新高考數(shù)學(xué)二輪復(fù)習(xí)圓錐曲線重難點(diǎn)提升專題2 直線與圓錐曲線的位置關(guān)系（原卷版）

專題2直線與圓錐曲線的位置關(guān)系一、考情分析直線與圓錐曲線的位置關(guān)系的常見題型,一是根據(jù)直線與圓錐曲線有兩個(gè)交點(diǎn),研究長度、面積、定點(diǎn)、定值等問題,二是判斷直線與圓錐曲線的公共點(diǎn)個(gè)數(shù),三是直線與圓錐曲線相切問題,其中第一類問題是高考考查頻率最高的問題.二、解題秘籍(一)根據(jù)直線與圓錐曲線有兩個(gè)交點(diǎn)研究圓錐曲線的性質(zhì)1.把直線l：SKIPIF1<0與橢圓C：SKIPIF1<0聯(lián)立,當(dāng)SKIPIF1<0時(shí)直線l與橢圓C有2個(gè)交點(diǎn)；2.直線l：SKIPIF1<0與雙曲線C：SKIPIF1<0聯(lián)立得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)直線l與雙曲線C有2個(gè)交點(diǎn)；當(dāng)SKIPIF1<0時(shí)直線l與雙曲線C的左右支各有一個(gè)交點(diǎn)；當(dāng)SKIPIF1<0時(shí)直線l與雙曲線C的右支有2個(gè)交點(diǎn)；3.直線l：SKIPIF1<0與拋物線C：SKIPIF1<0聯(lián)立,得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)直線l與拋物線C有2個(gè)交點(diǎn).【例1】（2023屆重慶市南開中學(xué)校高三上學(xué)期9月月考）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0,上頂點(diǎn)為D,斜率為k的直線l與橢圓C交于不同的兩點(diǎn)A,B,M為線段AB的中點(diǎn),當(dāng)點(diǎn)M的坐標(biāo)為SKIPIF1<0時(shí),直線l恰好經(jīng)過D點(diǎn).(1)求橢圓C的方程：(2)當(dāng)l不過點(diǎn)D時(shí),若直線DM與直線l的斜率互為相反數(shù),求k的取值范圍.【解析】（1）由題意知,離心率SKIPIF1<0,所以SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0兩式相減得SKIPIF1<0,所以SKIPIF1<0；所以直線為SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,橢圓方程為SKIPIF1<0；（2）設(shè)直線為SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0因?yàn)閘不過D點(diǎn),則SKIPIF1<0,即SKIPIF1<0則SKIPIF1<0,化簡得SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0或SKIPIF1<0.【例2】（2023屆廣東省部分學(xué)校高三上學(xué)期聯(lián)考）設(shè)直線SKIPIF1<0與雙曲線SKIPIF1<0：SKIPIF1<0的兩條漸近線分別交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),且三角形SKIPIF1<0的面積為SKIPIF1<0.(1)求SKIPIF1<0的值；(2)已知直線SKIPIF1<0與SKIPIF1<0軸不垂直且斜率不為0,SKIPIF1<0與SKIPIF1<0交于兩個(gè)不同的點(diǎn)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)為SKIPIF1<0,SKIPIF1<0為SKIPIF1<0的右焦點(diǎn),若SKIPIF1<0,SKIPIF1<0,SKIPIF1<0三點(diǎn)共線,證明：直線SKIPIF1<0經(jīng)過SKIPIF1<0軸上的一個(gè)定點(diǎn).【解析】（1）雙曲線SKIPIF1<0：SKIPIF1<0的漸近線方程為SKIPIF1<0,不妨設(shè)SKIPIF1<0,SKIPIF1<0因?yàn)槿切蜸KIPIF1<0的面積為SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0.