新高考數(shù)學(xué)三輪沖刺通關(guān)練習(xí)09 圓錐曲線大題（易錯(cuò)點(diǎn)+六大題型）（解析版）

秘籍09圓錐曲線大題目錄【高考預(yù)測(cè)】概率預(yù)測(cè)+題型預(yù)測(cè)+考向預(yù)測(cè)【應(yīng)試秘籍】總結(jié)?？键c(diǎn)及應(yīng)對(duì)的策略【誤區(qū)點(diǎn)撥】點(diǎn)撥常見的易錯(cuò)點(diǎn)易錯(cuò)點(diǎn)：解題規(guī)范【搶分通關(guān)】精選名校模擬題，講解通關(guān)策略【題型一】極點(diǎn)、極線【題型二】自極三角形與調(diào)和點(diǎn)列【題型三】齊次化法解決斜率相關(guān)問題【題型四】定比點(diǎn)差法【題型五】定點(diǎn)、定值【題型六】求軌跡方程型概率預(yù)測(cè)☆☆☆☆☆題型預(yù)測(cè)解答題☆☆☆☆☆考向預(yù)測(cè)極點(diǎn)、極線圓錐曲線大題和小題考察的類型不一致，但是肯定都是以基礎(chǔ)知識(shí)為前提的情況下進(jìn)行考察，所以一般第一問考察的大多還是求圓錐曲線的函數(shù)解析式，而第二問往往考察的是直線與圓錐曲線的位置關(guān)系，這里對(duì)于解析幾何的代數(shù)問題要求就比較高，題型也相應(yīng)較多，需要多加練習(xí)。一些固定題型解題方法的掌握還是需要熟練，并且理解圓錐曲線中解析幾何的解題思維，延伸知識(shí)點(diǎn)例如極點(diǎn)、極線，齊次化解法、定比點(diǎn)差法等等比較熱門的需要熟練于心。易錯(cuò)點(diǎn)一：解題規(guī)范圓錐曲線大題在遇到直線與曲線相交相關(guān)的問題是，極點(diǎn)、極線的思想只能輔助我們解題，不可出現(xiàn)在答題過程中，都需要設(shè)點(diǎn)或設(shè)線，寫出完整的證明過程。例（2023年全國乙卷）已知橢圓SKIPIF1<0的離心率是SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0上．(1)求SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0的直線交SKIPIF1<0于SKIPIF1<0兩點(diǎn)，直線SKIPIF1<0與SKIPIF1<0軸的交點(diǎn)分別為SKIPIF1<0，證明：線段SKIPIF1<0的中點(diǎn)為定點(diǎn)．【極線思維】記SKIPIF1<0，點(diǎn)B的極線SKIPIF1<0過點(diǎn)A，設(shè)極線與PQ交于點(diǎn)D，則B，P，D，Q為調(diào)和點(diǎn)列，AB，AP，AD，AQ為調(diào)和線束，而AB平行y軸，故MN的中點(diǎn)為y軸于極線的交點(diǎn)【詳解】（1）由題意可得SKIPIF1<0，解得SKIPIF1<0，所以橢圓方程為SKIPIF1<0.（2）由題意可知：直線SKIPIF1<0的斜率存在，設(shè)SKIPIF1<0，聯(lián)立方程SKIPIF1<0，消去y得：SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，可得SKIPIF1<0，因?yàn)镾KIPIF1<0，則直線SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0,同理可得SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，所以線段SKIPIF1<0的中點(diǎn)是定點(diǎn)SKIPIF1<0.變式1：（2024·湖南衡陽·二模）（多選）已知圓SKIPIF1<0是直線SKIPIF1<0上一動(dòng)點(diǎn)，過點(diǎn)SKIPIF1<0作直線SKIPIF1<0分別與圓SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則（

）A．圓SKIPIF1<0上恰有一個(gè)點(diǎn)到SKIPIF1<0的距離為SKIPIF1<0 B．直線SKIPIF1<0恒過點(diǎn)SKIPIF1<0C．SKIPIF1<0的最小值是SKIPIF1<0 D．四邊形SKIPIF1<0面積的最小值為SKIPIF1<0【答案】BCD【詳解】易知圓心SKIPIF1<0，半徑SKIPIF1<0，如下圖所示：對(duì)于A，圓心SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，可得圓SKIPIF1<0上的點(diǎn)到直線SKIPIF1<0距離的最小值為SKIPIF1<0,圓SKIPIF1<0上的點(diǎn)到直線SKIPIF1<0距離的最大值為SKIPIF1<0，所以圓SKIPIF1<0上恰有兩個(gè)點(diǎn)到SKIPIF1<0的距離為SKIPIF1<0，即A錯(cuò)誤；對(duì)于B，設(shè)SKIPIF1<0，可得SKIPIF1<0；易知SKIPIF1<0，由SKIPIF1<0，整理可得SKIPIF1<0，同理可得SKIPIF1<0，即可知SKIPIF1<0兩點(diǎn)在直線SKIPIF1<0上，所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