新高考數(shù)學(xué)一輪復(fù)習(xí)第8章 第05講 橢圓 精講（教師版）

第05講橢圓(精講）目錄第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第二部分：課前自我評(píng)估測(cè)試第三部分：典型例題剖析題型一：橢圓定義的應(yīng)用角度1：利用橢圓定義求軌跡方程角度2：利用橢圓定義解決焦點(diǎn)三角形問題角度3：利用橢圓定義求最值題型二：橢圓的標(biāo)準(zhǔn)方程題型三：橢圓的簡(jiǎn)單幾何性質(zhì)角度1：橢圓的長(zhǎng)軸、短軸、焦距角度2：求橢圓的離心率角度3：與橢圓幾何性質(zhì)有關(guān)的最值(范圍)問題第四部分：高考真題感悟第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶知識(shí)點(diǎn)一：橢圓的定義平面內(nèi)一個(gè)動(dòng)點(diǎn)SKIPIF1<0到兩個(gè)定點(diǎn)SKIPIF1<0、SKIPIF1<0的距離之和等于常數(shù)SKIPIF1<0，這個(gè)動(dòng)點(diǎn)SKIPIF1<0的軌跡叫橢圓.這兩個(gè)定點(diǎn)（SKIPIF1<0,SKIPIF1<0）叫橢圓的焦點(diǎn)，兩焦點(diǎn)的距離（SKIPIF1<0）叫作橢圓的焦距.說明：若SKIPIF1<0，SKIPIF1<0的軌跡為線段SKIPIF1<0；若SKIPIF1<0，SKIPIF1<0的軌跡無圖形定義的集合語(yǔ)言表述集合SKIPIF1<0.知識(shí)點(diǎn)二：橢圓的標(biāo)準(zhǔn)方程和幾何性質(zhì)1、橢圓的標(biāo)準(zhǔn)方程焦點(diǎn)位置焦點(diǎn)在SKIPIF1<0軸上焦點(diǎn)在SKIPIF1<0軸上標(biāo)準(zhǔn)方程SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）圖象焦點(diǎn)坐標(biāo)SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0的關(guān)系SKIPIF1<0范圍SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0頂點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0軸長(zhǎng)短軸長(zhǎng)＝SKIPIF1<0，長(zhǎng)軸長(zhǎng)＝SKIPIF1<0焦點(diǎn)SKIPIF1<0SKIPIF1<0焦距SKIPIF1<0對(duì)稱性對(duì)稱軸：SKIPIF1<0軸、SKIPIF1<0軸對(duì)稱中心：原點(diǎn)離心率SKIPIF1<0，SKIPIF1<0知識(shí)點(diǎn)三：常用結(jié)論1、與橢圓SKIPIF1<0SKIPIF1<0共焦點(diǎn)的橢圓方程可設(shè)為：SKIPIF1<0SKIPIF1<02、有相同離心率：SKIPIF1<0（SKIPIF1<0，焦點(diǎn)在SKIPIF1<0軸上）或SKIPIF1<0（SKIPIF1<0，焦點(diǎn)在SKIPIF1<0軸上）3、橢圓SKIPIF1<0的圖象中線段的幾何特征（如下圖）：（1）SKIPIF1<0；（2）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；（3）SKIPIF1<0,SKIPIF1<0，SKIPIF1<0；（4）橢圓通經(jīng)長(zhǎng)=SKIPIF1<0第二部分：課前自我評(píng)估測(cè)試第二部分：課前自我評(píng)估測(cè)試1．（2022·江蘇·高二）P是橢圓SKIPIF1<0上一點(diǎn)，SKIPIF1<0，SKIPIF1<0是該橢圓的兩個(gè)焦點(diǎn)，且SKIPIF1<0，則SKIPIF1<0（

）A．1 B．3 C．5 D．9【答案】A解：對(duì)橢圓方程SKIPIF1<0變形得SKIPIF1<0，易知橢圓長(zhǎng)半軸的長(zhǎng)為4，由橢圓的定義可得SKIPIF1<0,又SKIPIF1<0，故SKIPIF1<0.故選：A.2．（2022·湖南·周南中學(xué)高二期末）已知橢圓C：SKIPIF1<0的左右焦點(diǎn)分別為F1、F2，過左焦點(diǎn)F1，作直線交橢圓C于A、B兩點(diǎn)，則三角形ABF2的周長(zhǎng)為（

）A．10 B．15 C．20 D．25【答案】C由題意橢圓的長(zhǎng)軸為SKIPIF1<0，由橢圓定義知SKIPIF1<0∴SKIPIF1<0故選：C3．（2022·浙江紹興·模擬預(yù)測(cè)）已知橢圓SKIPIF1<0，則該橢圓的離心率SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：因?yàn)闄E圓SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，又SKIPIF1<0，故SKIPIF1<0.故選：C.4．（2022·上海靜安·二模）已知橢圓SKIPIF1<0SKIPIF1<0的一個(gè)焦點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0__________.