新高考數(shù)學(xué)三輪沖刺大題突破練習(xí)專題05 解析幾何（原卷版）

專題05解析幾何解析幾何一般作為解答題21題或者是22題形式出現(xiàn)。一般作為壓軸題或者是次壓軸題出現(xiàn)，難度較大。1與原有關(guān)問題（蒙日圓，阿氏圓等）2面積問題3齊次化解決直線定點(diǎn)問題4一般的定值問題5非對稱問題6探究性問題7切線問題與阿基米德三角形問題8極點(diǎn)極限與調(diào)和點(diǎn)列，蝴蝶模型問題9不聯(lián)立問題10與其他知識點(diǎn)交叉問題蒙日圓定理的內(nèi)容：橢圓的兩條切線互相垂直，則兩切線的交點(diǎn)位于一個(gè)與橢圓同心的圓上，該圓稱為蒙日圓，其半徑等于橢圓長半軸和短半軸平方和的算術(shù)平方根，具體結(jié)論及證明如下：結(jié)論一：曲線SKIPIF1<0的兩條互相垂直的切線的交點(diǎn)SKIPIF1<0的軌跡是圓：SKIPIF1<0．結(jié)論二：雙曲線SKIPIF1<0的兩條互相垂直的切線的交點(diǎn)的軌跡是圓SKIPIF1<0SKIPIF1<0．結(jié)論三：拋物線SKIPIF1<0的兩條互相垂直的切線的交點(diǎn)在該拋物線的準(zhǔn)線上．題型一：與原有關(guān)問題（蒙日圓，協(xié)同圓等）例題1已知橢圓SKIPIF1<00)．稱圓心在原點(diǎn)，半徑為SKIPIF1<0的圓為橢圓SKIPIF1<0的“準(zhǔn)圓”．若橢圓SKIPIF1<0的一個(gè)焦點(diǎn)為SKIPIF1<0SKIPIF1<0，其短軸上的一個(gè)端點(diǎn)到SKIPIF1<0的距離為SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程及其“準(zhǔn)圓”方程．(2)點(diǎn)SKIPIF1<0是橢圓SKIPIF1<0的“準(zhǔn)圓”上的動(dòng)點(diǎn)，過點(diǎn)SKIPIF1<0作橢圓的切線SKIPIF1<0交“準(zhǔn)圓”于點(diǎn)SKIPIF1<0．①當(dāng)點(diǎn)SKIPIF1<0為“準(zhǔn)圓”與SKIPIF1<0軸正半軸的交點(diǎn)時(shí)，求直線SKIPIF1<0的方程并證明SKIPIF1<0．②求證：線段SKIPIF1<0的長為定值．【解析】(1)依題意可得SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．SKIPIF1<0．(2)證明：①由(1)題可得SKIPIF1<0，設(shè)切線方程為：SKIPIF1<0．聯(lián)立SKIPIF1<0，消去SKIPIF1<0可得SKIPIF1<0，整理可得SKIPIF1<0．∴SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0．∴設(shè)直線PM：SKIPIF1<0，直線SKIPIF1<0SKIPIF1<0．∴SKIPIF1<0，即SKIPIF1<0．②設(shè)SKIPIF1<0，直線SKIPIF1<0SKIPIF1<0．則SKIPIF1<0，消去SKIPIF1<0可得SKIPIF1<0．即SKIPIF1<0SKIPIF1<0．∴SKIPIF1<0．整理得SKIPIF1<0．同理，設(shè)切線SKIPIF1<0的斜率為SKIPIF1<0，則有SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0在“準(zhǔn)圓”上．∴SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0為“準(zhǔn)圓”的直徑．∴SKIPIF1<0為定值，SKIPIF1<0．1．公元前3世紀(jì)，古希臘數(shù)學(xué)家阿波羅尼斯SKIPIF1<0在《平面軌跡》一書中，研究了眾多的平面軌跡問題，其中有如下著名結(jié)果：平面內(nèi)到兩個(gè)定點(diǎn)SKIPIF1<0，SKIPIF1<0距離之比為SKIPIF1<0且SKIPIF1<0的點(diǎn)SKIPIF1<0的軌跡為圓，此圓稱為阿波羅尼斯圓．(1)已知兩定點(diǎn)SKIPIF1<0，SKIPIF1<0，若動(dòng)點(diǎn)SKIPIF1<0滿足SKIPIF1<0，求點(diǎn)SKIPIF1<0的軌跡方程；(2)已知SKIPIF1<0，SKIPIF1<0是圓SKIPIF1<0上任意一點(diǎn)，在平面上是否存在點(diǎn)SKIPIF1<0，使得SKIPIF1<0恒成立？若存在，求出點(diǎn)SKIPIF1<0坐標(biāo)；若不存在，說明理由；(3)已知SKIPIF1<0是圓SKIPIF1<0上任意一點(diǎn)，在平面內(nèi)求出兩個(gè)定點(diǎn)SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0恒成立．只需寫出兩個(gè)定點(diǎn)SKIPIF1<0，SKIPIF1<0的坐標(biāo)，無需證明．題型二：面積問題1．已知M是平面直角坐標(biāo)系內(nèi)的一個(gè)動(dòng)點(diǎn)，直線SKIPIF1<0與直線SKIPIF1<0垂直，A為垂足且位于第一象限，直線SKIPIF1<0與直線SKIPIF1<0垂直，B為垂足且位于第四象限，四邊形SKIPIF1<0（O為原點(diǎn)）的面積為8，動(dòng)點(diǎn)M的軌跡為C.