得$0=-x^{2}-2x+3$,解得x=-3或x=1,$\therefore$B(1,0).由B(1,0),C(0,3)得直线BC的函数解析式为$y=-3x+3$.设P(m,$-3m+3$),D(n,$-n^{2}-2n+3$),过点P作PN⊥y轴于点N,过点D作DM⊥y轴于点M①$\because OA=OC=3$,$\therefore$当点P与点C重合,点D与点A重合时,$\triangle OPD$是等腰直角三角形,如答图②,

此时P(0,3);②当点P在第一象限,点D在第四象限时,如答图③,

$\because \triangle OPD$是以PD为斜边的等腰直角三角形,$\therefore OD=OP$,$\angle POD=90^{\circ}$,$\therefore \angle DOM=90^{\circ}-\angle PON=\angle OPN$.$\because \angle DMO=90^{\circ}=\angle PNO$,$\therefore \triangle DOM\cong \triangle OPN(AAS)$,$\therefore DM=ON$,$OM=PN$,$\therefore \left\{\begin{array}{l} n=-3m+3,\\ n^{2}+2n-3=m,\end{array}\right. $解得$\left\{\begin{array}{l} m=\frac{25+\sqrt{193}}{18},\\ n=\frac{-7-\sqrt{193}}{6}\end{array}\right. $(n小于0,舍去)或$\left\{\begin{array}{l} m=\frac{25-\sqrt{193}}{18},\\ n=\frac{-7+\sqrt{193}}{6},\end{array}\right. $$\therefore -3m+3=n=\frac{-7+\sqrt{193}}{6}$,$\therefore$点P的坐标为$(\frac{25-\sqrt{193}}{18},\frac{-7+\sqrt{193}}{6})$;③当点P在第四象限,点D在第三象限时,如答图④.

$\because \triangle OPD$是以PD为斜边的等腰直角三角形,$\therefore OD=OP$,$\angle POD=90^{\circ}$,$\therefore \angle DOM=90^{\circ}-\angle PON=\angle OPN$.$\because \angle DMO=90^{\circ}=\angle PNO$,$\therefore \triangle DOM\cong \triangle OPN(AAS)$,$\therefore PN=OM$,$ON=DM$,$\therefore \left\{\begin{array}{l} m=n^{2}+2n-3,\\ 3m-3=-n,\end{array}\right. $解得$\left\{\begin{array}{l} m=\frac{25+\sqrt{193}}{18},\\ n=\frac{-7-\sqrt{193}}{6}\end{array}\right. $或$\left\{\begin{array}{l} m=\frac{25-\sqrt{193}}{18},\\ n=\frac{-7+\sqrt{193}}{6}\end{array}\right. $(n大于0,舍去),$\therefore -3m+3=n=\frac{-7-\sqrt{193}}{6}$,$\therefore$点P的坐标为$(\frac{25+\sqrt{193}}{18},\frac{-7-\sqrt{193}}{6})$;④当点P在第四象限,点D在第一象限时,如答图⑤,

$\because \triangle OPD$是以PD为斜边的等腰直角三角形,$\therefore OD=OP$,$\angle POD=90^{\circ}$,$\therefore \angle DOM=90^{\circ}-\angle PON=\angle OPN$.$\because \angle DMO=90^{\circ}=\angle PNO$,$\therefore \triangle DOM\cong \triangle OPN(AAS)$,$\therefore PN=OM$,$ON=DM$,$\therefore \left\{\begin{array}{l} m=-n^{2}-2n+3,\\ 3m-3=n,\end{array}\right. $解得$\left\{\begin{array}{l} m=0,\\ n=-3\end{array}\right. $(舍去)或$\left\{\begin{array}{l} m=\frac{11}{9},\\ n=\frac{2}{3},\end{array}\right. $$\therefore -3m+3=-n=-\frac{2}{3}$,$\therefore$点P的坐标为$(\frac{11}{9},-\frac{2}{3})$.综上所述,点P的坐标为(0,3)或$(\frac{25-\sqrt{193}}{18},\frac{-7+\sqrt{193}}{6})$或$(\frac{25+\sqrt{193}}{18},\frac{-7-\sqrt{193}}{6})$或$(\frac{11}{9},-\frac{2}{3})$.