【答案】SKIPIF1<0由焦點(diǎn)坐標(biāo)SKIPIF1<0知焦點(diǎn)在SKIPIF1<0軸上，且SKIPIF1<0，解得SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<0.5．（2022·上海黃浦·模擬預(yù)測(cè)）已知橢圓SKIPIF1<0的左焦點(diǎn)為F，若A?B是橢圓上兩動(dòng)點(diǎn)，且SKIPIF1<0垂直于x軸，則SKIPIF1<0周長(zhǎng)的最大值為___________.【答案】12如圖.設(shè)SKIPIF1<0與x軸相交于點(diǎn)C，橢圓SKIPIF1<0右焦點(diǎn)為SKIPIF1<0，連接SKIPIF1<0，所以SKIPIF1<0周長(zhǎng)為SKIPIF1<0故SKIPIF1<0的周長(zhǎng)的最大值為12，故答案為：12.第三部分：典型例題剖析第三部分：典型例題剖析題型一：橢圓定義的應(yīng)用角度1：利用橢圓定義求軌跡方程典型例題例題1．（2022·全國(guó)·高二課時(shí)練習(xí)）SKIPIF1<0中，SKIPIF1<0為動(dòng)點(diǎn)，SKIPIF1<0，SKIPIF1<0且滿足SKIPIF1<0，則SKIPIF1<0點(diǎn)的軌跡方程為______．【答案】SKIPIF1<0.根據(jù)正弦定理，由SKIPIF1<0，所以點(diǎn)A點(diǎn)的軌跡是以SKIPIF1<0，SKIPIF1<0為焦點(diǎn)的橢圓，不包括兩點(diǎn)SKIPIF1<0，由SKIPIF1<0，所以A點(diǎn)的軌跡方程為SKIPIF1<0，故答案為：SKIPIF1<0.例題2．（2022·全國(guó)·高二專題練習(xí)）方程SKIPIF1<0化簡(jiǎn)的結(jié)果是___________．【答案】SKIPIF1<0解：∵SKIPIF1<0，故令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，∴方程表示的曲線是以SKIPIF1<0，SKIPIF1<0為焦點(diǎn)，長(zhǎng)軸長(zhǎng)SKIPIF1<0的橢圓，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴方程為SKIPIF1<0.故答案為：SKIPIF1<0.例題3．（2022·河北·深州長(zhǎng)江中學(xué)高二期末）已知SKIPIF1<0，SKIPIF1<0是圓SKIPIF1<0上一動(dòng)點(diǎn)，線段SKIPIF1<0的垂直平分線交SKIPIF1<0于SKIPIF1<0，則動(dòng)點(diǎn)SKIPIF1<0的軌跡方程為______________．【答案】SKIPIF1<0如圖所示，圓SKIPIF1<0的圓心坐標(biāo)為SKIPIF1<0，半徑SKIPIF1<0，因?yàn)镾KIPIF1<0是線段SKIPIF1<0的垂直平分線上的點(diǎn)，所以SKIPIF1<0，則SKIPIF1<0，根據(jù)橢圓的定義可知，點(diǎn)SKIPIF1<0的軌跡為以SKIPIF1<0為焦點(diǎn)的橢圓，其中SKIPIF1<0，SKIPIF1<0，則有SKIPIF1<0，故點(diǎn)P的軌跡方程為SKIPIF1<0．故答案為：SKIPIF1<0．例題4．（2022·山西·懷仁市第一中學(xué)校高二期中（文））已知兩圓SKIPIF1<0，動(dòng)圓SKIPIF1<0在圓SKIPIF1<0內(nèi)部且和圓SKIPIF1<0相內(nèi)切．和圓SKIPIF1<0相外切，則動(dòng)圓圓心SKIPIF1<0的軌跡方程為_________．【答案】SKIPIF1<0圓SKIPIF1<0，圓心SKIPIF1<0，圓SKIPIF1<0，圓心SKIPIF1<0，動(dòng)圓SKIPIF1<0設(shè)圓心SKIPIF1<0，半徑為r，動(dòng)圓M在圓SKIPIF1<0內(nèi)部，且動(dòng)圓M與圓SKIPIF1<0相內(nèi)切，與圓SKIPIF1<0相外切，所以SKIPIF1<0，①＋②可得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，則動(dòng)點(diǎn)M滿足橢圓定義，SKIPIF1<0，焦點(diǎn)SKIPIF1<0，SKIPIF1<0所以橢圓方程為SKIPIF1<0．故答案為：SKIPIF1<0同類題型歸類練1．（2022·廣東·廣州市第六十五中學(xué)高二期中）已知定圓SKIPIF1<0，動(dòng)圓C滿足與SKIPIF1<0外切且與SKIPIF1<0內(nèi)切，則動(dòng)圓圓心C的軌跡方程為__________.【答案】SKIPIF1<0由圓SKIPIF1<0：SKIPIF1<0可得圓心SKIPIF1<0，半徑SKIPIF1<0，由圓SKIPIF1<0：SKIPIF1<0可得圓心SKIPIF1<0，半徑SKIPIF1<0，設(shè)圓SKIPIF1<0的半徑為SKIPIF1<0，因?yàn)閯?dòng)圓SKIPIF1<0同時(shí)與圓SKIPIF1<0外切和圓SKIPIF1<0內(nèi)切，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0，SKIPIF1<0為焦點(diǎn)，SKIPIF1<0的橢圓，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以動(dòng)圓的圓心SKIPIF1<0的軌跡方程為：SKIPIF1<0，故答案為：SKIPIF1<0.2．