搜索
1. (2024·无锡模拟)如图，二次函数$y = x^{2} - 4x + 3$的图像交$x$轴于点$A,B$(点$A$在点$B$的左侧)，交$y$轴于点$C$，将线段$AC$绕着点$C$逆时针旋转$45^{\circ}$，其所在直线与二次函数图像交于点$E$，则点$E$的坐标为
($\frac{7}{2}$,$\frac{5}{4}$)
.
答案：
1.($\frac{7}{2}$,$\frac{5}{4}$) 点拨：令$y=0$，则$x^{2}-4x + 3 = 0$，
解得$x_1 = 1$，$x_2 = 3$，$\therefore A(1,0)$，$B(3,0)$，$\therefore OA = 1$，$OB = 3$.
令$x = 0$，则$y = 3$，$\therefore C(0,3)$，$\therefore OC = 3$.
过点$A$作$AC$的垂线，交$CE$于点$F$，过点$F$作$x$轴的垂线，交$x$轴于点$D$，如答图
$\because \angle ACF = 45^{\circ}$，$\angle CAF = 90^{\circ}$，$\therefore AC = AF$.
$\because \angle FAD + \angle CAO = 90^{\circ}$，$\angle ACO + \angle CAO = 90^{\circ}$，
$\therefore \angle FAD = \angle ACO$，$\therefore \triangle FAD \cong \triangle ACO(AAS)$，
$\therefore FD = OA = 1$，$AD = OC = 3$，
$\therefore OD = OA + AD = 1 + 3 = 4$，$\therefore F(4,1)$.
设直线$CE$的函数表达式为$y = kx + b$，
把$C(0,3)$，$F(4,1)$代入，得$\begin{cases}b = 3\\4k + b = 1\end{cases}$，
解得$\begin{cases}k = -\frac{1}{2}\\b = 3\end{cases}$，
$\therefore$直线$CE$的函数表达式为$y = -\frac{1}{2}x + 3$，
联立$\begin{cases}y = -\frac{1}{2}x + 3\\y = x^{2} - 4x + 3\end{cases}$，解得$\begin{cases}x = 0\\y = 3\end{cases}$或$\begin{cases}x = \frac{7}{2}\\y = \frac{5}{4}\end{cases}$
$\therefore E(\frac{7}{2},\frac{5}{4})$.
1.($\frac{7}{2}$,$\frac{5}{4}$) 点拨：令$y=0$，则$x^{2}-4x + 3 = 0$，
解得$x_1 = 1$，$x_2 = 3$，$\therefore A(1,0)$，$B(3,0)$，$\therefore OA = 1$，$OB = 3$.
令$x = 0$，则$y = 3$，$\therefore C(0,3)$，$\therefore OC = 3$.
过点$A$作$AC$的垂线，交$CE$于点$F$，过点$F$作$x$轴的垂线，交$x$轴于点$D$，如答图
$\because \angle ACF = 45^{\circ}$，$\angle CAF = 90^{\circ}$，$\therefore AC = AF$.
$\because \angle FAD + \angle CAO = 90^{\circ}$，$\angle ACO + \angle CAO = 90^{\circ}$，
$\therefore \angle FAD = \angle ACO$，$\therefore \triangle FAD \cong \triangle ACO(AAS)$，
$\therefore FD = OA = 1$，$AD = OC = 3$，
$\therefore OD = OA + AD = 1 + 3 = 4$，$\therefore F(4,1)$.
设直线$CE$的函数表达式为$y = kx + b$，
把$C(0,3)$，$F(4,1)$代入，得$\begin{cases}b = 3\\4k + b = 1\end{cases}$，
解得$\begin{cases}k = -\frac{1}{2}\\b = 3\end{cases}$，
$\therefore$直线$CE$的函数表达式为$y = -\frac{1}{2}x + 3$，
联立$\begin{cases}y = -\frac{1}{2}x + 3\\y = x^{2} - 4x + 3\end{cases}$，解得$\begin{cases}x = 0\\y = 3\end{cases}$或$\begin{cases}x = \frac{7}{2}\\y = \frac{5}{4}\end{cases}$
$\therefore E(\frac{7}{2},\frac{5}{4})$.
