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14 圓錐曲線中的證明問題（原卷版）

專題14圓錐曲線中的證明問題一、考情分析圓錐曲線中的證明問題在高考時(shí)有出現(xiàn),主要有兩大類：一是證明點(diǎn)線位置關(guān)系,如直線或曲線過某個(gè)點(diǎn)、直線平行與垂直、直線對(duì)稱等問題,二是證明直線與圓錐曲線中的一些數(shù)量關(guān)系,如相等與不相等.二、解題秘籍(一)證明直線或圓過定點(diǎn)證明直線過定點(diǎn),通常是設(shè)出直線方程SKIPIF1<0,由已知條件確定SKIPIF1<0的關(guān)系.若SKIPIF1<0,則SKIPIF1<0,則直線過定點(diǎn)SKIPIF1<0,證明圓過定點(diǎn),常見題型是證明以AB為直徑的圓過定點(diǎn)P,只需證明SKIPIF1<0.【例1】（2023屆重慶市南開中學(xué)校高三上學(xué)期質(zhì)量檢測(cè)）已知橢圓C：SKIPIF1<0的離心率為SKIPIF1<0,橢圓的上頂點(diǎn)B到兩焦點(diǎn)的距離之和為4．(1)求橢圓C的標(biāo)準(zhǔn)方程；(2)若直線l：SKIPIF1<0與橢圓C交于異于點(diǎn)B的兩點(diǎn)P,Q,直線BP,BQ與x軸相交于SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,求證：直線SKIPIF1<0過一定點(diǎn),并求出定點(diǎn)坐標(biāo)．【解析】（1）∵SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．故橢圓方程為SKIPIF1<0；（2）聯(lián)立直線和橢圓可得SKIPIF1<0,解得SKIPIF1<0,于是有：SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．由題意BP：SKIPIF1<0,BQ：SKIPIF1<0,分別和SKIPIF1<0聯(lián)立得,SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0整理得SKIPIF1<0,整理得SKIPIF1<0,解得SKIPIF1<0或者SKIPIF1<0．當(dāng)SKIPIF1<0時(shí),直線SKIPIF1<0過點(diǎn)B,與題意矛盾,應(yīng)舍去.故直線SKIPIF1<0的方程為：SKIPIF1<0,過定點(diǎn)為SKIPIF1<0．【例2】（2023屆福建省福州華僑中學(xué)高三上學(xué)期第二次考試）在平面直角坐標(biāo)系SKIPIF1<0中,已知點(diǎn)SKIPIF1<0,直線SKIPIF1<0,點(diǎn)M到l的距離為d,若點(diǎn)M滿足SKIPIF1<0,記M的軌跡為C．(1)求C的方程；(2)過點(diǎn)SKIPIF1<0且斜率不為0的直線與C交于P,Q兩點(diǎn),設(shè)SKIPIF1<0,證明：以P,Q為直徑的圓經(jīng)過點(diǎn)A．【解析】（1）設(shè)點(diǎn)SKIPIF1<0,則SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,兩邊平方整理得SKIPIF1<0,則所求曲線SKIPIF1<0的方程為SKIPIF1<0.（2）設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立方程SKIPIF1<0,消去SKIPIF1<0并整理得SKIPIF1<0,因?yàn)橹本€SKIPIF1<0與SKIPIF1<0交于兩點(diǎn),故SKIPIF1<0,此時(shí)SKIPIF1<0,所以SKIPIF1<0,而SKIPIF1<0.又SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0所以SKIPIF1<0,即以P,Q為直徑的圓經(jīng)過點(diǎn)A.(二)證明與斜率有關(guān)的定值問題證明與斜率有關(guān)的定值問題通常是證明斜率之和或斜率之積為定值問題,此類問題通常是把斜率之和或斜率之積用點(diǎn)的坐標(biāo)表示,再通過化簡(jiǎn)使結(jié)果為定值,此外證明垂直問題可轉(zhuǎn)化為斜率之積為SKIPIF1<0,證明兩直線關(guān)于直線SKIPIF1<0或SKIPIF1<0對(duì)稱,可轉(zhuǎn)化為證明斜率之和為0.【例3】（2023屆河南省安陽(yáng)市高三上學(xué)期10月月考）已知橢圓SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,面積為SKIPIF1<0的正方形ABCD的頂點(diǎn)都在SKIPIF1<0上.