（2022·安徽·六安一中高二期中）已知圓SKIPIF1<0：SKIPIF1<0和圓SKIPIF1<0：SKIPIF1<0，動(dòng)圓SKIPIF1<0同時(shí)與圓SKIPIF1<0外切和圓SKIPIF1<0內(nèi)切，則動(dòng)圓的圓心SKIPIF1<0的軌跡方程為________．【答案】SKIPIF1<0由圓SKIPIF1<0：SKIPIF1<0可得圓心SKIPIF1<0，半徑SKIPIF1<0，由圓SKIPIF1<0：SKIPIF1<0可得圓心SKIPIF1<0，半徑SKIPIF1<0，設(shè)圓SKIPIF1<0的半徑為SKIPIF1<0，因?yàn)閯?dòng)圓SKIPIF1<0同時(shí)與圓SKIPIF1<0外切和圓SKIPIF1<0內(nèi)切，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0，SKIPIF1<0為焦點(diǎn)，SKIPIF1<0的橢圓，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以動(dòng)圓的圓心SKIPIF1<0的軌跡方程為：SKIPIF1<0，故答案為：SKIPIF1<0.3．（2022·浙江·金華市江南中學(xué)高二期中）已知點(diǎn)SKIPIF1<0是圓SKIPIF1<0：SKIPIF1<0上動(dòng)點(diǎn)，SKIPIF1<0.若線段SKIPIF1<0的中垂線交SKIPIF1<0于點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0的軌跡方程為____________.【答案】SKIPIF1<0如圖所示，圓SKIPIF1<0：SKIPIF1<0，可得圓心SKIPIF1<0，半徑SKIPIF1<0，因?yàn)榫€段SKIPIF1<0的中垂線交SKIPIF1<0于點(diǎn)SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，根據(jù)橢圓的定義，可得N是以SKIPIF1<0,SKIPIF1<0為焦點(diǎn)的橢圓，且SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，所以點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0.故答案為：SKIPIF1<0.4．（2022·全國(guó)·高二課時(shí)練習(xí)）已知三角形ABC的周長(zhǎng)是8，頂點(diǎn)B，C的坐標(biāo)分別為SKIPIF1<0，（1，0），則頂點(diǎn)A的軌跡方程為________．【答案】SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即點(diǎn)SKIPIF1<0是以頂點(diǎn)SKIPIF1<0為焦點(diǎn)的橢圓，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以橢圓方程SKIPIF1<0，因?yàn)槿c(diǎn)SKIPIF1<0不能共線，所以SKIPIF1<0，則頂點(diǎn)A的軌跡方程為SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<0SKIPIF1<0角度2：利用橢圓定義解決焦點(diǎn)三角形問題典型例題例題1．（2022·安徽省亳州市第一中學(xué)高二期末）設(shè)SKIPIF1<0是橢圓SKIPIF1<0的兩個(gè)焦點(diǎn)，SKIPIF1<0是橢圓上一點(diǎn)，且SKIPIF1<0.則SKIPIF1<0的面積為（

）A．6 B．SKIPIF1<0 C．8 D．SKIPIF1<0【答案】B解：由橢圓SKIPIF1<0的方程可得SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，由余弦定理可得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，而SKIPIF1<0，所以，SKIPIF1<0，又因?yàn)?，SKIPIF1<0，所以SKIPIF1<0，所以，SKIPIF1<0故選：B例題2．（2022·全國(guó)·高二專題練習(xí)）已知橢圓SKIPIF1<0的左、右焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0為橢圓上動(dòng)點(diǎn)，則SKIPIF1<0的值是______；SKIPIF1<0的取值范圍是______．【答案】

SKIPIF1<0