（2）雙曲線SKIPIF1<0的方程為SKIPIF1<0：SKIPIF1<0,所以右焦點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,若直線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0,故可設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,聯(lián)立SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0且SKIPIF1<0,化簡得SKIPIF1<0且SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,因?yàn)橹本€SKIPIF1<0的斜率存在,所以直線SKIPIF1<0的斜率也存在,因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0三點(diǎn)共線,所以SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,化簡得SKIPIF1<0,所以SKIPIF1<0經(jīng)過SKIPIF1<0軸上的定點(diǎn)SKIPIF1<0.【例3】（2023屆福建省漳州市高三上學(xué)期第一次教學(xué)質(zhì)量檢測(cè)）已知拋物線SKIPIF1<0：SKIPIF1<0,直線SKIPIF1<0過點(diǎn)SKIPIF1<0.(1)若SKIPIF1<0與SKIPIF1<0有且只有一個(gè)公共點(diǎn),求直線SKIPIF1<0的方程；(2)若SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),點(diǎn)SKIPIF1<0在線段SKIPIF1<0上,且SKIPIF1<0,求點(diǎn)SKIPIF1<0的軌跡方程.【解析】（1）當(dāng)直線SKIPIF1<0斜率不存在時(shí),其方程為SKIPIF1<0,符合題意；當(dāng)直線SKIPIF1<0斜率存在時(shí),設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),直線SKIPIF1<0符合題意；當(dāng)SKIPIF1<0時(shí),令SKIPIF1<0,解得SKIPIF1<0,∴直線SKIPIF1<0的方程為SKIPIF1<0,即SKIPIF1<0.綜上,直線SKIPIF1<0的方程為SKIPIF1<0,或SKIPIF1<0,或SKIPIF1<0.（2）設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,不妨令SKIPIF1<0,∵直線SKIPIF1<0與拋物線SKIPIF1<0有兩個(gè)交點(diǎn),∴SKIPIF1<0,∴SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.由SKIPIF1<0,得SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0.∵SKIPIF1<0,且SKIPIF1<0,∴SKIPIF1<0,且SKIPIF1<0,∴點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0（SKIPIF1<0,且SKIPIF1<0）.(二)根據(jù)直線與圓錐曲線有一個(gè)公共點(diǎn)研究圓錐曲線的性質(zhì)1.直線與橢圓有一個(gè)公共點(diǎn),則直線與橢圓相切,可把直線方程與橢圓方程聯(lián)立,整理成關(guān)于x或y的一元二次方程,由SKIPIF1<0求解；2.直線l：SKIPIF1<0與雙曲線C：SKIPIF1<0聯(lián)立得SKIPIF1<0,當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)直線l與雙曲線C有1個(gè)交點(diǎn),即直線與雙曲線相切或與漸近線平行時(shí)與雙曲線有1個(gè)公共點(diǎn)；3.當(dāng)直線l：SKIPIF1<0與拋物線C：SKIPIF1<0聯(lián)立,得SKIPIF1<0,當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)直線l與拋物線C有1個(gè)交點(diǎn),即直線與拋物線相切或與拋物線準(zhǔn)線垂直時(shí)直線與拋物線有1個(gè)公共點(diǎn).【例4】（2023屆湖北省荊荊宜三校高三上學(xué)期9月聯(lián)考）設(shè)橢圓SKIPIF1<0：SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0的左、右焦點(diǎn),點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上,點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0外,且SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)若SKIPIF1<0,點(diǎn)SKIPIF1<0為橢圓SKIPIF1<0上橫坐標(biāo)大于1的一點(diǎn),過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與橢圓有且僅有一個(gè)交點(diǎn),并與直線SKIPIF1<0,SKIPIF1<0交于M,N兩點(diǎn),SKIPIF1<0為坐標(biāo)原點(diǎn),記SKIPIF1<0,SKIPIF1<0的面積分別為SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0的最小值．