【答案】
分析:(1)已知點D(0,3)和點E(0,-1),可以得到圓的直徑,連接AC,根據(jù)垂徑定理,以及勾股定理就可以求出OB,OE,OC的長度,得到三點的坐標,根據(jù)待定系數(shù)法就可以求出二次函數(shù)的解析式.
(2)過點P作PF⊥y軸于F,過點Q作QN⊥y軸于N,易證△PFA≌△QNA,則FA=NA,即|t-1|=|1-y|,即可得到函數(shù)解析式.
(3)當(dāng)y=0時,Q點與C點重合,連接PB,由PC為⊙A的直徑可以得到PB⊥x軸,就可以求出P點的坐標.求出直線PM的解析式,求出切線PM與拋物線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/0.png)
x
2-1交點坐標,橫坐標x的范圍就在兩個交點之間.
解答:解:(1)解法一:連接AC
∵DE為⊙A的直徑,DE⊥BC
∴BO=CO
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/images1.png)
∵D(0,3),E(0,-1)
∴DE=|3-(-1)|=4,OE=1
∴AO=1,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/1.png)
DE=2
在Rt△AOC中,AC
2=AO
2+OC
2
∴OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/2.png)
∴C(
,0),B(
,0)
設(shè)經(jīng)過B、E、C三點的拋物線的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/5.png)
,
則-1=a(0-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/6.png)
)(0+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/7.png)
)
解得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/8.png)
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/9.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/10.png)
)(x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/11.png)
)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/12.png)
x
2-1(2分).
解法二:∵DE為⊙A的直徑,DE⊥BC
∴BO=CO
∴OC
2=OD•OE
∵D(0,3),E(0,-1)
∴DO=3,OE=1
∴OC2=3×1=3
∴OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/13.png)
∴C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/14.png)
,0),B(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/15.png)
,0)
以下同解法一;
(2)解法一:過點P作PF⊥y軸于F,過點Q作QN⊥y軸于N
∴∠PFA=∠QNA=90°,F(xiàn)點的縱坐標為t
N點的縱坐標為y
∵∠PAF=∠QAN,PA=QA
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/images17.png)
∴△PFA≌△QNA
∴FA=NA
∵AO=1
∴A(0,1)
∴|t-1|=|1-y|
∵動切線PM經(jīng)過第一、二、三象限
觀察圖形可得1<t<3,-1<y<1.
∴t-1=1-y.
即y=-t+2.
∴y關(guān)于t的函數(shù)關(guān)系式為y=-t+2(1<t<3)(5分)
解法二:(i)當(dāng)經(jīng)過一、二、三象限的切線PM運動到使得Q點與C點重合時,y=0
連接PB
∵PC是直徑
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/images18.png)
∴∠PBC=90°
∴PB⊥x軸,
∴PB=t.
∵PA=AC,BO=OC,AO=1,
∴PB=2AO=2,
∴t=2.
即t=2時,y=0.
(ii)當(dāng)經(jīng)過一、二、三象限的切線
PM運動使得Q點在x軸上方時,y>0
觀察圖形可得1<t<2
過P作PS⊥x軸于S,過Q作QT⊥x軸于T
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/images19.png)
則PS∥AO∥QT
∵點A為線段PQ的中點
∴點O為線段ST的中點
∴AO為梯形QTSP的中位線
∴AO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/16.png)
∴1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/17.png)
∴y=-t+2.
∴y=-t+2(1<t<2).
(iii)當(dāng)經(jīng)過一、二、三象限的切線PM運動使得Q點在x軸下方時,y<0,觀察圖形可得2<t<3
過P作PS⊥x軸于S,過Q作QT⊥x軸于T,設(shè)PQ交x軸于R
則QT∥PS
∴△QRT∽△PRS
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/18.png)
設(shè)AR=m,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/19.png)
&&(1)
又∵AO⊥x軸,
∴AO∥PS
∴△ROA∽△RSP
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/20.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/21.png)
&&(2)
由(1)、(2)得y=-t+2
∴y=-t+2(2<t<3)
綜上所述:y與t的函數(shù)關(guān)系式為y=-t+2(1<t<3)(5分)
(3)解法一:當(dāng)y=0時,Q點與C點重合,連接PB
∵PC為⊙A的直徑
∴∠PBC=90°
即PB⊥x軸
∴s=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/22.png)
將y=0代入y=-t+2(1<t<3),得0=-t+2
∴t=2∴P(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/23.png)
,2)
設(shè)切線PM與y軸交于點I,則AP⊥PI
∴∠API=9
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/images28.png)
0°
在△API與△AOC中
∵∠API=∠AOC=90°,∠PAI=∠OAC
∴△API∽△AOC
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/24.png)
∴I點坐標為(0,5)
設(shè)切線PM的解析式為y=kx+5(k≠0),
∵P點的坐標為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/25.png)
,
∴2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/26.png)
3 k+5.
解得k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/27.png)
,
∴切線PM的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/28.png)
x+5(7分)
設(shè)切線PM與拋物線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/29.png)
x
2-1交于G、H兩點
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/30.png)
可得x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/31.png)
因此,G、H的橫坐標分別為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/32.png)
根據(jù)圖象可得拋物線在切線PM下方的點的橫坐標x的取值范圍是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/33.png)
(9分)
解法二:同(3)解法一
可得P(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/34.png)
,2)
∵直線PM為⊙A的切線,PC為⊙A的直徑
∴PC⊥PM
在Rt△CPM與Rt△CBP中
cos∠PCM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/35.png)
∵CB=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/36.png)
,PC=4
∴CM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/37.png)
設(shè)M點的坐標為(m,0),
則CM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/38.png)
-m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/39.png)
∴m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/40.png)
.
即M(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/41.png)
,0).
設(shè)切線PM的解析式為y=kx+b(k≠0),
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/42.png)
k+b
2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/43.png)
k+b.
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/44.png)
∴切線PM的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193535181609896/SYS201311011935351816098027_DA/45.png)
x+5(7分)
以下同解法一.
點評:本題是圓與函數(shù)相結(jié)合的題目,主要考查了垂徑定理以及勾股定理.待定系數(shù)法求函數(shù)的解析式,是一個比較難的題目.