全国中考数学压轴题精选-解析几何详细解析VIP免费

全国中考数学压轴题精选-解析几何71.（中考江苏镇江28题）（本小题满分8分）探索研究如图，在直角坐标系xOy中，点P为函数214yx在第一象限内的图象上的任一点，点A的坐标为(01)，，直线l过(01)B，且与x轴平行，过P作y轴的平行线分别交x轴，l于CQ，，连结AQ交x轴于H，直线PH交y轴于R．（1）求证：H点为线段AQ的中点；（2）求证：①四边形APQR为平行四边形；②平行四边形APQR为菱形；（3）除P点外，直线PH与抛物线214yx有无其它公共点？并说明理由．（中考江苏镇江28题解析）（1）法一：由题可知1AOCQ．90AOHQCHoQ，AHOQHC，AOHQCH△≌△．·············································································（1分）OHCH，即H为AQ的中点．····························································（2分）法二：(01)AQ，，(01)B，，OAOB．··················································（1分）又BQx∥轴，HAHQ．····································································（2分）（2）①由（1）可知AHQH，AHRQHP，ARPQQ∥，RAHPQH，RAHPQH△≌△．·············································································（3分）ARPQ，又ARPQ∥，四边形APQR为平行四边形．·············································（4分）②设214Pmm，，PQyQ∥轴，则(1)Qm，，则2114PQm．过P作PGy轴，垂足为G，在RtAPG△中，xlQCPAOBHRy22222222111111444APAGPGmmmmPQ．平行四边形APQR为菱形．····································································（6分）（3）设直线PR为ykxb，由OHCH，得22mH，，214Pmm，代入得：2021.4mkbkmbm，221.4mkbm，直线PR为2124myxm．·····················（7分）设直线PR与抛物线的公共点为214xx，，代入直线PR关系式得：22110424mxxm，21()04xm，解得xm．得公共点为214mm，．所以直线PH与抛物线214yx只有一个公共点P．·······································（8分）72（中考黑龙江齐齐哈尔28题）（本小题满分10分）如图，在平面直角坐标系中，点(30)C，，点AB，分别在x轴，y轴的正半轴上，且满足2310OBOA．（1）求点A，点B的坐标．（2）若点P从C点出发，以每秒1个单位的速度沿射线CB运动，连结AP．设ABP△的面积为S，点P的运动时间为t秒，求S与t的函数关系式，并写出自变量的取值范围．（3）在（2）的条件下，是否存在点P，使以点ABP，，为顶点的三角形与AOB△相似？若存在，请直接写出点P的坐标；若不存在，请说明理由．yxAOCB（中考黑龙江齐齐哈尔28题解析）解：（1）2310OBOAQ230OB，10OA·······································································（1分）3OB，1OAQ点A，点B分别在x轴，y轴的正半轴上(10)(03)AB，，，·················································································（2分）（2）求得90ABCo············································································（3分）23(023)23(23)ttStt≤（每个解析式各1分，两个取值范围共1分）················································（6分...