新高考數(shù)學(xué)一輪復(fù)習(xí)第8章 第06講 雙曲線(xiàn) 精講（教師版）

第06講雙曲線(xiàn)(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：雙曲線(xiàn)的定義及其應(yīng)用題型二：雙曲線(xiàn)的標(biāo)準(zhǔn)方程題型三：雙曲線(xiàn)的簡(jiǎn)單幾何性質(zhì)角度1：漸近線(xiàn)角度2：離心率題型四：與雙曲線(xiàn)有關(guān)的最值和范圍問(wèn)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：雙曲線(xiàn)的定義1、定義:一般地,我們把平面內(nèi)與兩個(gè)定點(diǎn)SKIPIF1<0,SKIPIF1<0的距離的差的絕對(duì)值等于非零常數(shù)(小于SKIPIF1<0)的點(diǎn)的軌跡叫做雙曲線(xiàn).這兩個(gè)定點(diǎn)叫做雙曲線(xiàn)的焦點(diǎn),兩焦點(diǎn)間的距離叫做雙曲線(xiàn)的焦距.2、集合語(yǔ)言表達(dá)式雙曲線(xiàn)就是下列點(diǎn)的集合:SKIPIF1<0.3、說(shuō)明若將定義中差的絕對(duì)值中的絕對(duì)值符號(hào)去掉,則點(diǎn)SKIPIF1<0的軌跡為雙曲線(xiàn)的一支,具體是哪一支,取決于SKIPIF1<0與SKIPIF1<0的大小.(1)若SKIPIF1<0,則SKIPIF1<0,點(diǎn)SKIPIF1<0的軌跡是靠近定點(diǎn)SKIPIF1<0的那一支;(2)若SKIPIF1<0,則SKIPIF1<0,點(diǎn)SKIPIF1<0的軌跡是靠近定點(diǎn)SKIPIF1<0的那一支.知識(shí)點(diǎn)二：雙曲線(xiàn)的標(biāo)準(zhǔn)方程和簡(jiǎn)單幾何性質(zhì)標(biāo)準(zhǔn)方程SKIPIF1<0（SKIPIF1<0）SKIPIF1<0（SKIPIF1<0）圖形性質(zhì)范圍SKIPIF1<0或SKIPIF1<0SKIPIF1<0或SKIPIF1<0對(duì)稱(chēng)性對(duì)稱(chēng)軸：坐標(biāo)軸；對(duì)稱(chēng)中心：原點(diǎn)頂點(diǎn)坐標(biāo)SKIPIF1<0，SKIPIF1<0SKIPIF1<0,SKIPIF1<0漸近線(xiàn)SKIPIF1<0SKIPIF1<0離心率SKIPIF1<0，SKIPIF1<0，SKIPIF1<0間的關(guān)系SKIPIF1<0知識(shí)點(diǎn)三：等軸雙曲線(xiàn)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）當(dāng)SKIPIF1<0時(shí)稱(chēng)雙曲線(xiàn)為等軸雙曲線(xiàn)①SKIPIF1<0；②離心率SKIPIF1<0；③兩漸近線(xiàn)互相垂直，分別為SKIPIF1<0；④等軸雙曲線(xiàn)的方程SKIPIF1<0，SKIPIF1<0；知識(shí)點(diǎn)四：雙曲線(xiàn)與漸近線(xiàn)的關(guān)系1、若雙曲線(xiàn)方程為SKIPIF1<0SKIPIF1<0漸近線(xiàn)方程：SKIPIF1<0SKIPIF1<02、若雙曲線(xiàn)方程為SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）SKIPIF1<0漸近線(xiàn)方程：SKIPIF1<0SKIPIF1<03、若漸近線(xiàn)方程為SKIPIF1<0，則雙曲線(xiàn)方程可設(shè)為SKIPIF1<0，4、若雙曲線(xiàn)與SKIPIF1<0有公共漸近線(xiàn)，則雙曲線(xiàn)的方程可設(shè)為SKIPIF1<0（SKIPIF1<0，焦點(diǎn)在SKIPIF1<0軸上，SKIPIF1<0，焦點(diǎn)在SKIPIF1<0軸上）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·海南·瓊海市嘉積第三中學(xué)高三階段練習(xí)）雙曲線(xiàn)SKIPIF1<0的離心率為SKIPIF1<0，且過(guò)SKIPIF1<0，則雙曲線(xiàn)方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D解：由雙曲線(xiàn)離心率為SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，所以雙曲線(xiàn)方程為SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0SKIPIF1<0.所以雙曲線(xiàn)的方程為SKIPIF1<0.故選：D2．（2022·四川甘孜·高二期末（文））雙曲線(xiàn)的方程為SKIPIF1<0?，則該雙曲線(xiàn)的離心率為（

）A．SKIPIF1<0? B．SKIPIF1<0?C．SKIPIF1<0? D．SKIPIF1<0?【答案】D由雙曲線(xiàn)方程SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，則雙曲線(xiàn)的離心率為SKIPIF1<0.故選：D.3．（多選）（2022·廣東·佛山市南海區(qū)藝術(shù)高級(jí)中學(xué)模擬預(yù)測(cè)）若方程SKIPIF1<0所表示的曲線(xiàn)為SKIPIF1<0，則下面四個(gè)命題中正確的是（

