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. (1) ∵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

 

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(2013年四川资阳3分)在一个不透明的盒子里,装有4个黑球和若干个白球,它们除颜色外没有任何其他区别,摇匀后从中随机摸出一个球记下颜色,再把它放回盒子中,不断重复,共摸球40次,其中10次摸到黑球,则估计盒子中大约有白球【    】

A.12个       B.16个       C.20个      D.30个

 

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(2013年四川资阳3分)资阳市2012年财政收入取得重大突破,地方公共财政收入用四舍五入取近似值后为27.39亿元,那么这个数值【    】

A.精确到亿位       B.精确到百分位       C.精确到千万位      D.精确到百万位

 

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(2013年四川资阳3分)在函数中,自变量x的取值范围是【    】

A.x≤1       B.x≥1       C.x<1      D.x>1

 

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(2013年四川资阳9分)钓鱼岛历来是中国领土,以它为圆心在周围12海里范围内均属于禁区,不允许它国船只进入,如图,今有一中国海监船在位于钓鱼岛A正南方距岛60海里的B处海域巡逻,值班人员发现在钓鱼岛的正西方向52海里的C处有一艘日本渔船,正以9节的速度沿正东方向驶向钓鱼岛,中方立即向日本渔船发出警告,并沿北偏西30°的方向以12节的速度前往拦截,期间多次发出警告,2小时候海监船到达D处,与此同时日本渔船到达E处,此时海监船再次发出严重警告.

(1)当日本渔船受到严重警告信号后,必须沿北偏东转向多少度航行,才能恰好避免进入钓鱼岛12海里禁区?

(2)当日本渔船不听严重警告信号,仍按原速度,原方向继续前进,那么海监船必须尽快到达距岛12海里,且位于线段AC上的F处强制拦截渔船,问海监船能否比日本渔船先到达F处?(注:①中国海监船的最大航速为18节,1节=1海里/小时;②参考数据:sin26.3°≈0.44,sin20.5°≈0.35,sin18.1°≈0.31,≈1.4,≈1.7)

 

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(2013年四川资阳9分)如图,已知直线l分别与x轴、y轴交于A,B两点,与双曲线(a≠0,x>0)分别交于D、E两点.

(1)若点D的坐标为(4,1),点E的坐标为(1,4):

①分别求出直线l与双曲线的解析式;

②若将直线l向下平移m(m>0)个单位,当m为何值时,直线l与双曲线有且只有一个交点?

(2)假设点A的坐标为(a,0),点B的坐标为(0,b),点D为线段AB的n等分点,请直接写出b的值.

 

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