2023高考一轮复习课时作业52 椭圆（有答案）.doc

第页共页高考一轮复习课时作业椭圆一选择题共小题椭圆11990923119910241的焦点坐标为1010202001010202设定点119865103119865203动点119875满足条件1198751198651119875119865211988691198861198860则点119875的轨迹是椭圆线段不存在椭圆或线段已知椭圆的方程为119910216119909291则椭圆的长轴长为482710椭圆119909225119910291的焦点11986511198652119875为椭圆上一点已知11987511986511198751198652则11986511198751198652的面积为912108二填空题共小题巳知119860120119861是圆11986511990912211991024119865为圆心上一动点线段119860119861的垂直平分线交119861119865于119875则动点119875的轨迹方程为双曲线119909211991021的渐近线方程为椭圆1199092119898119910241的焦距等于2则119898的值为若椭圆1199092119898119910241的焦点在119909轴上焦距为2则实数119898的值为焦点的坐标分别是0202并且经过点3252的椭圆的方程为设119860119861是椭圆119864的长轴点119862在119864上且1198621198611198604若11986011986141198611198622则119864的两个焦点之间的距离为已知椭圆1199092251199102161上一点119875到椭圆一个焦点的距离是3则119875点到另一个焦点的距离为已知椭圆11986211199092119886121199102119887121119886111988710和椭圆11986221199092119886221199102119887221119886211988720的焦点相同且11988611198862给出如下四个结论椭圆1198621和椭圆1198622一定没有公共点11988612119886221198871211988722119886111988621198871119887211988611198862其中所有正确的结论是填序号若一个椭圆长轴的长度短轴的长度和焦距成等差数列则该椭圆的离心率是第页共页已知点1198721011987310若直线119897的图象上存在点119875使得1198751198721198751198734成立则说直线119897是119879型直线给出下列直线11198971199102021198972119909503119897119909211991040411989731199091199103051198972119898111990911991010常数119898119825其中代表119879型直线的序号是要求写出所有119879型直线的序号三解答题共小题如图已知椭圆11986211990924119910231的左焦点为1198651点119875是椭圆119862上位于第一象限的点119872119873是119910轴上的两个动点点119872位于119909轴上方满足119875119872119875119873且11986511198721198651119873线段119875119873交119909轴于点119876若119865111987552求点119875的坐标若四边形119865119872119875119873为矩形求点119872的坐标求证119875119876119876119873为定值已知椭圆1198661199092411991021过点1198980作圆119909211991021的切线119897交椭圆119866于119860119861两点求椭圆119866的焦点坐标将119860119861表示为119898的函数并求119860119861的最大值已知椭圆119862119909211988621199102119887211198861198870的两个焦点和短轴的两个端点都在圆119909211991021上求椭圆119862的方程若斜率为119896的直线经过点11987220且与椭圆119862相交于119860119861两点试探讨119896为何值时119874119860119874119861第页共页答案解析因为椭圆方程为119910216119909291所以119886216所以1198864长轴为2119886所以2119886811990924119910231119910119909解析由双曲线11990921198862119910211988721的渐近线方程为119910119887119886119909则双曲线119909211991021的渐近线方程为1199101199095或35119910210119909261436解析设椭圆标准方程为11990921198862119910211988721由题意知211988641198862因为11986211986111986041198611198622所以点11986211因为点119862在椭圆上所以124111988721所以119887243所以11988821198862119887244383所以119888236则椭圆119864的两个焦点之间的距离为4367解析由已知条件可得11988612119887121198862211988722所以11988612119886221198871211988722第页共页而11988611198862可知两椭圆无公共点则正确因为119886121198871211988821198862211988722119888211988611198862所以1198882119886121119887121198861211988712119886121198872211988622即11988611198862不正确因为119886111988710119886211988720所以11988611198862119887111988720而又由11988611198862119886111988621198871119887211988711198872可得11988611198862则正确综上正确的结论序号为35345解析由椭圆的定义可知点119875的轨迹是以119872119873为焦点的椭圆其中2119886421198882所以11988621198881所以1198872119886211988823所以椭圆的方程为11990924119910231对于1由方程组1199102011990924119910231得1199092413不成立所以方程组无解所以直线11989711991020不是119879型直线对于2由方程组21199095011990924119910231得11991023916不成立所以方程组无解所以直线119897211990950不是119879型直线对于3由方程组11990921199104011990924119910231得411991021211991090由1205491224490所以方程组有解所以直线119897119909211991040是119879型直线对于4由方程组31199091199103011990924119910231得1311990922411990980由12054924241381600所以方程组有解所以直线119897311990911991030是119879型直线对于5因为1198972119898111990911991010常数119898119825过定点01且点01在椭圆11990924119910231的内部第页共页所以直线1198972119898111990911991010与椭圆有交点所以直线1198972119898111990911991010常数119898119825是119879型直线设119875119909111991011199091011991010由题意119865110所以11986511198751199091121199101252又1199091241199101231所以11990911119910132所以点119875坐标为132连接1198651119875交119872119873于点119877则119877为1198651119875中点且119877为119872119873中点所以119875132119877034设119872011989811989801198730119899则11989811989932又119865111987211986511198731119898111989911198981198990所以1198982故点119872的坐标是02由2知119865111987211986511198731119898111989911198981198990所以1198991119898由题意1198721198751198731198751199091119910111989811990911199101119899119909121199101211989811989911991011198981198990又1199091241199101231所以1199101231198981119898119910190所以11991013119898或11991013119898舍去所以1198751198761198761198731199101119910119873311989811198983为定值由已知椭圆1198661199092411991021得11988621198871所以1198883所以椭圆119866的焦点坐标为3030由题意椭圆1198661199092411991021过点1198980作圆119909211991021的切线119897交椭圆119866于119860119861两点知1198981当1198981时切线119897的方程为1199091点119860119861的坐标分别为132132此时1198601198613当1198981时同理可得1198601198613当1198981时设切线方程为119910119896119909119898由1199101198961199091198981199092411991021得14119896211990928119896211989811990941198962119898240第页共页设119860119861两点的坐标分别为1199091119910111990921199102则1199091119909281198962119898141198962119909111990924119896211989824141198962又由119897与圆119909211991021相切得119896119898111989621即1198982119896211989621所以11986011986111198962811989621198981411989622441198962119898241411989624311989811989823由于当1198981时1198601198613所以119860119861431198981198982311989811因为11986011986143119898119898234311989831198982当且仅当1198983时1198601198612所以119860119861的最大值为2依题意椭圆的两个焦点和短轴的两个端点都在圆119909211991021上可得11988711198881所以11988622所以椭圆119862的方程为1199092211991021设1198601199091119910111986111990921199102直线119860119861的方程为1199101198961199092由11991011989611990921199092211991021消去119910得1211989621199092811989621199098119896220所以119909111990928119896212119896211990911199092811989622121198962因为119874119860119874119861所以11991011199102119909111990921即11990911199092119910111991020而1199101119910211989621199091211990922所以11990911199092119896211990912119909220所以11198962811989622121198962161198964121198962411989620解得119896215此时1205490所以11989655