）A．若SKIPIF1<0為橢圓，則SKIPIF1<0 B．若SKIPIF1<0為雙曲線(xiàn)，則SKIPIF1<0或SKIPIF1<0C．曲線(xiàn)SKIPIF1<0可能是圓 D．若SKIPIF1<0為橢圓，且長(zhǎng)軸在SKIPIF1<0軸上，則SKIPIF1<0【答案】BC若SKIPIF1<0為橢圓，則SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，故A錯(cuò)誤若SKIPIF1<0為雙曲線(xiàn)，則SKIPIF1<0，SKIPIF1<0，故B正確若SKIPIF1<0為圓，則SKIPIF1<0，SKIPIF1<0，故C正確若SKIPIF1<0為橢圓，且長(zhǎng)軸在SKIPIF1<0軸上，則SKIPIF1<0，SKIPIF1<0，故D錯(cuò)誤故選：BC4．（2022·貴州遵義·高二期末（理））過(guò)點(diǎn)SKIPIF1<0且與雙曲線(xiàn)：SKIPIF1<0的漸近線(xiàn)垂直的直線(xiàn)方程為_(kāi)_________．【答案】SKIPIF1<0，SKIPIF1<0由雙曲線(xiàn)：SKIPIF1<0可得其漸近線(xiàn)方程為SKIPIF1<0，∴過(guò)點(diǎn)SKIPIF1<0且與雙曲線(xiàn)：SKIPIF1<0的漸近線(xiàn)垂直的直線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0.5．（2022·上?！とA東師范大學(xué)附屬東昌中學(xué)高三階段練習(xí)）若雙曲線(xiàn)SKIPIF1<0的焦距等于虛軸長(zhǎng)的3倍，則SKIPIF1<0的值為_(kāi)_____．【答案】SKIPIF1<0SKIPIF1<0化為標(biāo)準(zhǔn)方程：SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，則可得：SKIPIF1<0，解得：SKIPIF1<0，故答案為：SKIPIF1<0第三部分：典型例題剖析第三部分：典型例題剖析題型一：雙曲線(xiàn)的定義及其應(yīng)用典型例題例題1．（2022·河南許昌·高二期末（理））已知雙曲線(xiàn)SKIPIF1<0的左右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，其一條漸近線(xiàn)傾斜角為SKIPIF1<0，若點(diǎn)P在雙曲線(xiàn)上，且SKIPIF1<0，則SKIPIF1<0______．【答案】13由題意，SKIPIF1<0，故SKIPIF1<0，雙曲線(xiàn)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0小于SKIPIF1<0到右頂點(diǎn)的距離，故SKIPIF1<0在雙曲線(xiàn)的左支上，由雙曲線(xiàn)的定義可得SKIPIF1<0，解得SKIPIF1<0故答案為：13例題2．（2022·安徽師范大學(xué)附屬中學(xué)模擬預(yù)測(cè)（文））設(shè)SKIPIF1<0為橢圓SKIPIF1<0和雙曲線(xiàn)SKIPIF1<0的一個(gè)公共點(diǎn)，且SKIPIF1<0在第一象限，SKIPIF1<0是SKIPIF1<0的左焦點(diǎn)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A由橢圓SKIPIF1<0方程知其焦點(diǎn)為SKIPIF1<0；由雙曲線(xiàn)SKIPIF1<0方程知其焦點(diǎn)為SKIPIF1<0；SKIPIF1<0橢圓SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0共焦點(diǎn)，設(shè)其右焦點(diǎn)為SKIPIF1<0，SKIPIF1<0為橢圓SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0在第一象限內(nèi)的交點(diǎn)，SKIPIF1<0由橢圓和雙曲線(xiàn)定義知：SKIPIF1<0，解得：SKIPIF1<0.故選：A.例題3．（2022·全國(guó)·高二專(zhuān)題練習(xí)）雙曲線(xiàn)SKIPIF1<0的左?右焦點(diǎn)分別是SKIPIF1<0?SKIPIF1<0，過(guò)SKIPIF1<0的弦AB與其右支交于SKIPIF1<0?SKIPIF1<0兩點(diǎn)，SKIPIF1<0，則SKIPIF1<0的周長(zhǎng)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C由題可得SKIPIF1<0，則SKIPIF1<0的周長(zhǎng)為SKIPIF1<0.故選：C．例題4．（2022·江蘇·高二）已知SKIPIF1<0?SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0的兩個(gè)焦點(diǎn)，點(diǎn)SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0上一點(diǎn)，且SKIPIF1<0，求SKIPIF1<0的面積.【答案】SKIPIF1<0因?yàn)镾KIPIF1<0?SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0的兩個(gè)焦點(diǎn)，所以SKIPIF1<0，所以SKIPIF1<0；設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)辄c(diǎn)M是雙曲線(xiàn)上一點(diǎn)，且SKIPIF1<0，所以SKIPIF1<0；在△SKIPIF1<0中，由余弦定理可得：SKIPIF1<0；聯(lián)立上述兩式可得：SKIPIF1<0，所以SKIPIF1<0的面積SKIPIF1<0.