【答案】
分析:(1)由已知在等腰直角三角形中解出OH的長,因直線過頂點和OH長等于點到直線距離,聯(lián)立方程求出k,b;
(2)思維要嚴密,分兩類情況:①若DN為等腰直角三角形的直角邊;②若DN為等腰直角三角形的斜邊.
根據(jù)相似的比例關(guān)系和幾何關(guān)系,作適合的輔助線,構(gòu)造垂直從而驗證相似比例關(guān)系是否成立.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/images0.png)
解:(1)∵直線y=kx+b過P(-2,0)?-2k+b=0…①
∵AO=BO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/0.png)
,AO⊥BO?三角形AOB為等腰直角三角形,
AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/1.png)
=2?∠OAB=45°?OH=OA×sin45°=1,
∵OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/2.png)
=1…②
由①②方程解得:k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/3.png)
,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/4.png)
,OH=1.
(2)設(shè)存在實數(shù)a,使拋物線y=a(x+1)(x-5)上有一點E,滿足以D,N,E為頂點的三角形與等腰直角△AOB相似.
∴以D,N,E為頂點的三角形為等腰直角三角形,且這樣的三角形最多只有兩類,一類是以DN為直角邊的等腰直角三角形,另一類是以DN為斜邊的等腰直角三角形.
①若DN為等腰直角三角形的直角邊,則ED⊥DN.
在拋物線y=a(x+1)(x-5)中,令y=0,解得x=-1或5,則得:M(-1,0),N(5,0).
∴D(2,0),
∴ED=DN=3.
∴E的坐標為(2,3).
把E(2,3)代入拋物線解析式y(tǒng)=a(x+1)(x-5),得:a(2+1)(2-5)=3,解得a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/5.png)
.
∴拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/6.png)
(x+1)(x-5).
即y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/7.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/8.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/9.png)
.
②若DN為等腰直角三角形的斜邊,
則DE⊥EN,DE=EN.
∴E的坐標為(3.5,1.5).
把E(3.5,1.5)代入拋物線解析式y(tǒng)=a(x+1)(x-5)得:a(3.5+1)(3.5-5)=1.5,解得a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/10.png)
.
∴拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/11.png)
(x+1)(x-5),
即y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/12.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/13.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/14.png)
.
當a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/15.png)
時,在拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/16.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/17.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/18.png)
上存在一點E(2,3)滿足條件,
如果此拋物線上還有滿足條件的E點,不妨設(shè)為E′點,那么只有可能△DE′N是以DN為斜邊的等腰直角三角形,
由此得E′(3.5,1.5),顯然E′不在拋物線.
y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/19.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/20.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/21.png)
上,
因此拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/22.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/23.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/24.png)
上沒有符合條件的其他的E點.
當a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/25.png)
時,同理可得拋物線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/26.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/27.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/28.png)
上沒有符合條件的其他的E點.
當E的坐標為(2,3),對應(yīng)的拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/29.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/30.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/31.png)
時,
∵△EDN和△ABO都是等腰直角三角形,
∴∠GNP=∠PBO=45°.
又∵∠NPG=∠BPO,
∴△NPG∽△BPO.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/32.png)
,
∴PB•PG=PO•PN=2×7=14,
∴總滿足PB•PG<10
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/33.png)
.
當E的坐標為(3.5,1.5),解得對應(yīng)的拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/34.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/35.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/36.png)
時,
同理可證得:PB•PG=PO•PN=2×7=14,
∴總滿足PB•PG<10
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129574212983/SYS201311031011295742129029_DA/37.png)
.
點評:此題考查在直角三角形中解題技巧,通過解方程組來求拋物線解析式,第二問探究三角形相似問題,考查思維的嚴密性,不要漏掉其它情況,學會分類討論.