2. 如图，在平面直角坐标系$xOy$中，抛物线$y = ax^{2} + bx - 6(a \neq 0)$与$x$轴交于点$A(-2,0),B(6,0)$，与$y$轴交于点$C$，顶点为$D$，连接$BC$.
(1)抛物线的函数表达式为
(2)连接$AC$并延长，交$BD$的延长线于点$E$，求$\angle CEB$的度数.
(1)抛物线的函数表达式为
$y = \frac{1}{2}x^{2} - 2x - 6$
；(2)连接$AC$并延长，交$BD$的延长线于点$E$，求$\angle CEB$的度数.
答案：
2.
(1)$y = \frac{1}{2}x^{2} - 2x - 6$
(2)解：设直线$AC$的函数表达式为$y = k_1x + b_1$，
$\because A(-2,0)$，$C(0,-6)$，$\therefore \begin{cases}-2k_1 + b_1 = 0\\b_1 = -6\end{cases}$，解得$\begin{cases}k_1 = -3\\b_1 = -6\end{cases}$，
$\therefore$直线$AC$的函数表达式为$y = -3x - 6$.
同理，由点$D(2,-8)$，$B(6,0)$，可得直线$BD$的函数表达式为$y = 2x - 12$，令$-3x - 6 = 2x - 12$，解得$x = \frac{6}{5}$，$\therefore$点$E$的坐标为$(\frac{6}{5},-\frac{48}{5})$，
由题意，$OA = 2$，$OB = OC = 6$，$AB = 8$，
$\therefore AC = \sqrt{OA^{2} + OC^{2}} = \sqrt{2^{2} + 6^{2}} = 2\sqrt{10}$.
如答图，过点$E$作$EF \perp x$轴于点$F$，
$\therefore AE = \sqrt{AF^{2} + EF^{2}} = \sqrt{(2 + \frac{6}{5})^{2} + (\frac{48}{5})^{2}} = \frac{16\sqrt{10}}{5}$，
$\because \frac{AC}{AB} = \frac{\sqrt{10}}{4}$，$\frac{AB}{AE} = \frac{8}{\frac{16\sqrt{10}}{5}} = \frac{\sqrt{10}}{4}$，$\therefore \frac{AC}{AB} = \frac{AB}{AE}$
$\because \angle BAC = \angle EAB$，
$\therefore \triangle ABC \backsim \triangle AEB$，$\therefore \angle ABC = \angle AEB$.
$\because OB = OC$，$\angle COB = 90^{\circ}$，$\therefore \angle ABC = 45^{\circ}$，
$\therefore \angle CEB = 45^{\circ}$.
2.
(1)$y = \frac{1}{2}x^{2} - 2x - 6$
(2)解：设直线$AC$的函数表达式为$y = k_1x + b_1$，
$\because A(-2,0)$，$C(0,-6)$，$\therefore \begin{cases}-2k_1 + b_1 = 0\\b_1 = -6\end{cases}$，解得$\begin{cases}k_1 = -3\\b_1 = -6\end{cases}$，
$\therefore$直线$AC$的函数表达式为$y = -3x - 6$.
同理，由点$D(2,-8)$，$B(6,0)$，可得直线$BD$的函数表达式为$y = 2x - 12$，令$-3x - 6 = 2x - 12$，解得$x = \frac{6}{5}$，$\therefore$点$E$的坐标为$(\frac{6}{5},-\frac{48}{5})$，
由题意，$OA = 2$，$OB = OC = 6$，$AB = 8$，
$\therefore AC = \sqrt{OA^{2} + OC^{2}} = \sqrt{2^{2} + 6^{2}} = 2\sqrt{10}$.