同類(lèi)題型歸類(lèi)練1．（2022·湖北·宜城市第一中學(xué)高三階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0分別是雙曲線(xiàn)SKIPIF1<0的左、右焦點(diǎn)，動(dòng)點(diǎn)SKIPIF1<0在雙曲線(xiàn)SKIPIF1<0的右支上，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B因?yàn)閯?dòng)點(diǎn)SKIPIF1<0在雙曲線(xiàn)SKIPIF1<0的右支上，由雙曲線(xiàn)定義可得：SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0.故選：B．2．（2022·陜西·西北工業(yè)大學(xué)附屬中學(xué)高二階段練習(xí)（文））已知雙曲線(xiàn)SKIPIF1<0的左焦點(diǎn)為F，點(diǎn)M在雙曲線(xiàn)C的右支上，SKIPIF1<0，當(dāng)SKIPIF1<0的周長(zhǎng)最小時(shí)，SKIPIF1<0的面積為（

）A．2 B．4 C．8 D．12【答案】D解：設(shè)雙曲線(xiàn)SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，由雙曲線(xiàn)的定義知，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的周長(zhǎng)為SKIPIF1<0，即當(dāng)SKIPIF1<0在SKIPIF1<0即SKIPIF1<0與雙曲線(xiàn)的交點(diǎn)處時(shí)，SKIPIF1<0的周長(zhǎng)最小，此時(shí)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0的面積為SKIPIF1<0．故選：D．3．（2022·寧夏·銀川一中模擬預(yù)測(cè)（文））已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，一條漸近線(xiàn)方程為SKIPIF1<0，若點(diǎn)SKIPIF1<0在雙曲線(xiàn)SKIPIF1<0上，且SKIPIF1<0，則SKIPIF1<0________.【答案】9由雙曲線(xiàn)C的方程可得其漸近線(xiàn)方程為SKIPIF1<0，由已知可得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，由雙曲線(xiàn)定義可知SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，又因?yàn)镾KIPIF1<0，故SKIPIF1<0，故答案為：9.4．（2022·河北·衡水市第二中學(xué)高二期中）已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0.雙曲線(xiàn)SKIPIF1<0上有一點(diǎn)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0______.【答案】1或13##13或1因?yàn)殡p曲線(xiàn)SKIPIF1<0：SKIPIF1<0，所以a=3，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，故答案為：1或13.題型二：雙曲線(xiàn)的標(biāo)準(zhǔn)方程典型例題例題1．（2022·江蘇·高二課時(shí)練習(xí)）求適合下列條件的雙曲線(xiàn)的標(biāo)準(zhǔn)方程：(1)頂點(diǎn)在SKIPIF1<0軸上，焦距為10，離心率是SKIPIF1<0；(2)一個(gè)頂點(diǎn)的坐標(biāo)為SKIPIF1<0，一個(gè)焦點(diǎn)的坐標(biāo)為SKIPIF1<0；(3)焦點(diǎn)在SKIPIF1<0軸上，一條漸近線(xiàn)方程為SKIPIF1<0，實(shí)軸長(zhǎng)為12；(4)漸近線(xiàn)方程為SKIPIF1<0，焦點(diǎn)坐標(biāo)為SKIPIF1<0和SKIPIF1<0．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；(4)SKIPIF1<0.(1)由題設(shè)，SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，又頂點(diǎn)在x軸上，故雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0.(2)由題設(shè)，SKIPIF1<0，則SKIPIF1<0，又一個(gè)焦點(diǎn)為SKIPIF1<0，故雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0.(3)由題設(shè)，SKIPIF1<0，又焦點(diǎn)在y軸上，令雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0，又一條漸近線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，所以雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0.