(1)求軌跡C的方程；(2)已知SKIPIF1<0是軌跡C上一點(diǎn)，直線l交軌跡C于P，Q兩點(diǎn)，直線SKIPIF1<0，SKIPIF1<0的斜率之和為1，SKIPIF1<0，求SKIPIF1<0的面積.【答案】(1)SKIPIF1<0（SKIPIF1<0）(2)SKIPIF1<0【詳解】（1）設(shè)動(dòng)點(diǎn)SKIPIF1<0，由題意知M只能在直線SKIPIF1<0與直線SKIPIF1<0所夾的范圍內(nèi)活動(dòng).SKIPIF1<0，SKIPIF1<0，動(dòng)點(diǎn)SKIPIF1<0在SKIPIF1<0右側(cè)，有SKIPIF1<0，同理有SKIPIF1<0，∵四邊形SKIPIF1<0的面積為8，∴SKIPIF1<0，即SKIPIF1<0，所以所求軌跡C方程為SKIPIF1<0（SKIPIF1<0）.（2）如圖，設(shè)直線SKIPIF1<0的傾斜角為SKIPIF1<0，斜率為k，直線SKIPIF1<0傾斜角為SKIPIF1<0，則SKIPIF1<0斜率為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在曲線C上，過點(diǎn)T直線與曲線C有兩個(gè)交點(diǎn)，則SKIPIF1<0或SKIPIF1<0，同時(shí)SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.

SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去）.SKIPIF1<0時(shí)，直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，消y得：SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，得SKIPIF1<0.直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，消y得：SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，點(diǎn)Q到直線SKIPIF1<0的距離SKIPIF1<0

，SKIPIF1<0.方法二：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.1已知橢圓SKIPIF1<0離心率為SKIPIF1<0，經(jīng)過SKIPIF1<0的左焦點(diǎn)SKIPIF1<0斜率為1的直線與SKIPIF1<0軸正半軸相交于點(diǎn)SKIPIF1<0，且SKIPIF1<0.(1)求SKIPIF1<0的方程；(2)設(shè)M，N是SKIPIF1<0上異于SKIPIF1<0的兩點(diǎn)，若SKIPIF1<0，求SKIPIF1<0面積的最大值.題型三：齊次化解決定值定點(diǎn)問題1已知橢圓C：SKIPIF1<0（a>b>0），四點(diǎn)P1（1,1），P2（0,1），P3（–1，SKIPIF1<0），P4（1，SKIPIF1<0）中恰有三點(diǎn)在橢圓C上.（Ⅰ）求C的方程；（Ⅱ）設(shè)直線l不經(jīng)過P2點(diǎn)且與C相交于A，B兩點(diǎn).若直線P2A與直線P2B的斜率的和為–1，證明：l過定點(diǎn).【答案】(1)SKIPIF1<0.(2)證明見解析.解題方法一：試題解析：（1）由于SKIPIF1<0，SKIPIF1<0兩點(diǎn)關(guān)于y軸對稱，故由題設(shè)知C經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn).又由SKIPIF1<0知，C不經(jīng)過點(diǎn)P1，所以點(diǎn)P2在C上.因此SKIPIF1<0，解得SKIPIF1<0.故C的方程為SKIPIF1<0.（2）設(shè)直線P2A與直線P2B的斜率分別為k1，k2，如果l與x軸垂直，設(shè)l：x=t，由題設(shè)知SKIPIF1<0，且SKIPIF1<0，可得A，B的坐標(biāo)分別為（t，SKIPIF1<0），（t，SKIPIF1<0）.則SKIPIF1<0，得SKIPIF1<0，不符合題設(shè).從而可設(shè)l：SKIPIF1<0（SKIPIF1<0）.將SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0由題設(shè)可知SKIPIF1<0.設(shè)A（x1，y1），B（x2，y2），則x1+x2=SKIPIF1<0，x1x2=SKIPIF1<0.而SKIPIF1<0SKIPIF1<0SKIPIF1<0.由題設(shè)SKIPIF1<0，故SKIPIF1<0.即SKIPIF1<0.解得SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，欲使l：SKIPIF1<0，即SKIPIF1<0，所以l過定點(diǎn)（2，SKIPIF1<0）解題方法二：齊次化處理：1．已知橢圓C：SKIPIF1<0的離心率為SKIPIF1<0，且過點(diǎn)SKIPIF1<0．