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專(zhuān)題16妙解離心率問(wèn)題目錄01頂角為直角的焦點(diǎn)三角形求解離心率的取值范圍問(wèn)題 202焦點(diǎn)三角形頂角范圍與離心率 603共焦點(diǎn)的橢圓與雙曲線問(wèn)題 804橢圓與雙曲線的4a通徑體 1105橢圓與雙曲線的4a直角體 1406橢圓與雙曲線的等腰三角形問(wèn)題 1907雙曲線的4a底邊等腰三角形 2108焦點(diǎn)到漸近線距離為b 2509焦點(diǎn)到漸近線垂線構(gòu)造的直角三角形 2910以兩焦點(diǎn)為直徑的圓與漸近線相交問(wèn)題 3211漸近線平行線與面積問(wèn)題 3612數(shù)形結(jié)合轉(zhuǎn)化長(zhǎng)度角度 3801頂角為直角的焦點(diǎn)三角形求解離心率的取值范圍問(wèn)題1.(2024·安徽宣城·高三統(tǒng)考期末)已知橢圓SKIPIF1<0上一點(diǎn)SKIPIF1<0關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn)為點(diǎn)SKIPIF1<0,SKIPIF1<0為其右焦點(diǎn),若SKIPIF1<0,設(shè)SKIPIF1<0,且SKIPIF1<0,則該橢圓的離心率SKIPIF1<0的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】由題意設(shè)橢圓的左焦點(diǎn)為N,連接AN,BN,因?yàn)锳F⊥BF,所以四邊形AFBN為長(zhǎng)方形,再根據(jù)橢圓的定義化簡(jiǎn)得SKIPIF1<0,得到離心率關(guān)于SKIPIF1<0的函數(shù)表達(dá)式,再利用輔助角公式和三角函數(shù)的單調(diào)性求得離心率的范圍.由題意橢圓SKIPIF1<0SKIPIF1<0上一點(diǎn)A關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn)為點(diǎn)B,F(xiàn)為其右焦點(diǎn),設(shè)左焦點(diǎn)為N,連接AN,BN,因?yàn)锳F⊥BF,所以四邊形AFBN為長(zhǎng)方形.根據(jù)橢圓的定義:SKIPIF1<0,由題∠ABF=α,則∠ANF=α,所以SKIPIF1<0,利用SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,即橢圓離心率SKIPIF1<0的取值范圍是SKIPIF1<0,故選B.2.(2024·河北唐山·高三統(tǒng)考期末)已知橢圓SKIPIF1<0上一點(diǎn)SKIPIF1<0關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn)為點(diǎn)SKIPIF1<0,SKIPIF1<0為其右焦點(diǎn),若SKIPIF1<0,設(shè)SKIPIF1<0,且SKIPIF1<0,則該橢圓的離心率SKIPIF1<0的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】設(shè)橢圓SKIPIF1<0的左焦點(diǎn)為:SKIPIF1<0,根據(jù)SKIPIF1<0,得到四邊形為SKIPIF1<0為矩形,再由SKIPIF1<0,結(jié)合橢圓的定義得到SKIPIF1<0,然后由SKIPIF1<0求解.設(shè)橢圓SKIPIF1<0的左焦點(diǎn)為:SKIPIF1<0,因?yàn)镾KIPIF1<0,所以四邊形為SKIPIF1<0為矩形,所以SKIPIF1<0因?yàn)镾KIPIF1<0,所以SKIPIF1<0由橢圓的定義得:SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故選:B3.(2024·江西南昌·高三南昌十中??计谀┮阎獧E圓SKIPIF1<0上一點(diǎn)SKIPIF1<0關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn)為SKIPIF1<0點(diǎn),SKIPIF1<0為其右焦點(diǎn),若SKIPIF1<0,設(shè)SKIPIF1<0,且SKIPIF1<0,則該橢圓的離心率的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】SKIPIF1<0SKIPIF1<0和SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱(chēng),SKIPIF1<0SKIPIF1<0也在橢圓上,設(shè)左焦點(diǎn)為SKIPIF1<0,根據(jù)橢圓的定義:SKIPIF1<0,又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0

(1)又原點(diǎn)是SKIPIF1<0的斜邊中點(diǎn),SKIPIF1<0SKIPIF1<0,又SKIPIF1<0

(2)SKIPIF1<0

