24．如图10.已知点A的坐标是.点B的坐标是(9.0).以AB为直径作⊙O′.交y轴的负半轴于点C.连接AC.BC.过A.B.C三点作抛物线． (1)求抛物线的解析式, (2)点E是AC延长——青夏教育精英家教网—

24．如图10.已知点A的坐标是.点B的坐标是(9.0).以AB为直径作⊙O′.交y轴的负半轴于点C.连接AC.BC.过A.B.C三点作抛物线． (1)求抛物线的解析式, (2)点E是AC延长线上一点.∠BCE的平分线CD交⊙O′于点D.连结BD.求直线BD的解析式, 的条件下.抛物线上是否存在点P.使得∠PDB＝∠CBD?如果存在.请求出点P的坐标,如果不存在.请说明理由． ∵以AB为直径作⊙O′.交y轴的负半轴于点C. ∴∠OCA+∠OCB=90°. 又∵∠OCB+∠OBC=90°. ∴∠OCA=∠OBC. 又∵∠AOC= ∠COB=90°. ∴ΔAOC∽ ΔCOB.························································································ 1分 ∴．又∵A. ∴.解得OC=3． ∴C. ······················································································································ 3分设抛物线解析式为y=a. ∴–3=a.解得a=. ∴二次函数的解析式为y=.即y=x2–x–3．···························· 4分 (2) ∵AB为O′的直径.且A. ∴OO′=4.O′(4.0).····················································································· 5分 ∵点E是AC延长线上一点.∠BCE的平分线CD交⊙O′于点D. ∴∠BCD=∠BCE=×90°=45°. 连结O′D交BC于点M.则∠BO′D=2∠BCD=2×45°=90°.OO′=4.O′D=AB=5． ∴D．································································································· 6分 ∴设直线BD的解析式为y=kx+b ∴··························································· 7分解得 ∴直线BD的解析式为y=x–9.····································· 8分 (3) 假设在抛物线上存在点P.使得∠PDB=∠CBD. 解法一:设射线DP交⊙O′于点Q.则．分两种情况: ①∵O′.C． ∴把点C.D绕点O′逆时针旋转90°.使点D与点B重合.则点C与点Q1重合. 因此.点Q1符合. ∵D.Q1. ∴用待定系数法可求出直线DQ1解析式为y=x–．··································· 9分解方程组得 ∴点P1坐标为(.).[坐标为(.)不符合题意.舍去]． ······················································································································ 10分 ②∵Q1. ∴点Q1关于x轴对称的点的坐标为Q2(7.4)也符合． ∵D.Q2(7.4)． ∴用待定系数法可求出直线DQ2解析式为y=3x–17．······································ 11分解方程组得 ∴点P2坐标为不符合题意.舍去]． ······················································································································ 12分 ∴符合条件的点P有两个:P1(.).P2．解法二:分两种情况: ①当DP1∥CB时.能使∠PDB=∠CBD． ∵B． ∴用待定系数法可求出直线BC解析式为y=x–3．又∵DP1∥CB.∴设直线DP1的解析式为y=x+n．把D代入可求n= –. ∴直线DP1解析式为y=x–．························· 9分解方程组得 ∴点P1坐标为(.).[坐标为(.)不符合题意.舍去]． ······················································································································ 10分 ②在线段O′B上取一点N.使BN=DM时.得ΔNBD≌ΔMDB(SAS).∴∠NDB=∠CBD．由①知.直线BC解析式为y=x–3．取x=4.得y= –.∴M(4.–).∴O′N=O′M=.∴N(.0). 又∵D. ∴直线DN解析式为y=3x–17．······································································ 11分解方程组得 ∴点P2坐标为不符合题意.舍去]． ······················································································································ 12分 ∴符合条件的点P有两个:P1(.).P2．解法三:分两种情况: ①求点P1坐标同解法二．··············································································· 10分 ②过C点作BD的平行线,交圆O′于G, 此时.∠GDB=∠GCB=∠CBD．由(2)题知直线BD的解析式为y=x–9, 又∵ C ∴可求得CG的解析式为y=x–3, 设G.作GH⊥x轴交与x轴与H. 连结O′G,在Rt△O′GH中.利用勾股定理可得.m=7. 由D可得. DG的解析式为.··········································································· 11分解方程组得 ∴点P2坐标为不符合题意.舍去]．························ 12分 ∴符合条件的点P有两个:P1(.).P2．说明:本题解法较多.如有不同的正确解法.请按此步骤给分. 【查看更多】

题目列表(包括答案和解析)

（1）(

)-2-(

)2
（2）(-9

)×

（3）如图1：AD⊥BC，EF⊥BC，∠1=∠2，说明AB∥DM．
（4）如图2：已知点A（-4，4），B（-2，-2），求△AOB的面积．

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28、（1）如图1，已知点P在正三角形ABC的边BC上，以AP为边作正三角形APQ，连接CQ．
①求证：△ABP≌△ACQ；
②若AB=6，点D是AQ的中点，直接写出当点P由点B运动到点C时，点D运动路线的长．
（2）已知，△EFG中，EF=EG=13，FG=10．如图2，把△EFG绕点E旋转到△EF′G′的位置，点M是边EF′与边FG的交点，点N在边EG′上且EN=EM，连接GN．求点E到直线GN的距离．

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（2011•下关区一模）（1）如图1，已知点P在正三角形ABC的边BC上，以AP为边作正三角形APQ，连接CQ．
①求证：△ABP≌△ACQ；
②若AB=6，点D是AQ的中点，直接写出当点P由点B运动到点C时，点D运动路线的长．
（2）已知，△EFG中，EF=EG=13，FG=10．如图2，把△EFG绕点E旋转到△EF'G'的位置，点M是边EF'与边FG的交点，点N在边EG'上且EN=EM，连接GN．求点E到直线GN的距离．

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．(10分)(1)如图1，已知点P在正三角形ABC的边BC上，以AP为边作正三角形APQ，连接CQ．

①求证：△ABP≌△ACQ；

②若AB＝6，点D是AQ的中点，直接写出当点P由点B运动到点C时，点D运动路线的长．

(2)已知，△EFG中，EF＝EG＝13，FG＝10．如图2，把△EFG绕点E旋转到△EF＇G＇的位置，点M是边EF＇与边FG的交点，点N在边EG＇上且EN＝EM，连接GN．求点E到直线GN的距离．

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．(10分)(1)如图1，已知点P在正三角形ABC的边BC上，以AP为边作正三角形APQ，连接CQ．
①求证：△ABP≌△ACQ；
②若AB＝6，点D是AQ的中点，直接写出当点P由点B运动到点C时，点D运动路线的长．
(2)已知，△EFG中，EF＝EG＝13，FG＝10．如图2，把△EFG绕点E旋转到△EF＇G＇的位置，点M是边EF＇与边FG的交点，点N在边EG＇上且EN＝EM，连接GN．求点E到直线GN的距离．

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