(4)由題設(shè)，SKIPIF1<0且焦點(diǎn)在x軸上，令SKIPIF1<0又漸近線(xiàn)方程為SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，故雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0例題2．（2022·全國(guó)·高二課時(shí)練習(xí)）求適合下列條件的雙曲線(xiàn)的標(biāo)準(zhǔn)方程：(1)SKIPIF1<0，SKIPIF1<0，焦點(diǎn)在SKIPIF1<0軸上；(2)焦點(diǎn)為SKIPIF1<0?SKIPIF1<0，經(jīng)過(guò)點(diǎn)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(1)由題設(shè)知，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0.因?yàn)殡p曲線(xiàn)的焦點(diǎn)在x軸上，所以所求雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0；(2)由已知得SKIPIF1<0，且焦點(diǎn)在y軸上.因?yàn)辄c(diǎn)SKIPIF1<0在雙曲線(xiàn)上，所以點(diǎn)A與兩焦點(diǎn)的距離的差的絕對(duì)值是常數(shù)2a，即SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.因此，所求雙曲線(xiàn)的標(biāo)準(zhǔn)方程是SKIPIF1<0.同類(lèi)題型歸類(lèi)練1．（2022·全國(guó)·高二課時(shí)練習(xí)）已知雙曲線(xiàn)的焦點(diǎn)與橢圓SKIPIF1<0的左、右頂點(diǎn)相同，且經(jīng)過(guò)橢圓的右焦點(diǎn)，求該雙曲線(xiàn)的方程.【答案】SKIPIF1<0解：橢圓SKIPIF1<0的左頂點(diǎn)為SKIPIF1<0，右頂點(diǎn)為SKIPIF1<0，右焦點(diǎn)為SKIPIF1<0，所以雙曲線(xiàn)中SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以雙曲線(xiàn)方程為SKIPIF1<0；2．（2022·全國(guó)·高二課時(shí)練習(xí)）求適合下列條件的雙曲線(xiàn)的標(biāo)準(zhǔn)方程：(1)焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，且雙曲線(xiàn)上的一點(diǎn)到兩個(gè)焦點(diǎn)距離之差為2；(2)焦點(diǎn)在y軸上，焦距為10，且經(jīng)過(guò)點(diǎn)SKIPIF1<0；(3)經(jīng)過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0.(1)因?yàn)殡p曲線(xiàn)的焦點(diǎn)在SKIPIF1<0軸上，故可設(shè)方程為：SKIPIF1<0，又焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，故可得SKIPIF1<0，又雙曲線(xiàn)上的一點(diǎn)到兩個(gè)焦點(diǎn)距離之差為2，即SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0.故雙曲線(xiàn)方程為：SKIPIF1<0.(2)因?yàn)殡p曲線(xiàn)焦點(diǎn)在SKIPIF1<0軸上，故可設(shè)雙曲線(xiàn)方程為SKIPIF1<0，又其焦距為10，故可得SKIPIF1<0；又該雙曲線(xiàn)過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，故雙曲線(xiàn)方程為：SKIPIF1<0.(3)不妨設(shè)雙曲線(xiàn)方程為：SKIPIF1<0，因其過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0，故可得SKIPIF1<0，聯(lián)立方程組可得：SKIPIF1<0，故所求雙曲線(xiàn)方程為：SKIPIF1<0.題型三：雙曲線(xiàn)的簡(jiǎn)單幾何性質(zhì)角度1：漸近線(xiàn)典型例題例題1．（2022·四川·威遠(yuǎn)中學(xué)校高二階段練習(xí)（文））設(shè)雙曲線(xiàn)SKIPIF1<0的虛軸長(zhǎng)為2，焦距為SKIPIF1<0，則雙曲線(xiàn)的漸近線(xiàn)方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：依題意SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以雙曲線(xiàn)方程為SKIPIF1<0，所以雙曲線(xiàn)的漸近線(xiàn)方程為SKIPIF1<0；故選：C例題2．（2022·天津市第一中學(xué)濱海學(xué)校高二開(kāi)學(xué)考試）雙曲線(xiàn)SKIPIF1<0的離心率SKIPIF1<0，則其漸近線(xiàn)方程為_(kāi)_____.