（1）求SKIPIF1<0的方程：（2）點(diǎn)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為垂足．證明：存在定點(diǎn)SKIPIF1<0，使得SKIPIF1<0為定值．題型四：一般的定值定點(diǎn)問題1．已知SKIPIF1<0為雙曲線SKIPIF1<0的左?右焦點(diǎn)，SKIPIF1<0的一條漸近線方程為SKIPIF1<0為SKIPIF1<0上一點(diǎn)，且SKIPIF1<0．(1)求SKIPIF1<0的方程；(2)設(shè)點(diǎn)SKIPIF1<0在坐標(biāo)軸上，直線SKIPIF1<0與SKIPIF1<0交于異于SKIPIF1<0的SKIPIF1<0兩點(diǎn)，SKIPIF1<0為SKIPIF1<0的中點(diǎn)，且SKIPIF1<0，過SKIPIF1<0作SKIPIF1<0，垂足為SKIPIF1<0，是否存在點(diǎn)SKIPIF1<0，使得SKIPIF1<0為定值？若存在，求出點(diǎn)SKIPIF1<0的坐標(biāo)以及SKIPIF1<0的長度；若不存在，請說明理由．【答案】(1)SKIPIF1<0(2)存在；點(diǎn)SKIPIF1<0，SKIPIF1<0為定值SKIPIF1<0【詳解】（1）由題意，在雙曲線SKIPIF1<0中，漸近線方程為SKIPIF1<0，由條件可知SKIPIF1<0．根據(jù)雙曲線的定義可知，SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0．（2）由題意及（1）得，在SKIPIF1<0中，SKIPIF1<0，∴點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0的左支上，當(dāng)點(diǎn)SKIPIF1<0在坐標(biāo)軸上，則點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0的斜率存在時(shí)，設(shè)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，整理得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0為SKIPIF1<0的中點(diǎn)，且SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，驗(yàn)證均滿足SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0的方程為SKIPIF1<0，則直線SKIPIF1<0過點(diǎn)SKIPIF1<0，不合題意，舍去；當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0的方程為SKIPIF1<0，則直線SKIPIF1<0過定點(diǎn)SKIPIF1<0，符合題意．當(dāng)直線SKIPIF1<0的斜率不存在時(shí)，由SKIPIF1<0，可設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，所以直線SKIPIF1<0的方程為：SKIPIF1<0，則直線SKIPIF1<0過定點(diǎn)SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0是以SKIPIF1<0為斜邊的直角三角形，∴點(diǎn)SKIPIF1<0在以SKIPIF1<0為直徑的圓上，則當(dāng)SKIPIF1<0為該圓的圓心SKIPIF1<0時(shí)，SKIPIF1<0為該圓的半徑，即SKIPIF1<0，故存在點(diǎn)SKIPIF1<0，使得SKIPIF1<0為定值SKIPIF1<0．1已知雙曲線SKIPIF1<0過點(diǎn)SKIPIF1<0，且SKIPIF1<0與SKIPIF1<0的兩個(gè)頂點(diǎn)連線的斜率之和為4．(1)求SKIPIF1<0的方程；(2)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)（異于點(diǎn)SKIPIF1<0）．設(shè)直線SKIPIF1<0與SKIPIF1<0軸垂直且交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，若線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0，證明：直線SKIPIF1<0的斜率為定值，并求該定值．類型五非對稱問題1已知橢圓SKIPIF1<0的長軸長為6，離心率為SKIPIF1<0.（1）求橢圓C的標(biāo)準(zhǔn)方程；（2）設(shè)橢圓C的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，左、右頂點(diǎn)分別為A，B，點(diǎn)M，N為橢圓C上位于x軸上方的兩點(diǎn)，且SKIPIF1<0，記直線AM，BN的斜率分別為SKIPIF1<0，且SKIPIF1<0,求直線SKIPIF1<0的方程.（1）SKIPIF1<0（2）SKIPIF1<0（1）由題意，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，聯(lián)立解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.