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）（原卷版）

秘籍09圓錐曲線大題目錄【高考預(yù)測】概率預(yù)測+題型預(yù)測+考向預(yù)測【應(yīng)試秘籍】總結(jié)?？键c(diǎn)及應(yīng)對的策略【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☆☆☆☆☆題型預(yù)測解答題☆☆☆☆☆考向預(yù)測極點(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)：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)．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【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【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________．【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．【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．【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).【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)．【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),連接對角線EH,FG交于N,連接對邊EG,FH交于M,則直線MN為點(diǎn)P對應(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對應(yīng)的極線,PN為點(diǎn)M所對應(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). 【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)．【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. 【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． 【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.【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(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),【例2】已知橢圓C:x2a(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.【變式1】（2024·全國·模擬預(yù)測）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，請說明理由．【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．【變式3】（2024·全國·模擬預(yù)測）已知曲線SKIPIF1<0與曲線SKIPIF1<0關(guān)于直線SKIPIF1<0對稱．(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，請說明理由．【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．【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．【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)的絕對值最大.【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．【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.【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.【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）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，請說明理由．【例2】（2023·河南焦作·模擬預(yù)測）已知橢圓SKIPIF1<0的長軸為4，直線SKIPIF1<0與圓SKIPIF1<0相切于點(diǎn)SKIPIF1<0，與SKIPIF1<0相交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.(1)記SKIPIF1<0的離心率為SKIPIF1<0，證明：SKIPIF1<0；(2)若SKIPIF1<0軸右側(cè)的點(diǎn)SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0軸，SKIPIF1<0，SKIPIF1<0是圓SKIPIF1<0的兩條切線，切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0（SKIPIF1<0在SKIPIF1<0上方），求SKIPIF1<0的值.【例3】（2024·上海奉賢·二模）已知曲線SKIPIF1<0SKIPIF1<0，SKIPIF1<0是坐標(biāo)原點(diǎn)，過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與曲線SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0兩點(diǎn).(1)當(dāng)SKIPIF1<0與SKIPIF1<0軸垂直時(shí)，求SKIPIF1<0的面積；(2)過圓SKIPIF1<0上任意一點(diǎn)SKIPIF1<0作直線SKIPIF1<0，SKIPIF1<0，分別與曲線SKIPIF1<0切于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，求證：SKIPIF1<0；

(3)過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與雙曲線SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)（SKIPIF1<0，SKIPIF1<0不與SKIPIF1<0軸重合）.記直線SKIPIF1<0的斜率為SKIPIF1<0，直線SKIPIF1<0斜率為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，求證：SKIPIF1<0與SKIPIF1<0都是定值.

