2019-2020年中考数学压轴题精选（九）及答案资料81.（08广东茂名25题）（本题满分10分）如图，在平面直角坐标系中，抛物线y =－32x 2+b x +c 经过A （0，－4）、B （x 1，0）、 C （x 2，0）三点，且x 2-x 1=5．（1）求b 、c 的值；（4分）（2）在抛物线上求一点D ，使得四边形BDCE 是以BC 为对角线的菱形；（3分）（3）在抛物线上是否存在一点P ，使得四边形B P O H 是以OB 为对角线的菱形？若存在，求出点P 的坐标，并判断这个菱形是否为正方形？若不存在，请说明理由．（3分）解：（08广东茂名25题解析）解：（1）解法一： ∵抛物线y =－32x 2+b x +c 经过点A （0，－4）， ∴c =－4 ……1分又由题意可知，x 1、x 2是方程－32x 2+b x +c =0的两个根， ∴x 1+x 2=23b ， x 1x 2=－23c =6 ·························································· 2分 由已知得（x 2-x 1）2=25 又（x 2-x 1）2=（x 2+x 1）2－4x1x 2=49b 2－24 ∴49b 2－24=25 解得b =±314··························································································· 3分当b =314时，抛物线与x 轴的交点在x 轴的正半轴上，不合题意，舍去．∴b =－314． ·························································································· 4分 解法二：∵x 1、x 2是方程－32x 2+b x +c=0的两个根， 即方程2x 2－3b x +12=0的两个根．（第25题图）x∴x =4969b 32-±b ， ································································ 2分∴x 2－x 1=2969b 2-=5，解得 b =±314 ·················································································· 3分 （以下与解法一相同．）（2）∵四边形BDCE 是以BC 为对角线的菱形，根据菱形的性质，点D 必在抛物线的对称轴上，····································································································· 5分又∵y =－32x 2－314x －4=－32（x +27）2+625···························· 6分 ∴抛物线的顶点（－27，625）即为所求的点D ． ································· 7分（3）∵四边形BPOH 是以OB 为对角线的菱形，点B 的坐标为（－6，0），根据菱形的性质，点P 必是直线x =-3与抛物线y =－32x 2-314x -4的交点， ··················································· 8分∴当x =－3时，y =－32×（－3）2－314×（－3）－4=4，∴在抛物线上存在一点P （－3，4），使得四边形BPOH 为菱形． ··············· 9分四边形BPOH 不能成为正方形，因为如果四边形BPOH 为正方形，点P 的坐标只能是（－3，3），但这一点不在抛物线上． ································································· 10分82.（08广东肇庆25题）（本小题满分10分）已知点A （a ，1y ）、B （2a ，y 2）、C （3a ，y 3）都在抛物线x x y 1252+=上. （1）求抛物线与x 轴的交点坐标； （2）当a =1时，求△ABC 的面积；（3）是否存在含有1y 、y 2、y 3，且与a 无关的等式？如果存在，试给出一个，并加以证明；如果不存在，说明理由.（08广东肇庆25题解析）（本小题满分10分）解：（1）由5x x 122+=0， ··································································· （1分）得01=x ，5122-=x ． ······································································ （2分） ∴抛物线与x 轴的交点坐标为（0，0）、（512-，0）． ································· （3分）（2）当a =1时，得A （1，17）、B （2，44）、C （3，81），·························· （4分）分别过点A 、B 、C 作x 轴的垂线，垂足分别为D 、E 、F ，则有ABC S ∆=S ADFC 梯形 -ADEB S 梯形 -BEFC S 梯形 ············································ （5分）=22)8117(⨯+-21)4417(⨯+-21)8144(⨯+ ······························· （6分）=5（个单位面积） ······························································ （7分）（3）如：)(3123y y y -=． ······························································· （8分）事实上，)3(12)3(523a a y ⨯+⨯= =45a 2+36a ．