新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練10.4 雙曲線（基礎(chǔ)版）（解析版）

10.4雙曲線（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一雙曲線的定義及應(yīng)用【例1-1】（2022·潮州二模）若點(diǎn)P是雙曲線SKIPIF1<0上一點(diǎn)，SKIPIF1<0，SKIPIF1<0分別為SKIPIF1<0的左、右焦點(diǎn)，則“SKIPIF1<0”是“SKIPIF1<0”的（）．A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】由題意可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0或1（舍去），若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0或13，故“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件．故答案為：A．【例1-2】（2022·成都模擬）設(shè)SKIPIF1<0，SKIPIF1<0是雙曲線SKIPIF1<0的左，右焦點(diǎn)，點(diǎn)P在雙曲線C的右支上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0面積為（）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】∵雙曲線SKIPIF1<0，∴SKIPIF1<0，又點(diǎn)P在雙曲線C的右支上，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0面積為SKIPIF1<0.故答案為：B.【例1-3】（2022常州期中）已知雙曲線SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，SKIPIF1<0為雙曲線左支上一點(diǎn)，點(diǎn)SKIPIF1<0，則SKIPIF1<0周長(zhǎng)的最小值為（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【解析】曲線SKIPIF1<0右焦點(diǎn)為SKIPIF1<0SKIPIF1<0，SKIPIF1<0周長(zhǎng)SKIPIF1<0要使SKIPIF1<0周長(zhǎng)最小，只需SKIPIF1<0最小，如圖：當(dāng)SKIPIF1<0三點(diǎn)共線時(shí)取到，故l=2|AF|+2a=SKIPIF1<0故答案為：B【例1-4】（2021河北月考）已知方程SKIPIF1<0表示雙曲線，則實(shí)數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C【解析】因?yàn)榉匠蘏KIPIF1<0表示雙曲線，所以當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0.綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0或SKIPIF1<0。故答案為：C【一隅三反】1．（2022高三上·廣東開學(xué)考）“k<2”是“方程SKIPIF1<0表示雙曲線”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】∵方程SKIPIF1<0為雙曲線，∴SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0，∴“SKIPIF1<0”是“方程SKIPIF1<0為雙曲線”的充分不必要條件，故答案為：A.2．（2022·運(yùn)城模擬）已知雙曲線SKIPIF1<0的左右焦點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是雙曲線上一點(diǎn)，SKIPIF1<0，則SKIPIF1<0（）A．1或13 B．1 C．13 D．9【答案】C【解析】根據(jù)雙曲線定義可得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故答案為：C3．（2022紅塔月考）已知SKIPIF1<0是雙曲線SKIPIF1<0的左焦點(diǎn)，點(diǎn)SKIPIF1<0，SKIPIF1<0是雙曲線右支上的動(dòng)點(diǎn)，則SKIPIF1<0的最小值為（）A．9 B．5 C．8 D．4【答案】A【解析】設(shè)右焦點(diǎn)為F'，則F'(4，0)，依題意，有PF|=|PF'|+4，|PF|+|PA|=|PF'|+|PA|+4≥|AF'|+4=5+4=9(當(dāng)P在線段AF'上時(shí)，取等號(hào))故|PF|+|PA|的最小值為9.故答案為：A4（2022廣東）已知SKIPIF1<0，SKIPIF1<0分別是雙曲線SKIPIF1<0的左?