以直線SKIPIF1<0恒過定點(diǎn)SKIPIF1<0，即B正確；對(duì)于C，由直線SKIPIF1<0恒過定點(diǎn)SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0與圓心SKIPIF1<0的連線垂直于SKIPIF1<0時(shí)，SKIPIF1<0的值最小，點(diǎn)SKIPIF1<0與圓心SKIPIF1<0之間的距離為SKIPIF1<0，所以SKIPIF1<0，故C正確；對(duì)于D，四邊形SKIPIF1<0的面積為SKIPIF1<0，根據(jù)切線長(zhǎng)公式可知SKIPIF1<0，當(dāng)SKIPIF1<0最小值，SKIPIF1<0最小，，所以，故四邊形的面積為，即D正確；故選：BCD【題型一】極點(diǎn)、極線二次曲線的極點(diǎn)極線（1）.二次曲線SKIPIF1<0極點(diǎn)SKIPIF1<0對(duì)應(yīng)的極線為SKIPIF1<0SKIPIF1<0（半代半不代）（2）圓錐曲線的三類極點(diǎn)極線（以橢圓為例）：橢圓方程SKIPIF1<0①極點(diǎn)SKIPIF1<0在橢圓外，SKIPIF1<0為橢圓的切線，切點(diǎn)為SKIPIF1<0則極線為切點(diǎn)弦SKIPIF1<0；②極點(diǎn)SKIPIF1<0在橢圓上，過點(diǎn)SKIPIF1<0作橢圓的切線SKIPIF1<0，則極線為切線SKIPIF1<0；③極點(diǎn)SKIPIF1<0在橢圓內(nèi)，過點(diǎn)SKIPIF1<0作橢圓的弦SKIPIF1<0，分別過SKIPIF1<0作橢圓切線，則切線交點(diǎn)軌跡為極線SKIPIF1<0；（3）圓錐曲線的焦點(diǎn)為極點(diǎn)，對(duì)應(yīng)準(zhǔn)線為極線.【例1】過點(diǎn)SKIPIF1<0作圓SKIPIF1<0的兩條切線，切點(diǎn)分別為SKIPIF1<0、SKIPIF1<0則直線SKIPIF1<0的方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0解析：直線SKIPIF1<0是點(diǎn)SKIPIF1<0對(duì)應(yīng)的極線，則方程為SKIPIF1<0，即SKIPIF1<0．故選A．【例2】已知點(diǎn)SKIPIF1<0為SKIPIF1<0上一動(dòng)點(diǎn)．過點(diǎn)SKIPIF1<0作橢圓SKIPIF1<0的兩條切線，切點(diǎn)分別SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0運(yùn)動(dòng)時(shí)，直線SKIPIF1<0過定點(diǎn)，該定點(diǎn)的坐標(biāo)是________．解析：設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)是SKIPIF1<0，則切點(diǎn)弦SKIPIF1<0的方程為SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，故直線SKIPIF1<0過定點(diǎn)SKIPIF1<0．【例3】（2024·廣東湛江·一模）已知點(diǎn)P為直線上的動(dòng)點(diǎn)，過P作圓的兩條切線，切點(diǎn)分別為A，B，若點(diǎn)M為圓上的動(dòng)點(diǎn)，則點(diǎn)M到直線AB的距離的最大值為．【答案】【詳解】設(shè)，則滿足；易知圓的圓心為，半徑；圓的圓心為，半徑，如下圖所示：易知，所以，即，整理可得；同理可得，即是方程的兩組解，可得直線的方程為，聯(lián)立，即；令，可得，即時(shí)等式與無關(guān)，所以直線恒過定點(diǎn)，可得；又在圓內(nèi)，當(dāng)，且點(diǎn)為的延長(zhǎng)線與圓的交點(diǎn)時(shí)，點(diǎn)到直線的距離最大；最大值為【變式1】（2024·陜西西安·一模）已知橢圓SKIPIF1<0的左，右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0與短軸的一個(gè)端點(diǎn)SKIPIF1<0構(gòu)成一個(gè)等腰直角三角形，點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0，過點(diǎn)SKIPIF1<0作互相垂直且與SKIPIF1<0軸不重合的兩直線SKIPIF1<0，SKIPIF1<0分別交橢圓SKIPIF1<0于SKIPIF1<0，SKIPIF1<0和點(diǎn)SKIPIF1<0，SKIPIF1<0，且點(diǎn)SKIPIF1<0，SKIPIF1<0分別是弦SKIPIF1<0，SKIPIF1<0的中點(diǎn)．