3（12y y -）=3[5×（2a ）2+12×2a -（5a 2+12a ）] =45a 2+36a ． ·········· （9分） ∴)(3123y y y -=． ········································································ （10分）83.（08辽宁沈阳26题）（本题14分）26．如图所示，在平面直角坐标系中，矩形ABOC 的边BO 在x 轴的负半轴上，边OC 在y 轴的正半轴上，且1AB =，OB =ABOC 绕点O 按顺时针方向旋转60后得到矩形EFOD ．点A 的对应点为点E ，点B 的对应点为点F ，点C 的对应点为点D ，抛物线2y ax bx c =++过点A E D ，，． （1）判断点E 是否在y 轴上，并说明理由； （2）求抛物线的函数表达式；（3）在x 轴的上方是否存在点P ，点Q ，使以点O B P Q ，，，为顶点的平行四边形的面积是矩形ABOC 面积的2倍，且点P 在抛物线上，若存在，请求出点P ，点Q 的坐标；若不存在，请说明理由．（08辽宁沈阳26题解析）解：（1）点E 在y 轴上 ··········································· 1分 理由如下：连接AO ，如图所示，在Rt ABO △中，1AB =，BO =，2AO ∴=1sin 2AOB ∴∠=，30AOB ∴∠= 由题意可知：60AOE ∠=306090BOE AOB AOE ∴∠=∠+∠=+=点B 在x 轴上，∴点E 在y 轴上． ······························································ 3分 （2）过点D 作DM x ⊥轴于点M第26题图1OD =，30DOM ∠=∴在Rt DOM △中，12DM =，OM =点D 在第一象限，∴点D 的坐标为12⎫⎪⎪⎝⎭， ············································································· 5分 由（1）知2EO AO ==，点E 在y 轴的正半轴上∴点E 的坐标为(02)，∴点A的坐标为( ··············································································· 6分抛物线2y ax bx c =++经过点E ，2c ∴=由题意，将(A，122D ⎛⎫ ⎪ ⎪⎝⎭，代入22y ax bx =++中得321312422a a ⎧-+=⎪⎨++=⎪⎩解得899a b ⎧=-⎪⎪⎨⎪=-⎪⎩∴所求抛物线表达式为：28299y x x =--+ ··············································· 9分（3）存在符合条件的点P ，点Q ． ······························································· 10分 理由如下：矩形ABOC 的面积3AB BO ==∴以O B P Q ，，，为顶点的平行四边形面积为由题意可知OB 为此平行四边形一边， 又3OB =OB ∴边上的高为2 ····················································································· 11分 依题意设点P 的坐标为(2)m ，点P在抛物线28299y x x =--+上28229m ∴--+=x解得，10m =，2m=1(02)P ∴，，228P ⎛⎫- ⎪ ⎪⎝⎭以O B P Q ，，，为顶点的四边形是平行四边形，PQ OB ∴∥，PQ OB == ∴当点1P的坐标为(02)，时，点Q的坐标分别为1(2)Q ，2Q ；当点2P 的坐标为2⎛⎫⎪ ⎪⎝⎭时，点Q 的坐标分别为32Q ⎛⎫⎪ ⎪⎝⎭，42Q ⎫⎪⎪⎝⎭． ········································· 14分84.（08辽宁12市26题）（本题14分）26．如图16，在平面直角坐标系中，直线y =-x 轴交于点A ，与y 轴交于点C ，抛物线2(0)3y ax x c a =-+≠经过A B C ，，三点． （1）求过A B C ，，三点抛物线的解析式并求出顶点F 的坐标； （2）在抛物线上是否存在点P ，使ABP △为直角三角形，若存在，直接写出P 点坐标；若不存在，请说明理由； （3）试探究在直线AC 上是否存在一点M ，使得MBF △的周长最小，若存在，求出M 点的坐标；若不存在，请说明理由．（08辽宁12市26题解析）解：（1）直线y =x 轴交于点A ，与y 轴交于点C．(10)A ∴-，，(0C -， (1)分 点AC ，都在抛物线上，03a c c ⎧=++⎪∴⎨⎪=⎩ 3a c ⎧=⎪∴⎨⎪=⎩∴抛物线的解析式为2y x x =··············································· 3分 xx∴顶点13F ⎛- ⎝⎭， ················································································· 4分（2）存在································································································ 5分 1(0P ····························································································· 7分 2(2P ····························································································· 9分 （3）存在······························································································· 10分理由： 解法一：延长BC 到点B '，使B C BC '=，连接B F '交直线AC 于点M ，则点M 就是所求的点． ········································································· 11分 过点B '作B H AB '⊥于点H ．