(3)將(2)(3)代入(1)SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,所以橢圓的離心率的取值范圍為SKIPIF1<0,故選:A4.(2024·黑龍江大慶·高三鐵人中學(xué)??计谀┮阎p曲線SKIPIF1<0:SKIPIF1<0(SKIPIF1<0,SKIPIF1<0)右支上非頂點(diǎn)的一點(diǎn)SKIPIF1<0關(guān)于原點(diǎn)SKIPIF1<0的對(duì)稱(chēng)點(diǎn)為SKIPIF1<0,SKIPIF1<0為其右焦點(diǎn),若SKIPIF1<0,設(shè)SKIPIF1<0,且SKIPIF1<0,則雙曲線SKIPIF1<0離心率的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】如圖所示,設(shè)雙曲線的左焦點(diǎn)為SKIPIF1<0,連接SKIPIF1<0,SKIPIF1<0.SKIPIF1<0,SKIPIF1<0四邊形SKIPIF1<0為矩形.所以SKIPIF1<0.則SKIPIF1<0,SKIPIF1<0.SKIPIF1<0.SKIPIF1<0.即SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,故雙曲線離心率的取值范圍是SKIPIF1<0,故選:D.02焦點(diǎn)三角形頂角范圍與離心率5.(2024·河南南陽(yáng)·高三鄭州一中階段練習(xí))已知SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0的左右兩個(gè)焦點(diǎn),P為橢圓上的一點(diǎn),且SKIPIF1<0,則橢圓的離心率的取值范圍為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】設(shè)點(diǎn)SKIPIF1<0,則SKIPIF1<0,由SKIPIF1<0得:SKIPIF1<0,而SKIPIF1<0,即SKIPIF1<0,因此有SKIPIF1<0,即SKIPIF1<0,因SKIPIF1<0,于是得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以橢圓的離心率的取值范圍為SKIPIF1<0.故選:D6.(2024·黑龍江·校聯(lián)考)已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,是雙曲線SKIPIF1<0的兩個(gè)焦點(diǎn),若點(diǎn)Р為橢圓SKIPIF1<0上的動(dòng)點(diǎn),當(dāng)P為橢圓的短軸端點(diǎn)時(shí),SKIPIF1<0取最小值,則橢圓SKIPIF1<0離心率的取值范圍為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】假設(shè)點(diǎn)SKIPIF1<0在SKIPIF1<0軸上方,設(shè)SKIPIF1<0,則SKIPIF1<0,由已知得SKIPIF1<0,SKIPIF1<0,設(shè)直線SKIPIF1<0的傾斜角為SKIPIF1<0,直線SKIPIF1<0的傾斜角為SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0考慮對(duì)勾函數(shù)SKIPIF1<0,由于SKIPIF1<0為橢圓的短軸端點(diǎn)時(shí),SKIPIF1<0,SKIPIF1<0取最小值,即SKIPIF1<0取最小值,SKIPIF1<0也取最小值,此時(shí)SKIPIF1<0,∵函數(shù)在SKIPIF1<0上單調(diào)遞減,∴SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0.即橢圓SKIPIF1<0離心率的取值范圍為SKIPIF1<0.故選:SKIPIF1<0.7.(2024·貴州·高三凱里一中??计谀┮阎獧E圓SKIPIF1<0,SKIPIF1<0,SKIPIF1<0分別為橢圓的左右焦點(diǎn),若橢圓SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0,則橢圓的離心率的取值范圍為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】設(shè)SKIPIF1<0,SKIPIF1<0,若橢圓SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0即SKIPIF1<0,SKIPIF1<0SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0SKIPIF1<0.故選D8.(2024·全國(guó)·高三專(zhuān)題練習(xí))已知橢圓SKIPIF1<0,SKIPIF1<0,SKIPIF1<0分別為橢圓的左右焦點(diǎn),若橢圓C上存在點(diǎn)SKIPIF1<0(SKIPIF1<0)使得SKIPIF1<0,則橢圓的離心率的取值范圍為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】根據(jù)題意作圖如下:由圖可得:當(dāng)點(diǎn)P在橢圓的上(下)頂點(diǎn)處時(shí),SKIPIF1<0最大,要滿足橢圓C上存在點(diǎn)SKIPIF1<0(SKIPIF1<0)使得SKIPIF1<0,則SKIPIF1<0,∴SKIPIF1<0,即:SKIPIF1<0,整理得:SKIPIF1<0,又SKIPIF1<0,∴得到:SKIPIF1<0,∴SKIPIF1<0,∴橢圓離心率的取值范圍為SKIPIF1<0,故選:B.03共焦點(diǎn)的橢圓與雙曲線問(wèn)題9.(2024·安徽·校聯(lián)考)已知中心在原點(diǎn)的橢圓與雙曲線有公共焦點(diǎn),左右焦點(diǎn)分別為SKIPIF1<0、SKIPIF1<0,且兩條曲線在第一象限的交點(diǎn)為SKIPIF1<0,SKIPIF1<0是以SKIPIF1<0為底邊的等腰三角形,若SKIPIF1<0,橢圓與雙曲線的離心率分別為SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0與SKIPIF1<0滿足的關(guān)系是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】由橢圓與雙曲線定義得SKIPIF1<0,所以SKIPIF1<0,選B.10.