(1)求SKIPIF1<0的方程；(2)已知P為橢圓SKIPIF1<0上一點(diǎn),過點(diǎn)P作SKIPIF1<0的兩條切線SKIPIF1<0和SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0的斜率分別為SKIPIF1<0,SKIPIF1<0,求證：SKIPIF1<0為定值.【解析】（1）根據(jù)對(duì)稱性,不妨設(shè)正方形的一個(gè)頂點(diǎn)為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,整理得SKIPIF1<0.①又SKIPIF1<0,②由①②解得SKIPIF1<0,SKIPIF1<0,故所求橢圓方程為SKIPIF1<0.（2）由已知及（1）可得SKIPIF1<0,設(shè)點(diǎn)SKIPIF1<0,則SKIPIF1<0.設(shè)過點(diǎn)P與SKIPIF1<0相切的直線l的方程為SKIPIF1<0,與SKIPIF1<0聯(lián)立消去y整理可得SKIPIF1<0,令SKIPIF1<0,整理可得SKIPIF1<0,③根據(jù)題意SKIPIF1<0和SKIPIF1<0為方程③的兩個(gè)不等實(shí)根,所以SKIPIF1<0,即SKIPIF1<0為定值SKIPIF1<0.【例4】（2023屆天津市第四十七中學(xué)高三上學(xué)期測(cè)試）已知橢圓SKIPIF1<0：SKIPIF1<0的右焦點(diǎn)和上頂點(diǎn)均在直線SKIPIF1<0上.(1)求橢圓SKIPIF1<0的方程；(2)已知點(diǎn)SKIPIF1<0,若過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與橢圓SKIPIF1<0交于不同的兩點(diǎn)SKIPIF1<0,SKIPIF1<0.直線SKIPIF1<0和直線SKIPIF1<0的斜率分別為SKIPIF1<0和SKIPIF1<0,求證：SKIPIF1<0為定值.【解析】（1）對(duì)于直線SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,因?yàn)闄E圓的右焦點(diǎn)和上頂點(diǎn)均在直線SKIPIF1<0上,所以SKIPIF1<0,所以SKIPIF1<0,所以橢圓方程為SKIPIF1<0,（2）因?yàn)镾KIPIF1<0在橢圓外,過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與橢圓SKIPIF1<0交于不同的兩點(diǎn),所以直線SKIPIF1<0的斜率一定存在,所以設(shè)直線SKIPIF1<0方程為SKIPIF1<0,設(shè)SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0(三)證明與線段長(zhǎng)度有關(guān)的等式證明與線段長(zhǎng)度有關(guān)的等式問題,一般是利用距離公式或弦長(zhǎng)公式寫出長(zhǎng)度表達(dá)式,再借助根與系數(shù)之間的關(guān)系或斜率、截距等證明等式兩邊相等.【例5】（2023屆江蘇省高三上學(xué)期起航調(diào)研）在平面直角坐標(biāo)系xOy中,拋物線SKIPIF1<0．SKIPIF1<0,SKIPIF1<0為C上兩點(diǎn),且SKIPIF1<0,SKIPIF1<0分別在第一、四象限．直線SKIPIF1<0與x正半軸交于SKIPIF1<0,與y負(fù)半軸交于SKIPIF1<0．(1)若SKIPIF1<0,求SKIPIF1<0橫坐標(biāo)的取值范圍；(2)記SKIPIF1<0的重心為G,直線SKIPIF1<0,SKIPIF1<0的斜率分別為SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0．若SKIPIF1<0,證明：λ為定值．【解析】（1）設(shè)SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,直線SKIPIF1<0的方程為：SKIPIF1<0,整理可得,SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0橫坐標(biāo)的取值范圍SKIPIF1<0；（2）SKIPIF1<0的重心為SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,又SKIPIF1<0,且SKIPIF1<0,∴SKIPIF1<0,化簡(jiǎn)得,SKIPIF1<0,∵SKIPIF1<0,SKIPIF1<0∴SKIPIF1<0,SKIPIF1<0.即SKIPIF1<0,所以λ為定值．【例6】已知雙曲線SKIPIF1<0的離心率是SKIPIF1<0,點(diǎn)SKIPIF1<0是雙曲線SKIPIF1<0的一個(gè)焦點(diǎn),且點(diǎn)SKIPIF1<0到雙曲線SKIPIF1<0的一條漸近線的距離是2.