右焦點(diǎn)，若P是雙曲線左支上的點(diǎn)，且SKIPIF1<0.則SKIPIF1<0的面積為（）A．8 B．SKIPIF1<0 C．16 D．SKIPIF1<0【答案】C【解析】因?yàn)镻是雙曲線左支上的點(diǎn)，所以SKIPIF1<0，兩邊平方得SKIPIF1<0，所以SKIPIF1<0.在SKIPIF1<0中，由余弦定理得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0。故答案為：C考點(diǎn)二雙曲線的離心率及漸近線【例2-1】（2022龍崗期中）雙曲線SKIPIF1<0的漸近線方程是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0表示焦點(diǎn)在y軸上的雙曲線，且a2=3，b2=1，則SKIPIF1<0則漸近線為SKIPIF1<0，即SKIPIF1<0.故答案為為：A【例2-2】（2022長(zhǎng)春月考）在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0.若以A，B為焦點(diǎn)的雙曲線經(jīng)過(guò)點(diǎn)C，則該雙曲線的離心率為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)雙曲線的實(shí)半軸長(zhǎng)，半焦距分別為a，c，因?yàn)镾KIPIF1<0，所以AC>BC，因?yàn)橐訟，B為焦點(diǎn)的雙曲線經(jīng)過(guò)點(diǎn)C所以AC-BC=2a，AB=BC=2c，在三角形ABC中由余弦定理得SKIPIF1<0，即SKIPIF1<0，解得AC2=12c2，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：C【例2-3】（2022·重慶市模擬）已知雙曲線C：SKIPIF1<0的左右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，SKIPIF1<0為等邊三角形，且線段SKIPIF1<0的中點(diǎn)恰在雙曲線C上，則雙曲線C的離心率為（）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】如圖所示，設(shè)SKIPIF1<0，SKIPIF1<0，設(shè)線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0，則SKIPIF1<0在雙曲線C的右支上，又SKIPIF1<0為等邊三角形，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0連接SKIPIF1<0，則在等邊三角形SKIPIF1<0中SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即雙曲線SKIPIF1<0的離心率為SKIPIF1<0.故答案為：C.【一隅三反】1．（2022·湖南模擬）已知O是坐標(biāo)原點(diǎn)，F(xiàn)是雙曲線SKIPIF1<0的右焦點(diǎn)，過(guò)雙曲線C的右頂點(diǎn)且垂直于x軸的直線與雙曲線C的一條漸近線交于A點(diǎn)，若以F為圓心的圓經(jīng)過(guò)點(diǎn)A，O，則雙曲線C的漸近線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由已知，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，故SKIPIF1<0，因?yàn)橐訤為圓心的圓經(jīng)過(guò)點(diǎn)A，O，所以SKIPIF1<0SKIPIF1<0，則△SKIPIF1<0為等邊三角形，所以SKIPIF1<0，則SKIPIF1<0，所以雙曲線C的漸近線方程為SKIPIF1<0.故答案為：A2．（2022高三上·廣西開學(xué)考）已知SKIPIF1<0，SKIPIF1<0是雙曲線C的兩個(gè)焦點(diǎn)，P為雙曲線上的一點(diǎn)，且SKIPIF1<0；則C的離心率為（）A．1 B．2 C．3 D．4【答案】B【解析】SKIPIF1<0。故答案為：B3．（202懷仁期末）設(shè)SKIPIF1<0，SKIPIF1<0分別是雙曲線SKIPIF1<0的左、右焦點(diǎn)，若雙曲線右支上存在一點(diǎn)SKIPIF1<0，使SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)），且SKIPIF1<0，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】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，故答案為：D4（2022德州月考）已知雙曲線SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0，曲線SKIPIF1<