B点在抛物线2y x x =(30)B ∴， 在Rt BOC △中，tan OBC ∠=，30OBC ∴∠=，BC =在Rt BB H '△中，12BH BB ''== 6BH H '==，3OH ∴=，(3B '∴--， ········································· 12分设直线BF '的解析式为y kx b=+3k b k b ⎧-=-+⎪∴⎨=+⎪⎩解得2k b ⎧=⎪⎪⎨⎪=-⎪⎩62y x ∴=- ················································································ (1)3分y y ⎧=⎪∴⎨=⎪⎩ 解得377x y ⎧=⎪⎪⎨⎪=-⎪⎩37M ⎛∴ ⎝⎭ ∴在直线AC 上存在点M ，使得MBF △的周长最小，此时37M ⎛ ⎝⎭．··· 14分解法二：过点F 作AC 的垂线交y 轴于点H ，则点H 为点F 关于直线AC 的对称点．连接BH 交AC 于点M ，则点M 即为所求． ············································· 11分 过点F 作FG y ⊥轴于点G ，则OB FG ∥，BC FH ∥．90BOC FGH ∴∠=∠=，BCO FHG ∠=∠HFG CBO ∴∠=∠同方法一可求得(30)B ，．在Rt BOC △中，tan 3OBC ∠=，30OBC ∴∠=，可求得3GH GC ==， GF ∴为线段CH 的垂直平分线，可证得CFH △为等边三角形，AC ∴垂直平分FH ．即点H 为点F 关于AC的对称点．03H ⎛⎫∴- ⎪ ⎪⎝⎭， ······································· 12分设直线BH 的解析式为y kx b =+，由题意得03k b b =+⎧⎪⎨=⎪⎩解得k b ⎧=⎪⎪⎨⎪=⎪⎩y ∴=··················································································· 13分y y ⎧=⎪∴⎨⎪=⎩解得37x y ⎧=⎪⎪⎨⎪=⎪⎩37M ⎛∴ ⎝⎭， ∴在直线AC 上存在点M ，使得MBF △的周长最小，此时377M ⎛- ⎝⎭，．··· 14分85.（08内蒙古赤峰25题）（本题满分14分）在平面直角坐标系中给定以下五个点17(30)(14)(03)(10)24A B C D E ⎛⎫-- ⎪⎝⎭，，，，，，，，，． （1）请从五点中任选三点，求一条以平行于y 轴的直线为对称轴的抛物线的解析式； （2）求该抛物线的顶点坐标和对称轴，并画出草图； （3）已知点1514F ⎛⎫- ⎪⎝⎭，在抛物线的对称轴上，直线174y =x过点1714G ⎛⎫- ⎪⎝⎭，且垂直于对称轴．验证：以(10)E ，为圆心，EF 为半径的圆与直线174y =相切．请你进一步验证，以抛物线上的点1724D ⎛⎫ ⎪⎝⎭，为圆心DF 为半径的圆也与直线174y =相切．由此你能猜想到怎样的结论．（08内蒙古赤峰25题解析）25．解：（1）设抛物线的解析式为2y ax bx c =++， 且过点(30)(03)(10)A C E -，，，，，， 由(03)，在2y ax bx c =++H ．则3c =． ·························································································· （2分）得方程组93300a b a b c -+=⎧⎨++=⎩，解得12a b =-=-，．∴抛物线的解析式为223y x x =--+ ·············· （4分）（2）由2223(1)4y x x x =--+=-++ ··········· （6分） 得顶点坐标为(14)-，，对称轴为1x =-． ········· （8分） （3）①连结EF ，过点E 作直线174y =的垂线，垂足为N ， 则174EN HG ==． 在Rt FHE △中，2HE =，154HF =，174EF ∴==， EF EN ∴=，∴以E 点为圆心，EF 为半径的E 与直线174y =相切． ························· （10分） ②连结DF 过点D 作直线174y =的垂线，垂足为M ．过点D 作DQ GH ⊥垂足为Q ， 则1771054442DM QG ==-==． 在Rt FQD △中，32QD =，15782444QF =-==．52FD ==．∴以D 点为圆心DF 为半径的D 与直线174y =相切． ··························· （12分）x③以抛物线上任意一点P 为圆心，以PF 为半径的圆与直线174y =相切． ···· （14分）86.（08青海西宁28题）如图14，已知半径为1的1O 与x 轴交于A B ，两点，OM 为1O 的切线，切点为M ，圆心1O 的坐标为(20)，，二次函数2y x bx c =-++的图象经过A B ，两点． （1）求二次函数的解析式；（2）求切线OM 的函数解析式；（3）线段OM 上是否存在一点P ，使得以P O A ，，为顶点的三角形与1OO M △相似．若存在，请求出所有符合条件的点P 的坐标；若不存在，请说明理由．