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24． 如圖10.已知點A的坐標是.點B的坐標是(9.0).以AB為直徑作⊙O′.交y軸的負半軸于點C.連接AC.BC.過A.B.C三點作拋物線． (1)求拋物線的解析式, (2)點E是AC延長線上一點.∠BCE的平分線CD交⊙O′于點D.連結(jié)BD.求直線BD的解析式, 的條件下.拋物線上是否存在點P.使得∠PDB＝∠CBD?如果存在.請求出點P的坐標,如果不存在.請說明理由． (1) ∵以AB為直徑作⊙O′.交y軸的負半軸于點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 設拋物線解析式為y=a. ∴–3=a.解得a=. ∴二次函數(shù)的解析式為y=.即y=x2–x–3．···························· 4分 (2) ∵AB為O′的直徑.且A. ∴OO′=4.O′(4.0).····················································································· 5分 ∵點E是AC延長線上一點.∠BCE的平分線CD交⊙O′于點D. ∴∠BCD=∠BCE=×90°=45°. 連結(jié)O′D交BC于點M.則∠BO′D=2∠BCD=2×45°=90°.OO′=4.O′D=AB=5． ∴D．································································································· 6分 ∴設直線BD的解析式為y=kx+b ∴··························································· 7分 解得 ∴直線BD的解析式為y=x–9.····································· 8分 (3) 假設在拋物線上存在點P.使得∠PDB=∠CBD. 解法一:設射線DP交⊙O′于點Q.則． 分兩種情況: ①∵O′.C． ∴把點C.D繞點O′逆時針旋轉(zhuǎn)90°.使點D與點B重合.則點C與點Q1重合. 因此.點Q1符合. ∵D.Q1. ∴用待定系數(shù)法可求出直線DQ1解析式為y=x–．··································· 9分 解方程組得 ∴點P1坐標為(.).[坐標為(.)不符合題意.舍去]． ······················································································································ 10分 ②∵Q1. ∴點Q1關于x軸對稱的點的坐標為Q2(7.4)也符合． ∵D.Q2(7.4)． ∴用待定系數(shù)法可求出直線DQ2解析式為y=3x–17．······································ 11分 解方程組得 ∴點P2坐標為不符合題意.舍去]． ······················································································································ 12分 ∴符合條件的點P有兩個:P1(.).P2． 解法二:分兩種情況: ①當DP1∥CB時.能使∠PDB=∠CBD． ∵B． ∴用待定系數(shù)法可求出直線BC解析式為y=x–3． 又∵DP1∥CB.∴設直線DP1的解析式為y=x+n． 把D代入可求n= –. ∴直線DP1解析式為y=x–．························· 9分 解方程組得 ∴點P1坐標為(.).[坐標為(.)不符合題意.舍去]． ······················································································································ 10分 ②在線段O′B上取一點N.使BN=DM時.得ΔNBD≌ΔMDB(SAS).∴∠NDB=∠CBD． 由①知.直線BC解析式為y=x–3． 取x=4.得y= –.∴M(4.–).∴O′N=O′M=.∴N(.0). 又∵D. ∴直線DN解析式為y=3x–17．······································································ 11分 解方程組得 ∴點P2坐標為不符合題意.舍去]． ······················································································································ 12分 ∴符合條件的點P有兩個:P1(.).P2． 解法三:分兩種情況: ①求點P1坐標同解法二．··············································································· 10分 ②過C點作BD的平行線,交圓O′于G, 此時.∠GDB=∠GCB=∠CBD． 由(2)題知直線BD的解析式為y=x–9, 又∵ C ∴可求得CG的解析式為y=x–3, 設G.作GH⊥x軸交與x軸與H. 連結(jié)O′G,在Rt△O′GH中.利用勾股定理可得.m=7. 由D可得. DG的解析式為.··········································································· 11分 解方程組得 ∴點P2坐標為不符合題意.舍去]．························ 12分 ∴符合條件的點P有兩個:P1(.).P2． 說明:本題解法較多.如有不同的正確解法.請按此步驟給分. 【查看更多】

 

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（2013年四川資陽3分）資陽市2012年財政收入取得重大突破，地方公共財政收入用四舍五入取近似值后為27.39億元，那么這個數(shù)值【    】

A．精確到億位       B．精確到百分位       C．精確到千萬位      D．精確到百萬位

 

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（2013年四川資陽7分）解方程：．

 

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（2013年四川資陽3分）已知直線上有n（n≥2的正整數(shù)）個點，每相鄰兩點間距離為1，從左邊第1個點起跳，且同時滿足以下三個條件：

①每次跳躍均盡可能最大；

②跳n次后必須回到第1個點；

③這n次跳躍將每個點全部到達，

設跳過的所有路程之和為Sn，則S25=　  　．

 

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（2013年四川資陽3分）從所給出的四個選項中，選出適當?shù)囊粋�填入問號所在位置，使之呈現(xiàn)相同的特征【    】

A．       B．       C．      D．

 

 

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（2013年四川資陽9分）如圖，已知直線l分別與x軸、y軸交于A，B兩點，與雙曲線（a≠0，x＞0）分別交于D、E兩點．

（1）若點D的坐標為（4，1），點E的坐標為（1，4）：

①分別求出直線l與雙曲線的解析式；

②若將直線l向下平移m（m＞0）個單位，當m為何值時，直線l與雙曲線有且只有一個交點？

（2）假設點A的坐標為（a，0），點B的坐標為（0，b），點D為線段AB的n等分點，請直接寫出b的值．

 

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