【解析】（1）因?yàn)辄c(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上,所以SKIPIF1<0,①因?yàn)辄c(diǎn)SKIPIF1<0在橢圓SKIPIF1<0外,且SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,②由①②解得SKIPIF1<0,SKIPIF1<0,故橢圓SKIPIF1<0的方程為SKIPIF1<0．（2）設(shè)點(diǎn)SKIPIF1<0,SKIPIF1<0,設(shè)直線SKIPIF1<0：SKIPIF1<0,由橢圓性質(zhì)以及點(diǎn)SKIPIF1<0的橫坐標(biāo)大于1可知,SKIPIF1<0,將直線SKIPIF1<0代入方程SKIPIF1<0并化簡可得,SKIPIF1<0,即SKIPIF1<0,因?yàn)橹本€SKIPIF1<0與橢圓有且僅有一個(gè)交點(diǎn),所以SKIPIF1<0,即SKIPIF1<0．直線SKIPIF1<0的方程為：SKIPIF1<0；直線SKIPIF1<0的方程為SKIPIF1<0：SKIPIF1<0,聯(lián)立方程SKIPIF1<0得SKIPIF1<0,同理得SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),不等式取等號(hào),故當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值SKIPIF1<0．【例5】已知雙曲線C：SKIPIF1<0的焦距為4,且過點(diǎn)SKIPIF1<0.(1)求雙曲線方程；(2)若直線SKIPIF1<0與雙曲線C有且只有一個(gè)公共點(diǎn),求實(shí)數(shù)SKIPIF1<0的值.【解析】（1）由題意可知雙曲線的焦點(diǎn)為SKIPIF1<0和SKIPIF1<0,根據(jù)定義有SKIPIF1<0．SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．SKIPIF1<0所求雙曲線SKIPIF1<0的方程為SKIPIF1<0．（2）因?yàn)殡p曲線SKIPIF1<0的方程為SKIPIF1<0,所以漸近線方程為SKIPIF1<0；由SKIPIF1<0,消去SKIPIF1<0整理得SKIPIF1<0．①當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),此時(shí)直線SKIPIF1<0與雙曲線的漸近線平行,此時(shí)直線與雙曲線相交于一點(diǎn),符合題意；②當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),由SKIPIF1<0,解得SKIPIF1<0,此時(shí)直線SKIPIF1<0雙曲線相切于一個(gè)公共點(diǎn),符合題意．綜上所述：符合題意的SKIPIF1<0的所有取值為SKIPIF1<0,SKIPIF1<0．【例6】已知頂點(diǎn)在原點(diǎn),焦點(diǎn)在SKIPIF1<0軸上的拋物線過點(diǎn)SKIPIF1<0．(1)求拋物線的標(biāo)準(zhǔn)方程；(2)過點(diǎn)P作直線l與拋物線有且只有一個(gè)公共點(diǎn),求直線l的方程；(3)過點(diǎn)SKIPIF1<0作直線交拋物線于A、B兩點(diǎn),使得Q恰好平分線段AB,求直線AB的方程．【解析】（1）因?yàn)轫旤c(diǎn)在原點(diǎn),焦點(diǎn)在y軸上的拋物線過點(diǎn)SKIPIF1<0,所以拋物線的焦點(diǎn)在y軸正半軸,設(shè)其方程為SKIPIF1<0,將點(diǎn)SKIPIF1<0代入可得SKIPIF1<0,所以SKIPIF1<0,所以拋物線的標(biāo)準(zhǔn)方程為SKIPIF1<0,（2）當(dāng)直線斜率不存在時(shí),過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與拋物線SKIPIF1<0有一個(gè)交點(diǎn)；當(dāng)直線斜率存在時(shí),設(shè)直線斜率為SKIPIF1<0,直線方程為SKIPIF1<0由SKIPIF1<0得SKIPIF1<0,直線與拋物線只有一個(gè)交點(diǎn),所以SKIPIF1<0,解得SKIPIF1<0,所以直線方程為SKIPIF1<0綜上,過點(diǎn)SKIPIF1<0與拋物線SKIPIF1<0有且只有一個(gè)交點(diǎn)的直線方程為SKIPIF1<0和SKIPIF1<0；（3）設(shè)點(diǎn)SKIPIF1<0,直線SKIPIF1<0斜率為SKIPIF1<0點(diǎn)SKIPIF1<0在拋物線上,所以SKIPIF1<0所以SKIPIF1<0,即SKIPIF1<0,所以直線方程為SKIPIF1<0經(jīng)檢驗(yàn),直線SKIPIF1<0符合題意.