(多選題)(2024·重慶渝中·高三重慶巴蜀中學(xué)??茧A段練習(xí))已知橢圓SKIPIF1<0:SKIPIF1<0SKIPIF1<0與雙曲線SKIPIF1<0:SKIPIF1<0(SKIPIF1<0,SKIPIF1<0)有公共焦點(diǎn)SKIPIF1<0,SKIPIF1<0,且兩條曲線在第一象限的交點(diǎn)為SKIPIF1<0,若SKIPIF1<0是以SKIPIF1<0為底邊的等腰三角形,SKIPIF1<0,SKIPIF1<0的離心率分別為SKIPIF1<0和SKIPIF1<0,則(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】AD【解析】設(shè)SKIPIF1<0,SKIPIF1<0的焦距為SKIPIF1<0,由SKIPIF1<0,SKIPIF1<0共焦點(diǎn)知SKIPIF1<0,故SKIPIF1<0正確;△SKIPIF1<0是以SKIPIF1<0為底邊的等腰三角形知SKIPIF1<0,由SKIPIF1<0在第一象限知:SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0錯(cuò);由SKIPIF1<0,得SKIPIF1<0,又SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,從而SKIPIF1<0,故SKIPIF1<0正確.故選:SKIPIF1<0.11.(2024·湖北孝感·高三統(tǒng)考期末)已知橢圓和雙曲線有共同的焦點(diǎn)SKIPIF1<0、SKIPIF1<0,SKIPIF1<0是它們的一個(gè)交點(diǎn),且SKIPIF1<0,記橢圓和雙曲線的離心率分別為SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0的最大值為.【答案】SKIPIF1<0/SKIPIF1<0【解析】不妨設(shè)SKIPIF1<0為第一象限的點(diǎn),SKIPIF1<0為左焦點(diǎn),設(shè)橢圓的長(zhǎng)半軸長(zhǎng)為SKIPIF1<0,雙曲線的實(shí)半軸長(zhǎng)為SKIPIF1<0,則根據(jù)橢圓及雙曲線的定義可得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,在△SKIPIF1<0中,SKIPIF1<0,由余弦定理得SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0,即SKIPIF1<0.所以SKIPIF1<0,從而SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,且SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0時(shí)等號(hào)成立.故答案為:SKIPIF1<012.(2024·江蘇蘇州·高三江蘇省蘇州第十中學(xué)校??茧A段練習(xí))已知橢圓和雙曲線有共同的焦點(diǎn)SKIPIF1<0分別是它們?cè)诘谝幌笙藓偷谌笙薜慕稽c(diǎn),且SKIPIF1<0,記橢圓和雙曲線的離心率分別為SKIPIF1<0,則SKIPIF1<0等于.【答案】SKIPIF1<0【解析】設(shè)橢圓長(zhǎng)半軸長(zhǎng)為SKIPIF1<0,雙曲線實(shí)半軸長(zhǎng)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0為兩曲線在第一象限的交點(diǎn),SKIPIF1<0為兩曲線在第三象限的交點(diǎn).由橢圓和雙曲線定義知:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,由橢圓和雙曲線對(duì)稱(chēng)性可知:四邊形SKIPIF1<0為平行四邊形,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0.故答案為:SKIPIF1<0.13.(2024·重慶沙坪壩·高三重慶一中校考期末)已知橢圓和雙曲線有共同的焦點(diǎn)SKIPIF1<0、SKIPIF1<0,SKIPIF1<0是它們的一個(gè)交點(diǎn),SKIPIF1<0,記橢圓和雙曲線的離心率分別為SKIPIF1<0、SKIPIF1<0,則SKIPIF1<0的最小值是.【答案】SKIPIF1<0【解析】不妨設(shè)橢圓與雙曲線的標(biāo)準(zhǔn)方程分別為SKIPIF1<0,SKIPIF1<0,設(shè)兩曲線的焦距為SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,所以,SKIPIF1<0,SKIPIF1<0,化為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),取等號(hào),則SKIPIF1<0的最小值是SKIPIF1<0.故答案為:SKIPIF1<0.04橢圓與雙曲線的4a通徑體14.