【答案】
分析:(1)由于A(-1,m)與B(2,m+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/0.png)
)是反比例函數(shù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/1.png)
圖象上的兩個(gè)點(diǎn),根據(jù)反比例函數(shù)性質(zhì)可知:坐標(biāo)之積相等,可列方程求k的值;
(2)判斷是不是梯形,就要判定一組對(duì)邊平行且不相等.求出坐標(biāo),既能求線段長(zhǎng)度,又能判別平行,即解.
解答:解:(1)將A(-1,m)與B(2,m+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/2.png)
)代入反比例函數(shù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/3.png)
中,
得:m=-k,m+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/5.png)
,
∴(-1)•m=2•(m+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/6.png)
),解得:m=-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/7.png)
,
則k=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/8.png)
;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/images9.png)
(2)如圖1,作BE⊥x軸,E為垂足,
則CE=3,BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/9.png)
,BC=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/10.png)
,
∵Rt△CBE中,BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/11.png)
BC,
∴∠BCE=30°,
又點(diǎn)C與點(diǎn)A的橫坐標(biāo)相同,
∴CA⊥x軸,
∴∠ACB=120°,
當(dāng)AC為底時(shí),由于過(guò)點(diǎn)B且平行于AC的直線與雙曲線只有一個(gè)公共點(diǎn)B,故不符題意;
當(dāng)BC為底時(shí),過(guò)點(diǎn)A作BC的平行線,交雙曲線于點(diǎn)D,
過(guò)點(diǎn)A,D分別作x軸,y軸的平行線,交于點(diǎn)F,
由于∠DAF=30°,設(shè)DF=m
1(m
1>0),則AF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/12.png)
m
1,AD=2m
1,
由點(diǎn)A(-1,-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/13.png)
),得點(diǎn)D(-1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/14.png)
m
1,-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/15.png)
+m
1),
因此(-1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/16.png)
m
1)•(-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/17.png)
+m
1)=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/18.png)
,
解之得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/19.png)
(m
1=0舍去),
因此點(diǎn)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/20.png)
,
此時(shí)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/21.png)
,與BC的長(zhǎng)度不等,故四邊形ADBC是梯形,
如圖2,當(dāng)AB為底時(shí),過(guò)點(diǎn)C作AB的平行線,與雙曲線在第一象限內(nèi)的交點(diǎn)為D,
由于AC=BC,因此∠CAB=30°,從而∠ACD=150°,作DH⊥x軸,H為垂足,
則∠DCH=60°,設(shè)CH=m
2(m
2>0),則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/22.png)
,CD=2m
2,
由點(diǎn)C(-1,0),得點(diǎn)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/23.png)
,
因此
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/24.png)
,
解之得m
2=2(m
2=-1舍去),因此點(diǎn)D(1,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/25.png)
,
此時(shí)CD=4,與AB的長(zhǎng)度不相等,故四邊形ABDC是梯形,
如圖3,當(dāng)過(guò)點(diǎn)C作AB的平行線,與雙曲線在第三象限內(nèi)的交點(diǎn)為D時(shí),
同理可得,點(diǎn)D(-2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/26.png)
),四邊形ABCD是梯形,
綜上所述,函數(shù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/27.png)
圖象上存在點(diǎn)D,使得以A,B,C,D四點(diǎn)為頂點(diǎn)的四邊形為梯形,
點(diǎn)D的坐標(biāo)為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/28.png)
或D(1,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/29.png)
或D(-2,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101209909027352/SYS201311031012099090273029_DA/30.png)
).
點(diǎn)評(píng):此題難度中等,考查了反比例函數(shù)的圖象和性質(zhì)與四邊形性質(zhì)的結(jié)合,綜合性較強(qiáng),同學(xué)們要熟練掌握.