(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)若SKIPIF1<0，求以SKIPIF1<0為直徑的圓的方程；(3)直線SKIPIF1<0是否過SKIPIF1<0軸上的一個(gè)定點(diǎn)？若是，求出該定點(diǎn)坐標(biāo)；若不是，說明理由．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【詳解】（1）解：因?yàn)闄E圓SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0與短軸的一個(gè)端點(diǎn)SKIPIF1<0構(gòu)成一個(gè)等腰直角三角形，可得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以橢圓SKIPIF1<0的標(biāo)準(zhǔn)分別為SKIPIF1<0.（2）解：由（1）得SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立方程組SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0，則CD的中點(diǎn)為SKIPIF1<0且SKIPIF1<0，故以SKIPIF1<0為直徑的圓的方程為SKIPIF1<0.（3）解：設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，且SKIPIF1<0，則直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立方程組SKIPIF1<0，整理得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，由中點(diǎn)坐標(biāo)公式得SKIPIF1<0，將SKIPIF1<0的坐標(biāo)中的用SKIPIF1<0代換，可得SKIPIF1<0的中點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，則直線SKIPIF1<0過定點(diǎn)SKIPIF1<0.【變式2】（2024·上海徐匯·二模）已知橢圓SKIPIF1<0，SKIPIF1<0分別為橢圓SKIPIF1<0的左、右頂點(diǎn)，SKIPIF1<0分別為左、右焦點(diǎn)，直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn)（SKIPIF1<0不過點(diǎn)SKIPIF1<0）.(1)若SKIPIF1<0為橢圓SKIPIF1<0上（除SKIPIF1<0外）任意一點(diǎn)，求直線SKIPIF1<0和SKIPIF1<0的斜率之積；(2)若SKIPIF1<0，求直線SKIPIF1<0的方程；(3)若直線SKIPIF1<0與直線SKIPIF1<0的斜率分別是SKIPIF1<0，且SKIPIF1<0，求證：直線SKIPIF1<0過定點(diǎn).【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)證明見解析【詳解】（1）在橢圓SKIPIF1<0中，左、右頂點(diǎn)分別為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0.（2）設(shè)SKIPIF1<0，由已知可得SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0，聯(lián)立SKIPIF1<0解得SKIPIF1<0由SKIPIF1<0得直線SKIPIF1<0過點(diǎn)SKIPIF1<0，SKIPIF1<0，所以，所求直線方程為SKIPIF1<0.（3）設(shè)SKIPIF1<0，易知直線SKIPIF1<0的斜率不為SKIPIF1<0，設(shè)其方程為SKIPIF1<0（SKIPIF1<0），聯(lián)立SKIPIF1<0，可得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0.由韋達(dá)定理，得SKIPIF1<0.SKIPIF1<0，SKIPIF1<0.可化為SKIPIF1<0，整理即得SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，進(jìn)一步得SKIPIF1<0，化簡(jiǎn)可得SKIPIF1<0，解得SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，恒過定點(diǎn)SKIPIF1<0.【變式3】（2024·新疆喀什·二模）已知橢圓SKIPIF1<0的左焦點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上，過點(diǎn)SKIPIF1<0的兩條直線SKIPIF1<0分別與橢圓SKIPIF1<0交于另一點(diǎn)SKIPIF1<0，且直線SKIPIF1<0的斜率滿足SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)證明直線SKIPIF1<0過定點(diǎn)．【答案】(1)SKIPIF1<0；(2)證明見解析.【詳解】（1）由點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上，得SKIPIF1<0，由SKIPIF1<0為橢圓SKIPIF1<0的左焦點(diǎn)，得SKIPIF1<0，所以橢圓SKIPIF1<0的方程為SKIPIF1<0.