SKIPIF1<0對(duì)橢圓SKIPIF1<0，其SKIPIF1<0，焦點(diǎn)坐標(biāo)分別為SKIPIF1<0，由橢圓定義可得：SKIPIF1<0SKIPIF1<0；設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，故SKIPIF1<0SKIPIF1<0，又SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0的取值范圍為：SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0.例題3．（2022·青海青?！じ叨谀ㄎ模┮阎獧E圓的方程為SKIPIF1<0，過橢圓中心的直線交橢圓于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，SKIPIF1<0是橢圓的右焦點(diǎn)，則SKIPIF1<0的周長(zhǎng)的最小值為______．【答案】10解：橢圓的方程為SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，則由橢圓的中心對(duì)稱性可得SKIPIF1<0的周長(zhǎng)SKIPIF1<0，當(dāng)AB位于短軸的端點(diǎn)時(shí)，SKIPIF1<0取最小值，最小值為SKIPIF1<0，SKIPIF1<0．故答案為：10同類題型歸類練1．（2022·江蘇·高二）已知橢圓SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，點(diǎn)SKIPIF1<0在橢圓上，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：由題意，橢圓方程SKIPIF1<0，可得SKIPIF1<0，所以焦點(diǎn)SKIPIF1<0，又由橢圓的定義，可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，在SKIPIF1<0中，由余弦定理可得SKIPIF1<0,所以SKIPIF1<0，解得SKIPIF1<0，又由SKIPIF1<0，所以SKIPIF1<0.故選:C．2．(多選）（2022·廣東·仲元中學(xué)高二期中）雙曲線SKIPIF1<0的左，右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)P在C上.若SKIPIF1<0是直角三角形，則SKIPIF1<0的面積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．4 D．2【答案】AC解：由雙曲線SKIPIF1<0可得SKIPIF1<0.根據(jù)雙曲線的對(duì)稱性只需考慮SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，將SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0的面積為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，由雙曲線的定義可知，SKIPIF1<0，由勾股定理可得SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，此時(shí)SKIPIF1<0的面積為SKIPIF1<0綜上所述，SKIPIF1<0的面積為4或SKIPIF1<0.故選：SKIPIF1<0.3．（2022·重慶八中模擬預(yù)測(cè)）已知SKIPIF1<0分別為橢圓SKIPIF1<0的左、右焦點(diǎn)，直線SKIPIF1<0與橢圓交于P，Q兩點(diǎn)，則SKIPIF1<0的周長(zhǎng)為______．【答案】SKIPIF1<0解：橢圓SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0、SKIPIF1<0，直線SKIPIF1<0過左焦點(diǎn)SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0；故答案為：SKIPIF1<04（2022·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0分別為橢圓SKIPIF1<0的左右焦點(diǎn)，傾斜角為SKIPIF1<0的直線SKIPIF1<0經(jīng)過SKIPIF1<0，且與橢圓交于SKIPIF1<0兩點(diǎn)，則△SKIPIF1<0的周長(zhǎng)為___．【答案】20由橢圓方程知：SKIPIF1<0，而SKIPIF1<0SKIPIF1<0，又△ABF2的周長(zhǎng)是SKIPIF1<0SKIPIF1<0.故答案為：20．7．（2022·全國(guó)·高二專題練習(xí)）已知點(diǎn)SKIPIF1<0在焦點(diǎn)為SKIPIF1<0、SKIPIF1<0的橢圓SKIPIF1<0上，若SKIPIF1<0，則SKIPIF1<0的值為______．【答案】SKIPIF1<0在橢圓SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由橢圓的定義可得SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0.故答案為：SKIPIF1<0.角度3：利用橢圓定義求最值典型例題例題1．（2022·四川遂寧·高二期末（理））已知SKIPIF1<0是橢圓SKIPIF1<0的左焦點(diǎn)，SKIPIF1<0為橢圓SKIPIF1<0上任意一點(diǎn)，點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．3 B．5 C．SKIPIF1<0 D．13【答案】B因?yàn)闄E圓SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則橢圓的右焦點(diǎn)為SKIPIF1<0，由橢圓的定義得：SKIPIF1<0，當(dāng)點(diǎn)P在點(diǎn)SKIPIF1<0處，取等號(hào)，所以SKIPIF1<0的最大值為5，故選：B.例題2．（2022·全國(guó)·高三專題練習(xí)（文））已知點(diǎn)SKIPIF1<0，且SKIPIF1<0是橢圓SKIPIF1<0的左焦點(diǎn)，SKIPIF1<0是橢圓上任意一點(diǎn)，則SKIPIF1<0的最小值是（