【答案】SKIPIF1<0或SKIPIF1<0由題意得：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，此時(shí)雙曲線(xiàn)方程為：SKIPIF1<0漸近線(xiàn)方程為：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，此時(shí)雙曲線(xiàn)方程為：SKIPIF1<0，漸近線(xiàn)方程為：SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0同類(lèi)題型歸類(lèi)練1．（2022·全國(guó)·高三專(zhuān)題練習(xí)（文））已知雙曲線(xiàn)SKIPIF1<0的離心率為SKIPIF1<0，則雙曲線(xiàn)E的兩條漸近線(xiàn)的夾角為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】B【詳解】當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0此時(shí)漸近線(xiàn)的斜率為SKIPIF1<0，所以漸近線(xiàn)的傾斜角為SKIPIF1<0和SKIPIF1<0，所以雙曲線(xiàn)E的兩條漸近線(xiàn)的夾角為SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0此時(shí)漸近線(xiàn)的斜率為SKIPIF1<0，所以漸近線(xiàn)的傾斜角為SKIPIF1<0和SKIPIF1<0，所以雙曲線(xiàn)E的兩條漸近線(xiàn)的夾角為SKIPIF1<0；故選：B.2．（2022·北京·高三專(zhuān)題練習(xí)）已知雙曲線(xiàn)SKIPIF1<0的一個(gè)焦點(diǎn)為SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0的一條漸近線(xiàn)方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A根據(jù)題意，SKIPIF1<0，故可得SKIPIF1<0，則SKIPIF1<0，故雙曲線(xiàn)的漸近線(xiàn)方程為SKIPIF1<0，故SKIPIF1<0是其中一條漸近線(xiàn).故選：A.3．（2022·上海理工大學(xué)附屬中學(xué)高二期中）雙曲線(xiàn)SKIPIF1<0的兩條漸近線(xiàn)的夾角為_(kāi)_____．【答案】SKIPIF1<0##SKIPIF1<0因?yàn)殡p曲線(xiàn)SKIPIF1<0，則SKIPIF1<0，所以漸近線(xiàn)方程為SKIPIF1<0，又直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，所以?xún)蓷l漸近線(xiàn)的夾角為SKIPIF1<0.故答案為：SKIPIF1<0角度2：離心率典型例題例題1．（2022·江蘇南通·高二期中）若SKIPIF1<0是1和4的等比中項(xiàng)，則曲線(xiàn)SKIPIF1<0的離心率為（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】Am是1和4的等比中項(xiàng),所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，曲線(xiàn)SKIPIF1<0化為SKIPIF1<0是焦點(diǎn)在SKIPIF1<0軸上的橢圓，離心率為：SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，曲線(xiàn)SKIPIF1<0化為SKIPIF1<0是焦點(diǎn)在SKIPIF1<0軸上的雙曲線(xiàn)，離心率為：SKIPIF1<0.故選：A.例題2．（2022·四川省瀘縣第二中學(xué)模擬預(yù)測(cè)（理））已知SKIPIF1<0為雙曲線(xiàn)SKIPIF1<0的左、右焦點(diǎn)，以線(xiàn)段SKIPIF1<0為直徑的圓與雙曲線(xiàn)SKIPIF1<0的右支交于SKIPIF1<0兩點(diǎn)，若SKIPIF1<0為等邊三角形，則SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】DSKIPIF1<0雙曲線(xiàn)SKIPIF1<0與以SKIPIF1<0為直徑的圓均關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，SKIPIF1<0為等邊三角形，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；由雙曲線(xiàn)定義知：SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0雙曲線(xiàn)離心率SKIPIF1<0.故選：D例題3．（2022·山東泰安·三模）已知雙曲線(xiàn)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的右焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0為雙曲線(xiàn)虛軸的上端點(diǎn)，SKIPIF1<0為雙曲線(xiàn)的左頂點(diǎn)，若SKIPIF1<0，則雙曲線(xiàn)的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由已知雙曲線(xiàn)SKIPIF1<0的右焦點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，虛軸的上端點(diǎn)B的坐標(biāo)為SKIPIF1<0，左頂點(diǎn)A的坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0又SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以雙曲線(xiàn)的離心率SKIPIF1<0，故選：D.