【答案】
分析:(1)利用拋物線y=(x-2)
2+1的與y軸交于點(diǎn)A(0,5),它的頂點(diǎn)為點(diǎn)B(2,1),求出直線解析式即可;
(2)首先得出點(diǎn)A的坐標(biāo)為(0,-3),以及點(diǎn)C的坐標(biāo)為(0,3),進(jìn)而求出BE=2,得出頂點(diǎn)B的坐標(biāo)求出解析式即可;
(3)①由已知可得A坐標(biāo)為(0,b),C點(diǎn)坐標(biāo)為(0,-b),以及n=-2m+b,即點(diǎn)B點(diǎn)的坐標(biāo)為(m,-2m+b),利用勾股定理求出;
②利用①中B點(diǎn)坐標(biāo),以及BD的長度即可得出P點(diǎn)的坐標(biāo).
解答:解:(1)由拋物線y=a(x-m)
2+n與y軸交于點(diǎn)A,它的頂點(diǎn)為點(diǎn)B,
∴拋物線y=(x-2)
2+1的與y軸交于點(diǎn)A(0,5),它的頂點(diǎn)為點(diǎn)B(2,1),
設(shè)所求直線解析式為y=kx+b,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/0.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/1.png)
,
∴所求直線解析式為y=-2x+5;
(2)如圖,作BE⊥AC于點(diǎn)E,由題意得四邊形ABCD是平行四邊形,點(diǎn)A的坐標(biāo)為(0,-3),
點(diǎn)C的坐標(biāo)為(0,3),
可得:AC=6,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/images2.png)
∵平行四邊形ABCD的面積為12,
∴S
△ABC=6即S
△ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/2.png)
AC•BE=6,
∴BE=2,
∵m>0,即頂點(diǎn)B在y軸的右側(cè),且在直線y=x-3上,
∴頂點(diǎn)B的坐標(biāo)為(2,-1),
又拋物線經(jīng)過點(diǎn)A(0,-3),
∴a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/3.png)
,
∴y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/4.png)
(x-2)
2-1;
(3)①如圖,作BF⊥x軸于點(diǎn)F,
由已知可得A坐標(biāo)為(0,b),C點(diǎn)坐標(biāo)為(0,-b),
∵頂點(diǎn)B(m,n)在直線y=-2x+b(b>0)上,
∴n=-2m+b,即點(diǎn)B點(diǎn)的坐標(biāo)為(m,-2m+b),
在矩形ABCD中,CO=BO.
∴b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/5.png)
,
∴b
2=m
2+4m
2-4mb+b
2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/images7.png)
∴m=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/6.png)
b,
n=-2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/7.png)
b+b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/8.png)
b,
②∵B點(diǎn)坐標(biāo)為(m,n),即(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/9.png)
b,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/10.png)
b),
∴BO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/11.png)
=b,
∴BD=2b,
當(dāng)BD=BP,
∴PF=2b-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/12.png)
b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/13.png)
b,
∴P點(diǎn)的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/14.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/15.png)
b);
如圖3,當(dāng)DP=PB時(shí),
過點(diǎn)D作DE⊥PB,于點(diǎn)E,
∵B點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/16.png)
b,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/17.png)
b),
∴D點(diǎn)坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/18.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/19.png)
b),
∴DE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/20.png)
b,BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/21.png)
b,設(shè)PE=x,
∴DP=PB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/22.png)
b+x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/images25.png)
∴DE
2+PE
2=DP
2,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/23.png)
+x
2=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/24.png)
b+x)
2,
解得:x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/25.png)
b,
∴PF=PE+EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/26.png)
b+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/27.png)
b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/28.png)
b,
∴此時(shí)P點(diǎn)坐標(biāo)為:(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/29.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/30.png)
b);
同理P可以為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/31.png)
b,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/32.png)
b);(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/33.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/34.png)
b),
故P點(diǎn)坐標(biāo)為:(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/35.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/36.png)
b);(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/37.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/38.png)
b);(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/39.png)
b,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/40.png)
b);(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/41.png)
b,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191544632735308/SYS201311011915446327353023_DA/42.png)
b).
點(diǎn)評(píng):此題主要考查了二次函數(shù)的綜合應(yīng)用以及勾股定理和點(diǎn)的坐標(biāo)性質(zhì),二次函數(shù)的綜合應(yīng)用是初中階段的重點(diǎn)題型特別注意利用數(shù)形結(jié)合是這部分考查的重點(diǎn)也是難點(diǎn)同學(xué)們應(yīng)重點(diǎn)掌握.