(1)求雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程.(2)設(shè)點(diǎn)SKIPIF1<0在直線SKIPIF1<0上,過點(diǎn)SKIPIF1<0作兩條直線SKIPIF1<0,直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn),直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn).若直線SKIPIF1<0與直線SKIPIF1<0的傾斜角互補(bǔ),證明：SKIPIF1<0.【解析】根據(jù)雙曲線的對(duì)稱性,不妨設(shè)SKIPIF1<0,其漸近線方程為SKIPIF1<0,因?yàn)榻裹c(diǎn)SKIPIF1<0到雙曲線SKIPIF1<0的一條漸近線的距離是2.所以SKIPIF1<0,因?yàn)殡p曲線SKIPIF1<0的離心率是SKIPIF1<0,所以,SKIPIF1<0,解得SKIPIF1<0所以,雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）證明：由題意可知直線SKIPIF1<0的斜率存在,設(shè)SKIPIF1<0,直線SKIPIF1<0.聯(lián)立SKIPIF1<0整理得SKIPIF1<0,所以,SKIPIF1<0.故SKIPIF1<0.設(shè)直線SKIPIF1<0的斜率為SKIPIF1<0,同理可得SKIPIF1<0.因?yàn)橹本€SKIPIF1<0與直線SKIPIF1<0的傾斜角互補(bǔ),所以SKIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0.(四)證明代數(shù)式的值為定值或證明與代數(shù)式有關(guān)的恒等式證明此類問題一般是把代數(shù)式用點(diǎn)的坐標(biāo)表示后化簡(jiǎn)，或構(gòu)造方程求解【例7】（2023屆甘肅省張掖市重點(diǎn)校高三上學(xué)期檢測(cè)）橢圓SKIPIF1<0的方程為SKIPIF1<0,過橢圓左焦點(diǎn)SKIPIF1<0且垂直于SKIPIF1<0軸的直線在第二象限與橢圓相交于點(diǎn)SKIPIF1<0,橢圓的右焦點(diǎn)為SKIPIF1<0,已知SKIPIF1<0,橢圓過點(diǎn)SKIPIF1<0．(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)過橢圓SKIPIF1<0的右焦點(diǎn)SKIPIF1<0作直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn),交SKIPIF1<0軸于SKIPIF1<0點(diǎn),若SKIPIF1<0,SKIPIF1<0,求證：SKIPIF1<0為定值．【解析】（1）依題可知：SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0又∵橢圓SKIPIF1<0過點(diǎn)SKIPIF1<0,則SKIPIF1<0聯(lián)立SKIPIF1<0可得SKIPIF1<0,橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0．（2）設(shè)點(diǎn)SKIPIF1<0、SKIPIF1<0,SKIPIF1<0,由題意可知,直線SKIPIF1<0的斜率存在,可設(shè)直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立SKIPIF1<0,可得SKIPIF1<0,由于點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0的內(nèi)部,直線SKIPIF1<0與橢圓SKIPIF1<0必有兩個(gè)交點(diǎn),由韋達(dá)定理可得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．【例8】（2023屆廣東省揭陽(yáng)市高三上學(xué)期8月調(diào)研）已知SKIPIF1<0?SKIPIF1<0是橢圓SKIPIF1<0：SKIPIF1<0的左?右焦點(diǎn),點(diǎn)SKIPIF1<0SKIPIF1<0是橢圓上的動(dòng)點(diǎn)．(1)求SKIPIF1<0的重心SKIPIF1<0的軌跡方程；(2)設(shè)點(diǎn)SKIPIF1<0是SKIPIF1<0的內(nèi)切圓圓心,求證：SKIPIF1<0．