搜索
1. 如图，抛物线$y = - \frac{1}{2}x^{2} + \frac{3}{2}x + 2$与$x$轴交于点$A,B$，与$y$轴交于点$C$，顶点为$M$，连接$BC$. 已知$P$是抛物线上的一点，连接$CP$，若$\angle PCB = \angle ABC$，则点$P$的坐标为
(3,2)或($\frac{17}{3}$,−$\frac{50}{9}$)
.
答案：
1.(3,2)或($\frac{17}{3}$,−$\frac{50}{9}$) 点拨:由$y = -\frac{1}{2}x^{2} + \frac{3}{2}x + 2$易知$A(-1,0),B(4,0),C(0,2)$.
如答图①,当点$P$在$BC$上方时,
∵$\angle PCB = \angle ABC$，
∴$CP// AB$.又
∵$C(0,2)$，
∴$y_{P} = 2$，
∴$-\frac{1}{2}x^{2} + \frac{3}{2}x + 2 = 2$，解得$x = 0$(舍去)或$x = 3$，
则点$P(3,2)$.
如答图②,当点$P$在$BC$下方时,$CP$交$x$轴于点$H$.
∵$\angle PCB = \angle ABC$，
∴$CH = BH$.设$H(u,0)$，
在$Rt\triangle COH$中，$OH^{2} + OC^{2} = CH^{2}$，
∴$u^{2} + 2^{2} = (4 - u)^{2}$，解得$u = \frac{3}{2}$，
∴$H(\frac{3}{2},0)$.设直线
$CP$的函数表达式为$y = kx + 2$，把点$H$的坐标代入，得
$0 = \frac{3}{2}k + 2$，解得$k = -\frac{4}{3}$，所以直线$CP$的函数表达式为$y = -\frac{4}{3}x + 2$.联立上式和抛物线的函数表达式，得
$-\frac{4}{3}x + 2 = -\frac{1}{2}x^{2} + \frac{3}{2}x + 2$，解得$x = 0$(舍去)或$x = \frac{17}{3}$，
∴$P(\frac{17}{3}, -\frac{50}{9})$.
综上所述，点$P$的坐标为$(3,2)$或$(\frac{17}{3}, -\frac{50}{9})$.
1.(3,2)或($\frac{17}{3}$,−$\frac{50}{9}$) 点拨:由$y = -\frac{1}{2}x^{2} + \frac{3}{2}x + 2$易知$A(-1,0),B(4,0),C(0,2)$.
如答图①,当点$P$在$BC$上方时,
∵$\angle PCB = \angle ABC$，
∴$CP// AB$.又
∵$C(0,2)$，
∴$y_{P} = 2$，
∴$-\frac{1}{2}x^{2} + \frac{3}{2}x + 2 = 2$，解得$x = 0$(舍去)或$x = 3$，
则点$P(3,2)$.
如答图②,当点$P$在$BC$下方时,$CP$交$x$轴于点$H$.
∵$\angle PCB = \angle ABC$，
∴$CH = BH$.设$H(u,0)$，
在$Rt\triangle COH$中，$OH^{2} + OC^{2} = CH^{2}$，
∴$u^{2} + 2^{2} = (4 - u)^{2}$，解得$u = \frac{3}{2}$，
∴$H(\frac{3}{2},0)$.设直线
$CP$的函数表达式为$y = kx + 2$，把点$H$的坐标代入，得
$0 = \frac{3}{2}k + 2$，解得$k = -\frac{4}{3}$，所以直线$CP$的函数表达式为$y = -\frac{4}{3}x + 2$.联立上式和抛物线的函数表达式，得
$-\frac{4}{3}x + 2 = -\frac{1}{2}x^{2} + \frac{3}{2}x + 2$，解得$x = 0$(舍去)或$x = \frac{17}{3}$，
∴$P(\frac{17}{3}, -\frac{50}{9})$.