（2）依題意，直線SKIPIF1<0不垂直于坐標(biāo)軸，設(shè)其方程為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0消去y并整理得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，整理得SKIPIF1<0，即有SKIPIF1<0，而SKIPIF1<0，解得SKIPIF1<0，滿足SKIPIF1<0，直線SKIPIF1<0：SKIPIF1<0過定點(diǎn)SKIPIF1<0，所以直線SKIPIF1<0過定點(diǎn)SKIPIF1<0.【題型二】自極三角形與調(diào)和點(diǎn)列一、調(diào)和點(diǎn)列的充要條件 如圖，若SKIPIF1<0四點(diǎn)構(gòu)成調(diào)和點(diǎn)列，則有（一般前2個(gè)出現(xiàn)較多）SKIPIF1<0二、調(diào)和點(diǎn)列與極點(diǎn)極線的聯(lián)系如圖，過極點(diǎn)SKIPIF1<0作任意直線，與橢圓交于SKIPIF1<0，與極線交點(diǎn)SKIPIF1<0則點(diǎn)SKIPIF1<0成調(diào)和點(diǎn)列，若點(diǎn)SKIPIF1<0的極線通過另一點(diǎn)SKIPIF1<0，則SKIPIF1<0的極線也通過SKIPIF1<0．一般稱SKIPIF1<0、SKIPIF1<0互為共軛點(diǎn)．三、自極三角形如圖,設(shè)P是不在圓雉曲線上的一點(diǎn),過P點(diǎn)引兩條割線依次交二次曲線于E,F,G,H四點(diǎn),連接對(duì)角線EH,FG交于N,連接對(duì)邊EG,FH交于M,則直線MN為點(diǎn)P對(duì)應(yīng)的極線.若P為圓雉曲線上的點(diǎn),則過P點(diǎn)的切線即為極線.同理,PM為點(diǎn)N對(duì)應(yīng)的極線,PN為點(diǎn)M所對(duì)應(yīng)的極線.因而將△MNP稱為自極三點(diǎn)形.設(shè)直線MN交圓錐曲線于點(diǎn)A,B兩點(diǎn),則PA,PB恰為圓錐曲線的兩條切線.從直線SKIPIF1<0上任意一點(diǎn)SKIPIF1<0向橢圓SKIPIF1<0的左右頂點(diǎn)SKIPIF1<0引兩條割線SKIPIF1<0與橢圓交于SKIPIF1<0兩點(diǎn)，則直線SKIPIF1<0恒過定點(diǎn)SKIPIF1<0．【例1】已知A、B分別為橢圓E：SKIPIF1<0（a>1）的左、右頂點(diǎn)，G為E的上頂點(diǎn)，SKIPIF1<0，P為直線x=6上的動(dòng)點(diǎn)，PA與E的另一交點(diǎn)為C，PB與E的另一交點(diǎn)為D．（1）求E的方程； （2）證明：直線CD過定點(diǎn). 【自極三角形思路】延長(zhǎng)CB,AD交于點(diǎn)Q，SKIPIF1<0，則△EPG為自極三角形，故x=6為E點(diǎn)的極線，則E為SKIPIF1<0【詳解】（1）依據(jù)題意作出如下圖象：由橢圓方程SKIPIF1<0可得：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0橢圓方程為：SKIPIF1<0（2）證明：設(shè)SKIPIF1<0，則直線SKIPIF1<0的方程為：SKIPIF1<0，即：SKIPIF1<0聯(lián)立直線SKIPIF1<0的方程與橢圓方程可得：SKIPIF1<0，整理得：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0將SKIPIF1<0代入直線SKIPIF1<0可得：SKIPIF1<0所以點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0.同理可得：點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0直線SKIPIF1<0的方程為：SKIPIF1<0，整理可得：SKIPIF1<0整理得：SKIPIF1<0所以直線SKIPIF1<0過定點(diǎn)SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0：SKIPIF1<0，直線過點(diǎn)SKIPIF1<0．故直線CD過定點(diǎn)SKIPIF1<0．【例2】（2022·全國乙卷高考真題）已知橢圓E的中心為坐標(biāo)原點(diǎn)，對(duì)稱軸為x軸、y軸，且過SKIPIF1<0兩點(diǎn)．(1)求E的方程； (2)設(shè)過點(diǎn)SKIPIF1<0的直線交E于M，N兩點(diǎn)，過M且平行于x軸的直線與線段AB交于點(diǎn)T，點(diǎn)H滿足SKIPIF1<0．證明：直線HN過定點(diǎn)．【調(diào)和點(diǎn)列思路】AB為P所對(duì)應(yīng)的極線，故P，M，C，N四點(diǎn)成調(diào)和點(diǎn)列，故AP，AM，AC，AN四條線成調(diào)和線束，因?yàn)橹本€HM平行AP，且T為HM中點(diǎn)，由調(diào)和線束平行性質(zhì)(平行于一組調(diào)和線束中的其中一條直線交另外三條直線的三個(gè)交點(diǎn)，其中一個(gè)點(diǎn)為另外兩個(gè)點(diǎn)的中點(diǎn))，故H點(diǎn)必然在直線AN上，故直線HN過定SKIPIF1<0【詳解】（I）解：設(shè)橢圓SKIPIF1<0的方程為SKIPIF1<0，過SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，所以橢圓SKIPIF1<0的方程為：SKIPIF1<0．