(2024·河南·高三統(tǒng)考階段練習(xí))已知橢圓SKIPIF1<0的離心率為SKIPIF1<0,左、右焦點(diǎn)分別為SKIPIF1<0、SKIPIF1<0,過(guò)SKIPIF1<0的直線與橢圓SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),若SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】設(shè)SKIPIF1<0,橢圓的焦距為SKIPIF1<0,由題意得出SKIPIF1<0,橢圓的離心率為SKIPIF1<0,SKIPIF1<0.由橢圓的定義可得SKIPIF1<0,由余弦定理得SKIPIF1<0,設(shè)SKIPIF1<0,由橢圓的定義可得SKIPIF1<0,由余弦定理得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0.所以,SKIPIF1<0,SKIPIF1<0,因此,SKIPIF1<0.故選D.15.(2024·全國(guó)·高三校聯(lián)考階段練習(xí))已知橢圓SKIPIF1<0:SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0(如圖),過(guò)SKIPIF1<0的直線交SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn),且SKIPIF1<0軸,SKIPIF1<0,則SKIPIF1<0的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】由SKIPIF1<0,SKIPIF1<0,將SKIPIF1<0代入橢圓方程知SKIPIF1<0,解得:SKIPIF1<0,即SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸,則SKIPIF1<0,又SKIPIF1<0SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0所以點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,即SKIPIF1<0又點(diǎn)SKIPIF1<0在橢圓上,SKIPIF1<0,即SKIPIF1<0又SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0故選:D16.(2024·云南·校聯(lián)考模擬預(yù)測(cè))已知橢圓SKIPIF1<0:SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0(如圖),過(guò)SKIPIF1<0的直線交SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn),且SKIPIF1<0軸,SKIPIF1<0,則SKIPIF1<0的離心率為(

A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】設(shè)橢圓SKIPIF1<0的半焦距為SKIPIF1<0,由題意可得:SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0,且點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上,則SKIPIF1<0,整理得SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0.故選:A.17.(2024·山西太原·高三山西大附中校考階段練習(xí))已知橢圓E:SKIPIF1<0的左,右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0(如圖),過(guò)SKIPIF1<0的直線交E于P,Q兩點(diǎn),且SKIPIF1<0軸,SKIPIF1<0,則SKIPIF1<0的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】依題意設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0;由于SKIPIF1<0,所以SKIPIF1<0由SKIPIF1<0得SKIPIF1<0,化為SKIPIF1<0,所以SKIPIF1<0,得SKIPIF1<0故選:A05橢圓與雙曲線的4a直角體18.(2024·全國(guó)·高三校聯(lián)考階段練習(xí))已知橢圓SKIPIF1<0的左、右焦點(diǎn)為SKIPIF1<0,SKIPIF1<0,過(guò)SKIPIF1<0的直線交SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn),若SKIPIF1<0,且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【解析】運(yùn)用特殊值法進(jìn)行求解.不妨設(shè)SKIPIF1<0,利用勾股定理、余弦定理,結(jié)合橢圓的定義和離心率公式進(jìn)行求解即可.不妨設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0(舍),∴SKIPIF1<0,∴SKIPIF1<0,又由SKIPIF1<0得SKIPIF1<0,∴SKIPIF1<0.故選:C19.(2024·重慶·校聯(lián)考)已知雙曲線SKIPIF1<0的左右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,過(guò)SKIPIF1<0的直線交雙曲線C的左支于P,Q兩點(diǎn),若SKIPIF1<0,且SKIPIF1<0的周長(zhǎng)為SKIPIF1<0,則雙曲線C的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】由雙曲線定義知SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,∴SKIPIF1<0的周長(zhǎng)為SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,在SKIPIF1<0中,SKIPIF1<0,故SKIPIF1<0.