0上一點(diǎn)SKIPIF1<0到SKIPIF1<0軸的距離為SKIPIF1<0，且SKIPIF1<0，則雙曲線SKIPIF1<0的離心率為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】作SKIPIF1<0軸于SKIPIF1<0，如圖，依題意SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，由SKIPIF1<0得：SKIPIF1<0，由雙曲線定義知SKIPIF1<0，而SKIPIF1<0，在SKIPIF1<0中，由余弦定理得：SKIPIF1<0，解得：SKIPIF1<0，即SKIPIF1<0，又因?yàn)殡x心率SKIPIF1<0，于是有SKIPIF1<0，所以雙曲線SKIPIF1<0的離心率為SKIPIF1<0。故答案為：B5．（2022遼寧期中）（多選）已知雙曲線SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是雙曲線上一點(diǎn)，若SKIPIF1<0，則該雙曲線的離心率可以是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】A,B【解析】SKIPIF1<0是雙曲線右支上一點(diǎn)，SKIPIF1<0則有SKIPIF1<0，又SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0，則雙曲線的離心率取值范圍為SKIPIF1<0AB符合題意；CD不符合題意.故答案為：AB考點(diǎn)三雙曲線的標(biāo)準(zhǔn)方程【例3-1】（2022·海寧模擬）已知雙曲線C的漸近線方程為SKIPIF1<0，且焦距為10，則雙曲線C的標(biāo)準(zhǔn)方程是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C【解析】漸近線方程為SKIPIF1<0的雙曲線為SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故答案為：C．【例3-2】（2022·河西模擬）已知雙曲線的一個(gè)焦點(diǎn)與拋物線SKIPIF1<0的焦點(diǎn)重合，且雙曲線上的一點(diǎn)SKIPIF1<0到雙曲線的兩個(gè)焦點(diǎn)的距離之差的絕對(duì)值等于6，則雙曲線的標(biāo)準(zhǔn)方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因?yàn)镾KIPIF1<0的焦點(diǎn)為SKIPIF1<0，故雙曲線的焦點(diǎn)在SKIPIF1<0軸上，故設(shè)雙曲線方程為SKIPIF1<0，則SKIPIF1<0；由雙曲線定義知：SKIPIF1<0，解得SKIPIF1<0；故可得SKIPIF1<0；則雙曲線方程為：SKIPIF1<0.故答案為：C.【一隅三反】1．（2022·河南模擬）已知雙曲線SKIPIF1<0的一條漸近線過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的左焦點(diǎn)，且SKIPIF1<0，則雙曲線SKIPIF1<0的方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意可知，雙曲線SKIPIF1<0的漸近線方程為SKIPIF1<0，點(diǎn)SKIPIF1<0在一條漸近線上，如圖示：所以SKIPIF1<0，則SKIPIF1<0，且兩條漸近線的傾斜角分別為60°，120°，則SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)），所以SKIPIF1<0為等邊三角形，從而SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以雙曲線SKIPIF1<0的方程為SKIPIF1<0，故答案為：A.2（2021高三上·寧波期末）已知雙曲線SKIPIF1<0與雙曲線SKIPIF1<0有相同的漸近線，且它們的離心率不相同，則下列方程中有可能為雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】雙曲線SKIPIF1<0中，SKIPIF1<0，則漸近線方程為SKIPIF1<0，離心率為SKIPIF1<0。對(duì)于A，SKIPIF1<0，則離心率SKIPIF1<0，故A錯(cuò)誤；對(duì)于B，SKIPIF1<0，則漸近線方程為SKIPIF1<0，故B錯(cuò)誤；對(duì)于C，SKIPIF1<0，則離心率SKIPIF1<0，故C錯(cuò)誤；對(duì)于D，SKIPIF1<0，則漸近線方程為SKIPIF1<0，離心率SKIPIF1<0，故D正確。故選：D3（2022南山期末）已知雙曲線C過(guò)點(diǎn)SKIPIF1<0且漸近線為SKIPIF1<0，則雙曲線C的方程是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意可設(shè)雙曲線的方程為SKIPIF1<0，即3x2-y2=λ將點(diǎn)SKIPIF1<0代入上式，得SKIPIF1<0則雙曲線的方程為3x2-y2=1故答案為：A4．