（08青海西宁28题解析）解：（1）圆心1O 的坐标为(20)，，1O 半径为1，(10)A ∴，，(30)B ，……1分二次函数2y x bx c =-++的图象经过点A B ，，∴可得方程组10930b c b c -++=⎧⎨-++=⎩···································································· 2分解得：43b c =⎧⎨=-⎩∴二次函数解析式为243y x x =-+- ······································ 3分（2）过点M 作MF x ⊥轴，垂足为F ． ······················································ 4分 OM 是1O 的切线，M 为切点，1O M OM ∴⊥（圆的切线垂直于经过切点的半径）． 在1Rt OO M △中，1111sin 2O M O OM OO ∠== 1O OM ∠为锐角，130O OM ∴∠= ······················· 5分1cos3022OM OO ∴==⨯=，在Rt MOF △中，3cos30322OF OM ===． 1sin 3032MF OM ===． ∴点M 坐标为32⎛ ⎝⎭············································································ 6分图14设切线OM 的函数解析式为(0)y kx k =≠32k =，k ∴= ···· 7分 ∴切线OM的函数解析式为3y x =·························································· 8分 （3）存在． ···························································································· 9分①过点A 作1AP x ⊥轴，与OM 交于点1P ．可得11Rt Rt APO MO O △∽△（两角对应相等两三角形相似）113tan tan 303P A OA AOP =∠==113P ⎛∴ ⎝⎭，······································ 10分 ②过点A 作2AP OM ⊥，垂足为2P ，过2P 点作2P H OA ⊥，垂足为H ． 可得21Rt Rt AP O O MO △∽△（两角对应相等两三角开相似） 在2Rt OP A △中，1OA =，23cos302OP OA ∴==在2Rt OPH △中，223cos 4OHOP AOP =∠==， 2221sin 224P H OP AOP =∠=⨯=2344P ⎛⎫∴ ⎪ ⎪⎝⎭， ·································11分∴符合条件的P 点坐标有1⎛⎝⎭，34⎛ ⎝⎭··············································· 12分87.（08青海省卷28题）王亮同学善于改进学习方法，他发现对解题过程进行回顾反思，效果会更好．某一天他利用30分钟时间进行自主学习．假设他用于解题的时间x （单位：分钟）与学习收益量y 的关系如图甲所示，用于回顾反思的时间x （单位：分钟）与学习收益量y 的关系如图乙所示（其中OA 是抛物线的一部分，A 为抛物线的顶点），且用于回顾反思的时间不超过用于解题的时间． （1）求王亮解题的学习收益量y 与用于解题的时间x 之间的函数关系式，并写出自变量x 的取值范围； （2）求王亮回顾反思的学习收益量y 与用于回顾反思的时间x 之间的函数关系式； （3）王亮如何分配解题和回顾反思的时间，才能使这30分钟的学习收益总量最大？ （学习收益总量=解题的学习收益量+回顾反思的学习收益量）图甲图乙（08青海省卷28题解析）解：（1）设y kx =，把(24)，代入，得2k =．2y x ∴=． ······················································································· （1分） 自变量x 的取值范围是：030x ≤≤． ··················································· （2分）（2）当05x ≤≤时，设2(5)25y a x =-+， ········································································ （3分）把(00)，代入，得25250a +=，1a =-． 22(5)2510y x x x ∴=--+=-+．························································ （5分）当515x ≤≤时，25y = ····························································································· （6分） 即210(05)25(515)x x x y x ⎧-+=⎨⎩≤≤≤≤．（3）设王亮用于回顾反思的时间为(015)x x ≤≤分钟，学习效益总量为Z ，则他用于解题的时间为(30)x -分钟．当05x ≤≤时，222102(30)860(4)76Z x x x x x x =-++-=-++=--+． ····················· （7分）∴当4x =时，76Z =最大． ································································· （8分）当515x ≤≤时，252(30)285Z x x =+-=-+． ··························································· （9分） Z 随x 的增大而减小，∴当5x =时，75Z =最大．综合所述，当4x =时，76Z =最大，此时3026x -=． ··························· （10分）即王亮用于解题的时间为26分钟，用于回顾反思的时间为4分钟时，学习收益总量最大．······································································································ （11分）。