（II）證法一：定點(diǎn)為SKIPIF1<0，證明如下：點(diǎn)SKIPIF1<0對(duì)應(yīng)的極線為SKIPIF1<0，即SKIPIF1<0，即為直線SKIPIF1<0，則SKIPIF1<0為調(diào)和線束，過SKIPIF1<0作SKIPIF1<0//SKIPIF1<0，交SKIPIF1<0于SKIPIF1<0，由調(diào)和性質(zhì)可知SKIPIF1<0為SKIPIF1<0中點(diǎn)，故直線SKIPIF1<0過定點(diǎn)SKIPIF1<0．證法二：SKIPIF1<0，所以SKIPIF1<0，①若過點(diǎn)SKIPIF1<0的直線斜率不存在，直線SKIPIF1<0．代入SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，代入AB方程SKIPIF1<0，可得SKIPIF1<0，由SKIPIF1<0得到SKIPIF1<0．求得SKIPIF1<0方程：SKIPIF1<0，過點(diǎn)SKIPIF1<0．②若過點(diǎn)SKIPIF1<0的直線斜率存在，設(shè)SKIPIF1<0．聯(lián)立SKIPIF1<0得SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0聯(lián)立SKIPIF1<0可得SKIPIF1<0，可求得此時(shí)SKIPIF1<0，將SKIPIF1<0，代入整理得SKIPIF1<0，將SKIPIF1<0代入，得SKIPIF1<0，顯然成立．綜上，可得直線SKIPIF1<0過定點(diǎn)SKIPIF1<0．【變式1】（2024江南十校聯(lián)考）在平面直角坐標(biāo)系SKIPIF1<0中，已知雙曲線Ｃ的中心為坐標(biāo)原點(diǎn)，對(duì)稱軸是坐標(biāo)軸，右支與x軸的交點(diǎn)為SKIPIF1<0，其中一條漸近線的傾斜角為SKIPIF1<0.(1)求C的標(biāo)準(zhǔn)方程；(2)過點(diǎn)SKIPIF1<0作直線l與雙曲線C的左右兩支分別交于A，B兩點(diǎn)，在線段SKIPIF1<0上取一點(diǎn)E滿足SKIPIF1<0，證明：點(diǎn)E在一條定直線上. 【極線思路】顯然E在T的極線上，故E點(diǎn)軌跡為T的極線SKIPIF1<0【詳解】（1）根據(jù)題意，設(shè)雙曲線的方程為SKIPIF1<0，由題知SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0；所以雙曲線方程為SKIPIF1<0.（2）易知SKIPIF1<0為雙曲線的右焦點(diǎn)，如下圖所示：

由題知直線l斜率存在，根據(jù)對(duì)稱性，不妨設(shè)斜率為SKIPIF1<0，故直線的方程為SKIPIF1<0，代入雙曲線方程得SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，由韋達(dá)定理有SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，點(diǎn)E在線段SKIPIF1<0上，所以SKIPIF1<0由SKIPIF1<0可得SKIPIF1<0化簡(jiǎn)得SKIPIF1<0，代入SKIPIF1<0和SKIPIF1<0并化簡(jiǎn)可得SKIPIF1<0，即存在點(diǎn)E滿足條件，并且在定直線SKIPIF1<0上.【變式2】設(shè)橢圓SKIPIF1<0過點(diǎn)SKIPIF1<0，且左焦點(diǎn)為SKIPIF1<0．（1）求橢圓SKIPIF1<0的方程；（2）當(dāng)過點(diǎn)SKIPIF1<0的動(dòng)直線SKIPIF1<0與橢圓SKIPIF1<0相交于兩不同點(diǎn)SKIPIF1<0，SKIPIF1<0時(shí)，在線段SKIPIF1<0上取點(diǎn)SKIPIF1<0，滿足SKIPIF1<0，證明：點(diǎn)SKIPIF1<0總在某定直線上． 解析：（1）由題意得SKIPIF1<0，解得SKIPIF1<0，所求橢圓方程為SKIPIF1<0．（2）解法：已知SKIPIF1<0，說明點(diǎn)SKIPIF1<0關(guān)于橢圓調(diào)和共軛，根據(jù)定理3，點(diǎn)SKIPIF1<0在點(diǎn)SKIPIF1<0對(duì)應(yīng)的極線上，此極線方程為SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0．故點(diǎn)SKIPIF1<0總在直線SKIPIF1<0．【變式3】已知SKIPIF1<0、SKIPIF1<0分別為橢圓SKIPIF1<0:SKIPIF1<0的上、下焦點(diǎn)，其中SKIPIF1<0也是拋物線SKIPIF1<0的焦點(diǎn)，點(diǎn)SKIPIF1<0是SKIPIF1<0與SKIPIF1<0在第二象限的交點(diǎn)，且SKIPIF1<0.