（2022商丘）若方程SKIPIF1<0表示雙曲線，則實(shí)數(shù)m的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】方程SKIPIF1<0可化為SKIPIF1<0，它表示雙曲線，則SKIPIF1<0，解得SKIPIF1<0.故答案為：A．5．（2021肇東月考）以SKIPIF1<0，SKIPIF1<0為焦點(diǎn)且過(guò)點(diǎn)SKIPIF1<0的雙曲線方程是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意得SKIPIF1<0，則SKIPIF1<0，則雙曲線方程是SKIPIF1<0，故答案為：A考點(diǎn)四直線與雙曲線的位置關(guān)系【例4-1】（2022·全國(guó)·高三專題練習(xí)）過(guò)SKIPIF1<0且與雙曲線SKIPIF1<0有且只有一個(gè)公共點(diǎn)的直線有（

）A．1條 B．2條 C．3條 D．4條【答案】D【解析】當(dāng)斜率不存在時(shí)，過(guò)SKIPIF1<0的直線與雙曲線沒有公共點(diǎn)；當(dāng)斜率存在時(shí)，設(shè)直線為SKIPIF1<0，聯(lián)立SKIPIF1<0，得SKIPIF1<0①.當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，①式只有一個(gè)解；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，解得SKIPIF1<0；綜上可知過(guò)SKIPIF1<0且與雙曲線SKIPIF1<0有且只有一個(gè)公共點(diǎn)的直線有4條.故選：D.【例4-2】（2022·全國(guó)·專題練習(xí)）若過(guò)點(diǎn)SKIPIF1<0的直線SKIPIF1<0與雙曲線SKIPIF1<0:SKIPIF1<0的右支相交于不同兩點(diǎn)，則直線SKIPIF1<0斜率的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意可得直線SKIPIF1<0斜率存在，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，設(shè)交點(diǎn)SKIPIF1<0，聯(lián)立SKIPIF1<0可得SKIPIF1<0，由題意可得SKIPIF1<0解得：SKIPIF1<0，故選：D．【一隅三反】1．（2022·安徽）直線SKIPIF1<0與雙曲線SKIPIF1<0沒有公共點(diǎn)，則斜率k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】聯(lián)立直線SKIPIF1<0和雙曲線：SKIPIF1<0，消去SKIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，此時(shí)方程為SKIPIF1<0，解得SKIPIF1<0，此時(shí)直線與雙曲線有且只有一個(gè)交點(diǎn)；當(dāng)SKIPIF1<0，此時(shí)SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0時(shí)直線與雙曲線無(wú)交點(diǎn)；故選：A2．（2022·全國(guó)·高三專題練習(xí)）若雙曲線SKIPIF1<0的一個(gè)頂點(diǎn)為A，過(guò)點(diǎn)A的直線SKIPIF1<0與雙曲線只有一個(gè)公共點(diǎn)，則該雙曲線的焦距為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0斜率為SKIPIF1<0，過(guò)點(diǎn)A的直線SKIPIF1<0與雙曲線只有一個(gè)公共點(diǎn)，則該直線與雙曲線的漸近線SKIPIF1<0平行，且過(guò)雙曲線右頂點(diǎn)(a，0)，故SKIPIF1<0=SKIPIF1<0，且a-3=0，解得a=3，b=1，故c=SKIPIF1<0，故焦距為2c=SKIPIF1<0．故選：D．考點(diǎn)五弦長(zhǎng)與中點(diǎn)弦【例5-1】（2021·江西?。┮阎p曲線x2－y2＝a2（a＞0）與直線y＝SKIPIF1<0x交于A、B兩點(diǎn)，且｜AB｜＝2SKIPIF1<0，則a＝_____【答案】3【解析】由題設(shè)，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0.