【九年高考理科数学真题分类训练】专题十五 坐标系与参数方程第四十一讲坐标系与参数方程答案

一线名师凭借教学实践科学分类，高质量的解析，你能感受到名家不一样的解题思路专题十五坐标系与参数方程第四十一讲坐标系与参数方程 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 部分1．【解析】利用，，可得直线的方程为，圆的方程为，所以圆心，半径，由于直线与圆相切，故圆心到直线的距离等于半径，即，∴或，又，∴．2．1【解析】圆的普通方程为，即．设圆心为，所以．3．2【解析】直线的普通方程为，圆的普通方程为，因为圆心到直线的距离，所以有两个交点．4．2【解析】将化为直角坐标方程为，将ρ=2cosθ化为直角坐标方程为，圆心坐标为(1,0)，半径r=1，又(1，0)在直线上，所以|AB|=2r=2．5．【解析】由得，所以，故直线的直角坐标方程为，而点对应的直角坐标为，所以点到直线：的距离为．6．6【解析】圆即，化为直角坐标方程为，直线，则，化为直角坐标方程为，圆心到直线的距离为，所以圆上的点到直线距离的最大值为6．7．【解析】(1)由，得的直角坐标方程为．(2)由(1)知是圆心为，半径为的圆．由题设知，是过点且关于轴对称的两条射线．记轴右边的射线为，轴左边的射线为．由于在圆的外面，故与有且仅有三个公共点等价于与只有一个公共点且与有两个公共点，或与只有一个公共点且与有两个公共点．当与只有一个公共点时，到所在直线的距离为，所以，故或．经检验，当时，与没有公共点；当时，与只有一个公共点，与有两个公共点．当与只有一个公共点时，到所在直线的距离为，所以，故或．经检验，当时，与没有公共点；当时，与没有公共点．综上，所求的方程为．8．【解析】(1)曲线的直角坐标方程为．当时，的直角坐标方程为，当时，的直角坐标方程为．(2)将的参数方程代入的直角坐标方程，整理得关于的方程．①因为曲线截直线所得线段的中点在内，所以①有两个解，设为，，则．又由①得，故，于是直线的斜率．9．【解析】(1)的直角坐标方程为．当时，与交于两点．当时，记，则的方程为．与交于两点当且仅当，解得或，即或．综上，的取值范围是．(2)的参数方程为为参数，EMBEDEquation.DSMT4．设，，对应的参数分别为，，，则，且，满足．于是，．又点的坐标满足所以点的轨迹的参数方程是EMBEDEquation.DSMT4为参数，EMBEDEquation.DSMT4．10．C．【解析】因为曲线的极坐标方程为，所以曲线的圆心为，直径为4的圆．因为直线的极坐标方程为，则直线过，倾斜角为，所以A为直线与圆的一个交点．设另一个交点为B，则∠OAB=．连结OB，因为OA为直径，从而∠OBA=，所以．因此，直线被曲线截得的弦长为．11．【解析】（1）曲线的普通方程为．当时，直线的普通方程为．由解得或．从而与的交点坐标为，．（2）直线的普通方程为，故上的点到的距离为．当时，的最大值为．由题设得，所以；当时，的最大值为．由题设得，所以．综上，或．12．【解析】（1）设的极坐标为EMBEDEquation.DSMT4，的极坐标为EMBEDEquation.DSMT4．由椭圆知，．由得的极坐标方程EMBEDEquation.DSMT4．因此的直角坐标方程为．（2）设点的极坐标为EMBEDEquation.DSMT4．由题设知，，于是面积．当时，取得最大值．所以面积的最大值为．13．【解析】（1）消去参数得的普通方程；消去参数得的普通方程．设，由题设得，消去得．所以的普通方程为（2）的极坐标方程为EMBEDEquation.DSMT4联立得．故，从而代入得，所以交点的极径为．14．【解析】直线的普通方程为.因为点在曲线上，设，从而点到直线的的距离，当时，.因此当点的坐标为时，曲线上点到直线的距离取到最小值.15．【解析】(1)（均为参数）∴①∴为以为圆心，为半径的圆．方程为∵∴即为的极坐标方程(2)两边同乘得即②：化为普通方程为，由题意：和的公共方程所在直线即为①—②得：，即为∴，∴16．【解析】（Ⅰ）整理圆的方程得，由可知圆的极坐标方程为．(Ⅱ)记直线的斜率为，则直线的方程为，由垂径定理及点到直线距离公式知：，即，整理得，则．17．【解析】（Ⅰ）的普通方程为，的直角坐标方程为.（Ⅱ）由题意，可设点的直角坐标为，因为是直线，所以的最小值，即为到的距离的最小值，．当且仅当时，取得最小值，最小值为，此时的直角坐标为．18．【解析】椭圆的普通方程为，将直线的参数方程，代入，得，即，解得，.所以.19．【解析】（Ⅰ）因为，∴的极坐标方程为，的极坐标方程为．（Ⅱ）将代入，得，解得=，=，|MN|=－=，因为的半径为1，则的面积=．20．【解析】（Ⅰ）曲线的直角坐标方程为，曲线的直角坐标方程为．联立解得或所以与交点的直角坐标为和．（Ⅱ）曲线的极坐标方程为，其中．因此得到极坐标为，的极坐标为．所以EMBEDEquation.DSMT4，当时，取得最大值，最大值为．21．【解析】以极坐标系的极点为平面直角坐标系的原点，以极轴为轴的正半轴，建立直角坐标系．圆的极坐标方程为，化简，得．则圆的直角坐标方程为，即，所以圆的半径为．22．【解析】（Ⅰ）由，从而有．（Ⅱ）设，则，故当=0时，||取最小值，此时点的直角坐标为.23．【解析】……5分（Ⅱ）24．【解析】（I）C的普通方程为．可得C的参数方程为（t为参数，）（Ⅱ）设D.由（I）知C是以G（1,0）为圆心，1为半径的上半圆。因为C在点D处的切线与t垂直，所以直线GD与t的斜率相同，．故D的直角坐标为，即．25．【解析】将消去参数，化为普通方程，即：，将代入得，，∴的极坐标方程为；（Ⅱ）的普通方程为，由解得或，∴与的交点的极坐标分别为()，．26．【解析】（Ⅰ）由题意有因此的轨迹的参数方程为（）（Ⅱ）点到坐标原点的距离（）当时，，故的轨迹过坐标原点．27．【解析】（1）点的极坐标为点的直角坐标为（2）设；则28．【解析】（I）设，则由条件知M().由于M点在上，所以，即．从而的参数方程为（为参数），（Ⅱ）曲线的极坐标方程为，曲线的极坐标方程为．射线与的交点的极径为，射线与的交点的极径为．所以．