例題4．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0（SKIPIF1<0），以SKIPIF1<0的焦點(diǎn)為圓心，3為半徑的圓與SKIPIF1<0的漸近線(xiàn)相交，則雙曲線(xiàn)SKIPIF1<0的離心率的取值范圍是________________．【答案】SKIPIF1<0雙曲線(xiàn)C的漸近線(xiàn)方程為SKIPIF1<0，右焦點(diǎn)SKIPIF1<0，∵漸近線(xiàn)與圓相交，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，可得SKIPIF1<0，∴雙曲線(xiàn)C的離心率為：SKIPIF1<0，且SKIPIF1<0．∴SKIPIF1<0．故答案為：SKIPIF1<0同類(lèi)題型歸類(lèi)練1．（2022·江蘇連云港·模擬預(yù)測(cè)）已知雙曲線(xiàn)SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，一條漸近線(xiàn)被圓SKIPIF1<0截得的弦長(zhǎng)為SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A雙曲線(xiàn)的一條漸近線(xiàn)為SKIPIF1<0，即SKIPIF1<0過(guò)F作漸近線(xiàn)的垂線(xiàn)，垂足為B則右焦點(diǎn)SKIPIF1<0到漸近線(xiàn)的距離SKIPIF1<0由題可知SKIPIF1<0由勾股定理得，SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，得SKIPIF1<0故選：A2．（2022·河南商丘·三模（理））已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0，且SKIPIF1<0的實(shí)軸長(zhǎng)大于SKIPIF1<0，則SKIPIF1<0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由題意可知，SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0故選：D3．（2022·四川省內(nèi)江市第六中學(xué)高二階段練習(xí)（文））已知SKIPIF1<0，SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0的左?右焦點(diǎn)，過(guò)SKIPIF1<0作斜率為SKIPIF1<0的直線(xiàn)SKIPIF1<0，SKIPIF1<0分別交SKIPIF1<0軸和雙曲線(xiàn)右支于點(diǎn)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由SKIPIF1<0可得，SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0為SKIPIF1<0的中點(diǎn)，由于O為SKIPIF1<0的中點(diǎn)，故SKIPIF1<0，故SKIPIF1<0軸，將SKIPIF1<0代入SKIPIF1<0中得：SKIPIF1<0，故SKIPIF1<0，因?yàn)橹本€(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，故SKIPIF1<0,所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0（負(fù)值舍去），故SKIPIF1<0，故選：D4．（2022·四川雅安·三模（文））已知雙曲線(xiàn)SKIPIF1<0的右焦點(diǎn)為F，若過(guò)點(diǎn)F且傾斜角為SKIPIF1<0的直線(xiàn)與雙曲線(xiàn)的右支有且只有一個(gè)交點(diǎn)，則此雙曲線(xiàn)離心率的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由題可得漸近線(xiàn)SKIPIF1<0的斜率滿(mǎn)足SKIPIF1<0，所以離心率SKIPIF1<0.故選：D.5．（2022·廣西·昭平中學(xué)高二階段練習(xí)（理））已知雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0軸的垂線(xiàn)與雙曲線(xiàn)交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，且SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0的離心率的取值范圍是__________.【答案】SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，不等式兩邊同除以SKIPIF1<0得：SKIPIF1<0，解得：SKIPIF1<0故答案為：SKIPIF1<0題型四：與雙曲線(xiàn)有關(guān)的最值和范圍問(wèn)題典型例題例題1．