【答案】
分析:(1)根據(jù)C點(diǎn)的坐標(biāo)可以求出OC的長(zhǎng)度,根據(jù)CO的長(zhǎng)度和∠CBA=45°,tanA=3通過(guò)解直角三角形可以得出OB、OA的長(zhǎng)度從而求出A、B的坐標(biāo).
(2)根據(jù)A、B、C的坐標(biāo),利用待定系數(shù)法就可以求出拋物線的解析式,然后轉(zhuǎn)化為頂點(diǎn)式就可以求出頂點(diǎn)坐標(biāo)D.
(3)①通過(guò)勾股定理可以證明△BDC為直角三角形,當(dāng)直線EB與△BCD外切時(shí)EB⊥BD,利用三角形相似可以求出OE的長(zhǎng)來(lái)確定E點(diǎn)的坐標(biāo)而確定m的值.
②通過(guò)情況討論當(dāng)點(diǎn)E在C點(diǎn)的上方和下方來(lái)分別計(jì)算比較,∠DEC與∠DBC的大小,確定相應(yīng)的m的取值范圍.
解答:解:(1)∵C(0,3)
∴OC=3
∵∠CBA=45°
∴OC=OB=3
∵tanA=3
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/0.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/1.png)
∴OA=1
∴A(1,O),B(-3,0)
(2)設(shè)拋物線的解析式為:y=a(x-1)(x+3)
把C(0,3)代入得-3a=3
∴a=-1
∴y=-(x-1)(x+3)
y=-x
2-2x+3
∴-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/2.png)
=-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/3.png)
=4
∴D(-1,4)
(3)①作DH⊥y軸于H,則DH=1,CH=OH-OC=1
由勾股定理得:CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/4.png)
,CD
2=2
在△BOC中,由勾股定理得,BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/5.png)
OC
∴BC=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/6.png)
,BC
2=18
在Rt△BDF中,BF=BO-OF=2,DF=4,由勾股定理得;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/images7.png)
BD=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/7.png)
∴DB
2=20
在△BCD中∴CD
2+BC
2=DB
2
∴△BCD是直角三角形.
∴BD是△BCD的外接圓的直徑
∵BE與△BCD的外接圓相切
∴BE⊥BD
∴∠DBE=90°
∴∠EBO=∠BDF
∴△BDF∽△EBO
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/8.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/9.png)
∴OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/10.png)
∴E(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/11.png)
)
即m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/12.png)
②當(dāng)點(diǎn)E在C點(diǎn)的上方時(shí),當(dāng)∠DEC=∠DBC時(shí),
∵∠DHE=∠DCB=90°
∴△DEH∽△DBC
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163426498683674/SYS201310221634264986836025_DA/13.png)
∴EH=3,OE=EH+HO=7
∴E(0,7)
∴當(dāng)m=7時(shí),∠DEC=∠DBC
當(dāng)m>時(shí),∠DEC<∠DBC
當(dāng)m<7時(shí),∠DEC>∠DBC
點(diǎn)E在C下方時(shí),同理可得當(dāng)∠DEC=∠DBC時(shí),EH=3
∴此時(shí)OE=4-3=1
∴E(0,1)
∴當(dāng)m=1時(shí),∠DEC=∠DBC
當(dāng)1<m<3時(shí),∠DEC>∠DBC
當(dāng)m<1時(shí),∠DEC<∠DBC
綜上所述得:m>7或m<1時(shí),∠DEC<∠DBC
m=7或m=1時(shí),∠DEC=∠DBC
1<m<7且m≠3時(shí),∠DEC>∠DBC
點(diǎn)評(píng):本題是一道二次函數(shù)的綜合試題,考查了利用解直角三角形求線段的長(zhǎng)度來(lái)求點(diǎn)的坐標(biāo),待定系數(shù)法求函數(shù)的解析式,頂點(diǎn)式的運(yùn)用,勾股定理及其逆定理的運(yùn)用,相似三角形的判定及性質(zhì),圓的切線的性質(zhì)等多個(gè)知識(shí)點(diǎn).