【解析】（1）連接SKIPIF1<0,由三角形重心性質(zhì)知SKIPIF1<0在SKIPIF1<0的三等分點(diǎn)處（靠近原點(diǎn)）設(shè)SKIPIF1<0,則有SKIPIF1<0又SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0SKIPIF1<0的重心SKIPIF1<0的軌跡方程為SKIPIF1<0；（2）根據(jù)對(duì)稱性,不妨設(shè)點(diǎn)SKIPIF1<0在第一象限內(nèi),易知圓SKIPIF1<0的半徑為等于SKIPIF1<0,利用等面積法有：SKIPIF1<0結(jié)合橢圓定義：SKIPIF1<0有SKIPIF1<0,解得SKIPIF1<0由SKIPIF1<0?SKIPIF1<0兩點(diǎn)的坐標(biāo)可知直線SKIPIF1<0的方程為SKIPIF1<0根據(jù)圓心SKIPIF1<0到直線SKIPIF1<0的距離等于半徑,有SKIPIF1<0∴SKIPIF1<0,∴SKIPIF1<0∴SKIPIF1<0,又SKIPIF1<0化簡(jiǎn)得SKIPIF1<0,即SKIPIF1<0∴SKIPIF1<0,即SKIPIF1<0由已知得SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0所以SKIPIF1<0,即SKIPIF1<0．三、跟蹤檢測(cè)1．（2023屆湖南省長(zhǎng)沙市一中等名校聯(lián)考聯(lián)合體高三上學(xué)期11月聯(lián)考）設(shè)橢圓SKIPIF1<0：SKIPIF1<0的左?右焦點(diǎn)分別為SKIPIF1<0,SKIPIF1<0.SKIPIF1<0,SKIPIF1<0是該橢圓SKIPIF1<0的下頂點(diǎn)和右頂點(diǎn),且SKIPIF1<0,若該橢圓的離心率為SKIPIF1<0.(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)經(jīng)過點(diǎn)SKIPIF1<0的直線SKIPIF1<0：SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn)（點(diǎn)SKIPIF1<0在點(diǎn)SKIPIF1<0下方）,過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線交直線SKIPIF1<0于點(diǎn)SKIPIF1<0,交直線SKIPIF1<0于點(diǎn)SKIPIF1<0,求證：SKIPIF1<0為定值.2．（2023屆河南省焦作市高三上學(xué)期期中）已知橢圓SKIPIF1<0：SKIPIF1<0的離心率為SKIPIF1<0,點(diǎn)SKIPIF1<0,SKIPIF1<0,橢圓SKIPIF1<0的右頂點(diǎn)SKIPIF1<0滿足SKIPIF1<0.(1)求橢圓SKIPIF1<0上一點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的最小距離；(2)若經(jīng)過SKIPIF1<0點(diǎn)的直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn),證明：當(dāng)直線SKIPIF1<0的傾斜角任意變化時(shí),總存在實(shí)數(shù)SKIPIF1<0,使得SKIPIF1<0.3．已知橢圓SKIPIF1<0的長(zhǎng)軸長(zhǎng)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0為SKIPIF1<0的左、右焦點(diǎn),點(diǎn)SKIPIF1<0在SKIPIF1<0上運(yùn)動(dòng),且SKIPIF1<0的最小值為SKIPIF1<0.連接SKIPIF1<0,SKIPIF1<0并延長(zhǎng)分別交橢圓SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn).(1)求SKIPIF1<0的方程；(2)證明：SKIPIF1<0為定值.4．（2022屆湖北省十堰市丹江口市高三下學(xué)期模擬）已知雙曲線SKIPIF1<0的左、右頂點(diǎn)分別為SKIPIF1<0,右焦點(diǎn)為SKIPIF1<0,點(diǎn)P為C上一動(dòng)點(diǎn)（異于SKIPIF1<0兩點(diǎn)）,直線SKIPIF1<0和直線SKIPIF1<0與直線SKIPIF1<0分別交于M,N兩點(diǎn),當(dāng)SKIPIF1<0垂直于x軸時(shí),SKIPIF1<0的面積為2．(1)求C的方程；(2)求證：SKIPIF1<0為定值,并求出該定值．5．（2023屆湖北省荊荊宜三校高三上學(xué)期10月聯(lián)考）記以坐標(biāo)原點(diǎn)為頂點(diǎn)、SKIPIF1<0為焦點(diǎn)的拋物線為SKIPIF1<0,過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與拋物線SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn).(1)已知點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,求SKIPIF1<0最大時(shí)直線SKIPIF1<0的傾斜角；(2)當(dāng)SKIPIF1<0的斜率為SKIPIF1<0時(shí),若平行SKIPIF1<0的直線SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0,SKIPIF1<0兩點(diǎn),且SKIPIF1<0與SKIPIF1<0相交于點(diǎn)SKIPIF1<0,證明：點(diǎn)SKIPIF1<0在定直線上.6．在平面直角坐標(biāo)系SKIPIF1<0中,點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,以線段SKIPIF1<0為直徑的圓與SKIPIF1<0軸相切．(1)求點(diǎn)SKIPIF1<0的軌跡SKIPIF1<0的方程；(2)設(shè)SKIPIF1<0是SKIPIF1<0上橫坐標(biāo)為2的點(diǎn),SKIPIF1<0的平行線SKIPIF1<0交SKIPIF1<0于SKIPIF1<0,SKIPIF1<0兩點(diǎn),交曲線SKIPIF1<0在SKIPIF1<0處的切線于點(diǎn)SKIPIF1<0,求證：SKIPIF1<0.7．已知雙曲線SKIPIF1<0,雙曲線SKIPIF1<0的右焦點(diǎn)為F,圓C的圓心在y軸正半軸上,且經(jīng)過坐標(biāo)原點(diǎn)O,圓C與雙曲線Γ的右支交于A、B兩點(diǎn)．(1)當(dāng)△OFA是以F為直角頂點(diǎn)的直角三角形,求△OFA的面積；(2)若點(diǎn)A的坐標(biāo)是SKIPIF1<0,求直線AB的方程；(3)求證：直線AB與圓x2+y2＝2相切．8．（2023屆湖北省重點(diǎn)高中智學(xué)聯(lián)盟高三上學(xué)期10月聯(lián)考）已知直線SKIPIF1<0：SKIPIF1<0與橢圓SKIPIF1<0：SKIPIF1<0相切于點(diǎn)SKIPIF1<0,與直線SKIPIF1<0：SKIPIF1<0相交于點(diǎn)SKIPIF1<0（異于點(diǎn)SKIPIF1<0）．(1)求點(diǎn)SKIPIF1<0的坐標(biāo)；(2)直線SKIPIF1<0交SKIPIF1<0于點(diǎn)SKIPIF1<0,SKIPIF1<0兩點(diǎn),證明：SKIPIF1<0．9．（2023屆重慶市巴蜀中學(xué)校2023屆高三上學(xué)期月考）已知橢圓SKIPIF1<0的左?右頂點(diǎn)分別為SKIPIF1<0,橢圓SKIPIF1<0的長(zhǎng)半軸的長(zhǎng)等于它的焦距,且過點(diǎn)SKIPIF1<0.(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)設(shè)橢圓SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0,過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與橢圓SKIPIF1<0相交于SKIPIF1<0兩點(diǎn)（不同于SKIPIF1<0）,直線SKIPIF1<0與直線SKIPIF1<0相交于點(diǎn)SKIPIF1<0,直線SKIPIF1<0與直線SKIPIF1<0相交于點(diǎn)SKIPIF1<0,證明：SKIPIF1<0軸.10．已知拋物線C：SKIPIF1<0,其焦點(diǎn)為F,O為坐標(biāo)原點(diǎn),直線l與拋物線C相交于不同的兩點(diǎn)A,B,M為AB的中點(diǎn).(1)若SKIPIF1<0,M的坐標(biāo)為SKIPIF1<0,求直線l的方程.(2)若直線l過焦點(diǎn)F,AB的垂直平分線交x軸于點(diǎn)N,求證：SKIPIF1<0為定值.11．（2023屆河北省邯鄲市大名縣第一中學(xué)高三月考）己知橢圓SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0,左頂點(diǎn)為SKIPIF1<0,離心率為SKIPIF1<0．(1)求SKIPIF1<0的方程；(2)若直線SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0,線段SKIPIF1<0的中點(diǎn)分別為SKIPIF1<0．設(shè)過點(diǎn)SKIPIF1<0且垂直于SKIPIF1<0軸的直線為SKIPIF1<0,若直線SKIPIF1<0與直線SKIPIF1<0交于點(diǎn)SKIPIF1<0,直線SKIPIF1<0與直線SKIPIF1<0交于點(diǎn)SKIPIF1<0,求證：SKIPIF1<0為定值．12．已知拋物線SKIPIF1<0的焦點(diǎn)到