（2022·全國(guó)·模擬預(yù)測(cè)（文））已知點(diǎn)SKIPIF1<0為雙曲線(xiàn)SKIPIF1<0的右焦點(diǎn)，過(guò)SKIPIF1<0作雙曲線(xiàn)的一條漸近線(xiàn)的垂線(xiàn)，垂足為SKIPIF1<0．若SKIPIF1<0（點(diǎn)SKIPIF1<0為坐標(biāo)原點(diǎn)）的面積為4，雙曲線(xiàn)的離心率SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B取雙曲線(xiàn)的一條漸近線(xiàn)為SKIPIF1<0，即SKIPIF1<0.則SKIPIF1<0到漸近線(xiàn)的距離即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0.又SKIPIF1<0，SKIPIF1<0，易得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.故選：B.例題2．（2022·全國(guó)·高二專(zhuān)題練習(xí)）直線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0沒(méi)有交點(diǎn)，則SKIPIF1<0的取值范圍為_(kāi)____.【答案】SKIPIF1<0由題意，雙曲線(xiàn)SKIPIF1<0的漸近線(xiàn)方程為：SKIPIF1<0，因?yàn)橹本€(xiàn)SKIPIF1<0過(guò)原點(diǎn)且與雙曲線(xiàn)SKIPIF1<0沒(méi)有交點(diǎn)，故需滿(mǎn)足SKIPIF1<0，故答案為：SKIPIF1<0例題3．（2022·全國(guó)·高二專(zhuān)題練習(xí)）已知SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0的左右焦點(diǎn)，以SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓與雙曲線(xiàn)的一條漸近線(xiàn)交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，若SKIPIF1<0，則雙曲線(xiàn)的離心率的取值范圍是______．【答案】SKIPIF1<0SKIPIF1<0，SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0的左右焦點(diǎn)，以SKIPIF1<0圓心，SKIPIF1<0為半徑的圓與雙曲線(xiàn)的一條漸近線(xiàn)SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，則焦點(diǎn)到漸近線(xiàn)的距離：SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，即：SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以雙曲線(xiàn)的離心率的取值范圍是：SKIPIF1<0．故答案為：SKIPIF1<0．同類(lèi)題型歸類(lèi)練1．（2022·安徽滁州·高二期末）已知雙曲線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0有相同的漸近線(xiàn)，過(guò)雙曲線(xiàn)SKIPIF1<0右焦點(diǎn)SKIPIF1<0的直線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0相交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，弦SKIPIF1<0的中點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0右支上的動(dòng)點(diǎn)，點(diǎn)SKIPIF1<0是以點(diǎn)SKIPIF1<0為圓心，SKIPIF1<0為半徑的圓上的動(dòng)點(diǎn)，點(diǎn)SKIPIF1<0是圓SKIPIF1<0上的動(dòng)點(diǎn)，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由雙曲線(xiàn)SKIPIF1<0知漸近線(xiàn)方程為SKIPIF1<0，又雙曲線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0有相同的漸近線(xiàn)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0雙曲線(xiàn)方程為SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又弦SKIPIF1<0的中點(diǎn)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，所以雙曲線(xiàn)的方程為SKIPIF1<0，由圓SKIPIF1<0的方程可得SKIPIF1<0，圓心為SKIPIF1<0，半徑為SKIPIF1<0，SKIPIF1<0．當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三點(diǎn)共線(xiàn)時(shí)取等號(hào)．故選：D．2．