（1）求橢圓SKIPIF1<0的方程; （2）已知點(diǎn)SKIPIF1<0和圓SKIPIF1<0:SKIPIF1<0，過點(diǎn)SKIPIF1<0的動(dòng)直線SKIPIF1<0與圓SKIPIF1<0相交于不同的兩點(diǎn)SKIPIF1<0，在線段SKIPIF1<0上取一點(diǎn)SKIPIF1<0，滿足:SKIPIF1<0，SKIPIF1<0，(SKIPIF1<0且SKIPIF1<0).求證:點(diǎn)SKIPIF1<0總在某定直線上.【極線思路】由題可知SKIPIF1<0，即SKIPIF1<0，故點(diǎn)Q在P點(diǎn)的極線上【詳解】（1）設(shè)SKIPIF1<0，因?yàn)辄c(diǎn)M在拋物線SKIPIF1<0上，且SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，又點(diǎn)M在拋物線SKIPIF1<0上，所以SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以橢圓SKIPIF1<0的方程SKIPIF1<0；（2）設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即有SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即有SKIPIF1<0，所以SKIPIF1<0得：SKIPIF1<0，又點(diǎn)A、B在圓SKIPIF1<0上，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，故點(diǎn)Q總在直線SKIPIF1<0上．【題型三】齊次化法解決斜率相關(guān)問題“齊次”從詞面上解釋是“次數(shù)相等”的意思.在代數(shù)里也有“齊次”的叫法，例如f=1：“齊次化”方法使用場(chǎng)景題目中出現(xiàn)了一個(gè)定點(diǎn)引出的兩條動(dòng)直線的斜率之和k?+k?2：用法：必須先把該定點(diǎn)平移至原點(diǎn)位置，然后將兩個(gè)動(dòng)點(diǎn)所成的直線假設(shè)為mx+3:方程為mx+【例1】如圖，橢圓E:x2a2+y(1):求橢圓E的方程;(2)：經(jīng)過點(diǎn)(1，1)，且斜率為k的直線與橢圓E交于不同的兩點(diǎn)P、Q(均異于點(diǎn)A),解:(1)由題意,b=1,又c所以，a=2,b=1,c=1,所以，橢圓E的方程為首先將橢圓向上平移1個(gè)單位，得橢圓.E'的方程：x22此時(shí),點(diǎn)(1,1)向上平移1個(gè)單位,變?yōu)?1.2),設(shè)直線方程為mx+my=1,直線過(l,2),則有m+2m=1,將直線方程代入x2+2y2?4y=0得x2+2y2?4y平移之后，點(diǎn)P、Q變?yōu)镻'變換得4n?2y所以k原問題成立.【例2】已知橢圓C:x2a2(1)求橢圓C的方程;(2)點(diǎn)M,N在橢圓C上,且.AM⊥AN,AD⊥MN,D為垂足.解:(1)由題意得:ca=224a橢圓的方程為x26+(2)當(dāng)直線A'將整個(gè)圖形平移一下，將點(diǎn)A平移到坐標(biāo)原點(diǎn)A相應(yīng)的其它點(diǎn)平移為M'橢圓方程為：x+22+2y+12=6,設(shè)MN':mx+ny=1,即:4n+2yx2所以，?43m?43n=1,再將整個(gè)圖形重新平移回去，直線MN經(jīng)過定點(diǎn)B由題意可知，△ADB為直角三角形，斜邊AB的中點(diǎn)Q4當(dāng)直線AM，AN的斜率有一個(gè)不存在時(shí)，不妨設(shè)N2?1,M?21,綜上所述，存在點(diǎn)Q431【變式1】（2024·全國·模擬預(yù)測(cè)）已知P為橢圓SKIPIF1<0上一點(diǎn)，過原點(diǎn)且斜率存在的直線SKIPIF1<0與橢圓C相交于A，B兩點(diǎn)，過原點(diǎn)且斜率存在的直線SKIPIF1<0（SKIPIF1<0與SKIPIF1<0不重合）與橢圓C相交于M，N兩點(diǎn)，且點(diǎn)P滿足到直線SKIPIF1<0和SKIPIF1<0的距離都等于SKIPIF1<0．(1)求直線SKIPIF1<0和SKIPIF1<0的斜率之積；(2)當(dāng)點(diǎn)P在C上運(yùn)動(dòng)時(shí)，SKIPIF1<0是否為定值？若是，求出該值；若不是，請(qǐng)說明理由．【答案】(1)SKIPIF1<0；(2)是，12.【詳解】（1）設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，SKIPIF1<0，則根據(jù)點(diǎn)到直線的距離公式可得SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，同理可得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0是一元二次方程SKIPIF1<0的兩實(shí)數(shù)根，SKIPIF1<0，則有SKIPIF1<0．又因?yàn)辄c(diǎn)SKIPIF1<0在C上，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0（定值）．（2）SKIPIF1<0是定值，且定值為12．理由如下：設(shè)SKIPIF1<0，SKIPIF1<0，聯(lián)立方程組SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，同理可得SKIPIF1<0．由橢圓的對(duì)稱性知SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．由（1）知SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0（定值）．