故選:A.20.(2024·廣西桂林·高三統(tǒng)考期末)設(shè)SKIPIF1<0,SKIPIF1<0分別是橢圓SKIPIF1<0的左、右焦點(diǎn),過(guò)點(diǎn)SKIPIF1<0的直線交橢圓SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn),SKIPIF1<0,若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】D【解析】設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∵SKIPIF1<0,在SKIPIF1<0中,由余弦定理,得:SKIPIF1<0,∴SKIPIF1<0,化簡(jiǎn)可得SKIPIF1<0,而SKIPIF1<0,故SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,且SKIPIF1<0,∴SKIPIF1<0是等腰直角三角形,SKIPIF1<0,∴SKIPIF1<0,∴橢圓的離心率SKIPIF1<0.故選:D.21.(2024·湖南·校聯(lián)考)已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是雙曲線SKIPIF1<0上的三個(gè)點(diǎn),直線SKIPIF1<0經(jīng)過(guò)原點(diǎn)SKIPIF1<0,SKIPIF1<0經(jīng)過(guò)右焦SKIPIF1<0,若SKIPIF1<0,且SKIPIF1<0,則該雙曲線的離心率為A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】如圖,因?yàn)镾KIPIF1<0,所以四邊形SKIPIF1<0為矩形,設(shè)SKIPIF1<0,則SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,又因?yàn)镾KIPIF1<0,即SKIPIF1<0,所以得離心率SKIPIF1<0,選擇A22.(2024·湖北·高三開(kāi)學(xué)考試)已知SKIPIF1<0是雙曲線SKIPIF1<0上的三個(gè)點(diǎn),SKIPIF1<0經(jīng)過(guò)原點(diǎn)SKIPIF1<0,SKIPIF1<0經(jīng)過(guò)右焦點(diǎn)SKIPIF1<0,若SKIPIF1<0且SKIPIF1<0,則該雙曲線的離心率是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】設(shè)左焦點(diǎn)為SKIPIF1<0,SKIPIF1<0

,連接SKIPIF1<0則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0因?yàn)镾KIPIF1<0,且SKIPIF1<0經(jīng)過(guò)原點(diǎn)SKIPIF1<0所以四邊形SKIPIF1<0為矩形在Rt△SKIPIF1<0中,SKIPIF1<0,代入SKIPIF1<0化簡(jiǎn)得SKIPIF1<0所以在Rt△SKIPIF1<0中,SKIPIF1<0,代入SKIPIF1<0化簡(jiǎn)得SKIPIF1<0,即SKIPIF1<0所以選B23.(2024·山東聊城·統(tǒng)考)已知A,B,C是雙曲線SKIPIF1<0上的三點(diǎn),直線AB經(jīng)過(guò)原點(diǎn)O,AC經(jīng)過(guò)右焦點(diǎn)F,若SKIPIF1<0,且SKIPIF1<0,則該雙曲線的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】設(shè)雙曲線的左焦點(diǎn)為SKIPIF1<0,連接SKIPIF1<0由題意知SKIPIF1<0∴四邊形SKIPIF1<0為矩形,令SKIPIF1<0∵SKIPIF1<0,SKIPIF1<0∴在SKIPIF1<0中,SKIPIF1<0將SKIPIF1<0帶入可得SKIPIF1<0∴SKIPIF1<0∴在SKIPIF1<0中,SKIPIF1<0即SKIPIF1<0可得SKIPIF1<0故選:D06橢圓與雙曲線的等腰三角形問(wèn)題24.(2024·江西上饒·高三階段練習(xí))已知雙曲線SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,過(guò)SKIPIF1<0的直線與雙曲線SKIPIF1<0的右支相交于SKIPIF1<0兩點(diǎn),若SKIPIF1<0,且SKIPIF1<0,則雙曲線的離心率SKIPIF1<0A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0且SKIPIF1<0,所以SKIPIF1<0為等腰直角三角形,所以SKIPIF1<0,由雙曲線定義可得SKIPIF1<0,SKIPIF1<0,兩式相加得SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,在SKIPIF1<0中,由余弦定理,可得SKIPIF1<0,解得SKIPIF1<0.故選:A.25.