（2）如圖，由（1）知SKIPIF1<0，SKIPIF1<0方程為SKIPIF1<0SKIPIF1<0，直線SKIPIF1<0與橢圓的另一個(gè)交點(diǎn)為SKIPIF1<0，∵SKIPIF1<0，根據(jù)對稱性可得SKIPIF1<0，聯(lián)立SKIPIF1<0，整理得SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0∵SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，聯(lián)立解得SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.1已知橢圓SKIPIF1<0過點(diǎn)SKIPIF1<0，且SKIPIF1<0．（Ⅰ）求橢圓C的方程：（Ⅱ）過點(diǎn)SKIPIF1<0的直線l交橢圓C于點(diǎn)SKIPIF1<0,直線SKIPIF1<0分別交直線SKIPIF1<0于點(diǎn)SKIPIF1<0．求SKIPIF1<0的值．類型六探究性問題1．已知雙曲線SKIPIF1<0的左右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上且不在SKIPIF1<0軸上，直線SKIPIF1<0與雙曲線的交點(diǎn)分別為A，B，直線SKIPIF1<0與雙曲線的交點(diǎn)分別為C，D．(1)設(shè)直線SKIPIF1<0和SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值；(2)問直線l上是否存在點(diǎn)P，使得直線OA，OB，OC，OD的斜率SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0？若存在，求出所有滿足條件的點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，請說明理由．解析：(1)SKIPIF1<0(2)存在，SKIPIF1<0或SKIPIF1<0（1）設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0；（2）假設(shè)直線l上存在點(diǎn)SKIPIF1<0，SKIPIF1<0．設(shè)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0∴SKIPIF1<0，同理SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由（1）SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由（1）SKIPIF1<0得SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0．所以SKIPIF1<0或SKIPIF1<0．1在直角坐標(biāo)平面中，SKIPIF1<0的兩個(gè)頂點(diǎn)的坐標(biāo)分別為SKIPIF1<0，兩動(dòng)點(diǎn)SKIPIF1<0滿足SKIPIF1<0，向量SKIPIF1<0與SKIPIF1<0共線.(1)求SKIPIF1<0的頂點(diǎn)SKIPIF1<0的軌跡方程；(2)若過點(diǎn)SKIPIF1<0的直線與（1）的軌跡相交于SKIPIF1<0兩點(diǎn)，求SKIPIF1<0的取值范圍.(3)若SKIPIF1<0為SKIPIF1<0點(diǎn)的軌跡在第一象限內(nèi)的任意一點(diǎn)，則是否存在常數(shù)SKIPIF1<0，使得SKIPIF1<0恒成立？若存在，求出SKIPIF1<0的值；若不存在，請說明理由.類型七切線問題與阿基米德三角形問題拋物線的弦與過弦的端點(diǎn)的兩條切線所圍的三角形，這個(gè)三角形又常被稱為阿基米德三角形．阿基米德三角形的得名，是因?yàn)榘⒒椎卤救俗钤缋帽平乃枷胱C明如下結(jié)論：拋物線與阿基米德三角形定理：拋物線的弦與拋物線所圍成的封閉圖形的面積，等于拋物線的弦與過弦的端點(diǎn)的兩條切線所圍成的三角形面積的三分之二．下面來逐一介紹阿基米德三角形的一些推論:如圖，已知SKIPIF1<0是拋物線SKIPIF1<0準(zhǔn)線上任意一點(diǎn)，過SKIPIF1<0作拋物線的切線SKIPIF1<0、SKIPIF1<0兩點(diǎn)，SKIPIF1<0中點(diǎn)，則：1.若過焦點(diǎn)，則的端點(diǎn)的兩條切線的交點(diǎn)SKIPIF1<0在其準(zhǔn)線上．2.阿基米德三角形底邊上的中線平行于坐標(biāo)軸，即SKIPIF1<0． 3.過拋物線的焦點(diǎn)4.SKIPIF1<05.阿基米德三角形面積的最小值為SKIPIF1<01如下圖，設(shè)拋物線方程為SKIPIF1<0，M為直線SKIPIF1<0上任意一點(diǎn)，過SKIPIF1<0引拋物線的切線，切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0．