中考数学圆和抛物线训练题(含答案)_.docx

中考数学圆和抛物线训练题(含答案)_中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)CxxyyAOBEDACBCDG图1图2圆和抛物线综合题专题训练1、如图1，在平面直角坐标系xOy中，二次函数yax2bxc(a0)的图象顶点为D，与y轴交于点C，与x轴交于点A、B，点A在原点的左侧，点B的坐标为(3，0)，OBOC，tanACO13(1)求这个二次函数的解析式；(2)若平行于x轴的直线与该抛物线交于点M、N，且以MN为直径的圆与x轴相切，求该圆的半径长度；(3)如图2，若点G(2，y)是该抛物线上一点，点P是直线AG下方的抛物线上的一动点，当点P运动到什么位置时，AGP的面积最大？求此时点P的坐标和AGP的最大面积解：1由OC=OB=3，知C(03)，连接AC，在RtAOC中，OA=OC×tanACO=1313，故A10，设所求二次函数的表达式为(1)(3)yaxx将C(03)，代入得3(01)(03)a，解得1a，这个二次函数的表达式为223yxx。2解法一：当直线MN在x轴上方时，设所求圆的半径为RR>0，设M在N的左侧，所求圆的圆心在抛物线的对称轴1x上，NR+1，R代入223yxx中得2(1)2(1)3RRR，解得1172R。当直线MN在x轴下方时，设所求圆的半径为(0)rr，由可知N(1)rr，代入抛物线方程可得1172r。中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)2解法二：当直线MN在x轴上方时，设所求的半径为RR>0,12()()MxRNxR，、，则1x和2x是方程223Rxx的两根=44(3)R121223xxxxR由122xxR得，221212()44xxxxR244(3)4RR。解得1172R。当直线MN在x轴下方时，设所求圆的半径为(0)rr，12()()MxrNxr，、，则1x和2x是方程223rxx的两根=44(3)0r，解得4r。121223xxxxr由122xxr得，221212()44xxxxr244(3)4rr。解得1172r。又11742，1172r。3过点P作y轴的平行线与AG交于点Q，把G2，y代入抛物线的解析式223yxx得G(23)，。由A(10)，可得直线AG的方程为1yx设2(23)Pxxx，则(1)Qxx，22PQxx，213(2)22APGAPQGPQSSSPQGAxx横坐标横坐标当12x时，APG的面积最大。此时P点的坐标为115()24，APG的面积最大值为278。中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)2、如图1，直线y43x1与抛物线y41x2交于A，B两点A在B的左侧，与y轴交于点C1求线段AB的长；2若以AB为直径的圆与直线xm有公共点，求m的取值范围；3如图2，把抛物线向右平移2个单位，再向上平移n个单位n0，抛物线与x轴交于P，Q两点，过C，P，Q三点的圆的面积能否存在最小值的情况？若存在，请求出这个最小值和此时n的值，若不存在，请讲明理由解：由题意：，解得：x2+3x-4=0，即x=-4或x=1代入求得y=-4或-，或，即点A-4，-4B1，-，则AB=；2由1可得A，B中点即圆的圆心点O为，则圆的方程式为：与x=m有公共点即有解，CPyOxQ图2ACyOxB图1中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)把代入断定判别式即可3抛物线平移后为：存在理由如下：抛物线平移后为：，其对称轴是x=2由于过P、Q的圆的圆心必在对称轴上，要使圆的面积最小，则圆的半径要最小，即点C到圆心的距离要最短，过C作CE垂直抛物线的对称轴，垂足为E，则符合条件的圆是以E为圆心，EC长为半径的圆，其面积为43、如图，在平面直角坐标系xOy中，半径为1的圆的圆心O在坐标原点，且与两坐标轴分别交于ABCD、四点抛物线2yaxbxc与y轴交于点D，与直线yx交于点MN、，且MANC、分别与圆O相切于点A和点C1求抛物线的解析式；2抛物线的对称轴交x轴于点E，连结DE，并延长DE交圆O于F，求EF的长3过点B作圆O的切线交DC的延长线于点P，判定点P能否在抛物线上，讲明理由解：1圆心O在坐标原点，圆O的半径为1，点ABCD、的坐标分别为(10)(01)(10)(01)ABCD，、，、，、，抛物线与直线yx交于点MN、，且MANC、分别与圆O相切于点A和点C，(11)(11)MN，、，·······························································································2分点DMN、在抛物线上，将(01)(11)(11)DMN，、，、，的坐标代入OxyNCDEFBMA中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)2yaxbxc，得：111cabcabc解之，得：111abc抛物线的解析式为：21yxx······································································4分22215124yxxx抛物线的对称轴为12x，1151242OEDE，····················6分连结90BFBFD，°，BFDEOD，DEODDBFD，又5122DEODDB，455FD，455355210EFFDDE·····································································8分3点P在抛物线上······························································································9分设过DC、点的直线为：ykxb，将点(10)(01)CD，、，的坐标代入ykxb，得：11kb，直线DC为：1yx·····················································································10分过点B作圆O的切线BP与x轴平行，P点的纵坐标为1y，将1y代入1yx，得：2xP点的坐标为(21)，··························································································11分当2x时，2212211yxx，所以，P点在抛物线21yxx上·····································································12分OxyNCDEFBMAP中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)4、如下图，抛物线与x轴交于点1030AB，、，两点，与y轴交于点03.C，以AB为直径作M，过抛物线上一点P作M的切线PD，切点为D，并与M的切线AE相交于点E，连结DM并延长交M于点N，连结.ANAD、1求抛物线所对应的函数关系式及抛物线的顶点坐标；2若四边形EAMD的面积为43，求直线PD的函数关系式；3抛物线上能否存在点P，使得四边形EAMD的面积等于DAN的面积？若存在，求出点P的坐标；若不存在，讲明理由.解：1由于抛物线与x轴交于点1030AB，、，两点，设抛物线的函数关系式为：13yaxx，抛物线与y轴交于点03C，3103a，1.a所以，抛物线的函数关系式为：223yxx，···························································2分又214yx，因而，抛物线的顶点坐标为14，············································································3分2连结EM，EAED、是M，的两条切线，EAEDEAAMEDMN，EAMEDM又四边形EAMD的面积为43，23EAMS，1232AMAE·，又2AM，23.AE因而，点E的坐标为1123E，或2123.E，····················································5分当E点在第二象限时，切点D在第一象限.在直角三角形EAM中，23tan32EAEMAAM，60EMA°，60DMB°过切点D作DFAB，垂足为点F，13MFDF，中考数学圆和抛物线训练题(含答案)中考数学圆和抛物线训练题(含答案)因而，切点D的坐标为23，····················································································6分设直线PD的函数关系式为ykxb，将12323ED，、，的坐标代入得3223kbkb解之，得33533kb所以，直线PD的函数关系式为353.33yx······················································7分当E点在第三象限时，切点D在第四象限.同理可求：切点D的坐标为23，-，直线PD的函数关系式为353.33yx因而，直线PD的函数关系式为35333yx或353.33yx·········································································8分3若四边形EAMD的面积等于DAN的面积又22EAMDANAMDEAMDSSSS四边形，AMDEAMSSED、两点到x轴的距离相等，PD与M相切，点D与点E在x轴同侧，切线PD与x轴平行，此时切线PD的函数关系式为2y或2.y·················································································9分当2y时，由223yxx得，16x；当2y时，由223yxx得，12x.·························································11分故知足条件的点P的位置有4个，分别是123162162122PPP，、，、，、4122.P，·········································································································12分