(2024·北京海淀·校考模擬預(yù)測(cè))雙曲線SKIPIF1<0:SKIPIF1<0SKIPIF1<0的左、右焦點(diǎn)分別為F1、F2,過(guò)F1的直線與雙曲線C的右支在第一象限的交點(diǎn)為A,與y軸的交點(diǎn)為B,且△ABF2為等邊三角形,則雙曲線的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】SKIPIF1<0為等邊三角形,則SKIPIF1<0,設(shè)SKIPIF1<0為原點(diǎn),SKIPIF1<0中,SKIPIF1<0,故SKIPIF1<0,故SKIPIF1<0.故選:B26.(2024·安徽·高三校聯(lián)考階段練習(xí))如圖,已知SKIPIF1<0,SKIPIF1<0分別為雙曲線SKIPIF1<0:SKIPIF1<0的左右焦點(diǎn),過(guò)SKIPIF1<0的直線與雙曲線SKIPIF1<0的左支交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),連接SKIPIF1<0,SKIPIF1<0,在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,則雙曲線的離心率為(

)A.2 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】設(shè)SKIPIF1<0,由雙曲線的定義可得SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0,即有SKIPIF1<0,因?yàn)镾KIPIF1<0為等腰三角形,所以SKIPIF1<0,解得SKIPIF1<0,在△SKIPIF1<0中,SKIPIF1<0,化為SKIPIF1<0,即有SKIPIF1<0.故選:SKIPIF1<0.07雙曲線的4a底邊等腰三角形27.(2024·四川成都·石室中學(xué)??迹┮阎猄KIPIF1<0,SKIPIF1<0是雙曲線SKIPIF1<0的左,右焦點(diǎn),過(guò)點(diǎn)SKIPIF1<0作斜率為SKIPIF1<0的直線SKIPIF1<0與雙曲線的左,右兩支分別交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),以SKIPIF1<0為圓心的圓過(guò)SKIPIF1<0,SKIPIF1<0,則雙曲線SKIPIF1<0的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.2 D.SKIPIF1<0【答案】B【解析】取MN中點(diǎn)A,連AF2,由已知令SKIPIF1<0,則SKIPIF1<0,如圖:因點(diǎn)M,N為雙曲線左右兩支上的點(diǎn),由雙曲線定義得SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,令雙曲線半焦距為c,SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0中,SKIPIF1<0,則有SKIPIF1<0,即SKIPIF1<0,因直線SKIPIF1<0的斜率為SKIPIF1<0,即SKIPIF1<0,而SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,于是有SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以雙曲線SKIPIF1<0的離心率為SKIPIF1<0.故選:B28.(2024·江西九江·統(tǒng)考)設(shè)雙曲線SKIPIF1<0的左、右焦點(diǎn)分別為F1,F(xiàn)2,過(guò)點(diǎn)F2的直線分別交雙曲線左、右兩支于點(diǎn)P,Q,點(diǎn)M為線段PQ的中點(diǎn),若P,Q,F(xiàn)1都在以M為圓心的圓上,且SKIPIF1<0,則雙曲線C的離心率為(

)A.SKIPIF1<0 B.2SKIPIF1<0 C.SKIPIF1<0 D.2SKIPIF1<0【答案】C【解析】以PQ為直徑的圓經(jīng)過(guò)點(diǎn)F1,則SKIPIF1<0,又SKIPIF1<0,可知PQ⊥MF1,則|PF1|=|QF1|,故三角形PF1Q是等腰直角三角形,設(shè)|PF1|=t,則|PQ|SKIPIF1<0t,由雙曲線的定義可知:|PF2|=t+2a,|QF2|=t﹣2a,可得|PQ|=4a,則SKIPIF1<0t=4a,即t=2SKIPIF1<0a,則:|PF2|SKIPIF1<0,在Rt△MF1F2中,|MF1|SKIPIF1<02a,|MF2|=|PF1|﹣|PM|=2SKIPIF1<0a,由勾股定理可知|F1F2|=2SKIPIF1<0a=2c,則雙曲線C的離心率為:eSKIPIF1<0.故選:C.29.(2024·安徽合肥·校聯(lián)考模擬預(yù)測(cè))設(shè)雙曲線SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,過(guò)SKIPIF1<0的直線SKIPIF1<0與雙曲線左右兩支交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),以SKIPIF1<0為直徑的圓過(guò)SKIPIF1<0,且SKIPIF1<0,則雙曲線C的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0即SKIPIF1<0所以SKIPIF1<0在三角形SKIPIF1<0中,有余弦定理可得:SKIPIF1<0所以SKIPIF1<0SKIPIF1<0即SKIPIF1<0因?yàn)橐訫N為直徑的圓經(jīng)過(guò)右焦點(diǎn)F2,所以SKIPIF1<0,又|MF2|=|NF2|,可得△MNF2為等腰直角三角形,設(shè)|MF2|=|NF2|=m,則|MN|SKIPIF1<0m,由|MF2|﹣|MF1|=2a,|NF1|﹣|NF2|=2a,兩式相加可得|NF1|﹣|MF1|=|MN|=4a,即有m=2SKIPIF1<0a,在直角三角形HF1F2中可得4c2=4a2+(2a+2SKIPIF1<0a﹣2a)2,化為c2=3a2,即eSKIPIF1<0.