综上所述，点$P$的坐标为$(3,2)$或$(\frac{17}{3}, -\frac{50}{9})$.
2. 已知抛物线$y = ax^{2} - 4ax + 4a - 1$，该抛物线与$x$轴交于点$A,B$，与$y$轴交于点$C$，点$A$的坐标为$(1,0)$，若此抛物线的对称轴上的点$P$满足$\angle APB ＜ \angle ACB$，则点$P$的纵坐标$n$的取值范围是
$n ＞ 2 + \sqrt{5}$或$n ＜ -2 - \sqrt{5}$
.
答案：
2.$n ＞ 2 + \sqrt{5}$或$n ＜ -2 - \sqrt{5}$ 点拨:将$A(1,0)$代入$y = ax^{2} - 4ax + 4a - 1$，得$a - 4a + 4a - 1 = 0$，解得$a = 1$，故抛物线的函数表达式为$y = x^{2} - 4x + 3$，则点$A,B,C$的坐标分别为$(1,0),(3,0),(0,3)$.过点$A,B,C$作$\triangle ABC$的外接圆，则圆心$M$的坐标为$(2,m)$，则$MA = MC$，即$4 + (m - 3)^{2} = 1 + m^{2}$，解得$m = 2$.
则$\odot M$的半径为$\sqrt{5}$，如答图，当点$P$在圆上时，$\angle APB = \angle ACB$，点$P$在圆外时，$\angle APB ＜ \angle ACB$，点$P$的坐标为$(2,2 + \sqrt{5})$，则点$P$关于$x$轴的对称点$P'$的坐标为$(2,-2 - \sqrt{5})$.
∵$\angle APB ＜ \angle ACB$，
∴$n ＞ 2 + \sqrt{5}$或$n ＜ -2 - \sqrt{5}$.
2.$n ＞ 2 + \sqrt{5}$或$n ＜ -2 - \sqrt{5}$ 点拨:将$A(1,0)$代入$y = ax^{2} - 4ax + 4a - 1$，得$a - 4a + 4a - 1 = 0$，解得$a = 1$，故抛物线的函数表达式为$y = x^{2} - 4x + 3$，则点$A,B,C$的坐标分别为$(1,0),(3,0),(0,3)$.过点$A,B,C$作$\triangle ABC$的外接圆，则圆心$M$的坐标为$(2,m)$，则$MA = MC$，即$4 + (m - 3)^{2} = 1 + m^{2}$，解得$m = 2$.
则$\odot M$的半径为$\sqrt{5}$，如答图，当点$P$在圆上时，$\angle APB = \angle ACB$，点$P$在圆外时，$\angle APB ＜ \angle ACB$，点$P$的坐标为$(2,2 + \sqrt{5})$，则点$P$关于$x$轴的对称点$P'$的坐标为$(2,-2 - \sqrt{5})$.
∵$\angle APB ＜ \angle ACB$，
∴$n ＞ 2 + \sqrt{5}$或$n ＜ -2 - \sqrt{5}$.
3. 如图，抛物线$y = ax^{2} + bx - 3$经过点$A(-1,0),B(3,0)$，与$y$轴交于点$C$，$P$为第四象限内抛物线上一点，过点$P$作$PM \perp x$轴于点$M$，连接$AC,AP$，$AP$与$y$轴交于点$D$.
(1)求抛物线的函数表达式；
(2)当$\angle MPA = 2\angle PAC$时，求直线$AP$的函数表达式及点$P$的坐标.
(1)求抛物线的函数表达式；
(2)当$\angle MPA = 2\angle PAC$时，求直线$AP$的函数表达式及点$P$的坐标.
答案：
3.解:
(1)
∵抛物线$y = ax^{2} + bx - 3$经过$A(-1,0),B(3,0)$两点，把点$A,B$的坐标代入$y = ax^{2} + bx - 3$，得$\begin{cases}a - b - 3 = 0,\\9a + 3b - 3 = 0.\end{cases}$解得$\begin{cases}a = 1,\\b = -2.\end{cases}$
∴抛物线的函数表达式为$y = x^{2} - 2x - 3$.