如答图，过点$E$作$EF \perp x$轴于点$F$，
$\therefore AE = \sqrt{AF^{2} + EF^{2}} = \sqrt{(2 + \frac{6}{5})^{2} + (\frac{48}{5})^{2}} = \frac{16\sqrt{10}}{5}$，
$\because \frac{AC}{AB} = \frac{\sqrt{10}}{4}$，$\frac{AB}{AE} = \frac{8}{\frac{16\sqrt{10}}{5}} = \frac{\sqrt{10}}{4}$，$\therefore \frac{AC}{AB} = \frac{AB}{AE}$
$\because \angle BAC = \angle EAB$，
$\therefore \triangle ABC \backsim \triangle AEB$，$\therefore \angle ABC = \angle AEB$.
$\because OB = OC$，$\angle COB = 90^{\circ}$，$\therefore \angle ABC = 45^{\circ}$，
$\therefore \angle CEB = 45^{\circ}$.
3. 如图，抛物线$y = ax^{2} - 2ax + 3$与$x$轴交于点$A(-1,0)$和点$B$，与$y$轴交于点$C$，连接$AC$，过$B,C$两点作直线.
(1)求$a$的值；
(2)抛物线上是否存在点$P$，使$\angle PBC + \angle ACO = 45^{\circ}$，若存在，请求出直线$BP$的函数表达式；若不存在，请说明理由.
(1)求$a$的值；
(2)抛物线上是否存在点$P$，使$\angle PBC + \angle ACO = 45^{\circ}$，若存在，请求出直线$BP$的函数表达式；若不存在，请说明理由.
答案：
3.解：
(1)$\because$抛物线$y = ax^{2} - 2ax + 3$与$x$轴交于点$A(-1,0)$，$\therefore a + 2a + 3 = 0$，$\therefore a = -1$.
(2)存在.由
(1)知，$y = -x^{2} + 2x + 3$，则易知$B(3,0)$，$C(0,3)$.当$\angle PBC$在$BC$的下方时，在$y$轴正半轴上取点$M(0,1)$，连接$BM$交抛物线于点$P$，如答图①.$\because A(-1,0)$，$B(3,0)$，$C(0,3)$，$M(0,1)$，$\therefore OB = OC = 3$，$OM = OA = 1$，$\angle BOM = \angle COA = 90^{\circ}$，$\therefore \triangle BOM \cong \triangle COA(SAS)$，
$\therefore \angle MBO = \angle ACO$.$\because \angle CBO = 45^{\circ}$，$\therefore \angle CBP + \angle MBO = 45^{\circ}$，$\therefore \angle CBP + \angle ACO = 45^{\circ}$.设直线$BM$的函数表达式为$y = k^{\prime}x + b^{\prime}$，则$\begin{cases}3k^{\prime} + b^{\prime} = 0\\b^{\prime} = 1\end{cases}$，解得$\begin{cases}k^{\prime} = -\frac{1}{3}\\b^{\prime} = 1\end{cases}$，
$\therefore$直线$BM$的函数表达式为$y = -\frac{1}{3}x + 1$，
联立$\begin{cases}y = -\frac{1}{3}x + 1\\y = -x^{2} + 2x + 3\end{cases}$，
解得$\begin{cases}x_1 = 3\\y_1 = 0\end{cases}$(舍去)，$\begin{cases}x_2 = -\frac{2}{3}\\y_2 = \frac{11}{9}\end{cases}$，$\therefore P(-\frac{2}{3},\frac{11}{9})$.
当$\angle PBC$在$BC$的上方时，作点$M$关于直线$BC$的对称点$M^{\prime}$，如答图②，连接$MM^{\prime}$，$CM^{\prime}$，直线$BM^{\prime}$交抛物线于点$P$.