故答案為：3【例5-2】（2022·河南·模擬預(yù)測(cè)（文））已知雙曲線SKIPIF1<0的離心率為SKIPIF1<0，直線SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，則SKIPIF1<0與SKIPIF1<0的斜率的乘積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，兩式作差，并化簡(jiǎn)得，SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0為線段SKIPIF1<0的中點(diǎn)，即SKIPIF1<0所以SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0.故選：B.【一隅三反】1．（2021·全國(guó)·高二課時(shí)練習(xí)）已知雙曲線SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線l與雙曲線C交于M?N兩點(diǎn)，若P為線段MN的中點(diǎn)，則弦長(zhǎng)|MN|等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題設(shè)，直線l的斜率必存在，設(shè)過(guò)SKIPIF1<0的直線MN為SKIPIF1<0，聯(lián)立雙曲線：SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.弦長(zhǎng)|MN|SKIPIF1<0.故選：D.2．（2022·全國(guó)·高三專題練習(xí)）已知雙曲線SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0過(guò)左焦點(diǎn)SKIPIF1<0作斜率為2的直線與雙曲線交于A，B兩點(diǎn)，P是AB的中點(diǎn)，O為坐標(biāo)原點(diǎn)，若直線OP的斜率為SKIPIF1<0，則b的值是（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】設(shè)SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，兩式相減可得SKIPIF1<0，SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故選：D.3．（2022·山東煙臺(tái)·三模）過(guò)雙曲線SKIPIF1<0：SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的焦點(diǎn)且斜率不為0的直線交SKIPIF1<0于A，SKIPIF1<0兩點(diǎn)，SKIPIF1<0為SKIPIF1<0中點(diǎn)，若SKIPIF1<0，則SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】妨設(shè)過(guò)雙曲線SKIPIF1<0的焦點(diǎn)且斜率不為0的直線為SKIPIF1<0，令SKIPIF1<0由SKIPIF1<0，整理得SKIPIF1<0則SKIPIF1<0，SKIPIF1<0則SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0則有SKIPIF1<0，即SKIPIF1<0，則雙曲線SKIPIF1<0的離心率SKIPIF1<0故選：D4．（2023·全國(guó)·高三專題練習(xí)）已知雙曲線C的中心在坐標(biāo)原點(diǎn)，其中一個(gè)焦點(diǎn)為SKIPIF1<0，過(guò)F的直線l與雙曲線C交于A、B兩點(diǎn)，且AB的中點(diǎn)為SKIPIF1<0，則C的離心率為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由F、N兩點(diǎn)的坐標(biāo)得直線l的斜率SKIPIF1<0．∵雙曲線一個(gè)焦點(diǎn)為(-2，0)，∴c=2．設(shè)雙曲線C的方程為SKIPIF1<0，則SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，易得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴雙曲線C的離心率SKIPIF1<0．故選：B．5．（2022·四川省）已知中心在原點(diǎn)的雙曲線C的右焦點(diǎn)為（2，0），右頂點(diǎn)為（SKIPIF1<0，0）．(1)求雙曲線C的方程；(2)若直線l：y＝x+2與雙曲線交于A，B兩點(diǎn)，求弦長(zhǎng)|AB|．【答案】(1)SKIPIF1<0y2＝1(2)2SKIPIF1<0【解析】(1)由已知得aSKIPIF1<0，c＝2，再由c2＝a2+b2，得b2＝1，所以雙曲線C的方程為SKIPIF1<0y2＝1．(2)由直線與雙曲線聯(lián)立得2x2+12x+15＝0，解得x＝﹣3±SKIPIF1<0,SKIPIF1<0,∴|AB|SKIPIF1<02SKIPIF1<0．10.4雙曲線（精練）（基礎(chǔ)版）題組一題組一雙曲線的定義及應(yīng)用1．