【答案】
分析:(1)由點(diǎn)A的坐標(biāo)可求得OA的長(zhǎng),將線段OA繞原點(diǎn)O順時(shí)針旋轉(zhuǎn)120°后,恰好落在x軸上,由此得出B點(diǎn)的坐標(biāo).
(2)利用待定系數(shù)法求拋物線的解析式即可.
(3)過P作y軸的平行線交線段AB于D,首先求出直線AB的解析式,結(jié)合直線和拋物線的解析式先表達(dá)出P、D點(diǎn)的坐標(biāo),進(jìn)而能得出線段PD的長(zhǎng),以PD為底,A、B橫坐標(biāo)差的絕對(duì)值為高即可求出△ABP的面積函數(shù)關(guān)系式,根據(jù)函數(shù)的性質(zhì)進(jìn)行判斷即可.
(4)欲求反比例函數(shù)的解析式,必須先求出點(diǎn)Q的坐標(biāo);點(diǎn)Q、A關(guān)于M對(duì)稱,那么點(diǎn)Q的橫坐標(biāo)必為3;已知線段AB為Rt△QAB的直角邊,那么需要分兩種情況討論:
①BQ為直角邊,即BQ⊥AB,那么這兩天直線的斜率乘積為-1,即:kAB×kBQ=-1,結(jié)合點(diǎn)B的坐標(biāo)即可求出直線BQ的解析式,進(jìn)而能求出點(diǎn)Q的坐標(biāo)以及反比例函數(shù)的解析式;
②AQ為直角邊,解題方法和①完全相同.
解答:解:(1)∵A(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/0.png)
),
∴OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/1.png)
=2;
∵OA繞O順時(shí)針旋轉(zhuǎn)120°得OB,
∴OB=OA=2,且B在x軸正半軸上,
∴B(2,0).
(2)由于拋物線過原點(diǎn),可設(shè)其解析式為y=ax
2+bx,代入A(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/2.png)
)、B(2,0),得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/3.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/4.png)
∴拋物線的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/5.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/6.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/images7.png)
(3)設(shè)P(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/7.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/8.png)
)(0<x<2),過P作PD∥y軸交線段AB于D;
設(shè)直線AB:y=kx+b(k≠0),將A(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/9.png)
)、B(2,0)代入,得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/10.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/11.png)
∴直線AB:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/12.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/13.png)
,則點(diǎn)D的坐標(biāo)(x,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/14.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/15.png)
);
∴PD=(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/16.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/17.png)
)-(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/18.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/19.png)
)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/20.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/21.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/22.png)
,
∴S
△APB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/23.png)
×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/24.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/25.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/26.png)
)×3=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/27.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/28.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/29.png)
;
S是關(guān)于x的二次函數(shù),且開口向下,對(duì)稱軸x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/30.png)
在0<x<2的范圍內(nèi),因此當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/31.png)
時(shí),△PAB的面積最大,且最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/32.png)
;
此時(shí)P點(diǎn)的坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/33.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/34.png)
).
(4)點(diǎn)Q與拋物線上的點(diǎn)A(-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/35.png)
)關(guān)于點(diǎn)M(1,t)成中心對(duì)稱,所以點(diǎn)Q的橫坐標(biāo)必為3;
①BQ為Rt△QAB的直角邊時(shí),BQ⊥AB,即:k
AB×k
BQ=-1,解得:k
BQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/36.png)
;
可設(shè)直線BQ:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/37.png)
x+b,代入B(2,0),得:b=-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/38.png)
,
∴直線BQ:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/39.png)
x-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/40.png)
,當(dāng)x=3時(shí),y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/41.png)
,即 Q(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/42.png)
);
將點(diǎn)Q的坐標(biāo)代入反比例函數(shù)的解析式中,得:k
1=xy=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/43.png)
;
②AQ為Rt△AOB的直角邊時(shí),AQ⊥AB,同①可求得:k
2=15
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/44.png)
;
綜上,符合條件的反比例函數(shù)解析式為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/45.png)
或y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193744542660478/SYS201311011937445426604021_DA/46.png)
.
點(diǎn)評(píng):此題主要考查了函數(shù)解析式的確定、直角三角形的性質(zhì)以及圖形面積的求法等重要知識(shí);最后一題中,互相垂直的兩條直線斜率的乘積為-1,這個(gè)結(jié)論需要記住;這個(gè)小題也可以分別過A、Q作坐標(biāo)軸的垂線,通過構(gòu)建相似三角形來(lái)求點(diǎn)Q的坐標(biāo),不過這樣的計(jì)算過程會(huì)稍微復(fù)雜一些.