【變式2】（2024·安徽合肥·二模）已知橢圓SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，左頂點(diǎn)為SKIPIF1<0，短軸長(zhǎng)為SKIPIF1<0，且經(jīng)過點(diǎn)SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0的直線SKIPIF1<0（不與SKIPIF1<0軸重合）與SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)分別為SKIPIF1<0，記直線SKIPIF1<0的斜率分別為SKIPIF1<0，證明：SKIPIF1<0為定值．【答案】(1)SKIPIF1<0；(2)證明見解析.【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，再將點(diǎn)SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0，解得SKIPIF1<0，故橢圓SKIPIF1<0的方程為SKIPIF1<0；（2）由題意可設(shè)SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，易知SKIPIF1<0恒成立，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，同理SKIPIF1<0，從而SKIPIF1<0，故SKIPIF1<0為定值．【變式3】（2024·全國·模擬預(yù)測(cè)）已知曲線SKIPIF1<0與曲線SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱．(1)求曲線SKIPIF1<0的方程．(2)若過原點(diǎn)的兩條直線分別交曲線SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)），則四邊形SKIPIF1<0的面積是否為定值？若為定值，求四邊形SKIPIF1<0的面積；若不為定值，請(qǐng)說明理由．【答案】(1)SKIPIF1<0(2)是定值，SKIPIF1<0【詳解】（1）設(shè)點(diǎn)SKIPIF1<0為曲線SKIPIF1<0上任一點(diǎn)，則點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0在曲線SKIPIF1<0上．根據(jù)對(duì)稱性，得SKIPIF1<0解得SKIPIF1<0將SKIPIF1<0代入曲線SKIPIF1<0并整理，得SKIPIF1<0．故曲線SKIPIF1<0的方程為SKIPIF1<0．（2）四邊形SKIPIF1<0的面積為定值．理由如下：當(dāng)直線SKIPIF1<0的斜率不存在時(shí)，直線SKIPIF1<0軸，則SKIPIF1<0．因?yàn)镾KIPIF1<0，所以不妨設(shè)SKIPIF1<0，則SKIPIF1<0，此時(shí)取SKIPIF1<0，SKIPIF1<0，根據(jù)對(duì)稱性可知四邊形SKIPIF1<0為平行四邊形，則四邊形SKIPIF1<0的面積SKIPIF1<0，為定值．當(dāng)直線SKIPIF1<0的斜率存在時(shí)，設(shè)SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0．聯(lián)立SKIPIF1<0得SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0．因?yàn)镾KIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0．因?yàn)樵c(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離SKIPIF1<0，由于四邊形SKIPIF1<0為平行四邊形，所以四邊形SKIPIF1<0的面積SKIPIF1<0．綜上，四邊形SKIPIF1<0的面積為定值SKIPIF1<0．【題型四】定比點(diǎn)差法直線與圓雉曲線相交時(shí),中點(diǎn)(定比分點(diǎn))問題通常運(yùn)用韋達(dá)定理和點(diǎn)差法兩種方式.點(diǎn)差法(定比點(diǎn)差)是從設(shè)點(diǎn)的視角,將點(diǎn)的坐標(biāo)代人曲線方程,通過系數(shù)調(diào)配后進(jìn)行兩式作差.一般地,設(shè)橢圓SKIPIF1<0上兩點(diǎn)SKIPIF1<0,若定點(diǎn)SKIPIF1<0滿足SKIPIF1<0,則得到SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0由SKIPIF1<0得SKIPIF1<0兩式相減得SKIPIF1<0.把SKIPIF1<0代人,得SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0.特別地,如果SKIPIF1<0(或SKIPIF1<0),則可以得到方程組SKIPIF1<0繼而能相對(duì)快捷地求出交點(diǎn)坐標(biāo),避免暴求交點(diǎn).橢圓、雙曲線中的多點(diǎn)共線的倍值問題,也可類似解決,其實(shí)質(zhì)就是一種降維處理.此外,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0是SKIPIF1<0的中點(diǎn)即轉(zhuǎn)化為中點(diǎn)弦問題.【例1】直線SKIPIF1<0與橢圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0與SKIPIF1<0軸、SKIPIF1<0軸分別交于點(diǎn)SKIPIF1<0.如果SKIPIF1<0是線段SKIPIF1<0的兩個(gè)三等分點(diǎn)，則直線SKIPIF1<0的斜率為．【解析】設(shè)點(diǎn)SKIPIF1<0.由SKIPIF1<0得SKIPIF1<0；由SKIPIF1<0得SKIPIF1<0.所以SKIPIF1<0.(*)由SKIPIF1<0兩式相減得SKIPIF1<0.把(*)代入，知SKIPIF1<0，故SKIPIF1<0，所以點(diǎn)SKIPIF1<0，所以SKIPIF1<0.【例2】設(shè)SKIPIF1<0分別為橢圓SKIPIF1<0的左右兩個(gè)焦點(diǎn)，點(diǎn)SKIPIF1<0在橢圓上.若SKIPIF1<0SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)是．