（Ⅰ）設(shè)線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0；（ⅰ）求證：SKIPIF1<0平行于SKIPIF1<0軸；（ⅱ）已知當(dāng)SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0時(shí)，SKIPIF1<0，求此時(shí)拋物線的方程；（Ⅱ）是否存在點(diǎn)SKIPIF1<0，使得點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對稱點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，其中，點(diǎn)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)）．若存在，求出所有適合題意的點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，請說明理由．【答案】（Ⅰ）（?。┳C明見解析；（ⅱ）SKIPIF1<0或SKIPIF1<0；（Ⅱ）僅存在一點(diǎn)SKIPIF1<0適合題意.【分析】（Ⅰ）（?。┰O(shè)出SKIPIF1<0的坐標(biāo)，利用導(dǎo)數(shù)求得切線SKIPIF1<0的方程，結(jié)合SKIPIF1<0是線段SKIPIF1<0的中點(diǎn)進(jìn)行化簡，得到SKIPIF1<0兩點(diǎn)的橫坐標(biāo)相等，由此證得SKIPIF1<0平行于SKIPIF1<0軸.（ⅱ）利用SKIPIF1<0列方程，解方程求得SKIPIF1<0，進(jìn)而求得拋物線方程.（Ⅱ）設(shè)出SKIPIF1<0點(diǎn)坐標(biāo)，由SKIPIF1<0點(diǎn)坐標(biāo)求得線段SKIPIF1<0中點(diǎn)的坐標(biāo)，由直線SKIPIF1<0的方程和拋物線的方程，求得SKIPIF1<0點(diǎn)的坐標(biāo)，由此進(jìn)行分類討論求得SKIPIF1<0點(diǎn)的坐標(biāo).【詳解】（Ⅰ）（ⅰ）證明：由題意設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．因此直線SKIPIF1<0的方程為SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0．所以SKIPIF1<0，①SKIPIF1<0．②由①、②得SKIPIF1<0，因此SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0.所以SKIPIF1<0平行于SKIPIF1<0軸．（ⅱ）解：由（ⅰ）知，當(dāng)SKIPIF1<0時(shí)，將其代入①、②并整理得：SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0是方程SKIPIF1<0的兩根，因此SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0．由弦長公式的SKIPIF1<0．又SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，因此所求拋物線方程為SKIPIF1<0或SKIPIF1<0．（Ⅱ）解：設(shè)SKIPIF1<0，由題意得SKIPIF1<0，則SKIPIF1<0的中點(diǎn)坐標(biāo)為SKIPIF1<0，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，由點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，并注意到點(diǎn)SKIPIF1<0也在直線SKIPIF1<0上，代入得SKIPIF1<0．若SKIPIF1<0在拋物線上，則SKIPIF1<0，因此SKIPIF1<0或SKIPIF1<0．即SKIPIF1<0或SKIPIF1<0．（1）當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，此時(shí)，點(diǎn)SKIPIF1<0適合題意．（2）當(dāng)SKIPIF1<0，對于SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，矛盾．對于SKIPIF1<0，因?yàn)镾KIPIF1<0，此時(shí)直線SKIPIF1<0平行于SKIPIF1<0軸，又SKIPIF1<0，所以直線SKIPIF1<0與直線SKIPIF1<0不垂直，與題設(shè)矛盾，所以SKIPIF1<0時(shí)，不存在符合題意得SKIPIF1<0點(diǎn)．綜上所述，僅存在一點(diǎn)SKIPIF1<0適合題目如圖，設(shè)拋物線方程為SKIPIF1<0，SKIPIF1<0為直線SKIPIF1<0上任意一點(diǎn)，過SKIPIF1<0引拋物線的切線，切點(diǎn)分別為SKIPIF1<0．（Ⅰ）求證：SKIPIF1<0三點(diǎn)的橫坐標(biāo)成等差數(shù)列；（Ⅱ）已知當(dāng)SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0時(shí)，SKIPIF1<0．