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1 圓錐曲線的方程與軌跡方程（原卷版）

專題1圓錐曲線的方程與軌跡方程一、考情分析求圓錐曲線的方程,一般出現(xiàn)在圓錐曲線解答題的第（1）問,多用待定系數(shù)法,通過解方程確定待定系數(shù),考查頻率非常高,也比較容易得分；求圓錐曲線的軌跡方程一般用定義法,有時(shí)可用到直接法、相關(guān)點(diǎn)法、交軌法等,難度一般中等或中等以下.二、解題秘籍(一)用待定系數(shù)法求圓錐曲線的方程1.求橢圓標(biāo)準(zhǔn)方程的基本方法是待定系數(shù)法,具體過程是先定形,再定量,即首先確定焦點(diǎn)所在位置,然后再根據(jù)條件建立關(guān)于a,b的方程組．如果焦點(diǎn)位置不確定,要考慮是否有兩解,有時(shí)為了解題方便,也可把橢圓方程設(shè)為mx2＋ny2＝1(m>0,n>0,m≠n)的形式．2.雙曲線標(biāo)準(zhǔn)方程的形式,注意焦點(diǎn)F1,F2的位置是雙曲線定位的條件,它決定了雙曲線標(biāo)準(zhǔn)方程的類型.“焦點(diǎn)跟著正項(xiàng)走”,若x2項(xiàng)的系數(shù)為正,則焦點(diǎn)在x軸上;若y2項(xiàng)的系數(shù)為正,那么焦點(diǎn)在y軸上.確定方程的形式后,然后再根據(jù)a,b,c,e及漸近線之間的關(guān)系,求出a,b的值,當(dāng)雙曲線焦點(diǎn)的位置不確定時(shí),為了避免討論焦點(diǎn)的位置,常設(shè)雙曲線方程為Ax2＋By2＝1(A·B＜0),這樣可以簡化運(yùn)算.3.如果已知雙曲線的漸近線方程SKIPIF1<0,求雙曲線的標(biāo)準(zhǔn)方程,可設(shè)雙曲線方程為eq\f(x2,a2)－eq\f(y2,b2)＝λ(λ≠0),再由條件求出λ的值即可．與雙曲線eq\f(x2,a2)－eq\f(y2,b2)＝1(a>0,b>0)有共同漸近線的方程可表示eq\f(x2,a2)－eq\f(y2,b2)＝λ(λ≠0)．4.利用待定系數(shù)法求拋物線的標(biāo)準(zhǔn)方程的步驟(1)依據(jù)條件設(shè)出拋物線的標(biāo)準(zhǔn)方程的類型.(2)求參數(shù)p的值.(3)確定拋物線的標(biāo)準(zhǔn)方程.【例1】（2023屆山西省長治市高三上學(xué)期質(zhì)量檢測）已知點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）上,且點(diǎn)SKIPIF1<0到橢圓右頂點(diǎn)SKIPIF1<0的距離為SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)若點(diǎn)SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0上不同的兩點(diǎn)（均異于SKIPIF1<0）且滿足直線SKIPIF1<0與SKIPIF1<0斜率之積為SKIPIF1<0．試判斷直線SKIPIF1<0是否過定點(diǎn),若是,求出定點(diǎn)坐標(biāo),若不是,說明理由．【解析】（1）點(diǎn)SKIPIF1<0,在橢圓SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）上代入得：SKIPIF1<0,點(diǎn)SKIPIF1<0到橢圓右頂點(diǎn)SKIPIF1<0的距離為SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0,故橢圓SKIPIF1<0的方程為SKIPIF1<0．（2）由題意,直線SKIPIF1<0的斜率存在,可設(shè)直線SKIPIF1<0的方程為SKIPIF1<0（SKIPIF1<0）,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．聯(lián)立SKIPIF1<0得SKIPIF1<0．SKIPIF1<0．∴SKIPIF1<0,SKIPIF1<0,∵直線SKIPIF1<0與直線SKIPIF1<0斜率之積為SKIPIF1<0．