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第05講橢圓(精講)目錄第一部分:知識點精準(zhǔn)記憶第二部分:課前自我評估測試第三部分: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:橢圓的簡單幾何性質(zhì)角度1:橢圓的長軸、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:知識點精準(zhǔn)記憶第一部分:知識點精準(zhǔn)記憶知識點一:橢圓的定義平面內(nèi)一個動點SKIPIF1<0到兩個定點SKIPIF1<0、SKIPIF1<0的距離之和等于常數(shù)SKIPIF1<0,這個動點SKIPIF1<0的軌跡叫橢圓.這兩個定點(SKIPIF1<0,SKIPIF1<0)叫橢圓的焦點,兩焦點的距離(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言表述集合SKIPIF1<0.知識點二:橢圓的標(biāo)準(zhǔn)方程和幾何性質(zhì)1、橢圓的標(biāo)準(zhǔn)方程焦點位置焦點在SKIPIF1<0軸上焦點在SKIPIF1<0軸上標(biāo)準(zhǔn)方程SKIPIF1<0(SKIPIF1<0)SKIPIF1<0(SKIPIF1<0)圖象焦點坐標(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SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0軸長短軸長=SKIPIF1<0,長軸長=SKIPIF1<0焦點SKIPIF1<0SKIPIF1<0焦距SKIPIF1<0對稱性對稱軸:SKIPIF1<0軸、SKIPIF1<0軸對稱中心:原點離心率SKIPIF1<0,SKIPIF1<0知識點三:常用結(jié)論1、與橢圓SKIPIF1<0SKIPIF1<0共焦點的橢圓方程可設(shè)為:SKIPIF1<0SKIPIF1<02、有相同離心率:SKIPIF1<0(SKIPIF1<0,焦點在SKIPIF1<0軸上)或SKIPIF1<0(SKIPIF1<0,焦點在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=SKIPIF1<0第二部分:課前自我評估測試第二部分:課前自我評估測試1.(2022·江蘇·高二)P是橢圓SKIPIF1<0上一點,SKIPIF1<0,SKIPIF1<0是該橢圓的兩個焦點,且SKIPIF1<0,則SKIPIF1<0(

)A.1 B.3 C.5 D.9【答案】A解:對橢圓方程SKIPIF1<0變形得SKIPIF1<0,易知橢圓長半軸的長為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的左右焦點分別為F1、F2,過左焦點F1,作直線交橢圓C于A、B兩點,則三角形ABF2的周長為(

)A.10 B.15 C.20 D.25【答案】C由題意橢圓的長軸為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)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:因為橢圓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的一個焦點坐標(biāo)為SKIPIF1<0,則SKIPIF1<0__________.【答案】SKIPIF1<0由焦點坐標(biāo)SKIPIF1<0知焦點在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)已知橢圓SKIPIF1<0的左焦點為F,若A?B是橢圓上兩動點,且SKIPIF1<0垂直于x軸,則SKIPIF1<0周長的最大值為___________.【答案】12如圖.設(shè)SKIPIF1<0與x軸相交于點C,橢圓SKIPIF1<0右焦點為SKIPIF1<0,連接SKIPIF1<0,所以SKIPIF1<0周長為SKIPIF1<0故SKIPIF1<0的周長的最大值為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·高二課時練習(xí))SKIPIF1<0中,SKIPIF1<0為動點,SKIPIF1<0,SKIPIF1<0且滿足SKIPIF1<0,則SKIPIF1<0點的軌跡方程為______.【答案】SKIPIF1<0.根據(jù)正弦定理,由SKIPIF1<0,所以點A點的軌跡是以SKIPIF1<0,SKIPIF1<0為焦點的橢圓,不包括兩點SKIPIF1<0,由SKIPIF1<0,所以A點的軌跡方程為SKIPIF1<0,故答案為:SKIPIF1<0.例題2.(2022·全國·高二專題練習(xí))方程SKIPIF1<0化簡的結(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,長軸長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·深州長江中學(xué)高二期末)已知SKIPIF1<0,SKIPIF1<0是圓SKIPIF1<0上一動點,線段SKIPIF1<0的垂直平分線交SKIPIF1<0于SKIPIF1<0,則動點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,因為SKIPIF1<0是線段SKIPIF1<0的垂直平分線上的點,所以SKIPIF1<0,則SKIPIF1<0,根據(jù)橢圓的定義可知,點SKIPIF1<0的軌跡為以SKIPIF1<0為焦點的橢圓,其中SKIPIF1<0,SKIPIF1<0,則有SKIPIF1<0,故點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,動圓SKIPIF1<0設(shè)圓心SKIPIF1<0,半徑為r,動圓M在圓SKIPIF1<0內(nèi)部,且動圓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,則動點M滿足橢圓定義,SKIPIF1<0,焦點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,動圓C滿足與SKIPIF1<0外切且與SKIPIF1<0內(nèi)切,則動圓圓心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