(三)判斷直線與圓錐曲線公共點(diǎn)個(gè)數(shù)判斷直線l：Ax＋By＋C＝0與圓錐曲線C：F(x,y)＝0公共點(diǎn)個(gè)數(shù)時(shí),通常將直線l的方程Ax＋By＋C＝0(A、B不同時(shí)為0)代入圓錐曲線C的方程F(x,y)＝0,消去y(也可以消去x)得到一個(gè)關(guān)于變量x(或變量y)的一元方程．消去y后得ax2＋bx＋c＝0.(1)當(dāng)a≠0時(shí),設(shè)一元二次方程ax2＋bx＋c＝0的判別式為Δ,則Δ>0?直線與圓錐曲線C有2個(gè)公共點(diǎn)；Δ＝0?直線與圓錐曲線C有1個(gè)公共點(diǎn)；Δ<0?直線與圓錐曲線C沒有公共點(diǎn).(2)當(dāng)a＝0,b≠0時(shí),即得到一個(gè)一次方程,則直線l與圓錐曲線C相交,且只有一個(gè)交點(diǎn),此時(shí),若C為雙曲線,則直線l與雙曲線的漸近線平行；若C為拋物線,則直線l與拋物線的對(duì)稱軸重合或平行．【例7】（2022屆浙江省溫州市樂清市高三下學(xué)期5月仿真）已知SKIPIF1<0分別是橢圓SKIPIF1<0的左、右焦點(diǎn),點(diǎn)SKIPIF1<0在直線SKIPIF1<0的同側(cè),且點(diǎn)SKIPIF1<0到直線l的距離分別為SKIPIF1<0．(1)若橢圓C的方程為SKIPIF1<0,直線l的方程為SKIPIF1<0,求SKIPIF1<0的值,并判斷直線與橢圓C的公共點(diǎn)的個(gè)數(shù)；(2)若直線l與橢圓C有兩個(gè)公共點(diǎn),試求SKIPIF1<0所需要滿足的條件；【解析】（1）橢圓C的方程為SKIPIF1<0,則SKIPIF1<0.又直線SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0所以SKIPIF1<0.聯(lián)立SKIPIF1<0,消去y可得：SKIPIF1<0.因?yàn)镾KIPIF1<0,所以直線SKIPIF1<0與橢圓C有1個(gè)公共點(diǎn).（2）聯(lián)立SKIPIF1<0,消去y可得：SKIPIF1<0.因?yàn)橹本€l與橢圓C有兩個(gè)公共點(diǎn),所以SKIPIF1<0,整理化簡得：SKIPIF1<0.又SKIPIF1<0,其中SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0.所以直線l與橢圓C有兩個(gè)公共點(diǎn),則SKIPIF1<0.【例8】如圖,F是拋物線SKIPIF1<0的焦點(diǎn),Q是準(zhǔn)線與x軸的交點(diǎn),斜率為k的直線l經(jīng)過點(diǎn)Q．(1)當(dāng)k取不同數(shù)值時(shí),求直線l與拋物線公共點(diǎn)的個(gè)數(shù)；(2)若直線l與拋物線相交于A、B兩點(diǎn),求證：SKIPIF1<0是定值．(3)在x軸上是否存在這樣的定點(diǎn)M,對(duì)任意的過點(diǎn)Q的直線l與拋物線相交于A、B兩點(diǎn),均能使得SKIPIF1<0為定值,若有,找出滿足條件的點(diǎn)M;若沒有,請(qǐng)說明理由．【解析】（1）拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0,SKIPIF1<0,設(shè)SKIPIF1<0,代入SKIPIF1<0并化簡得SKIPIF1<0①．當(dāng)SKIPIF1<0,直線SKIPIF1<0的方程為SKIPIF1<0,與SKIPIF1<0的交點(diǎn)為原點(diǎn)SKIPIF1<0,直線l與拋物線有SKIPIF1<0個(gè)公共點(diǎn)；當(dāng)SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,即SKIPIF1<0,直線l與拋物線有SKIPIF1<0個(gè)公共點(diǎn)；若SKIPIF1<0,即SKIPIF1<0時(shí),直線l與拋物線有SKIPIF1<0個(gè)公共點(diǎn)；若SKIPIF1<0,即SKIPIF1<0或SKIPIF1<0,直線l與拋物線沒有公共點(diǎn)．（2）由于直線SKIPIF1<0與拋物線有兩個(gè)交點(diǎn),由（1）得SKIPIF1<0.