∴SKIPIF1<0,∴SKIPIF1<0．化簡得SKIPIF1<0,∴SKIPIF1<0,化簡得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0．當(dāng)SKIPIF1<0時(shí),直線SKIPIF1<0方程為SKIPIF1<0,過定點(diǎn)SKIPIF1<0．SKIPIF1<0代入判別式大于零中,解得SKIPIF1<0（SKIPIF1<0）．當(dāng)SKIPIF1<0時(shí),直線SKIPIF1<0的方程為SKIPIF1<0,過定點(diǎn)SKIPIF1<0,不符合題意．綜上所述：直線SKIPIF1<0過定點(diǎn)SKIPIF1<0．【點(diǎn)評】利用待定系數(shù)法求橢圓的方程,一般需要兩個(gè)獨(dú)立的條件確定關(guān)于SKIPIF1<0的等式.【例2】（2023屆廣東省開平市忠源紀(jì)念中學(xué)高三階段性檢測）已知雙曲線SKIPIF1<0的離心率為SKIPIF1<0,點(diǎn)SKIPIF1<0在SKIPIF1<0上.(1)求雙曲線SKIPIF1<0的方程.(2)設(shè)過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn),問在SKIPIF1<0軸上是否存在定點(diǎn)SKIPIF1<0,使得SKIPIF1<0為常數(shù)？若存在,求出點(diǎn)SKIPIF1<0的坐標(biāo)以及該常數(shù)的值；若不存在,請說明理由.【解析】（1）因?yàn)殡p曲線SKIPIF1<0的離心率為SKIPIF1<0,所以SKIPIF1<0,化簡得SKIPIF1<0.將點(diǎn)SKIPIF1<0的坐標(biāo)代入SKIPIF1<0,可得SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0的方程為SKIPIF1<0.（2）設(shè)SKIPIF1<0,直線SKIPIF1<0的方程為SKIPIF1<0,聯(lián)立方程組SKIPIF1<0消去SKIPIF1<0得（1-SKIPIF1<0SKIPIF1<0,由題可知SKIPIF1<0且SKIPIF1<0,即SKIPIF1<0且SKIPIF1<0,所以SKIPIF1<0.設(shè)存在符合條件的定點(diǎn)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0,化簡得SKIPIF1<0.因?yàn)镾KIPIF1<0為常數(shù),所以SKIPIF1<0,解得SKIPIF1<0.此時(shí)該常數(shù)的值為SKIPIF1<0,所以,在SKIPIF1<0軸上存在點(diǎn)SKIPIF1<0,使得SKIPIF1<0為常數(shù),該常數(shù)為SKIPIF1<0.【點(diǎn)評】求雙曲線的標(biāo)準(zhǔn)方程的基本方法是待定系數(shù)法．具體過程是先定形,再定量,即先確定雙曲線標(biāo)準(zhǔn)方程的形式,然后再根據(jù)a,b,c,e及漸近線之間的關(guān)系,求出a,b的值．注意用待定系數(shù)法確定雙曲線的標(biāo)準(zhǔn)方程要注意方程的個(gè)數(shù)要與未知數(shù)的個(gè)數(shù)相等.【例3】（2023屆甘肅省張掖市高三上學(xué)期診斷）已知拋物線SKIPIF1<0上的點(diǎn)SKIPIF1<0到其焦點(diǎn)F的距離為SKIPIF1<0.(1)求拋物線C的方程；(2)點(diǎn)SKIPIF1<0在拋物線C上,過點(diǎn)SKIPIF1<0的直線l與拋物線C交于SKIPIF1<0,SKIPIF1<0SKIPIF1<0兩點(diǎn),點(diǎn)H與點(diǎn)A關(guān)于x軸對稱,直線AH分別與直線OE,OB交于點(diǎn)M,N(O為坐標(biāo)原點(diǎn)),求證：SKIPIF1<0.【解析】（1）由點(diǎn)SKIPIF1<0在拋物線上可得,SKIPIF1<0,解得SKIPIF1<0.