（2021·太原期末）已知SKIPIF1<0，SKIPIF1<0分別是雙曲線SKIPIF1<0的左右焦點(diǎn)，點(diǎn)P在該雙曲線上，若SKIPIF1<0，則SKIPIF1<0（）A．4 B．4或6 C．3 D．3或7【答案】D【解析】由雙曲線定義知：SKIPIF1<0，而SKIPIF1<0，又SKIPIF1<0且SKIPIF1<0，∴SKIPIF1<03或7，故答案為：D.2．（2022郫都期中）雙曲線SKIPIF1<0的兩個(gè)焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，雙曲線上一點(diǎn)SKIPIF1<0到SKIPIF1<0的距離為11，則點(diǎn)SKIPIF1<0到SKIPIF1<0的距離為（）A．1 B．21 C．1或21 D．2或21【答案】B【解析】不妨設(shè)SKIPIF1<0，SKIPIF1<0分別為雙曲線的左右焦點(diǎn)，當(dāng)P在雙曲線的左支時(shí)，由雙曲線的定義可知，SKIPIF1<0，又SKIPIF1<0=11，所以SKIPIF1<0，當(dāng)P在雙曲線的右支時(shí)，由雙曲線的定義可知，SKIPIF1<0，又SKIPIF1<0=11，所以SKIPIF1<0，又SKIPIF1<0，所以右支上不存在滿足條件的點(diǎn)P.故答案為：B.3．（2021懷仁期中）已知SKIPIF1<0，SKIPIF1<0是雙曲線SKIPIF1<0的左右焦點(diǎn)，過(guò)SKIPIF1<0的直線SKIPIF1<0與曲線SKIPIF1<0的右支交于SKIPIF1<0兩點(diǎn)，則SKIPIF1<0的周長(zhǎng)的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由雙曲線SKIPIF1<0可知：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0的周長(zhǎng)為SKIPIF1<0.SKIPIF1<0當(dāng)SKIPIF1<0軸時(shí)，SKIPIF1<0的周長(zhǎng)最小值為SKIPIF1<0故答案為：C4．（2022奉賢期中）已知SKIPIF1<0是雙曲線SKIPIF1<0右支上的一點(diǎn)，雙曲線的一條漸近線方程為SKIPIF1<0.設(shè)SKIPIF1<0分別為雙曲線的左、右焦點(diǎn).若SKIPIF1<0，則SKIPIF1<0.【答案】5【解析】因?yàn)殡p曲線的漸近線方程為3x-y=0，即y=3x=SKIPIF1<0，所以SKIPIF1<0，解得a=1，根據(jù)雙曲線定義P是雙曲線SKIPIF1<0右支上的一點(diǎn)，滿足|PF1|-|PF2|=2a=2，所以|PF1|=|PF2|+2=5.故答案為：55．（2022·開封模擬）若雙曲線SKIPIF1<0的焦距為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0．【答案】4或SKIPIF1<0【解析】當(dāng)焦點(diǎn)在x軸時(shí)，可得m>0，2m?4>0，2m+2m?4=42當(dāng)焦點(diǎn)在y軸時(shí)，可得m<0，2m?4<0，2?m+4?2m=42所以SKIPIF1<0或SKIPIF1<0．故答案為：4或SKIPIF1<06．（2022·岳普湖模擬）已知雙曲線SKIPIF1<0，F(xiàn)1，F(xiàn)2是雙曲線的左右兩個(gè)焦點(diǎn)，P在雙曲線上且在第一象限，圓M是△F1PF2的內(nèi)切圓.則M的橫坐標(biāo)為，若F1到圓M上點(diǎn)的最大距離為SKIPIF1<0，則△F1PF2的面積為.【答案】1；SKIPIF1<0【解析】雙曲線的方程為SKIPIF1<0，則SKIPIF1<0.設(shè)圓SKIPIF1<0分別與SKIPIF1<0相切于SKIPIF1<0，根據(jù)雙曲線的定義可知SKIPIF1<0，根據(jù)內(nèi)切圓的性質(zhì)可知SKIPIF1<0①，而SKIPIF1<0②.由①②得：SKIPIF1<0，所以SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0.設(shè)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0到圓M上點(diǎn)的最大距離為SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，解得SKIPIF1<0.所以線SKIPIF1<0的方程為SKIPIF1<0.由SKIPIF1<0且SKIPIF1<0在第一象限，解得SKIPIF1<0.所以SKIPIF1<0，SKIPIF1<0.所以△F1PF2的面積為SKIPIF1<0SKIPIF1<0SKIPIF1<0.故答案為：1；SKIPIF1<07．（2021溫州期中）已知雙曲線x2-y2=1，點(diǎn)F1，F(xiàn)2為其兩個(gè)焦點(diǎn)，點(diǎn)P為雙曲線上一點(diǎn)，若PF1⊥PF2，則∣PF1∣+∣PF2∣的值為.