高考真题专项分类（理科数学）第11页—共11页_1234568017.unknown_1234568145.unknown_1234568209.unknown_1234568241.unknown_1234568257.unknown_1234568273.unknown_1234568281.unknown_1234568289.unknown_1234568293.unknown_1234568297.unknown_1234568301.unknown_1234568303.unknown_1234568304.unknown_1234568305.unknown_1234568302.unknown_1234568299.unknown_1234568300.unknown_1234568298.unknown_1234568295.unknown_1234568296.unknown_1234568294.unknown_1234568291.unknown_1234568292.unknown_1234568290.unknown_1234568285.unknown_1234568287.unknown_1234568288.unknown_1234568286.unknown_1234568283.unknown_1234568284.unknown_1234568282.unknown_1234568277.unknown_1234568279.unknown_1234568280.unknown_1234568278.unknown_1234568275.unknown_1234568276.unknown_1234568274.unknown_1234568265.unknown_1234568269.unknown_1234568271.unknown_1234568272.unknown_1234568270.unknown_1234568267.unknown_1234568268.unknown_1234568266.unknown_1234568261.unknown_1234568263.unknown_1234568264.unknown_1234568262.unknown_1234568259.unknown_1234568260.unknown_1234568258.unknown_1234568249.unknown_1234568253.unknown_1234568255.unknown_1234568256.unknown_1234568254.unknown_1234568251.unknown_1234568252.unknown_1234568250.unknown_1234568245.unknown_1234568247.unknown_1234568248.unknown_1234568246.unknown_1234568243.unknown_1234568244.unknown_1234568242.unknown_1234568225.unknown_1234568233.unknown_1234568237.unknown_1234568239.unknown_1234568240.unknown_1234568238.unknown_1234568235.unknown_1234568236.unknown_1234568234.unknown_1234568229.unknown_1234568231.unknown_1234568232.unknown_1234568230.unknown_1234568227.unknown_1234568228.unknown_1234568226.unknown_1234568217.unknown_1234568221.unknown_1234568223.unknown_1234568224.unknown_1234568222.unknown_1234568219.unknown_1234568220.unknown_1234568218.unknown_1234568213.unknown_1234568215.unknown_1234568216.unknown_1234568214.unknown_1234568211.unknown_1234568212.unknown_1234568210.unknown_1234568177.unknown_1234568193.unknown_1234568201.unknown_1234568205.unknown_1234568207.unknown_1234568208.unknown_1234568206.unknown_1234568203.unknown_1234568204.unknown_1234568202.unknown_1234568197.unknown_1234568199.unknown_1234568200.unknown_1234568198.unknown_1234568195.unknown_1234568196.unknown_1234568194.unknown_1234568185.unknown_1234568189.unknown_1234568191.unknown_1234568192.unknown_1234568190.unknown_1234568187.unknown_1234568188.unknown_1234568186.unknown_1234568181.unknown_1234568183.unknown_1234568184.unknown_1234568182.unknown_1234568179.unknown_1234568180.unknown_1234568178.unknown_1234568161.unknown_1234568169.unknown_1234568173.unknown_12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