由拋物線的定義可得SKIPIF1<0,整理得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0(舍去).故拋物線C的方程為SKIPIF1<0.（2）由SKIPIF1<0在拋物線C上可得SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,直線OE的方程為SKIPIF1<0,因?yàn)辄c(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸對稱,所以SKIPIF1<0,SKIPIF1<0均不為0.由題意知直線l的斜率存在且大于0,設(shè)直線l的方程為SKIPIF1<0,聯(lián)立SKIPIF1<0消去y,得SKIPIF1<0.則SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0.由直線OE的方程為SKIPIF1<0,得SKIPIF1<0.易知直線OB的方程為SKIPIF1<0,故SKIPIF1<0.要證SKIPIF1<0,即證SKIPIF1<0,即證SKIPIF1<0,即證SKIPIF1<0,即證SKIPIF1<0,則SKIPIF1<0,此等式顯然成立,所以SKIPIF1<0.【點(diǎn)評】用待定系數(shù)法求拋物線的標(biāo)準(zhǔn)方程,只需要確定p的值,因此只需要由已知條件整理出一個(gè)關(guān)于p的等式.(二)直接法求曲線軌跡方程1.直接法求曲線方程的關(guān)鍵就是把幾何條件或等量關(guān)系翻譯為代數(shù)方程,要注意翻譯的等價(jià)性．通常將步驟簡記為建系、設(shè)點(diǎn)、列式、代換、化簡、證明這幾個(gè)步驟,但最后的證明可以省略．2.求出曲線的方程后還需注意檢驗(yàn)方程的純粹性和完備性．3.對方程化簡時(shí),要保證前后方程解集相同,必要時(shí)可說明x,y的取值范圍．【例4】設(shè)動(dòng)點(diǎn)SKIPIF1<0在直線SKIPIF1<0和SKIPIF1<0上的射影分別為點(diǎn)SKIPIF1<0和SKIPIF1<0,已知SKIPIF1<0,其中SKIPIF1<0為坐標(biāo)原點(diǎn).（1）求動(dòng)點(diǎn)SKIPIF1<0的軌跡SKIPIF1<0的方程；（2）過直線SKIPIF1<0上的一點(diǎn)SKIPIF1<0作軌跡SKIPIF1<0的兩條切線SKIPIF1<0和SKIPIF1<0(SKIPIF1<0,SKIPIF1<0為切點(diǎn)),求證：直線SKIPIF1<0經(jīng)過定點(diǎn).【分析】（1）利用直接法求軌跡方程,設(shè)SKIPIF1<0,把SKIPIF1<0坐標(biāo)化,即可得到動(dòng)點(diǎn)SKIPIF1<0的軌跡SKIPIF1<0的方程；（2）利用導(dǎo)數(shù)的幾何意義,求得切線斜率,設(shè)SKIPIF1<0,可得切線SKIPIF1<0、SKIPIF1<0的方程,聯(lián)立可得切點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,又點(diǎn)SKIPIF1<0在直線SKIPIF1<0上,代入可得SKIPIF1<0,再代入到直線SKIPIF1<0的方程即可得解.【解析】（1）設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,由條件可得SKIPIF1<0,整理可得點(diǎn)SKIPIF1<0的軌方程為SKIPIF1<0；（2）由（1）知,SKIPIF1<0,求導(dǎo)可得SKIPIF1<0,設(shè)SKIPIF1<0,則切線SKIPIF1<0的方程為SKIPIF1<0,即SKIPIF1<0①,同理可得切線SKIPIF1<0的方程為SKIPIF1<0②,聯(lián)立①②,解得點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,因?yàn)辄c(diǎn)SKIPIF1<0在直線SKIPIF1<0上,所以SKIPIF1<0,即SKIPIF1<0,又直線SKIPIF1<0的斜率SKIPIF1<0,所以直線SKIPIF1<0的方程為：SKIPIF1<0,即SKIPIF1<0,又SKIPIF1<0,代入可得SKIPIF1<0,所以直線SKIPIF1<0過定點(diǎn)SKIPIF1<0.