【解析】延長(zhǎng)SKIPIF1<0交橢圓于點(diǎn)SKIPIF1<0.由SKIPIF1<0知，SKIPIF1<0，所以SKIPIF1<0把點(diǎn)SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0兩式相減，得SKIPIF1<0，化簡(jiǎn)，得SKIPIF1<0.與SKIPIF1<0聯(lián)立，解得SKIPIF1<0，代入橢圓求得點(diǎn)SKIPIF1<0.【例3】已知點(diǎn)SKIPIF1<0，橢圓SKIPIF1<0上兩點(diǎn)SKIPIF1<0滿足SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0橫坐標(biāo)的絕對(duì)值最大.【解析】設(shè)點(diǎn)SKIPIF1<0.SKIPIF1<0由SKIPIF1<0.所以SKIPIF1<0，所以SKIPIF1<0，代入SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)取得等號(hào).【變式1】已知SKIPIF1<0是雙曲線SKIPIF1<0的左焦點(diǎn)，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，直線SKIPIF1<0與雙曲線SKIPIF1<0的兩條漸進(jìn)線分別交于點(diǎn)SKIPIF1<0.若SKIPIF1<0SKIPIF1<0，則雙曲線SKIPIF1<0的離心率為．【答案】SKIPIF1<0.【解析】設(shè)點(diǎn)SKIPIF1<0.由于SKIPIF1<0,故SKIPIF1<0,得到SKIPIF1<0由點(diǎn)SKIPIF1<0均在漸近線SKIPIF1<00上可以知道,SKIPIF1<0則SKIPIF1<0兩式相減得SKIPIF1<0.把SKIPIF1<0代人,知SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0.結(jié)合SKIPIF1<0,解得SKIPIF1<0,故點(diǎn)SKIPIF1<0.由SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0.【變式2】已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，斜率為SKIPIF1<0的直線SKIPIF1<0與拋物線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0.(1)若SKIPIF1<0，求直線SKIPIF1<0的方程；(2)若SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0.【解析】(1)設(shè)直線SKIPIF1<0的方程為SKIPIF1<0.SKIPIF1<0設(shè)點(diǎn)SKIPIF1<0,故SKIPIF1<0.由拋物線定義可得SKIPIF1<0SKIPIF1<0,解得SKIPIF1<0.故直線SKIPIF1<0方程為SKIPIF1<0.(2)設(shè)直線SKIPIF1<0的方程為SKIPIF1<0SKIPIF1<0SKIPIF1<0設(shè)點(diǎn)SKIPIF1<0,故SKIPIF1<0由SKIPIF1<0可得SKIPIF1<0,可得SKIPIF1<0代入上式可得SKIPIF1<0.故直線SKIPIF1<0方程為SKIPIF1<0.解得點(diǎn)SKIPIF1<0,故SKIPIF1<0.【變式3】如圖，橢圓SKIPIF1<0.過點(diǎn)SKIPIF1<0作直線SKIPIF1<0分別交橢圓SKIPIF1<0于SKIPIF1<0，SKIPIF1<0四點(diǎn)，且直線SKIPIF1<0的斜率為SKIPIF1<0.試判斷直線SKIPIF1<0與直線SKIPIF1<0的位置關(guān)系.【答案】SKIPIF1<0.【解析】設(shè)點(diǎn)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.設(shè)SKIPIF1<0,則由定比分點(diǎn)得到SKIPIF1<0又點(diǎn)SKIPIF1<0,SKIPIF1<0在橢圓SKIPIF1<0上,所以SKIPIF1<0SKIPIF1<0又SKIPIF1<0三式相加得SKIPIF1<0.同理,設(shè)SKIPIF1<0,可得SKIPIF1<0.兩式相減得SKIPIF1<0.又直線SKIPIF1<0的斜率為SKIPIF1<0,則SKIPIF1<0.所以SKIPIF1<0,即SKIPIF1<0.所以SKIPIF1<0.【題型五】定點(diǎn)、定值求定值問題常見的方法有兩種：(1)從特殊入手，求出定值，再證明這個(gè)值與變量無關(guān)．(2)直接推理、計(jì)算，并在計(jì)算推理的過程中消去變量，從而得到定值．直線過定點(diǎn)問題或圓過定點(diǎn)問題，通常要設(shè)出直線方程，與圓錐曲線聯(lián)立，得到兩根之和，兩根之積，再表達(dá)出直線方程或圓的方程，結(jié)合方程特點(diǎn)，求出所過的定點(diǎn)坐標(biāo).【例1】（2024·全國·模擬預(yù)測(cè)）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0，SKIPIF1<0的左焦點(diǎn)與點(diǎn)SKIPIF1<0連線的斜率為SKIPIF1<0．(1)求SKIPIF1<0的方程．(2)已知點(diǎn)SKIPIF1<0，過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，直線SKIPIF1<0分別交SKIPIF1<0于SKIPIF1<0．試問：直線SKIPIF1<0的斜率是否為定值？若是，求出該定值；若不是，請(qǐng)說明理由．【答案】(1)SKIPIF1<0；(2)是，定值為SKIPIF1<0【詳解】（1）由題意可設(shè)橢圓SKIPIF1<0的半焦距為SKIPIF1<0，則橢圓SKIPIF1<0的左焦點(diǎn)為SKIPIF1<0．由題意得SKIPIF1<0，則SKIPIF1<0，所以橢圓SKIPIF1<