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07 圓錐曲線中的定值問題（原卷版）

專題7圓錐曲線中的定值問題一、考情分析求定值是圓錐曲線中頗有難度的一類問題,也是備受高考關(guān)注的一類問題,由于它在解題之前不知道定值的結(jié)果,因而更增添了題目的神秘色彩.解決這類問題時(shí),要善于運(yùn)用辯證的觀點(diǎn)去思考分析,在動(dòng)點(diǎn)的“變”中尋求定值的“不變”性,用特殊探索法（特殊值、特殊位置、特殊圖形等）先確定出定值,揭開神秘的面紗,這樣可將盲目的探索問題轉(zhuǎn)化為有方向有目標(biāo)的一般性證明題,從而找到解決問題的突破口.同時(shí)有許多定值問題,通過特殊探索法不但能夠確定出定值,還可以為我們提供解題的線索.二、解題秘籍(一)定值問題解題思路與策略定值問題肯定含有參數(shù),若要證明一個(gè)式子是定值,則意味著參數(shù)是不影響結(jié)果的,也就是說參數(shù)在解式子的過程中都可以消掉,因此解決定值問題的關(guān)鍵是設(shè)參數(shù)：

(1)在解析幾何中參數(shù)可能是點(diǎn)（注意如果設(shè)點(diǎn)是兩個(gè)參數(shù)時(shí),注意橫坐標(biāo)要滿足圓錐曲線方程）

(2)可能是角（這里的角常常是將圓錐曲線上的點(diǎn)設(shè)為三角函數(shù)角的形式）,

(3)也可能是斜率（這個(gè)是最常用的,但是既然設(shè)斜率了,就要考慮斜率是否存在的情況）

常用的參數(shù)就是以上三種,但是注意我們設(shè)參數(shù)時(shí)要遵循一個(gè)原則：參數(shù)越少越好.

因此定值問題的解題思路是：

(1)設(shè)參數(shù)；

(2)用參數(shù)來表示要求定值的式子；

(3)消參數(shù).