【點(diǎn)評】利用直接法求曲線的軌跡方程一般是根據(jù)題中的一個(gè)等量關(guān)系式,將其坐標(biāo)化,即可得到曲線的軌跡方程.(三)定義法求曲線軌跡方程1.運(yùn)用圓錐曲線的定義求軌跡方程,可從曲線定義出發(fā)直接寫出方程,或從曲線定義出發(fā)建立關(guān)系式,從而求出方程．2.定義法和待定系數(shù)法適用于已知曲線的軌跡類型,利用條件把待定系數(shù)求出來,使問題得解．3.平面內(nèi)與兩個(gè)定點(diǎn)F1,F2的距離之和等于常數(shù)(大于|F1F2|)的點(diǎn)的軌跡叫做橢圓．這兩個(gè)定點(diǎn)叫做橢圓的焦點(diǎn),兩焦點(diǎn)間的距離叫做橢圓的焦距．集合P＝{M||MF1|＋|MF2|＝2a},|F1F2|＝2c,其中a>0,c>0,且a,c為常數(shù)：(1)若a>c,則集合P為橢圓；(2)若a＝c,則集合P為線段；(3)若a<c,則集合P為空集．4.平面內(nèi)與兩個(gè)定點(diǎn)F1,F2的距離的差的絕對值等于常數(shù)(小于|F1F2|)的點(diǎn)的軌跡叫做雙曲線．這兩個(gè)定點(diǎn)叫做雙曲線的焦點(diǎn),兩焦點(diǎn)間的距離叫做雙曲線的焦距．集合P＝{M|||MF1|－|MF2||＝2a},|F1F2|＝2c,其中a,c為常數(shù)且a>0,c>0.(1)當(dāng)2a<|F1F2|時(shí),P點(diǎn)的軌跡是雙曲線；(2)當(dāng)2a＝|F1F2|時(shí),P點(diǎn)的軌跡是以F1,F2為端點(diǎn)的兩條射線；(3)當(dāng)2a>|F1F2|時(shí),P點(diǎn)不存在．5.平面內(nèi)與一個(gè)定點(diǎn)F和一條定直線l(l不經(jīng)過點(diǎn)F)的距離相等的點(diǎn)的軌跡叫做拋物線．點(diǎn)F叫做拋物線的焦點(diǎn),直線l叫做拋物線的準(zhǔn)線．注意：(1)定直線l不經(jīng)過定點(diǎn)F.(2)定義中包含三個(gè)定值,分別為一個(gè)定點(diǎn),一條定直線及一個(gè)確定的比值.【例5】（2023屆河北省示范性高中高三上學(xué)期調(diào)研）已知圓A：SKIPIF1<0,直線l（與x軸不重合）過點(diǎn)SKIPIF1<0交圓A于C、D兩點(diǎn),過點(diǎn)B作直線SKIPIF1<0的平行線交直線SKIPIF1<0于點(diǎn)E.(1)證明SKIPIF1<0為定值,并求點(diǎn)E的軌跡方程；(2)設(shè)點(diǎn)E的軌跡方程為SKIPIF1<0,直線l與曲線SKIPIF1<0交于M、N兩點(diǎn),線段SKIPIF1<0的垂直平分線交x軸于點(diǎn)P,是否存在實(shí)常數(shù)入,使得SKIPIF1<0,若存在,求出SKIPIF1<0的值；若不存在,請說明理由.【解析】（1）SKIPIF1<0,得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),如圖1所示,因?yàn)镈,C都在圓A上所以SKIPIF1<0,即SKIPIF1<0又因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,∴SKIPIF1<0,所以SKIPIF1<0當(dāng)SKIPIF1<0時(shí),如圖2所示,同理可得,SKIPIF1<0因此SKIPIF1<0,所以點(diǎn)E的軌跡是以A,B為焦點(diǎn)的雙曲線,故SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,∴SKIPIF1<0為定值2,且點(diǎn)E的軌跡方程為SKIPIF1<0.