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1、苏、锡、常、镇四市一模数学第18 题说题稿1/4 苏、锡、常、镇四市一模数学第18 题说题稿题目：18（本小题满分16 分）如图，在平面直角坐标系xOy 中，已知A，B，C是椭圆22221(0)xyabab上不同的三点，3 2(3 2,)2A，(3,3)B，C在第三象限，线段BC的中点在直线OA上（1）求椭圆的标准方程；（2）求点 C 的坐标；（3）设动点P在椭圆上（异于点A，B，C）且直线 PB，PC 分别交直线OA 于M，N两点，证明OMON 为定值并求出该定值一说命题立意：掌握直线方程，理解椭圆方程，感受运用方程研究曲线几何性质的思想方法,要求考生一定的计算能力。二说知识考点：直线方程、
2、椭圆方程。三 说如何分析讲解：1.说计算（1）由已知，得222291821,991,abab解得2227,27.2ab，所以椭圆的标准方程为22127272xy法 2：设)0(122nmnymx，代如两点的坐标即得。（2）设点(,)C m n(0,0)mn，则BC中点为33(,)22mn由已知，求得直线OA的方程为20 xy，从而23mn又点C在椭圆上，22227mn由，解得3n（舍），1n，从而5m 所以点C的坐标为(5,1)NMPCBA yxO（第 18精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 1 页，共 4 页苏、锡、常、镇四市一模数学第18 题说题稿2/4 法
3、2：22227mn可化为0)9(2)9(22nm，即)3)(3(2)3)(3(nnmm把232123mn代入上式后约去3m得：)3(3nm，于是得一个二元一次方程组。2说思想：用代数解决几何，故方法为把几何中的“动”用代数上00,yx的“变”表示，代如运算即可。（3）设00(,)P xy，11(2,)Myy，22(2,)Nyy,P B M 三点共线，011033233yyyx，整理，得001003()23yxyxy,P C N 三点共线，022011255yyyx，整理，得00200523yxyxy点C在椭圆上，2200227xy，2200272xy从而22200000001222200000
4、003(56)3(3627)393449241822xyx yyx yy yxyx yyx y所以124552OMONy y OMON 为定值，定值为452法 2：分析：原题叙述为由动点P带出动直线PCPBll,，再出两动点NM,，但也可叙述为动直线PBl带出动点MP,，在由动点P到动直线PCl，最后出现动点N，从而方法可为把几何中的“动”用代数上PBl的斜率k的“变”表示 当PBl的斜率不存在时，3:xlPB，由椭圆对称性可得)3,3(P，由xylAO21:可得)23,3(M从而)5(21:xylPC，于是)3,6(N故 O MO N=)23,3()3,6(=452精品w o r d 学习资
5、料 可编辑资料-精心整理-欢迎下载-第 2 页，共 4 页文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 H
6、I4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6
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8、P10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N
9、2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文
10、档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK
11、2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9苏、锡、常、镇四市一模数学第18 题说题稿3/4 当PBl的斜率存在时，设)3(3:xkylPB，结合xylA
12、O21:得),1266(MykkM把PBl与椭圆方程联立，得：093618)66(2)12(2222kkxkkxk故129361822kkkxxPB且3Bx，得：)12366,123123(2222kkkkkkP由CP,两点坐标可得)5(4621341:22xkkkkylPC，结合xylAO21:得：),1336(22NykkkNOMONNMNMNMxxyyxx45133612664522kkkkk245)1)(12()1)(12)(1(2452kkkkk几点解释：1两种方法都是同一个思想，用代数的“变”限定几何的“动”。2由法 2 最后一步的分母可以看出，法2 还要单独讨论21k和1k，其实
13、这与题目中的“异于点A，B，C”有关，也是一种几何和代数的对应，讲解时可提问学生，再次渗透解析几何思想。四说指导学生作答1对（1）（2）要有耐心拿全分，对（3）要有信心多拿分。2考试中对（3）可用特殊法先得到答案。五说拓展用代数的“变”限定几何的“动”其实就是解析几何的本质思想，本题除了“动中有定”外，是否可以考虑“动中有值”，如PMN的最值问题，这样代数中的“变”就成了函数中的自变量了。精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 3 页，共 4 页文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI
14、4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S
15、6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP
16、10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2
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18、编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2
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20、5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9苏、锡、常、镇四市一模数学第18 题说题稿4/4 精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 4 页，共 4 页文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N
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22、档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK2F7I10G5H4 HI4D3C6S6V6 ZP10C5N2D2G9文档编码:CK
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