,因為動圓SKIPIF1<0同時與圓SKIPIF1<0外切和圓SKIPIF1<0內(nèi)切,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以點SKIPIF1<0的軌跡是以SKIPIF1<0,SKIPIF1<0為焦點,SKIPIF1<0的橢圓,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以動圓的圓心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,動圓SKIPIF1<0同時與圓SKIPIF1<0外切和圓SKIPIF1<0內(nèi)切,則動圓的圓心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,因為動圓SKIPIF1<0同時與圓SKIPIF1<0外切和圓SKIPIF1<0內(nèi)切,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以點SKIPIF1<0的軌跡是以SKIPIF1<0,SKIPIF1<0為焦點,SKIPIF1<0的橢圓,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以動圓的圓心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SKIPIF1<0是圓SKIPIF1<0:SKIPIF1<0上動點,SKIPIF1<0.若線段SKIPIF1<0的中垂線交SKIPIF1<0于點SKIPIF1<0,則點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,因為線段SKIPIF1<0的中垂線交SKIPIF1<0于點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為焦點的橢圓,且SKIPIF1<0,即SKIPIF1<0,可得SKIPIF1<0,所以點SKIPIF1<0的軌跡方程為SKIPIF1<0.故答案為:SKIPIF1<0.4.(2022·全國·高二課時練習(xí))已知三角形ABC的周長是8,頂點B,C的坐標(biāo)分別為SKIPIF1<0,(1,0),則頂點A的軌跡方程為________.【答案】SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,即點SKIPIF1<0是以頂點SKIPIF1<0為焦點的橢圓,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,所以橢圓方程SKIPIF1<0,因為三點SKIPIF1<0不能共線,所以SKIPIF1<0,則頂點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的兩個焦點,SKIPIF1<0是橢圓上一點,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,又因為,SKIPIF1<0,所以SKIPIF1<0,所以,SKIPIF1<0故選:B例題2.(2022·全國·高二專題練習(xí))已知橢圓SKIPIF1<0的左、右焦點為SKIPIF1<0,SKIPIF1<0,點SKIPIF1<0為橢圓上動點,則SKIPIF1<0的值是______;SKIPIF1<0的取值范圍是______.【答案】

SKIPIF1<0

SKIPIF1<0對橢圓SKIPIF1<0,其SKIPIF1<0,焦點坐標(biāo)分別為SKIPIF1<0,由橢圓定義可得:SKIPIF1<0SKIPIF1<0;設(shè)點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,SKIPIF1<0是橢圓的右焦點,則SKIPIF1<0的周長的最小值為______.【答案】10解:橢圓的方程為SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,連接SKIPIF1<0,SKIPIF1<0,則由橢圓的中心對稱性可得SKIPIF1<0的周長SKIPIF1<0,當(dāng)AB位于短軸的端點時,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、右焦點分別為SKIPIF1<0,點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SKIPIF1<0,又由橢圓的定義,可得SKIPIF1<0,因為SKIPIF1<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,右焦點分別為SKIPIF1<0,SKIPIF1<0,點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)雙曲線的對稱性只需考慮SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時,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,由雙曲線的定義可知,SKIPIF1<0,由勾股定理可得SKIPIF1<0.因為SKIPIF1<0,所以SKIPIF1<0,此時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)已知SKIPIF1<0分別為橢圓SKIPIF1<0的左、右焦點,直線SKIPIF1<0與橢圓交于P,Q兩點,則SKIPIF1<0的周長為______.【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SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0;故答案為:SKIPIF1<04(2022·全國·高三專題練習(xí))已知SKIPIF1<0分別為橢圓SKIPIF1<0的左右焦點,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,則△SKIPIF1<0的周長為___.【答案】20由橢圓方程知:SKIPIF1<0,而SKIPIF1<0SKIPIF1<0,又△ABF2的周長是SKIPIF1<0SKIPIF1<0.故答案為:20.7.(2022·全國·高二專題練習(xí))已知點SKIPIF1<0在焦點為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,因為SKIPIF1<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,SKIPIF1<0為橢圓SKIPIF1<0上任意一點,點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因為橢圓SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,則橢圓的右焦點為SKIPIF1<0,由橢圓的定義得:SKIPIF1<0,當(dāng)點P在點SKIPIF1<0處,取等號,所以SKIPIF1<0的最大值為5,故選:B.例題2.(2022·全國·高三專題練習(xí)(文))已知點SKIPIF1<0,且SKIPIF1<0是橢圓SKIPIF1<0的左焦點,SKIPIF1<0是橢圓上任意一點,則SKIPIF1<0的最小值是(