（2）由題知,直線l的斜率不為0,設(shè)l：SKIPIF1<0,聯(lián)立SKIPIF1<0消去x得,SKIPIF1<0,于是SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0,則有SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,所以線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0,從而線段SKIPIF1<0的中垂線的方程為SKIPIF1<0令SKIPIF1<0得,SKIPIF1<0,∴SKIPIF1<0又SKIPIF1<0故SKIPIF1<0,于是SKIPIF1<0即存在SKIPIF1<0使得SKIPIF1<0.【點(diǎn)評】利用雙曲線定義求軌跡方程,關(guān)鍵是利用題中條件,確定動(dòng)點(diǎn)到兩定點(diǎn)距離之差的絕對值為定值.【例6】已知一定點(diǎn)SKIPIF1<0,及一定直線l：SKIPIF1<0,以動(dòng)點(diǎn)M為圓心的圓M過點(diǎn)F,且與直線l相切．（1）求動(dòng)點(diǎn)M的軌跡C的方程；（2）設(shè)P在直線l上,直線PA,PB分別與曲線C相切于A,B,N為線段AB的中點(diǎn)．求證：SKIPIF1<0,且直線AB恒過定點(diǎn)．【解析】（1）動(dòng)點(diǎn)M為圓心的圓M過點(diǎn)F,且與直線l相切,動(dòng)圓圓心到定點(diǎn)F（0,1）與定直線y=-1的距離相等,

∴動(dòng)圓圓心的軌跡為拋物線,其中F（0,1）為焦點(diǎn),y=-1為準(zhǔn)線,SKIPIF1<0,∴動(dòng)圓圓心軌跡方程為x2=4y．（2）依題意可設(shè)SKIPIF1<0,又SKIPIF1<0故切線SKIPIF1<0的斜率為SKIPIF1<0,故切線SKIPIF1<0同理可得到切線SKIPIF1<0又SKIPIF1<0,∴SKIPIF1<0且SKIPIF1<0,故方程SKIPIF1<0有兩根SKIPIF1<0∴SKIPIF1<0,SKIPIF1<0SKIPIF1<0又SKIPIF1<0為線段SKIPIF1<0的中點(diǎn),SKIPIF1<0又由SKIPIF1<0得到：SKIPIF1<0即SKIPIF1<0同理可得到SKIPIF1<0,故直線AB方程為：SKIPIF1<0,故直線過定點(diǎn)SKIPIF1<0.【點(diǎn)評】利用拋物線定義求軌跡方程關(guān)鍵是確定動(dòng)點(diǎn)到一定點(diǎn)與定直線距離相等.(四)相關(guān)點(diǎn)法求曲線軌跡方程“相關(guān)點(diǎn)法”求軌跡方程的基本步驟(1)設(shè)點(diǎn)：設(shè)被動(dòng)點(diǎn)坐標(biāo)為(x,y),主動(dòng)點(diǎn)坐標(biāo)為(x1,y1)；(2)求關(guān)系式：求出兩個(gè)動(dòng)點(diǎn)坐標(biāo)之間的關(guān)系式eq\b\lc\{\rc\(\a\vs4\al\co1(x1＝fx，y，,y1＝gx，y；))(3)代換：將上述關(guān)系式代入已知曲線方程,便可得到所求動(dòng)點(diǎn)的軌跡方程．【例7】（2023屆廣東省揭陽市高三上學(xué)期調(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．(五)交軌法求曲線軌跡方程求兩曲線的交點(diǎn)軌跡時(shí),可由方程直接消去參數(shù),或者先引入?yún)?shù)來建立這些動(dòng)曲線的聯(lián)系,然后消去參數(shù)來得到軌跡方程,稱之交軌法.若動(dòng)點(diǎn)是兩曲線的交點(diǎn),可以通過這兩曲線的方程直接求出交點(diǎn)的軌跡方程,也可以解方程組先求出交點(diǎn)坐標(biāo)的參數(shù)方程,再化為普通方程.【例8】（2022屆重慶市第八中學(xué)高三上學(xué)期月考）已知拋物線SKIPIF1<0,過點(diǎn)SKIPIF1<0的直線交拋物線SKIPIF1<0于SKIPIF1<0兩點(diǎn),以SKIPIF1<0為切點(diǎn)分別作拋物線SKIPIF1<0的兩條切線交于點(diǎn)SKIPIF1<0.（1）若線段SKIPIF1<0的中點(diǎn)SKIPIF1<0的縱坐標(biāo)為SKIPIF1<0,求直線SKIPIF1<0的方程；（2）求動(dòng)點(diǎn)SKIPIF1<0的軌跡.