由对称，得$MM^{\prime} \perp BC$，$CM = CM^{\prime} = 2$，$\angle BCM^{\prime} = \angle BCM = 45^{\circ}$，$\therefore \angle MCM^{\prime} = 90^{\circ}$，$\therefore M^{\prime}(2,3)$，则直线$BM^{\prime}$的函数表达式为$y = -3x + 9$，联立$\begin{cases}y = -3x + 9\\y = -x^{2} + 2x + 3\end{cases}$，
解得$\begin{cases}x_1 = 3\\y_1 = 0\end{cases}$(舍去)，$\begin{cases}x_2 = 2\\y_2 = 3\end{cases}$，$\therefore P(2,3)$.
综上所述，抛物线上存在点$P$，使$\angle PBC + \angle ACO = 45^{\circ}$，直线$BP$的函数表达式为$y = -\frac{1}{3}x + 1$或$y = -3x + 9$.
3.解：
(1)$\because$抛物线$y = ax^{2} - 2ax + 3$与$x$轴交于点$A(-1,0)$，$\therefore a + 2a + 3 = 0$，$\therefore a = -1$.
(2)存在.由
(1)知，$y = -x^{2} + 2x + 3$，则易知$B(3,0)$，$C(0,3)$.当$\angle PBC$在$BC$的下方时，在$y$轴正半轴上取点$M(0,1)$，连接$BM$交抛物线于点$P$，如答图①.$\because A(-1,0)$，$B(3,0)$，$C(0,3)$，$M(0,1)$，$\therefore OB = OC = 3$，$OM = OA = 1$，$\angle BOM = \angle COA = 90^{\circ}$，$\therefore \triangle BOM \cong \triangle COA(SAS)$，
$\therefore \angle MBO = \angle ACO$.$\because \angle CBO = 45^{\circ}$，$\therefore \angle CBP + \angle MBO = 45^{\circ}$，$\therefore \angle CBP + \angle ACO = 45^{\circ}$.设直线$BM$的函数表达式为$y = k^{\prime}x + b^{\prime}$，则$\begin{cases}3k^{\prime} + b^{\prime} = 0\\b^{\prime} = 1\end{cases}$，解得$\begin{cases}k^{\prime} = -\frac{1}{3}\\b^{\prime} = 1\end{cases}$，
$\therefore$直线$BM$的函数表达式为$y = -\frac{1}{3}x + 1$，
联立$\begin{cases}y = -\frac{1}{3}x + 1\\y = -x^{2} + 2x + 3\end{cases}$，
解得$\begin{cases}x_1 = 3\\y_1 = 0\end{cases}$(舍去)，$\begin{cases}x_2 = -\frac{2}{3}\\y_2 = \frac{11}{9}\end{cases}$，$\therefore P(-\frac{2}{3},\frac{11}{9})$.
当$\angle PBC$在$BC$的上方时，作点$M$关于直线$BC$的对称点$M^{\prime}$，如答图②，连接$MM^{\prime}$，$CM^{\prime}$，直线$BM^{\prime}$交抛物线于点$P$.
由对称，得$MM^{\prime} \perp BC$，$CM = CM^{\prime} = 2$，$\angle BCM^{\prime} = \angle BCM = 45^{\circ}$，$\therefore \angle MCM^{\prime} = 90^{\circ}$，$\therefore M^{\prime}(2,3)$，则直线$BM^{\prime}$的函数表达式为$y = -3x + 9$，联立$\begin{cases}y = -3x + 9\\y = -x^{2} + 2x + 3\end{cases}$，
解得$\begin{cases}x_1 = 3\\y_1 = 0\end{cases}$(舍去)，$\begin{cases}x_2 = 2\\y_2 = 3\end{cases}$，$\therefore P(2,3)$.
综上所述，抛物线上存在点$P$，使$\angle PBC + \angle ACO = 45^{\circ}$，直线$BP$的函数表达式为$y = -\frac{1}{3}x + 1$或$y = -3x + 9$.
查看更多完整答案，请扫码查看