求此時(shí)拋物線的方程；yxBAOMSKIPIF1<0（Ⅲ）是否存在點(diǎn)SKIPIF1<0，使得點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對稱點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0上，其中，點(diǎn)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)）．若存在，求出所有適合題意的點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，請說明理由．yxBAOMSKIPIF1<0類型八極點(diǎn)極限調(diào)和點(diǎn)列蝴蝶模型1.極點(diǎn)和極線的幾何定義如圖,SKIPIF1<0為不在圓錐曲線SKIPIF1<0上的點(diǎn),過點(diǎn)SKIPIF1<0引兩條割線依次交圓錐曲線于四點(diǎn)SKIPIF1<0,連接SKIPIF1<0交于SKIPIF1<0,連接SKIPIF1<0交于SKIPIF1<0,我們稱點(diǎn)SKIPIF1<0為直線SKIPIF1<0關(guān)于圓錐曲線SKIPIF1<0的極點(diǎn),稱直線SKIPIF1<0為點(diǎn)SKIPIF1<0關(guān)于圓錐曲線SKIPIF1<0的極線.直線SKIPIF1<0交圓錐曲線SKIPIF1<0于SKIPIF1<0兩點(diǎn),則SKIPIF1<0為圓錐曲線SKIPIF1<0的兩條切線.若SKIPIF1<0在圓錐曲線SKIPIF1<0上,則過點(diǎn)SKIPIF1<0的切線即為極線.(1)自極三角形：極點(diǎn)SKIPIF1<0一一極線SKIPIF1<0；極點(diǎn)SKIPIF1<0一一極線SKIPIF1<0極點(diǎn)SKIPIF1<0一一極線SKIPIF1<0;即SKIPIF1<0中,三個(gè)頂點(diǎn)和對邊分別為一對極點(diǎn)和極線,稱SKIPIF1<0為“自極三角形”.(2)極點(diǎn)和極線的兩種特殊情況(1)當(dāng)四邊形變成三角形時(shí)：曲線上的點(diǎn)SKIPIF1<0對應(yīng)的極線,就是切線SKIPIF1<0;

(2)當(dāng)四邊有一組對邊平行時(shí),如：當(dāng)SKIPIF1<0時(shí),SKIPIF1<0和SKIPIF1<0的交點(diǎn)SKIPIF1<0落在無窮遠(yuǎn)處;點(diǎn)SKIPIF1<0的極線SKIPIF1<0和點(diǎn)SKIPIF1<0的極線SKIPIF1<0滿足:SKIPIF1<02.極點(diǎn)和極線的代數(shù)定義對于定點(diǎn)SKIPIF1<0與非退化二次曲線SKIPIF1<0過點(diǎn)SKIPIF1<0作動(dòng)直線與曲線SKIPIF1<0交于點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0,那么點(diǎn)SKIPIF1<0關(guān)于線段SKIPIF1<0的調(diào)和點(diǎn)SKIPIF1<0的軌跡是什么?可以證明:點(diǎn)SKIPIF1<0在一條定直線SKIPIF1<0上，如下圖.我們稱點(diǎn)SKIPIF1<0為直線SKIPIF1<0關(guān)于曲線SKIPIF1<0的極點(diǎn);相應(yīng)地,稱直線SKIPIF1<0為點(diǎn)SKIPIF1<0關(guān)于曲線SKIPIF1<0的極線.一般地,對于圓錐曲線SKIPIF1<0設(shè)極點(diǎn)SKIPIF1<0,則對應(yīng)的極線為 SKIPIF1<0【注】替換規(guī)則為:SKIPIF1<0(1)橢圓SKIPIF1<0的三類極點(diǎn)極線(1)若極點(diǎn)SKIPIF1<0在橢圓外,過點(diǎn)SKIPIF1<0作橢圓的兩條?線,切點(diǎn)為SKIPIF1<0,則極線為切點(diǎn)弦所在直線 SKIPIF1<0(2)若極點(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)若極點(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由此可得橢圓極線的幾何作法:(2)對于雙曲線SKIPIF1<0,極點(diǎn)SKIPIF1<0對應(yīng)的極線為SKIPIF1<0(3)對于拋物線SKIPIF1<0,極點(diǎn)SKIPIF1<0對應(yīng)的極線為SKIPIF1<0.3.極點(diǎn)和極線的性質(zhì)(1)引理:已知橢圓方程為SKIPIF1<0,直線SKIPIF1<0的方程為SKIPIF1<0,點(diǎn)SKIPIF1<0不與原點(diǎn)重合.過點(diǎn)SKIPIF1<0作直線交橢圓于SKIPIF1<0兩點(diǎn),SKIPIF1<0點(diǎn)在直線SKIPIF1<0上，則“點(diǎn)SKIPIF1<0在直線SKIPIF1<0上"的充要條件是SKIPIF1<0調(diào)和分割SKIPIF1<0,即SKIPIF1<0.1設(shè)橢圓C:x2a2(1)求敉圓C的方程;(2)當(dāng)過點(diǎn)P(4,1)的動(dòng)直線l于橢圓C相交于兩不同點(diǎn)A,B時(shí),在線段AB上取點(diǎn)Q,滿足|【答案】(1)x2【解析】（1）由題意得：SKIPIF1<0，解得SKIPIF1<0，所求橢圓方程為x24+y22=1.