【答案】
分析:(1)依題意設(shè)直線BC的解析式為y=kx+3,把B點(diǎn)坐標(biāo)代入解析式求出直線BC的表達(dá)式.然后又已知拋物線y=x
2+bx+c過點(diǎn)B,C,代入求出解析式.
(2)由y=x
2-4x+3求出點(diǎn)D,A的坐標(biāo).得出三角形OBC是等腰直角三角形求出∠OBC,CB的值.過A點(diǎn)作AE⊥BC于點(diǎn)E,求出BE,CE的值.證明△AEC∽△AFP求出PF可得點(diǎn)P在拋物線的對(duì)稱軸,求出點(diǎn)P的坐標(biāo).
(3)本題要靠輔助線的幫助.作點(diǎn)A(1,0)關(guān)于y軸的對(duì)稱點(diǎn)A',則A'(-1,0),求出A'C=AC,由勾股定理可得CD,A'D的值.得出△A'DC是等腰三角形后可推出∠OCA+∠OCD=45度.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/images0.png)
解:(1)∵y=kx沿y軸向上平移3個(gè)單位長(zhǎng)度后經(jīng)過y軸上的點(diǎn)C,
∴C(0,3).
設(shè)直線BC的解析式為y=kx+3.
∵B(3,0)在直線BC上,
∴3k+3=0.
解得k=-1.
∴直線BC的解析式為y=-x+3.(1分)
∵拋物線y=x
2+bx+c過點(diǎn)B,C,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/0.png)
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/1.png)
,
∴拋物線的解析式為y=x
2-4x+3.(2分)
(2)由y=x
2-4x+3.
可得D(2,-1),A(1,0).
∴OB=3,OC=3,OA=1,AB=2.
可得△OBC是等腰直角三角形,
∴∠OBC=45°,CB=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/2.png)
.
如圖1,設(shè)拋物線對(duì)稱軸與x軸交于點(diǎn)F,
∴AF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/3.png)
AB=1.
過點(diǎn)A作AE⊥BC于點(diǎn)E.
∴∠AEB=90度.
可得BE=AE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/4.png)
,CE=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/5.png)
.
在△AEC與△AFP中,∠AEC=∠AFP=90°,∠ACE=∠APF,
∴△AEC∽△AFP.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/6.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/7.png)
.
解得PF=2.∵點(diǎn)P在拋物線的對(duì)稱軸上,
∴點(diǎn)P的坐標(biāo)為(2,2)或(2,-2).(5分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/images9.png)
(3)解法一:
如圖2,作點(diǎn)A(1,0)關(guān)于y軸的對(duì)稱點(diǎn)A',則A'(-1,0).
連接A'C,A'D,
可得A'C=AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/8.png)
,∠OCA'=∠OCA.
由勾股定理可得CD
2=20,A'D
2=10.
又∵A'C
2=10,
∴A'D
2+A'C
2=CD
2.
∴△A'DC是等腰直角三角形,∠CA'D=90°,
∴∠DCA'=45度.
∴∠OCA'+∠OCD=45度.
∴∠OCA+∠OCD=45度.
即∠OCA與∠OCD兩角和的度數(shù)為45度.(7分)
解法二:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/images11.png)
如圖3,連接BD.
同解法一可得CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/9.png)
,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/10.png)
.
在Rt△DBF中,∠DFB=90°,BF=DF=1,
∴DB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/11.png)
.
在△CBD和△COA中,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/13.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/14.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348010_DA/15.png)
.
∴△CBD∽△COA.
∴∠BCD=∠OCA.
∵∠OCB=45°,
∴∠OCA+∠OCD=45度.
即∠OCA與∠OCD兩角和的度數(shù)為45度.(9分)
點(diǎn)評(píng):本題設(shè)計(jì)得很精致,將幾何與函數(shù)完美的結(jié)合在一起,對(duì)學(xué)生綜合運(yùn)用知識(shí)的能力要求較高,本題3問之間層層遞進(jìn),后兩問集中研究角度問題.
中等層次的學(xué)生能夠做出第(1)問,中上層次的學(xué)生可能會(huì)作出第(2)問,但第(2)問中符合條件的P點(diǎn)有兩個(gè),此時(shí)學(xué)生易忽視其中某一個(gè),成績(jī)較好的學(xué)生才可能作出第(3)問,本題是拉開不同層次學(xué)生分?jǐn)?shù)的一道好題.
本題考點(diǎn):函數(shù)圖形的平移、一次函數(shù)解析式的確定、二次函數(shù)解析式的確定、相似三角形、等腰直角三角形的判定及性質(zhì)、勾股定理.