【分析】（1）聯(lián)立直線與拋物線,根據(jù)韋達(dá)定理及中點(diǎn)求出k即可；（2）寫出圓的切線方程,根據(jù)P是交點(diǎn)可得SKIPIF1<0是方程SKIPIF1<0的兩根,由（1）中SKIPIF1<0代入化簡即可求出.【解析】（1）依題意有：直線SKIPIF1<0的斜率必存在,故可設(shè)直線SKIPIF1<0的方程為SKIPIF1<0由SKIPIF1<0可得：SKIPIF1<0.設(shè)SKIPIF1<0,則有SKIPIF1<0于是：SKIPIF1<0,解得SKIPIF1<0,故直線SKIPIF1<0的方程為SKIPIF1<0（2）設(shè)SKIPIF1<0,對于拋物線SKIPIF1<0,SKIPIF1<0于是：SKIPIF1<0點(diǎn)處切線方程為SKIPIF1<0,點(diǎn)SKIPIF1<0在該切線上,故SKIPIF1<0,即SKIPIF1<0.同理：SKIPIF1<0點(diǎn)坐標(biāo)也滿足SKIPIF1<0于是：SKIPIF1<0是方程SKIPIF1<0的兩根,所以SKIPIF1<0又由（1）可知：SKIPIF1<0,于是SKIPIF1<0,消k得SKIPIF1<0,于是SKIPIF1<0的軌跡方程為SKIPIF1<0,點(diǎn)SKIPIF1<0的軌跡是一條直線.【點(diǎn)評】求兩條動(dòng)直線交點(diǎn)軌跡方程一般用交軌法三、跟蹤檢測1．（2023屆廣東省廣東廣雅中學(xué)高三上學(xué)期9月階段測試）已知橢圓SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）的離心率為SKIPIF1<0．圓SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)）在橢圓SKIPIF1<0的內(nèi)部,半徑為SKIPIF1<0．SKIPIF1<0,SKIPIF1<0分別為橢圓SKIPIF1<0和圓SKIPIF1<0上的動(dòng)點(diǎn),且SKIPIF1<0,SKIPIF1<0兩點(diǎn)的最小距離為SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)SKIPIF1<0,SKIPIF1<0是橢圓SKIPIF1<0上不同的兩點(diǎn),且直線SKIPIF1<0與以SKIPIF1<0為直徑的圓的一個(gè)交點(diǎn)在圓SKIPIF1<0上．求證：以SKIPIF1<0為直徑的圓過定點(diǎn)．2．（2023屆山西省忻州市高三上學(xué)期聯(lián)考）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<0.3．（2023屆廣東省茂名市高三上學(xué)期9月大聯(lián)考）如圖,平面直角坐標(biāo)系SKIPIF1<0中,點(diǎn)SKIPIF1<0為SKIPIF1<0軸上的一個(gè)動(dòng)點(diǎn),動(dòng)點(diǎn)SKIPIF1<0滿足SKIPIF1<0,又點(diǎn)SKIPIF1<0滿足SKIPIF1<0.(1)求動(dòng)點(diǎn)SKIPIF1<0的軌跡SKIPIF1<0的方程；(2)過曲線SKIPIF1<0上的點(diǎn)SKIPIF1<0（SKIPIF1<0）的直線SKIPIF1<0與SKIPIF1<0,SKIPIF1<0軸的交點(diǎn)分別為SKIPIF1<0和SKIPIF1<0,且SKIPIF1<0,過原點(diǎn)SKIPIF1<0的直線與SKIPIF1<0平行,且與曲線SKIPIF1<0交于SKIPIF1<0、SKIPIF1<0兩點(diǎn),求SKIPIF1<0面積的最大值.4．（2023屆湖南省永州市高三上學(xué)期適應(yīng)性考試）點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0上,離心率SKIPIF1<0.