(2)解法1:定比點(diǎn)差法設(shè)點(diǎn)Q、A由題設(shè)知|AP|,|PB|,|AQ|,|又A,P于是4=x從而:4x=x又點(diǎn)A、B在橢圓x1x2(1)+(2)×2,并結(jié)合(3)(4)得4x即點(diǎn)Q(x,解法2：構(gòu)造同構(gòu)式設(shè)點(diǎn)Q(由題設(shè)知|AP|,|PB又A,P于是x1=4?由于Ax1,y1,B整理得:x2x2(4)-(3)得:8(2x即點(diǎn)Q(x,解法3：極點(diǎn)極線由|AP|?|QB說明點(diǎn)P,Q關(guān)于桞圓調(diào)和共軛,點(diǎn)Q在點(diǎn)此極線方程為4?x4+故點(diǎn)Q總在直線2x如圖,過直線l:5x?7y?70=0上的點(diǎn)P作橢圓x225+y(1)當(dāng)點(diǎn)P在直線l上運(yùn)動(dòng)時(shí),證明：直線MN恒過定點(diǎn)Q;(2)當(dāng)MN//l時(shí),定點(diǎn)Q平分線段蝴蝶定理（ButterflyTheorem），是古代歐氏平面幾何中最精彩的結(jié)果之一.這個(gè)命題最早出現(xiàn)在1815年,由W.G.霍納提出證明.【蝴蝶定理】M是⊙O中弦AB的中點(diǎn),過點(diǎn)M的兩條弦CD,EF,連接DE,CF交AB于P問題中的圖形酷似圓中翩翩起舞的蝴蝶,因此而被冠之“蝴蝶定理".蝴蝶定理還可以推廣到橢圓,甚至雙曲線與拋物線中.例題.如圖,O為坐標(biāo)原點(diǎn),橢圓C:x2a2+y(1)求橢圓C的方程;(2)過點(diǎn)P(0,1)作直線l交橢圓C于異于M,N的A,B兩點(diǎn),直線AM,類型九不聯(lián)立問題1已知點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0上，直線SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，直線SKIPIF1<0，SKIPIF1<0的斜率之和為0．（1）求SKIPIF1<0的斜率；（2）若SKIPIF1<0，求SKIPIF1<0的面積．解析：（1）設(shè)SKIPIF1<0，由點(diǎn)SKIPIF1<0都在雙曲線SKIPIF1<0上，得，，所以，結(jié)合斜率公式，相減后變形，可得：，.因?yàn)橹本€SKIPIF1<0的斜率之和為SKIPIF1<0，即SKIPIF1<0，所以，由得.②由得.③由②-③，得，從而，即SKIPIF1<0的斜率為SKIPIF1<0.1已知橢圓C：SKIPIF1<0的離心率為SKIPIF1<0，且過點(diǎn)SKIPIF1<0．（1）求SKIPIF1<0的方程：（2）點(diǎn)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為垂足．證明：存在定點(diǎn)SKIPIF1<0，使得SKIPIF1<0為定值．類型十與其他知識點(diǎn)交叉問題某電廠冷卻塔的外形是由SKIPIF1<0雙曲線的一部分繞其虛軸所在的直線旋轉(zhuǎn)所形成的曲面.如圖所示，已知它的最小半徑為SKIPIF1<0，上口半徑為SKIPIF1<0，下口半徑為SKIPIF1<0，高為SKIPIF1<0，選擇適當(dāng)?shù)钠矫嬷苯亲鴺?biāo)系.(1)求此雙曲線SKIPIF1<0的方程；(2)定義：以（1）中求出的雙曲線SKIPIF1<0的實(shí)軸為虛軸，以SKIPIF1<0的虛軸為實(shí)軸的雙曲線SKIPIF1<0叫做SKIPIF1<0的共軛雙曲線，求雙曲線SKIPIF1<0的方程；(3)對于（2）中的雙曲線SKIPIF1<0?SKIPIF1<0的離心率分別為SKIPIF1<0?SKIPIF1<0，寫出SKIPIF1<0與SKIPIF1<0滿足的一個(gè)關(guān)系式，并證明.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0（1）以冷卻塔的軸截面的最窄處所在的直線為SKIPIF1<0軸，垂直平分線為SKIPIF1<0軸建立平面直角坐標(biāo)系，設(shè)雙曲線的方程為SKIPIF1<0，由題意知SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以雙曲線SKIPIF1<0的方程為SKIPIF1<0.（2）以（1）中求出的雙曲線SKIPIF1<0的實(shí)軸為虛軸，以SKIPIF1<0的虛軸為實(shí)軸的雙曲線SKIPIF1<0為SKIPIF1<0.（3）SKIPIF1<0與SKIPIF1<0滿足的一個(gè)關(guān)系式為SKIPIF1<0，證明如下，雙曲線SKIPIF1<0的半焦距SKIPIF1<0，所以雙曲線SKIPIF1<0的離心率為SKIPIF1<0，雙曲線SKIPIF1<0的半焦距SKIPIF1<0，所以雙曲線SKIPIF1<0的離心率為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0滿足的一個(gè)關(guān)系式為SKIPIF1<0.1在平面直角坐標(biāo)系SKIPIF1<0中，對于直線SKIPIF1<0和點(diǎn)SKIPIF1<0、SKIPIF1<0，記SKIPIF1<0，若SKIPIF1<0，則稱點(diǎn)SKIPIF1<0、SKIPIF1<0被直線SKIPIF1<0分隔，若曲線SKIPIF1<0與直線SKIPIF1<0沒有公共點(diǎn)，且曲線SKIPIF1<0上存在點(diǎn)SKIPIF1<0、SKIPIF1<0被直線SKIPIF1<0分隔，則稱直線SKIPIF1<0為曲線SKIPIF1<0的一條分隔線．