【答案】SKIPIF1<0【解析】∵PF1⊥PF2，SKIPIF1<0∴|PF1|2+|PF2|2=|F1F2|2．∵雙曲線方程為x2﹣y2=1，∴a2=b2=1，c2=a2+b2=2，可得F1F2=2∴|PF1|2+|PF2|2=|F1F2|2=8又∵P為雙曲線x2﹣y2=1上一點(diǎn)，∴|PF1|﹣|PF2|=±2a=±2，（|PF1|﹣|PF2|）2=4因此（|PF1|+|PF2|）2=2（|PF1|2+|PF2|2）﹣（|PF1|﹣|PF2|）2=12∴|PF1|+|PF2|的值為SKIPIF1<0故答案為SKIPIF1<0題組二題組二雙曲線的離心率及漸近線1．（2021高三上·南開期末）已知雙曲線SKIPIF1<0，過(guò)原點(diǎn)作一條傾斜角為SKIPIF1<0的直線分別交雙曲線左、右兩支于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，以線段SKIPIF1<0為直徑的圓過(guò)右焦點(diǎn)SKIPIF1<0，則雙曲線的離心率為（）.A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)雙曲線的左焦點(diǎn)為SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0，如下圖所示：由題意可知，點(diǎn)SKIPIF1<0為SKIPIF1<0的中點(diǎn)，也為SKIPIF1<0的中點(diǎn)，且SKIPIF1<0，則四邊形SKIPIF1<0為矩形，故SKIPIF1<0，由已知可知SKIPIF1<0，由直角三角形的性質(zhì)可得SKIPIF1<0，故SKIPIF1<0為等邊三角形，故SKIPIF1<0，所以，SKIPIF1<0，由雙曲線的定義可得SKIPIF1<0，所以，SKIPIF1<0.故答案為：A.2．（2022湖南月考）已知雙曲線的左焦點(diǎn)為SKIPIF1<0，右焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為雙曲線右支上一點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，滿足SKIPIF1<0，且SKIPIF1<0，則該雙曲線的離心率為（）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】B【解析】∵SKIPIF1<0，O為SKIPIF1<0的中點(diǎn)，∴△SKIPIF1<0為直角三角形，設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，∴e＝SKIPIF1<0.故答案為：B.3．（2021·全國(guó)甲卷）已知F1，F(xiàn)2是雙曲線C的兩個(gè)焦點(diǎn)，P為C上一點(diǎn)，且∠F1PF2=60°，|PF1|=3|PF2|，則C的離心率為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由|PF1|=3|PF2|，|PF1|-|PF2|=2a得|PF1|=3a，|PF2|=a在△F1PF2中，由|F1F2|2=|PF1|2+|PF2|2-2|PF1||PF2|cos∠F1PF2得(2c)2=(3a)2+a2-2×3a×a×cos60°解得SKIPIF1<0所以SKIPIF1<0故答案為：A4．（2022·靖遠(yuǎn)模擬）若雙曲線SKIPIF1<0的兩條漸近線與直線y＝2圍成了一個(gè)等邊三角形，則C的離心率為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】D【解析】由題意得：漸近線方程的斜率為SKIPIF1<0，又漸近線方程為SKIPIF1<0，所以SKIPIF1<0，所以C的離心率為SKIPIF1<0故答案為：D5．（2022·新鄉(xiāng)三模）已知雙曲線SKIPIF1<0的頂點(diǎn)到一條漸近線的距離為實(shí)軸長(zhǎng)的SKIPIF1<0，則雙曲線C的離心率為（）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．3【答案】B【解析】因?yàn)殡p曲線C的頂點(diǎn)SKIPIF1<0到一條漸近線SKIPIF1<0的距離為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，雙曲線C的離心率SKIPIF1<0.故答案為：B6．（2022·湘贛皖模擬）已知雙曲線SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，雙曲線C上一點(diǎn)P到x軸的距離為c，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】C【解析】作SKIPIF1<0軸于M，依題意SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0為等腰直角三角形，令SKIPIF1<0，則SKIPIF1<0，由雙曲線定義知SKIPIF1<0．