(1)求雙曲線SKIPIF1<0的方程；(2)SKIPIF1<0是雙曲線SKIPIF1<0上的兩個(gè)動(dòng)點(diǎn)（異于點(diǎn)SKIPIF1<0）,SKIPIF1<0分別表示直線SKIPIF1<0的斜率,滿足SKIPIF1<0,求證：直線SKIPIF1<0恒過一個(gè)定點(diǎn),并求出該定點(diǎn)的坐標(biāo).5．（2023屆福建師范大學(xué)附屬中學(xué)高三上學(xué)期月考）在平面直角坐標(biāo)系SKIPIF1<0中,設(shè)點(diǎn)SKIPIF1<0,點(diǎn)SKIPIF1<0與SKIPIF1<0兩點(diǎn)的距離之和為SKIPIF1<0為一動(dòng)點(diǎn),點(diǎn)SKIPIF1<0滿足向量關(guān)系式：SKIPIF1<0．(1)求點(diǎn)SKIPIF1<0的軌跡方程SKIPIF1<0；(2)設(shè)SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0(SKIPIF1<0在SKIPIF1<0的左側(cè)),點(diǎn)SKIPIF1<0為SKIPIF1<0上一動(dòng)點(diǎn)(且不與SKIPIF1<0重合)．設(shè)直線SKIPIF1<0軸與直線SKIPIF1<0分別交于點(diǎn)SKIPIF1<0,取SKIPIF1<0,連接SKIPIF1<0,證明：SKIPIF1<0為SKIPIF1<0的角平分線．6．（2023屆云南省大理市轄區(qū)高三統(tǒng)一檢測）已知SKIPIF1<0為橢圓C的左、右焦點(diǎn),點(diǎn)SKIPIF1<0為其上一點(diǎn),且SKIPIF1<0．(1)求橢圓C的標(biāo)準(zhǔn)方程；(2)過點(diǎn)SKIPIF1<0的直線l與橢圓C相交于P,Q兩點(diǎn),點(diǎn)P關(guān)于坐標(biāo)原點(diǎn)O的對稱點(diǎn)R,試問SKIPIF1<0的面積是否存在最大值？若存在,求出這個(gè)最大值；若不存在,請說明理由．7．（2022屆福建省福州第十八中學(xué)高三上學(xué)期考試）已知拋物線SKIPIF1<0的焦點(diǎn)SKIPIF1<0到準(zhǔn)線的距離為2．(1)求SKIPIF1<0的方程；(2)已知SKIPIF1<0為坐標(biāo)原點(diǎn),點(diǎn)SKIPIF1<0在SKIPIF1<0上,點(diǎn)SKIPIF1<0滿足SKIPIF1<0,求直線SKIPIF1<0斜率的最大值．8．（2023屆陜西師范大學(xué)附屬中學(xué)、渭北中學(xué)等高三上學(xué)期聯(lián)考）已知拋物線SKIPIF1<0,O是坐標(biāo)原點(diǎn),F是C的焦點(diǎn),M是C上一點(diǎn),SKIPIF1<0,SKIPIF1<0．(1)求拋物線C的標(biāo)準(zhǔn)方程；(2)設(shè)點(diǎn)SKIPIF1<0在C上,過Q作兩條互相垂直的直線SKIPIF1<0,分別交C于A,B兩點(diǎn)（異于Q點(diǎn)）．證明：直線SKIPIF1<0恒過定點(diǎn)．9．（2023屆廣東省潮陽實(shí)驗(yàn)、湛江一中、深圳實(shí)驗(yàn)三校高三上學(xué)期9月聯(lián)考）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0,橢圓上一動(dòng)點(diǎn)SKIPIF1<0與左?右焦點(diǎn)構(gòu)成的三角形面積最大值為SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)設(shè)橢圓SKIPIF1<0的左?右頂點(diǎn)分別為SKIPIF1<0,直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn),記直線SKIPIF1<0的斜率為SKIPIF1<0,直線SKIPIF1<0的斜率為SKIPIF1<0,已知SKIPIF1<0.①求證：直線SKIPIF1<0恒過定點(diǎn)；②設(shè)SKIPIF1<0和SKIPIF1<0的面積分別為SKIPIF1<0,求SKIPIF1<0的最大值.10．（2022屆云南省紅河州高三檢測）在平面直角坐標(biāo)系SKIPIF1<0中,點(diǎn)SK