（2022·全國(guó)·高二專(zhuān)題練習(xí)）設(shè)雙曲線(xiàn)SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0的直線(xiàn)SKIPIF1<0交雙曲線(xiàn)左支于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，則SKIPIF1<0的最小值為_(kāi)_____．【答案】22根據(jù)雙曲線(xiàn)SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，由雙曲線(xiàn)的定義可得：SKIPIF1<0①，SKIPIF1<0②，①+②可得：SKIPIF1<0，由于過(guò)雙曲線(xiàn)的左焦點(diǎn)SKIPIF1<0的直線(xiàn)交雙曲線(xiàn)的左支于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，可得SKIPIF1<0，即有SKIPIF1<0．則SKIPIF1<0，當(dāng)SKIPIF1<0是雙曲線(xiàn)的通徑時(shí)SKIPIF1<0最小，故SKIPIF1<0.故答案為：223．（2022·河南洛陽(yáng)·模擬預(yù)測(cè)（理））已知F是橢圓SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）的右焦點(diǎn)，A為橢圓SKIPIF1<0的下頂點(diǎn)，雙曲線(xiàn)SKIPIF1<0：SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）與橢圓SKIPIF1<0共焦點(diǎn)，若直線(xiàn)SKIPIF1<0與雙曲線(xiàn)SKIPIF1<0的一條漸近線(xiàn)平行，SKIPIF1<0，SKIPIF1<0的離心率分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_____．【答案】SKIPIF1<0解：設(shè)SKIPIF1<0的半焦距為c（SKIPIF1<0），則SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，又直線(xiàn)SKIPIF1<0與SKIPIF1<0的一條漸近線(xiàn)平行，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，即SKIPIF1<0的最小值為SKIPIF1<0．故答案為：SKIPIF1<04．（2022·河南·南陽(yáng)中學(xué)三模（文））已知雙曲線(xiàn)SKIPIF1<0的一條漸近線(xiàn)方程為SKIPIF1<0，左焦點(diǎn)為SKIPIF1<0，點(diǎn)P在雙曲線(xiàn)右支上運(yùn)動(dòng)，點(diǎn)Q在圓SKIPIF1<0上運(yùn)動(dòng)，則SKIPIF1<0的最小值為_(kāi)__________.【答案】8由雙曲線(xiàn)SKIPIF1<0的一條漸近線(xiàn)方程為SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0.所以SKIPIF1<0，雙曲線(xiàn)的左焦點(diǎn)坐標(biāo)SKIPIF1<0，右焦點(diǎn)坐標(biāo)為SKIPIF1<0，由雙曲線(xiàn)的定義，知SKIPIF1<0，即SKIPIF1<0，由圓SKIPIF1<0可得圓心SKIPIF1<0，半徑為SKIPIF1<0，SKIPIF1<0，問(wèn)題轉(zhuǎn)化為求點(diǎn)SKIPIF1<0到圓SKIPIF1<0上的最小值，即SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.第四部分：高考真題感悟第四部分：高考真題感悟1．（2022·天津·高考真題）SKIPIF1<0、SKIPIF1<0是雙曲線(xiàn)SKIPIF1<0的兩個(gè)焦點(diǎn)，拋物線(xiàn)SKIPIF1<0的準(zhǔn)線(xiàn)SKIPIF1<0過(guò)雙曲線(xiàn)的焦點(diǎn)SKIPIF1<0，準(zhǔn)線(xiàn)與漸近線(xiàn)交于點(diǎn)SKIPIF1<0，SKIPIF1<0，則雙曲線(xiàn)的標(biāo)準(zhǔn)方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C拋物線(xiàn)SKIPIF1<0的準(zhǔn)線(xiàn)方程為SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0、SKIPIF1<0，不妨設(shè)點(diǎn)SKIPIF1<0為第二象限內(nèi)的點(diǎn)，聯(lián)立SKIPIF1<0，可得SKIPIF1<0，即點(diǎn)SKIPIF1<0，因?yàn)镾KIPIF1<0且SKIPIF1<0，則SKIPIF1<0為等腰直角三角形，且SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0，因此，雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0.故選：C.2．（多選）（2022·全國(guó)·高考真題（理））雙曲線(xiàn)C的兩個(gè)焦點(diǎn)為SKIPIF1<0，以C的實(shí)軸為直徑的圓記為D，過(guò)SKIPIF1<0作D的切線(xiàn)與C交于M，N兩點(diǎn)，且SKIPIF1<0，則C的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．