【答案】
分析:(1)將點(diǎn)M(a+c,0)代入拋物線y=x
2-2ax+b
2,整理可得a
2=b
2+c
2,從而判斷出三角形為直角三角形;
(2)①根據(jù)S
△MNP=3S
△NOP,判斷出MN=3ON,即MO=4ON,求出N點(diǎn)坐標(biāo)的表達(dá)式,得到x=a+c和x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/0.png)
是方程x
2-2ax+b
2=0的兩根,求出a、c之間的關(guān)系,然后根據(jù)銳角三角函數(shù)的定義求出cosC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/2.png)
.
②過(guò)D作DE⊥x軸于點(diǎn)E,則NE=EM,DN=DM,要使以MN為直徑的圓恰好過(guò)拋物線y=x
2-2ax+b
2的頂點(diǎn),則使△MND為等腰直角三角形,只須ED=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/3.png)
MN=EM.據(jù)此進(jìn)行計(jì)算即可.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/images4.png)
解:(1)∵拋物線y=x
2-2ax+b
2經(jīng)過(guò)點(diǎn)M(a+c,0),
∴(a+c)
2-2a(a+c)+b
2=0,即a
2=b
2+c
2.
由勾股定理的逆定理,得△ABC為直角三角形.
(2)①如圖1所示?∵S
△MNP=3S
△NOP,
∴MN=3ON,即MO=4ON,又M(a+c,0),
∴N(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/4.png)
,0),
∴x=a+c和x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/5.png)
是方程x
2-2ax+b
2=0的兩根,
此時(shí)兩個(gè)為x
1,2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/6.png)
=a±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/7.png)
,
∴a+c+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/8.png)
=2a,
∴c=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/9.png)
a,由(1)知:在△ABC中,∠A=90°,由勾股定理得b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/10.png)
a,
∴cosC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/12.png)
.
②能,由(1)知:y=x
2-2ax+b
2=x
2-2ax+a
2-c
2=(x-a)
2-c
2,
∴頂點(diǎn)D(a,-c
2).
過(guò)D作DE⊥x軸于點(diǎn)E,則NE=EM,DN=DM,要使以MN為直徑的圓恰好過(guò)拋物線y=x
2-2ax+b
2的頂點(diǎn),則使△MND為等腰直角三角形,只須ED=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/13.png)
MN=EM.
∵M(jìn)(a+c,0),D(a,-c
2),
∴DE=c
2,EM=c,
∴c
2=c,又c>0,
∴c=1.
∵c=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/14.png)
a,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/15.png)
a,
∴a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/16.png)
,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/17.png)
;
∴當(dāng)a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/18.png)
,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192112218427379/SYS201311011921122184273031_DA/19.png)
,c=1時(shí),△MND為等腰直角三角形.
此時(shí),EM=ED=EN,以MN為直徑的圓恰好過(guò)拋物線y=x
2-2ax+b
2的頂點(diǎn).
點(diǎn)評(píng):本題考查了二次函數(shù)的性質(zhì),是一道探索題,是近年來(lái)中考命題的熱點(diǎn)問(wèn)題.在第(2)小題中要求同學(xué)們先猜想可能的結(jié)論,再進(jìn)行證明,這對(duì)同學(xué)們的確有較高的能力要求.而在探索結(jié)論前可以自己先畫幾個(gè)草圖,做到心中有數(shù)再去努力求證.總之這是一道新課標(biāo)形勢(shì)下的優(yōu)秀壓軸.