)A.6 B.5 C.4 D.3【答案】DSKIPIF1<0,設(shè)橢圓的右焦點為SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0在SKIPIF1<0的正上方時,等號成立.故選:D例題3.(2022·全國·高二專題練習(xí))設(shè)SKIPIF1<0是橢圓SKIPIF1<0上一點,SKIPIF1<0、SKIPIF1<0是橢圓的兩個焦點,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,等號成立,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·高二課時練習(xí))已知SKIPIF1<0是橢圓SKIPIF1<0的左焦點,P是此橢圓上的動點,SKIPIF1<0是一定點,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點運動到SKIPIF1<0的延長線和橢圓交點SKIPIF1<0時,SKIPIF1<0取得最大,此時SKIPIF1<0,所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為:SKIPIF1<02.(2022·全國·高二專題練習(xí))已知點SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0內(nèi)的兩個點,M是橢圓上的動點,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,設(shè)左焦點為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·高二專題練習(xí))已知定點SKIPIF1<0、SKIPIF1<0和動點SKIPIF1<0.(1)再從條件①、條件②這兩個條件中選擇一個作為已知,求:動點SKIPIF1<0的軌跡及其方程.條件①:SKIPIF1<0條件②:SKIPIF1<0(2)SKIPIF1<0,求:動點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,因為SKIPIF1<0,故點SKIPIF1<0的軌跡是以SKIPIF1<0為焦點的橢圓,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SKIPIF1<0的軌跡是橢圓,其方程為SKIPIF1<0;選擇條件②:SKIPIF1<0,因為SKIPIF1<0,故點SKIPIF1<0的軌跡是線段SKIPIF1<0,其方程為SKIPIF1<0.(2)因為SKIPIF1<0,當(dāng)SKIPIF1<0時,此時動點SKIPIF1<0不存在,沒有軌跡和方程;當(dāng)SKIPIF1<0時,此時SKIPIF1<0,由(1)可知,此時動點SKIPIF1<0的軌跡是線段SKIPIF1<0,其方程為SKIPIF1<0;當(dāng)SKIPIF1<0時,此時SKIPIF1<0,此時點SKIPIF1<0的軌跡是以SKIPIF1<0為焦點的橢圓,其方程為SKIPIF1<0.綜上所述:當(dāng)SKIPIF1<0時,動點SKIPIF1<0沒有軌跡和方程;當(dāng)SKIPIF1<0時,動點SKIPIF1<0的軌跡是線段SKIPIF1<0,其方程為SKIPIF1<0;當(dāng)SKIPIF1<0時,動點SKIPIF1<0的軌跡是以SKIPIF1<0為焦點的橢圓,其方程為SKIPIF1<0.例題2.(2022·四川省資中縣球溪高級中學(xué)高二階段練習(xí)(文))(1)求焦點在SKIPIF1<0軸上,長軸長為6,焦距為4的橢圓標(biāo)準(zhǔn)方程;(2)求離心率SKIPIF1<0,焦點在SKIPIF1<0軸,且經(jīng)過點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中學(xué)高二階段練習(xí)(理))(1)求焦點在SKIPIF1<0軸上,長軸長為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若雙曲線焦點在SKIPIF1<0軸上,設(shè)其方程為SKIPIF1<0,同理SKIPIF1<0,此時將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SKIPIF1<0,且與橢圓SKIPIF1<0有共同的焦點;(3)經(jīng)過SKIPIF1<0,SKIPIF1<0兩點.【答案】(1)答案見解析(2)SKIPIF1<0(3)SKIPIF1<0(1)解:當(dāng)SKIPIF1<0,SKIPIF1<0時,SKIPIF1<0,若焦點在SKIPIF1<0軸上,則標(biāo)準(zhǔn)方程為SKIPIF1<0;若焦點在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,故它的焦點為SKIPIF1<0.設(shè)所求橢圓的方程為SKIPIF1<0,SKIPIF1<0,把點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,設(shè)所求橢圓的方程為SKIPIF1<0,把點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)長軸長為4,短軸長為2,焦點在y軸上;(2)經(jīng)過點SKIPIF1<0,SKIPIF1<0;(3)一個焦點為SKIPIF1<0,一個頂點為SKIPIF1<0;(4)一個焦點為SKIPIF1<0,長軸長為4;(5)一個焦點為SKIPIF1<0,離心率為SKIPIF1<0;(6)一個焦點到長軸的兩個端點的距離分別為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:橢圓的簡單幾何性質(zhì)角度1:橢圓的長軸、短軸、焦距典型例題例題1.