）A．6 B．5 C．4 D．3【答案】DSKIPIF1<0，設(shè)橢圓的右焦點(diǎn)為SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0的正上方時(shí)，等號(hào)成立.故選：D例題3．（2022·全國(guó)·高二專題練習(xí)）設(shè)SKIPIF1<0是橢圓SKIPIF1<0上一點(diǎn)，SKIPIF1<0、SKIPIF1<0是橢圓的兩個(gè)焦點(diǎn)，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A在橢圓SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由橢圓定義可得SKIPIF1<0，SKIPIF1<0，由余弦定理可得SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故選：A.同類題型歸類練1．（2022·全國(guó)·高二課時(shí)練習(xí)）已知SKIPIF1<0是橢圓SKIPIF1<0的左焦點(diǎn)，P是此橢圓上的動(dòng)點(diǎn)，SKIPIF1<0是一定點(diǎn)，則SKIPIF1<0的最大值為______．【答案】SKIPIF1<0##SKIPIF1<0根據(jù)題意橢圓方程為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，如圖，根據(jù)橢圓定義可得：SKIPIF1<0，當(dāng)SKIPIF1<0點(diǎn)運(yùn)動(dòng)到SKIPIF1<0的延長(zhǎng)線和橢圓交點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0取得最大，此時(shí)SKIPIF1<0，所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<02．（2022·全國(guó)·高二專題練習(xí)）已知點(diǎn)SKIPIF1<0，SKIPIF1<0是橢圓SKIPIF1<0內(nèi)的兩個(gè)點(diǎn)，M是橢圓上的動(dòng)點(diǎn)，則SKIPIF1<0的最大值為______．【答案】SKIPIF1<0##SKIPIF1<0依題意，橢圓方程為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0是橢圓的右焦點(diǎn)，設(shè)左焦點(diǎn)為SKIPIF1<0，根據(jù)橢圓的定義可知SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<0題型二：橢圓的標(biāo)準(zhǔn)方程典型例題例題1．（2022·全國(guó)·高二專題練習(xí)）已知定點(diǎn)SKIPIF1<0、SKIPIF1<0和動(dòng)點(diǎn)SKIPIF1<0．(1)再?gòu)臈l件①、條件②這兩個(gè)條件中選擇一個(gè)作為已知，求：動(dòng)點(diǎn)SKIPIF1<0的軌跡及其方程．條件①：SKIPIF1<0條件②：SKIPIF1<0(2)SKIPIF1<0，求：動(dòng)點(diǎn)SKIPIF1<0的軌跡及其方程．【答案】(1)答案見解析；(2)答案見解析.(1)選擇條件①：SKIPIF1<0，因?yàn)镾KIPIF1<0，故點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0為焦點(diǎn)的橢圓，設(shè)其方程為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故其方程為：SKIPIF1<0.即選擇條件①，點(diǎn)SKIPIF1<0的軌跡是橢圓，其方程為SKIPIF1<0；選擇條件②：SKIPIF1<0，因?yàn)镾KIPIF1<0，故點(diǎn)SKIPIF1<0的軌跡是線段SKIPIF1<0，其方程為SKIPIF1<0.(2)因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，此時(shí)動(dòng)點(diǎn)SKIPIF1<0不存在，沒有軌跡和方程；當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，由（1）可知，此時(shí)動(dòng)點(diǎn)SKIPIF1<0的軌跡是線段SKIPIF1<0，其方程為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0為焦點(diǎn)的橢圓，其方程為SKIPIF1<0.綜上所述：當(dāng)SKIPIF1<0時(shí)，動(dòng)點(diǎn)SKIPIF1<0沒有軌跡和方程；當(dāng)SKIPIF1<0時(shí)，動(dòng)點(diǎn)SKIPIF1<0的軌跡是線段SKIPIF1<0，其方程為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，動(dòng)點(diǎn)SKIPIF1<0的軌跡是以SKIPIF1<0為焦點(diǎn)的橢圓，其方程為SKIPIF1<0.