(1)判斷點(diǎn)SKIPIF1<0是否被直線SKIPIF1<0分隔并證明；(2)若直線SKIPIF1<0是曲線SKIPIF1<0的分隔線，求實(shí)數(shù)SKIPIF1<0的取值范圍；(3)動(dòng)點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離與到SKIPIF1<0軸的距離之積為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0的軌跡為曲線SKIPIF1<0，求證：通過原點(diǎn)的直線中，有且僅有一條直線是SKIPIF1<0的分隔線．1．（2022·北京海淀·?？寄M預(yù)測）橢圓C：SKIPIF1<0的右頂點(diǎn)為SKIPIF1<0，離心率為SKIPIF1<0(1)求橢圓C的方程及短軸長；(2)已知：過定點(diǎn)SKIPIF1<0作直線l交橢圓C于D，E兩點(diǎn)，過E作AB的平行線交直線DB于點(diǎn)F，設(shè)EF中點(diǎn)為G，直線BG與橢圓的另一點(diǎn)交點(diǎn)為M，若四邊形BEMF為平行四邊形，求G點(diǎn)坐標(biāo)．2．（2022·北京·統(tǒng)考模擬預(yù)測）如圖所示，過原點(diǎn)O作兩條互相垂直的線OA，OB分別交拋物線SKIPIF1<0于A，B兩點(diǎn)，連接AB，交y軸于點(diǎn)P．(1)求點(diǎn)P的坐標(biāo)；(2)證明：存在相異于點(diǎn)P的定點(diǎn)T，使得SKIPIF1<0恒成立，請求出點(diǎn)T的坐標(biāo)，并求出SKIPIF1<0面積的最小值．3．（2023·湖北武漢·統(tǒng)考模擬預(yù)測）過坐標(biāo)原點(diǎn)SKIPIF1<0作圓SKIPIF1<0的兩條切線，設(shè)切點(diǎn)為SKIPIF1<0，直線SKIPIF1<0恰為拋物SKIPIF1<0的準(zhǔn)線.(1)求拋物線SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)設(shè)點(diǎn)SKIPIF1<0是圓SKIPIF1<0上的動(dòng)點(diǎn)，拋物線SKIPIF1<0上四點(diǎn)SKIPIF1<0滿足：SKIPIF1<0，設(shè)SKIPIF1<0中點(diǎn)為SKIPIF1<0.（i）求直線SKIPIF1<0的斜率；（ii）設(shè)SKIPIF1<0面積為SKIPIF1<0，求SKIPIF1<0的最大值.4．（2022·江蘇南京·模擬預(yù)測）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．5．（2023河北·校聯(lián)考三模）已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，過SKIPIF1<0且斜率為SKIPIF1<0的直線SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，斜率為SKIPIF1<0的直線SKIPIF1<0與SKIPIF1<0相切于點(diǎn)SKIPIF1<0，且SKIPIF1<0與SKIPIF1<0不垂直，SKIPIF1<0為SKIPIF1<0的中點(diǎn).（1）若SKIPIF1<0，求SKIPIF1<0；（2）若直線SKIPIF1<0過SKIPIF1<0，求SKIPIF1<0.6．（2023·山東泰安·統(tǒng)考一模）已知橢圓SKIPIF1<0：SKIPIF1<0的左，右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，離心率為SKIPIF1<0，SKIPIF1<0是橢圓SKIPIF1<0上不同的兩點(diǎn)，且點(diǎn)SKIPIF1<0在SKIPIF1<0軸上方，SKIPIF1<0，直線SKIPIF1<0，SKIPIF1<0交于點(diǎn)SKIPIF1<0.已知當(dāng)SKIPIF1<0軸時(shí)，SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)求證：點(diǎn)SKIPIF1<0在以SKIPIF1<0，SKIPIF1<0為焦點(diǎn)的定橢圓上..7．（2023·河北邢臺·校聯(lián)考模擬預(yù)測）已知雙曲線SKIPIF1<0過點(diǎn)SKIPIF1<0，且SKIPIF1<0與SKIPIF1<0的兩個(gè)頂點(diǎn)連線的斜率之和為4．(1)求SKIPIF1<0的方程；(2)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)（異于點(diǎn)SKIPIF1<0）．設(shè)直線SKIPIF1<0與SKIPIF1<0軸垂直且交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，若線段SKIPIF1<