(2022·全國·高二課時練習(xí))橢圓SKIPIF1<0的長軸長為______.【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,長軸的長為SKIPIF1<0=SKIPIF1<0;故答案為:SKIPIF1<0.例題2.(2022·全國·高二課時練習(xí))若橢圓SKIPIF1<0與橢圓SKIPIF1<0焦點相同,則實數(shù)SKIPIF1<0___________.【答案】SKIPIF1<0由SKIPIF1<0得:SKIPIF1<0,則SKIPIF1<0且焦點在SKIPIF1<0軸上由SKIPIF1<0得:SKIPIF1<0,SKIPIF1<0與SKIPIF1<0共焦點,SKIPIF1<0;SKIPIF1<0,解得:SKIPIF1<0.故答案為:SKIPIF1<0.例題3.(2022·河南·新蔡縣第一高級中學(xué)高二開學(xué)考試(文))已知橢圓SKIPIF1<0,點SKIPIF1<0,SKIPIF1<0為橢圓上一動點,則SKIPIF1<0的最大值為____.【答案】SKIPIF1<0設(shè)點SKIPIF1<0,則SKIPIF1<0,可得SKIPIF1<0,其中SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時,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·高三專題練習(xí))已知橢圓的長軸長為SKIPIF1<0,短軸長為SKIPIF1<0,則橢圓上任意一點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)橢圓的焦點在SKIPIF1<0軸上,則該橢圓的標(biāo)準(zhǔn)方程為SKIPIF1<0,設(shè)點SKIPIF1<0,則SKIPIF1<0,且有SKIPIF1<0,所以,SKIPIF1<0.故選:A.2.(2022·全國·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,則曲線C為橢圓,又SKIPIF1<0,所以SKIPIF1<0;當(dāng)SKIPIF1<0時,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·高二課時練習(xí))求下列橢圓的長軸長、短軸長、離心率、頂點坐標(biāo)和焦點坐標(biāo):(1)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0;(4)SKIPIF1<0.【答案】(1)長軸長為SKIPIF1<0,短軸長為SKIPIF1<0,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0;(2)長軸長為SKIPIF1<0,短軸長為SKIPIF1<0,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0;(3)長軸長為10,短軸長為8,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0;(4)長軸長為8,短軸長為4,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0.(1)由橢圓方程知,SKIPIF1<0,所以橢圓的長軸長為SKIPIF1<0,短軸長為SKIPIF1<0,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0;(2)由橢圓方程知,SKIPIF1<0,所以橢圓的長軸長為SKIPIF1<0,短軸長為SKIPIF1<0,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0;(3)橢圓方程可變形為SKIPIF1<0,所以SKIPIF1<0,所以橢圓的長軸長為10,短軸長為8,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(biāo)為SKIPIF1<0;(4)橢圓方程可變形為SKIPIF1<0,所以SKIPIF1<0,所以橢圓的長軸長為8,短軸長為4,離心率為SKIPIF1<0,頂點坐標(biāo)為SKIPIF1<0,焦點坐標(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:因為橢圓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的兩個焦點,SKIPIF1<0為SKIPIF1<0上一點,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,因為SKIPIF1<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·高二專題練習(xí))橢圓SKIPIF1<0的兩焦點為SKIPIF1<0,若橢圓SKIPIF1<0上存在點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,SKIPIF1<0為等腰直角三角形,則點SKIPIF1<0位于橢圓的上下頂點,則滿足:SKIPIF1<0,當(dāng)SKIPIF1<0或者SKIPIF1<0時,此時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,點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,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的兩個焦點,SKIPIF1<0為橢圓SKIPIF1<0上一點,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則橢圓SKIPIF1<0的離心率SKIPIF1<0_________.【答案】SKIPIF1<0或SKIPIF1<0因為SKIPIF1<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

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