例題2．（2022·四川省資中縣球溪高級(jí)中學(xué)高二階段練習(xí)（文））（1）求焦點(diǎn)在SKIPIF1<0軸上，長(zhǎng)軸長(zhǎng)為6，焦距為4的橢圓標(biāo)準(zhǔn)方程；（2）求離心率SKIPIF1<0，焦點(diǎn)在SKIPIF1<0軸，且經(jīng)過點(diǎn)SKIPIF1<0的雙曲線標(biāo)準(zhǔn)方程.【答案】（1）SKIPIF1<0；（2）SKIPIF1<0.（1）設(shè)橢圓的標(biāo)準(zhǔn)方程為SKIPIF1<0.由題意知：SKIPIF1<0；SKIPIF1<0.SKIPIF1<0.所以橢圓的標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）設(shè)雙曲線的標(biāo)準(zhǔn)方程為SKIPIF1<0.則SKIPIF1<0所以雙曲線的標(biāo)準(zhǔn)方程為SKIPIF1<0.例題3．（2022·四川省資中縣球溪高級(jí)中學(xué)高二階段練習(xí)（理））（1）求焦點(diǎn)在SKIPIF1<0軸上，長(zhǎng)軸長(zhǎng)為6，焦距為4的橢圓標(biāo)準(zhǔn)方程；（2）求離心率SKIPIF1<0，經(jīng)過點(diǎn)SKIPIF1<0的雙曲線標(biāo)準(zhǔn)方程．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0（1）由題意得SKIPIF1<0，故SKIPIF1<0，橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0（2）①若雙曲線焦點(diǎn)在x軸上，設(shè)其方程為SKIPIF1<0，由題意SKIPIF1<0，而SKIPIF1<0故SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，故雙曲線標(biāo)準(zhǔn)方程為SKIPIF1<0②若雙曲線焦點(diǎn)在SKIPIF1<0軸上，設(shè)其方程為SKIPIF1<0，同理SKIPIF1<0，此時(shí)將SKIPIF1<0代入后方程無解綜上，雙曲線標(biāo)準(zhǔn)方程為SKIPIF1<0同類題型歸類練1．（2022·江蘇·高二）求適合下列條件的橢圓的標(biāo)準(zhǔn)方程：(1)SKIPIF1<0，SKIPIF1<0；(2)經(jīng)過點(diǎn)SKIPIF1<0，且與橢圓SKIPIF1<0有共同的焦點(diǎn)；(3)經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn)．【答案】(1)答案見解析(2)SKIPIF1<0(3)SKIPIF1<0(1)解：當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，若焦點(diǎn)在SKIPIF1<0軸上，則標(biāo)準(zhǔn)方程為SKIPIF1<0；若焦點(diǎn)在SKIPIF1<0軸上，則標(biāo)準(zhǔn)方程為SKIPIF1<0．(2)解：橢圓SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故它的焦點(diǎn)為SKIPIF1<0．設(shè)所求橢圓的方程為SKIPIF1<0，SKIPIF1<0，把點(diǎn)SKIPIF1<0代入，SKIPIF1<0,求得SKIPIF1<0，或SKIPIF1<0（舍去），故要求的橢圓的方程為SKIPIF1<0．(3)解：SKIPIF1<0橢圓經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn)，設(shè)所求橢圓的方程為SKIPIF1<0，把點(diǎn)SKIPIF1<0、SKIPIF1<0代入得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0所求橢圓的方程為SKIPIF1<0．2．（2022·江蘇·高二）求適合下列條件的橢圓的標(biāo)準(zhǔn)方程：(1)長(zhǎng)軸長(zhǎng)為4，短軸長(zhǎng)為2，焦點(diǎn)在y軸上；(2)經(jīng)過點(diǎn)SKIPIF1<0，SKIPIF1<0；(3)一個(gè)焦點(diǎn)為SKIPIF1<0，一個(gè)頂點(diǎn)為SKIPIF1<0；(4)一個(gè)焦點(diǎn)為SKIPIF1<0，長(zhǎng)軸長(zhǎng)為4；(5)一個(gè)焦點(diǎn)為SKIPIF1<0，離心率為SKIPIF1<0；(6)一個(gè)焦點(diǎn)到長(zhǎng)軸的兩個(gè)端點(diǎn)的距離分別為6，2.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；(4)SKIPIF1<0；(5)SKIPIF1<0；(6)SKIPIF1<0或SKIPIF1<0.(1)由題設(shè)，SKIPIF1<0，又焦點(diǎn)在y軸上，故橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0；(2)設(shè)橢圓方程為SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0在橢圓上，所以SKIPIF1<0，即SKIPIF1<0，故橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0.