【變式1】（2024·上海崇明·二模）已知橢圓SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的上頂點(diǎn)，SKIPIF1<0是SKIPIF1<0上不同于點(diǎn)SKIPIF1<0的兩點(diǎn)．(1)求橢圓SKIPIF1<0的離心率；(2)若SKIPIF1<0是橢圓SKIPIF1<0的右焦點(diǎn)，SKIPIF1<0是橢圓下頂點(diǎn)，SKIPIF1<0是直線SKIPIF1<0上一點(diǎn).若SKIPIF1<0有一個(gè)內(nèi)角為SKIPIF1<0，求點(diǎn)SKIPIF1<0的坐標(biāo)；(3)作SKIPIF1<0，垂足為SKIPIF1<0．若直線SKIPIF1<0與直線SKIPIF1<0的斜率之和為SKIPIF1<0，是否存在SKIPIF1<0軸上的點(diǎn)SKIPIF1<0，使得SKIPIF1<0為定值？若存在，請求出點(diǎn)SKIPIF1<0的坐標(biāo)，若不存在，請說明理由．【變式2】（2024·全國·模擬預(yù)測）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0，且過點(diǎn)SKIPIF1<0．(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程．(2)設(shè)過點(diǎn)SKIPIF1<0且斜率不為0的直線SKIPIF1<0與橢圓SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)．問：在SKIPIF1<0軸上是否存在定點(diǎn)SKIPIF1<0，使直線SKIPIF1<0的斜率SKIPIF1<0與SKIPIF1<0的斜率SKIPIF1<0的積為定值？若存在，求出該定點(diǎn)坐標(biāo)；若不存在，請說明理由．【變式3】（2024·全國·模擬預(yù)測）已知離心率為SKIPIF1<0的橢圓SKIPIF1<0的左、右頂點(diǎn)分別為SKIPIF1<0，點(diǎn)SKIPIF1<0為橢圓SKIPIF1<0上的動(dòng)點(diǎn)，且SKIPIF1<0面積的最大值為SKIPIF1<0．直線SKIPIF1<0與橢圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，點(diǎn)SKIPIF1<0，直線SKIPIF1<0分別交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn)，過點(diǎn)SKIPIF1<0作直線SKIPIF1<0的垂線，垂足為SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程．(2)記直線SKIPIF1<0的斜率為SKIPIF1<0，證明：SKIPIF1<0為定值．(3)試問：是否存在定點(diǎn)SKIPIF1<0，使SKIPIF1<0為定值？若存在，求出定點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，說明理由．【題型六】求軌跡方程型求軌跡方程的常見方法有:①直接法，設(shè)出動(dòng)點(diǎn)的坐標(biāo)SKIPIF1<0，根據(jù)題意列出關(guān)于SKIPIF1<0的等式即可；②定義法，根據(jù)題意動(dòng)點(diǎn)符合已知曲線的定義，直接求出方程；③參數(shù)法，把SKIPIF1<0分別用第三個(gè)變量表示，消去參數(shù)即可；④逆代法，將SKIPIF1<0代入SKIPIF1<0.【例1】（2024·上海嘉定·二模）如圖：已知三點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0都在橢圓SKIPIF1<0上．(1)若點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0都是橢圓的頂點(diǎn)，求SKIPIF1<0的面積；(2)若直線SKIPIF1<0的斜率為1，求弦SKIPIF1<0中點(diǎn)SKIPIF1<0的軌跡方程；(3)若直線SKIPIF1<0的斜率為2，設(shè)直線SKIPIF1<0的斜率為SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，是否存在定點(diǎn)SKIPIF1<0，使得SKIPIF1<0恒成立？若存在，求出所有滿足條件的點(diǎn)SKIPIF1<0，若不存在，說明理由．【例2】（2024·安徽合肥·二模）在數(shù)學(xué)中，廣義距離是泛函分析中最基本的概念之一．對平面直角坐標(biāo)系中兩個(gè)點(diǎn)SKIPIF1<0和SKIPIF1<0，記SKIPIF1<0，稱SKIPIF1<0為點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0之間的“SKIPIF1<0距離”，其中SKIPIF1<0表示SKIPIF1<0中較大者．(1)計(jì)算點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0之間的“SKIPIF1<0距離”；(2)設(shè)SKIPIF1<0是平面中一定點(diǎn)，SKIPIF1<0．我們把平面上到點(diǎn)SKIPIF1<0的“SKIPIF1<0距離”為SKIPIF1<0的所有點(diǎn)構(gòu)成的集合叫做以點(diǎn)SKIPIF1<0為圓心，以SKIPIF1<0為半徑的“SKIPIF1<0圓”．求以原點(diǎn)SKIPIF1<0為圓心，以SKIPIF1<0為半徑的“SKIPIF1<0圓”的面積；(3)證明：對任意點(diǎn)SKIPIF1<0．【例3】（2024·河南開封·三模）已知SKIPIF1<0，SKIPIF1<0，對于平面內(nèi)一動(dòng)點(diǎn)SKIPIF1<0，SKIPIF1<0軸于點(diǎn)M，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列．(1)求點(diǎn)P的軌跡C的方程；(2)已知過點(diǎn)A的直線l與C交于M，N兩點(diǎn)，若SKIPIF1<0，求直線l的方程．【變式1】（2024·廣東韶關(guān)·二模）已知橢圓SKIPIF1<0的離心率為SKIPIF1<0，長軸長為4，SKIPIF1<0是其左、