設(shè)交點(diǎn)SKIPIF1<0、SKIPIF1<0,由①得SKIPIF1<0,SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0為定值0．（3）若存在滿足條件的點(diǎn)SKIPIF1<0,使得SKIPIF1<0為定值．則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0為定值SKIPIF1<0．三、跟蹤檢測(cè)1．已知,分別是橢圓的左、右焦點(diǎn),點(diǎn),在直線的同側(cè),且點(diǎn),到直線l的距離分別為,.(1)若橢圓C的方程為,直線l的方程為,求的值,并判斷直線l與橢圓C的公共點(diǎn)的個(gè)數(shù)；(2)若直線l與橢圓C有兩個(gè)公共點(diǎn),試求所需要滿足的條件；(3)結(jié)合（1）和（2）,試寫出一個(gè)能判斷直線l與橢圓C有公共點(diǎn)的充要條件（不需要證明）.2．已知拋物線的方程為,直線l繞點(diǎn)旋轉(zhuǎn),討論直線l與拋物線的公共點(diǎn)個(gè)數(shù),并回答下列問題：（1）畫出圖形表示直線l與拋物線的各種位置關(guān)系,從圖中你發(fā)現(xiàn)直線l與拋物線只有一個(gè)公共點(diǎn)時(shí)是什么情況？（2）與直線l的方程組成的方程組解的個(gè)數(shù)與公共點(diǎn)的個(gè)數(shù)是什么關(guān)系？3．（2023屆四川、云南部分學(xué)校高三上學(xué)期9月聯(lián)考）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0,其左右焦點(diǎn)分別為SKIPIF1<0,點(diǎn)SKIPIF1<0是橢圓上任意一點(diǎn),且滿足SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)過橢圓外一點(diǎn)SKIPIF1<0作橢圓的兩條切線SKIPIF1<0,滿足SKIPIF1<0,求動(dòng)點(diǎn)SKIPIF1<0的軌跡方程.4．（2023屆甘肅省金昌市永昌縣高三上學(xué)期模擬）已知橢圓SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0是SKIPIF1<0上一動(dòng)點(diǎn),SKIPIF1<0的最大面積為SKIPIF1<0.(1)求SKIPIF1<0的方程；(2)若直線SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0兩點(diǎn),SKIPIF1<0為SKIPIF1<0上兩點(diǎn),且SKIPIF1<0,求四邊形SKIPIF1<0面積的最大值.5．設(shè)SKIPIF1<0為雙曲線SKIPIF1<0的左?右頂點(diǎn),直線SKIPIF1<0過右焦點(diǎn)SKIPIF1<0且與雙曲線SKIPIF1<0的右支交于SKIPIF1<0兩點(diǎn),當(dāng)直線SKIPIF1<0垂直于SKIPIF1<0軸時(shí),SKIPIF1<0為等腰直角三角形.(1)求雙曲線SKIPIF1<0的離心率；(2)已知SKIPIF1<0,若直線SKIPIF1<0分別交直線SKIPIF1<0于SKIPIF1<0兩點(diǎn),當(dāng)直線SKIPIF1<0的傾斜角變化時(shí),以SKIPIF1<0為直徑的圓是否過定點(diǎn),若過定點(diǎn)求出定點(diǎn)的坐標(biāo)；若不過定點(diǎn),請(qǐng)說明理由.6．（2023屆安徽省部分校高三上學(xué)期開學(xué)摸底）已知SKIPIF1<0為坐標(biāo)原點(diǎn),橢圓SKIPIF1<0過點(diǎn)SKIPIF1<0,記線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0．(1)若直線SKIPIF1<0的斜率為3,求直線SKIPIF1<0的斜率;(2)若四邊形SKIPIF1<0為平行四邊形,求SKIPIF1<0的取值范圍．7．（2023屆浙江省A9協(xié)作體高三上學(xué)期聯(lián)考）已知直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩個(gè)不同的點(diǎn).(1)求SKIPIF1<0的取值范圍；(2)若SKIPIF1<0為雙曲線SKIPIF1<0的左頂點(diǎn),點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0的左支上,點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0的右支上,且直線SKIPIF1<0、SKIPIF1<0分別與SKIPIF1<0軸交于SKIPIF