(3)由題設(shè)，SKIPIF1<0，則SKIPIF1<0，又焦點(diǎn)為SKIPIF1<0所以橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0.(4)由題設(shè)，SKIPIF1<0，則SKIPIF1<0，又焦點(diǎn)為SKIPIF1<0所以橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0.(5)由題設(shè)，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又焦點(diǎn)為SKIPIF1<0所以橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0.(6)由題設(shè)，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，所以橢圓標(biāo)準(zhǔn)方程為SKIPIF1<0或SKIPIF1<0.題型三：橢圓的簡(jiǎn)單幾何性質(zhì)角度1：橢圓的長(zhǎng)軸、短軸、焦距典型例題例題1．（2022·全國(guó)·高二課時(shí)練習(xí)）橢圓SKIPIF1<0的長(zhǎng)軸長(zhǎng)為______．【答案】SKIPIF1<0依題意SKIPIF1<0是橢圓方程，即SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，長(zhǎng)軸的長(zhǎng)為SKIPIF1<0=SKIPIF1<0；故答案為：SKIPIF1<0.例題2．（2022·全國(guó)·高二課時(shí)練習(xí)）若橢圓SKIPIF1<0與橢圓SKIPIF1<0焦點(diǎn)相同，則實(shí)數(shù)SKIPIF1<0___________.【答案】SKIPIF1<0由SKIPIF1<0得：SKIPIF1<0，則SKIPIF1<0且焦點(diǎn)在SKIPIF1<0軸上由SKIPIF1<0得：SKIPIF1<0，SKIPIF1<0與SKIPIF1<0共焦點(diǎn)，SKIPIF1<0；SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<0.例題3．（2022·河南·新蔡縣第一高級(jí)中學(xué)高二開學(xué)考試（文））已知橢圓SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0為橢圓上一動(dòng)點(diǎn)，則SKIPIF1<0的最大值為____.【答案】SKIPIF1<0設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值SKIPIF1<0.故答案為：SKIPIF1<0.同類題型歸類練1．（2022·全國(guó)·高三專題練習(xí)）已知橢圓的長(zhǎng)軸長(zhǎng)為SKIPIF1<0，短軸長(zhǎng)為SKIPIF1<0，則橢圓上任意一點(diǎn)SKIPIF1<0到橢圓中心SKIPIF1<0的距離的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A不妨設(shè)橢圓的焦點(diǎn)在SKIPIF1<0軸上，則該橢圓的標(biāo)準(zhǔn)方程為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，且有SKIPIF1<0，所以，SKIPIF1<0.故選：A.2．（2022·全國(guó)·高三專題練習(xí)（理））已知曲線SKIPIF1<0的焦距為8，則SKIPIF1<0___________.【答案】25或SKIPIF1<0解：由題意知半焦距SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則曲線C為橢圓，又SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，曲線C為雙曲線，所以SKIPIF1<0，所以SKIPIF1<0.故a的值為25或SKIPIF1<0.故答案為：25或SKIPIF1<03．（2022·江蘇·高二課時(shí)練習(xí)）求下列橢圓的長(zhǎng)軸長(zhǎng)、短軸長(zhǎng)、離心率、頂點(diǎn)坐標(biāo)和焦點(diǎn)坐標(biāo)：(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；(4)SKIPIF1<0．【答案】(1)長(zhǎng)軸長(zhǎng)為SKIPIF1<0，短軸長(zhǎng)為SKIPIF1<0，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；(2)長(zhǎng)軸長(zhǎng)為SKIPIF1<0，短軸長(zhǎng)為SKIPIF1<0，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；(3)長(zhǎng)軸長(zhǎng)為10，短軸長(zhǎng)為8，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；(4)長(zhǎng)軸長(zhǎng)為8，短軸長(zhǎng)為4，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0.