【答案】
分析:(1)已知A、D、E三點的坐標(biāo),利用待定系數(shù)法可確定拋物線的解析式,進(jìn)而能得到頂點B的坐標(biāo).
(2)過B作BM⊥y軸于M,由A、B、E三點坐標(biāo),可判斷出△BME、△AOE都為等腰直角三角形,易證得∠BEA=90°,即△ABE是直角三角形,而AB是△ABE外接圓的直徑,因此只需證明AB與CB垂直即可.BE、AE長易得,能求出tan∠BAE的值,結(jié)合tan∠CBE的值,可得到∠CBE=∠BAE,由此證得∠CBA=∠CBE+∠ABE=∠BAE+∠ABE=90°,此題得證.
(3)△ABE中,∠AEB=90°,tan∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/0.png)
,即AE=3BE,若以D、E、P為頂點的三角形與△ABE相似,那么該三角形必須滿足兩個條件:①有一個角是直角、②兩直角邊滿足1:3的比例關(guān)系;然后分情況進(jìn)行求解即可.
(4)過E作EF∥x軸交AB于F,當(dāng)E點運動在EF之間時,△AOE與△ABE重疊部分是個四邊形;當(dāng)E點運動到F點右側(cè)時,△AOE與△ABE重疊部分是個三角形.按上述兩種情況按圖形之間的和差關(guān)系進(jìn)行求解.
解答:(1)解:由題意,設(shè)拋物線解析式為y=a(x-3)(x+1).
將E(0,3)代入上式,解得:a=-1.
∴y=-x
2+2x+3.
則點B(1,4).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/images1.png)
(2)證明:如圖1,過點B作BM⊥y于點M,則M(0,4).
在Rt△AOE中,OA=OE=3,
∴∠1=∠2=45°,AE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/1.png)
=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/2.png)
.
在Rt△EMB中,EM=OM-OE=1=BM,
∴∠MEB=∠MBE=45°,BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/4.png)
.
∴∠BEA=180°-∠1-∠MEB=90°.
∴AB是△ABE外接圓的直徑.
在Rt△ABE中,tan∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/6.png)
=tan∠CBE,
∴∠BAE=∠CBE.
在Rt△ABE中,∠BAE+∠3=90°,∴∠CBE+∠3=90°.
∴∠CBA=90°,即CB⊥AB.
∴CB是△ABE外接圓的切線.
(3)解:Rt△ABE中,∠AEB=90°,tan∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/7.png)
,sin∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/8.png)
,cos∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/9.png)
;
若以D、E、P為頂點的三角形與△ABE相似,則△DEP必為直角三角形;
①DE為斜邊時,P
1在x軸上,此時P
1與O重合;
由D(-1,0)、E(0,3),得OD=1、OE=3,即tan∠DEO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/10.png)
=tan∠BAE,即∠DEO=∠BAE
滿足△DEO∽△BAE的條件,因此 O點是符合條件的P
1點,坐標(biāo)為(0,0).
②DE為短直角邊時,P
2在x軸上;
若以D、E、P為頂點的三角形與△ABE相似,則∠DEP
2=∠AEB=90°,sin∠DP
2E=sin∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/11.png)
;
而DE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/13.png)
,則DP
2=DE÷sin∠DP
2E=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/14.png)
÷
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/15.png)
=10,OP
2=DP
2-OD=9
即:P
2(9,0);
③DE為長直角邊時,點P
3在y軸上;
若以D、E、P為頂點的三角形與△ABE相似,則∠EDP
3=∠AEB=90°,cos∠DEP
3=cos∠BAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/16.png)
;
則EP
3=DE÷cos∠DEP
3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/17.png)
÷
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/19.png)
,OP
3=EP
3-OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/20.png)
;
綜上,得:P
1(0,0),P
2(9,0),P
3(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/21.png)
).
(4)解:設(shè)直線AB的解析式為y=kx+b.
將A(3,0),B(1,4)代入,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/22.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/23.png)
.
∴y=-2x+6.
過點E作射線EF∥x軸交AB于點F,當(dāng)y=3時,得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/24.png)
,∴F(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/25.png)
,3).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/images27.png)
情況一:如圖2,當(dāng)0<t≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/26.png)
時,設(shè)△AOE平移到△GNM的位置,MG交AB于點H,MN交AE于點S.
則ON=AG=t,過點H作LK⊥x軸于點K,交EF于點L.
由△AHG∽△FHM,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/27.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/28.png)
.
解得HK=2t.
∴S
陰=S
△MNG-S
△SNA-S
△HAG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/29.png)
×3×3-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/30.png)
(3-t)
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/31.png)
t•2t=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/32.png)
t
2+3t.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/images35.png)
情況二:如圖3,當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/33.png)
<t≤3時,設(shè)△AOE平移到△PQR的位置,PQ交AB于點I,交AE于點V.
由△IQA∽△IPF,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/34.png)
.即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/35.png)
,
解得IQ=2(3-t).
∵AQ=VQ=3-t,
∴S
陰=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/36.png)
IV•AQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/37.png)
(3-t)
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/38.png)
t
2-3t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/39.png)
.
綜上所述:s=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185228789250661/SYS201311011852287892506024_DA/40.png)
.
點評:該題考查了二次函數(shù)的綜合題,涉及到二次函數(shù)解析式的確定、切線的判定、相似三角形的判定、圖形面積的解法等重點知識,綜合性強(qiáng),難度系數(shù)較大.此題的難點在于后兩個小題,它們都需要分情況進(jìn)行討論,容易出現(xiàn)漏解的情況.在解答動點類的函數(shù)問題時,一定不要遺漏對應(yīng)的自變量取值范圍.