(2)
∵$OC// MP$，
∴$\angle CDP = \angle APM$.
∵$\angle MPA = 2\angle PAC$，
∴$\angle CDP = 2\angle PAC$，
∴$\angle PAC = \angle DCA$，
∴$AD = CD$.
∵在$y = x^{2} - 2x - 3$中，当$x = 0$时，$y = -3$，
∴$C(0,-3)$.
设$D(0,n)$，
∴$1 + n^{2} = (n + 3)^{2}$，解得$n = -\frac{4}{3}$，
∴$D(0,-\frac{4}{3})$.
设直线$AP$的函数表达式为$y = kx + t(k \neq 0)$，
将$D(0,-\frac{4}{3}),A(-1,0)$代入，得$\begin{cases}t = -\frac{4}{3},\\-k + t = 0.\end{cases}$解得$\begin{cases}k = -\frac{4}{3},\\t = -\frac{4}{3}.\end{cases}$
∴$y = -\frac{4}{3}x - \frac{4}{3}$.
联立$\begin{cases}y = -\frac{4}{3}x - \frac{4}{3},\\y = x^{2} - 2x - 3.\end{cases}$解得$\begin{cases}x = -1,\\y = 0.\end{cases}$(舍去)或$\begin{cases}x = \frac{5}{3},\\y = -\frac{32}{9}.\end{cases}$
∴$P(\frac{5}{3}, -\frac{32}{9})$，
∴直线$AP$的函数表达式为$y = -\frac{4}{3}x - \frac{4}{3}$，点$P$的坐标为$(\frac{5}{3}, -\frac{32}{9})$.
(1)
∵抛物线$y = ax^{2} + bx - 3$经过$A(-1,0),B(3,0)$两点，把点$A,B$的坐标代入$y = ax^{2} + bx - 3$，得$\begin{cases}a - b - 3 = 0,\\9a + 3b - 3 = 0.\end{cases}$解得$\begin{cases}a = 1,\\b = -2.\end{cases}$
∴抛物线的函数表达式为$y = x^{2} - 2x - 3$.
(2)
∵$OC// MP$，
∴$\angle CDP = \angle APM$.
∵$\angle MPA = 2\angle PAC$，
∴$\angle CDP = 2\angle PAC$，
∴$\angle PAC = \angle DCA$，
∴$AD = CD$.
∵在$y = x^{2} - 2x - 3$中，当$x = 0$时，$y = -3$，
∴$C(0,-3)$.
设$D(0,n)$，
∴$1 + n^{2} = (n + 3)^{2}$，解得$n = -\frac{4}{3}$，
∴$D(0,-\frac{4}{3})$.
设直线$AP$的函数表达式为$y = kx + t(k \neq 0)$，
将$D(0,-\frac{4}{3}),A(-1,0)$代入，得$\begin{cases}t = -\frac{4}{3},\\-k + t = 0.\end{cases}$解得$\begin{cases}k = -\frac{4}{3},\\t = -\frac{4}{3}.\end{cases}$
∴$y = -\frac{4}{3}x - \frac{4}{3}$.
联立$\begin{cases}y = -\frac{4}{3}x - \frac{4}{3},\\y = x^{2} - 2x - 3.\end{cases}$解得$\begin{cases}x = -1,\\y = 0.\end{cases}$(舍去)或$\begin{cases}x = \frac{5}{3},\\y = -\frac{32}{9}.\end{cases}$
∴$P(\frac{5}{3}, -\frac{32}{9})$，
∴直线$AP$的函数表达式为$y = -\frac{4}{3}x - \frac{4}{3}$，点$P$的坐标为$(\frac{5}{3}, -\frac{32}{9})$.
查看更多完整答案，请扫码查看