(1)由橢圓方程知，SKIPIF1<0，所以橢圓的長(zhǎng)軸長(zhǎng)為SKIPIF1<0，短軸長(zhǎng)為SKIPIF1<0，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；(2)由橢圓方程知，SKIPIF1<0，所以橢圓的長(zhǎng)軸長(zhǎng)為SKIPIF1<0，短軸長(zhǎng)為SKIPIF1<0，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；(3)橢圓方程可變形為SKIPIF1<0，所以SKIPIF1<0，所以橢圓的長(zhǎng)軸長(zhǎng)為10，短軸長(zhǎng)為8，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0；(4)橢圓方程可變形為SKIPIF1<0，所以SKIPIF1<0，所以橢圓的長(zhǎng)軸長(zhǎng)為8，短軸長(zhǎng)為4，離心率為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0.角度2：求橢圓的離心率典型例題例題1．（2022·貴州黔西·高二期末（理））已知橢圓SKIPIF1<0的離心率為SKIPIF1<0，則橢圓SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：因?yàn)闄E圓SKIPIF1<0的離心率為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，則橢圓SKIPIF1<0的離心率SKIPIF1<0.故選：C.例題2．（2022·江西上饒·高二期末（理））已知SKIPIF1<0是橢圓SKIPIF1<0的兩個(gè)焦點(diǎn)，SKIPIF1<0為SKIPIF1<0上一點(diǎn)，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C在橢圓SKIPIF1<0中，由橢圓的定義可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，由余弦定理得SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0所以SKIPIF1<0的離心率SKIPIF1<0.故選：C例題3．（2022·全國(guó)·高二專題練習(xí)）橢圓SKIPIF1<0的兩焦點(diǎn)為SKIPIF1<0，若橢圓SKIPIF1<0上存在點(diǎn)SKIPIF1<0使SKIPIF1<0為等腰直角三角形，則橢圓SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為等腰直角三角形，則點(diǎn)SKIPIF1<0位于橢圓的上下頂點(diǎn)，則滿足：SKIPIF1<0,當(dāng)SKIPIF1<0或者SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，SKIPIF1<0為等腰直角三角形，則滿足SKIPIF1<0,故SKIPIF1<0，SKIPIF1<0故選：C例題4．（2022·重慶一中高一期末）已知SKIPIF1<0，SKIPIF1<0為橢圓SKIPIF1<0的左，右焦點(diǎn)，點(diǎn)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0為等腰三角形，且頂角為SKIPIF1<0，則SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D解：依題意設(shè)橢圓方程為SKIPIF1<0，①若SKIPIF1<0為等腰三角形SKIPIF1<0的頂角，則SKIPIF1<0在橢圓的上（下）頂點(diǎn)，如下圖所示：則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0；②若SKIPIF1<0（或SKIPIF1<0）為等腰三角形SKIPIF1<0的頂角，不妨取SKIPIF1<0為頂角，如下圖所示：即SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，由余弦定理SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去）綜上可得SKIPIF1<0或SKIPIF1<0.故選：D．例題5．（2022·廣東汕尾·高二期末）設(shè)SKIPIF1<0，SKIPIF1<0為橢圓SKIPIF1<0的兩個(gè)焦點(diǎn)，SKIPIF1<0為橢圓SKIPIF1<0上一點(diǎn)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則橢圓SKIPIF1<0的離心率SKIPIF1<0_________.【答案】SKIPIF1<0或SKIPIF1<0因?yàn)镾KIPIF1<0，且SKIPIF1<0，故SKIPIF1<0為銳角，所以SKIPIF1<0，由余弦定理SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0故答案為：SKIPIF1<0或SKIPIF1<0同