而SKIPIF1<0，在SKIPIF1<0中SKIPIF1<0，解得：SKIPIF1<0，雙曲線離心率SKIPIF1<0，則SKIPIF1<0．故答案為：C．7．（2022·濟(jì)南二模）已知SKIPIF1<0，SKIPIF1<0分別為雙曲線SKIPIF1<0的左?右焦點(diǎn)，點(diǎn)P在雙曲線上，若SKIPIF1<0，SKIPIF1<0，則雙曲線的離心率為.【答案】SKIPIF1<0【解析】不妨假設(shè)點(diǎn)P在雙曲線右支上，則SKIPIF1<0，由于SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，而SKIPIF1<0，故SKIPIF1<0，故答案為：SKIPIF1<08．（2022·汝州模擬）已知雙曲線SKIPIF1<0的兩條漸近線所夾銳角為SKIPIF1<0，則雙曲線的離心率為．【答案】SKIPIF1<0【解析】由于SKIPIF1<0，雙曲線的漸近線方程為SKIPIF1<0，SKIPIF1<0，所以雙曲線的漸近線與SKIPIF1<0軸夾角小于SKIPIF1<0，由，得SKIPIF1<0，則雙曲線的離心率SKIPIF1<0．故答案為：SKIPIF1<0題組三題組三雙曲線的標(biāo)準(zhǔn)方程1．（2022·安徽模擬）與橢圓SKIPIF1<0共焦點(diǎn)且過(guò)點(diǎn)SKIPIF1<0的雙曲線的標(biāo)準(zhǔn)方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】橢圓SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0，設(shè)雙曲線的標(biāo)準(zhǔn)方程為SKIPIF1<0，由雙曲線的定義可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因此，雙曲線的方程為SKIPIF1<0。故答案為：C.2．（2022合肥期末）已知點(diǎn)SKIPIF1<0分別是等軸雙曲線SKIPIF1<0的左、右焦點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0的面積為8，則雙曲線SKIPIF1<0的方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的中點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，所以雙曲線方程為SKIPIF1<0．故答案為：D．3．（2022資陽(yáng)期末）已知雙曲線SKIPIF1<0過(guò)三點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0中的兩點(diǎn)，則SKIPIF1<0的方程為.【答案】SKIPIF1<0【解析】根據(jù)雙曲線SKIPIF1<0的對(duì)稱性可知，點(diǎn)SKIPIF1<0，SKIPIF1<0在雙曲線圖像上，將其代入雙曲線方程，所以8a2=1，16a2?4b所以雙曲線C：SKIPIF1<0，故答案為：SKIPIF1<0.4．（2022徐匯期末）已知雙曲線經(jīng)過(guò)點(diǎn)SKIPIF1<0，其漸近線方程為SKIPIF1<0，則該雙曲線的方程為．【答案】SKIPIF1<0【解析】考慮到雙曲線的實(shí)軸可能在x軸，也可能在y軸，分別設(shè)雙曲線方程如下：實(shí)軸在x軸時(shí)，設(shè)雙曲線方程為：SKIPIF1<0，則有SKIPIF1<0…①其漸近線方程為SKIPIF1<0，即SKIPIF1<0…②聯(lián)立①②，解得SKIPIF1<0，雙曲線方程為SKIPIF1<0；實(shí)軸在y軸時(shí)，設(shè)雙曲線方程為SKIPIF1<0，則有SKIPIF1<0…③其漸近線方程為SKIPIF1<0，即SKIPIF1<0…④聯(lián)立③④，無(wú)解；故答案為：SKIPIF1<0.5．（2022河南月考）經(jīng)過(guò)點(diǎn)SKIPIF1<0且與雙曲線SKIPIF1<0有公共漸近線的雙曲線方程為．【答案】SKIPIF1<0【解析】由題意設(shè)所求雙曲線的方程為SKIPIF1<0，∵點(diǎn)SKIPIF1<0在雙曲線上，∴SKIPIF1<0，∴所求的雙曲線方程為SKIPIF1<0，即SKIPIF1<0。答案：SKIPIF1<0。6．（2022·湖北模擬）在平面直角坐標(biāo)系中，已知圓SKIPIF1<0：SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0是圓SKIPIF1<0上任意一點(diǎn)，線段SKIPIF1<0的垂直平分線與直線SKIPIF1<0相交于點(diǎn)SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0的軌跡為曲線SKIPIF1<0