故選:B.30.(2024·河北石家莊·統(tǒng)考)已知SKIPIF1<0,SKIPIF1<0分別為橢圓SKIPIF1<0的左、右焦點(diǎn),點(diǎn)SKIPIF1<0是橢圓上位于第一象限內(nèi)的點(diǎn),延長(zhǎng)SKIPIF1<0交橢圓于點(diǎn)SKIPIF1<0,若SKIPIF1<0,且SKIPIF1<0,則橢圓的離心率為A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,故SKIPIF1<0.因SKIPIF1<0,故SKIPIF1<0,整理得到SKIPIF1<0,即SKIPIF1<0,故選A.31.(2024·山東煙臺(tái)·統(tǒng)考)已知雙曲線SKIPIF1<0:SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,點(diǎn)SKIPIF1<0在SKIPIF1<0的右支上,SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0,若SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】由SKIPIF1<0且SKIPIF1<0知:△SKIPIF1<0為等腰直角三角形且SKIPIF1<0、SKIPIF1<0,即SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0,故SKIPIF1<0,則SKIPIF1<0,而在△SKIPIF1<0中,SKIPIF1<0,∴SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0.故選:B.08焦點(diǎn)到漸近線距離為b32.(2024·四川瀘州·高三統(tǒng)考期末)已知F1,F(xiàn)2為雙曲線C:SKIPIF1<0=1(a>0,b>0)的左,右焦點(diǎn),過(guò)F2作C的一條漸近線的垂線,垂足為P,且與C的右支交于點(diǎn)Q,若SKIPIF1<0(O為坐標(biāo)原點(diǎn)),則C的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.2 D.3【答案】A【解析】根據(jù)對(duì)稱(chēng)性不妨設(shè)P為第一象限的點(diǎn),∵O為F1F2的中點(diǎn),又SKIPIF1<0,∴Q為PF2的中點(diǎn),又F2(c,0)到SKIPIF1<0的距離SKIPIF1<0,∴|PF2|=b,∴|QF2|=SKIPIF1<0,連接SKIPIF1<0,所以SKIPIF1<0,又|F1F2|=2c,∵PO的斜率為SKIPIF1<0,又QF2⊥PO,∴QF2的斜率為SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,在△QF2F1中,由余弦定理可得:SKIPIF1<0,化簡(jiǎn)可得a=b,∴雙曲線C的離心率為SKIPIF1<0=SKIPIF1<0.故選:A.33.(2024·安徽滁州·高三統(tǒng)考期末)設(shè)F1,F(xiàn)2分別是雙曲線SKIPIF1<0的左、右焦點(diǎn),過(guò)F2作雙曲線的一條漸近線的垂線,垂足為H,若|HF1|=3|HF2|,則雙曲線的離心率為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】由題設(shè)條件推導(dǎo)出SKIPIF1<0,SKIPIF1<0,可得SKIPIF1<0的坐標(biāo),由兩點(diǎn)間的距離公式得SKIPIF1<0,計(jì)算求出離心率SKIPIF1<0.由題設(shè)知雙曲線C:SKIPIF1<0的一條漸近線方程為SKIPIF1<0:SKIPIF1<0,∵右焦點(diǎn)SKIPIF1<0,且SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,由SKIPIF1<0,解得SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,平方化簡(jiǎn)得SKIPIF1<0,又SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,故得SKIPIF1<0,故選:D.34.(2024·遼寧葫蘆島·統(tǒng)考)設(shè)F1,F(xiàn)2是雙曲線C:SKIPIF1<0(a>0,b>0)的左、右焦點(diǎn),O是坐標(biāo)原點(diǎn).過(guò)F2作C的一條漸近線的垂線,垂足為P.若|PF1|=3|OP|,則C的離心率為(

)A.SKIPIF1<0 B.2 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】不妨設(shè)雙曲線的一條漸近線方程為SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,

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