2.圓錐曲線中的定值問題的常見類型及解題策略(1)求代數(shù)式為定值．依題意設(shè)條件,得出與代數(shù)式參數(shù)有關(guān)的等式,代入代數(shù)式、化簡即可得出定值；(2)求點(diǎn)到直線的距離為定值．利用點(diǎn)到直線的距離公式得出距離的解析式,再利用題設(shè)條件化簡、變形求得；(3)求某線段長度為定值．利用長度公式求得解析式,再依據(jù)條件對解析式進(jìn)行化簡、變形即可求得．【例1】（2023屆湖湘名校教育聯(lián)合體高三上學(xué)期9月大聯(lián)考）已知橢圓SKIPIF1<0為右焦點(diǎn)，直線SKIPIF1<0與橢圓C相交于A，B兩點(diǎn)，取A點(diǎn)關(guān)于x軸的對稱點(diǎn)S，設(shè)線段SKIPIF1<0與線段SKIPIF1<0的中垂線交于點(diǎn)Q．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0是否為定值？若為定值，則求出定值；若不為定值，則說明理由．【解析】（1）設(shè)SKIPIF1<0，線段SKIPIF1<0的中點(diǎn)M坐標(biāo)為SKIPIF1<0，聯(lián)立得SKIPIF1<0消去y可得：SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，代入直線SKIPIF1<0方程，求得SKIPIF1<0，因?yàn)镼為SKIPIF1<0三條中垂線的交點(diǎn)，所以SKIPIF1<0，有SKIPIF1<0，直線SKIPIF1<0方程為SKIPIF1<0.令SKIPIF1<0，所以SKIPIF1<0.由橢圓SKIPIF1<0可得右焦點(diǎn)SKIPIF1<0，故SKIPIF1<0．（2）設(shè)SKIPIF1<0，中點(diǎn)M坐標(biāo)為SKIPIF1<0．SKIPIF1<0相減得SKIPIF1<0，SKIPIF1<0.又Q為SKIPIF1<0的外心，故SKIPIF1<0，所以SKIPIF1<0，直線SKIPIF1<0方程為SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0而SKIPIF1<0,所以SKIPIF1<0，SKIPIF1<0，同理SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以當(dāng)t變化時(shí)，SKIPIF1<0為定值SKIPIF1<0．【例2】（2023屆河南省濮陽市高三上學(xué)期測試）已知橢圓SKIPIF1<0：SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，圓SKIPIF1<0：SKIPIF1<0，過SKIPIF1<0且垂直于SKIPIF1<0軸的直線被橢圓SKIPIF1<0和圓SKIPIF1<0所截得的弦長分別為SKIPIF1<0和SKIPIF1<0.(1)求SKIPIF1<0的方程；(2)過圓SKIPIF1<0上一點(diǎn)SKIPIF1<0（不在坐標(biāo)軸上）作SKIPIF1<0的兩條切線SKIPIF1<0，SKIPIF1<0，記SKIPIF1<0，SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，證明：SKIPIF1<0為定值.【解析】（1）設(shè)橢圓SKIPIF1<0的半焦距為SKIPIF1<0，過SKIPIF1<0且垂直于SKIPIF1<0軸的直線被橢圓SKIPIF1<0所截得的弦長分別為SKIPIF1<0，則SKIPIF1<0；過SKIPIF1<0且垂直于SKIPIF1<0軸的直線被圓SKIPIF1<0所截得的弦長分別為SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的方程為SKIPIF1<0.（2）設(shè)SKIPIF1<0，則SKIPIF1<0.①設(shè)過點(diǎn)SKIPIF1<0與橢圓SKIPIF1<0相切的直線方程為SKIPIF1<0，聯(lián)立SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，整理得SKIPIF1<0.②由題意知SKIPIF1<0，SKIPIF1<0為方程②的兩根，由根與系數(shù)的關(guān)系及①可得SKIPIF1<0.又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0為定值SKIPIF1<0．(二)與線段長度有關(guān)的定值問題與線段長度有關(guān)的定值問題通常是先引入?yún)?shù),利用距離公式或弦長公式得到長度解析式,再對解析式化簡,得出結(jié)果為定值【例3】（2023屆遼寧省朝陽市高三上學(xué)期9月月考）已知雙曲線SKIPIF1<0的離心率為SKIPIF1<0，點(diǎn)SKIPIF1<0在雙曲線SKIPIF1<0上.(1)求雙曲線SKIPIF1<0的方程；(2)點(diǎn)SKIPIF1<0，SKIPIF1<0在雙曲線SKIPIF1<0上，直線SKIPIF1<0，SKIPIF1<0與SKIPIF1<0軸分別相交于SKIPIF1<0兩點(diǎn)，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，若坐標(biāo)原點(diǎn)SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，SKIPIF1<0，證明：存在定點(diǎn)SKIPIF1<0，使得SKIPIF1<0為定值.【解析】（1）由題意，雙曲線SKIPIF1<0的離心率為SKIPIF1<0，且SKIPIF1<0在雙曲線SKIPIF1<0上，可得SKIPIF1<0，解得SKIPIF1<0，所以雙曲線的方程為SKIPIF1<0.（2）由題意知，直線的SKIPIF1<0的斜率存在，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立方程組SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0,同理可得SKIPIF1<0，因?yàn)镾KIPIF1<0為SKIPIF1<0的中點(diǎn)，所以SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，若SKIPIF1<0，則直線方程為SKIPIF1<0，即SKIPIF1<0，此時(shí)直線SKIPIF1<0過點(diǎn)SKIPIF1<0，不合題意；若SKIPIF1<0時(shí)，則直線方程為SKIPIF1<0，恒過定點(diǎn)SKIPIF1<0，所以SKIPIF1<0為定值，又由SKIPIF1<0為直角三角形，且SKIPIF1<0為斜邊，所以當(dāng)SKIPIF1<0為SKIPIF1<0的中點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0.(三)與面積有關(guān)的定值問題與面積有關(guān)的定值問題通常是利用面積公式把面積表示成某些變量的表達(dá)式,再利用題中條件化簡.【例4】（2023屆河南省部分學(xué)校高三上學(xué)期9月聯(lián)考）已知橢圓SKIPIF1<0：SKIPIF1<0的左焦點(diǎn)為SKIPIF1<0，上、下頂點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)求橢圓SKIPIF1<0的方程；(2)若橢圓上有三點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，證明：四邊形SKIPIF1<0的面積為定值．【解析】（1）依題意SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以橢圓方程為SKIPIF1<0.（2）證明：設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以四邊形SKIPIF1<0為平行四邊形，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，若直線SKIPIF1<0的斜率不存在，SKIPIF1<0與左頂點(diǎn)或右頂點(diǎn)重合，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，若直線SKIPIF1<0的斜率存在，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，代入橢圓方程整理得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0所以SKIPIF1<0，整理得SKIPIF1<0，又SKIPIF1<0，又原點(diǎn)SKIPIF1<0到SKIPIF1<0的距離SKIPIF1<0，所以SKIPIF1<0，將SKIPIF1<0代入得SKIPIF1<0，所以SKIPIF1<0，綜上可得，四邊形SKIPIF1<0的面積為定值SKIPIF1<0.(四)與斜率有關(guān)的定值問題與斜率有關(guān)的定值問題常見類型是斜率之積商或斜率之和差為定值,求解時(shí)一般先利用斜率公式寫出表達(dá)式,再利用題中條件或韋達(dá)定理化簡.【例5】（2023屆江蘇省南通市高三上學(xué)期第一次質(zhì)量監(jiān)測）已知SKIPIF1<0分別是橢圓SKIPIF1<0的左?右頂點(diǎn)，SKIPIF1<0分別是SKIPIF1<0的上頂點(diǎn)和左焦點(diǎn).點(diǎn)SKIPIF1<0在SKIPIF1<0上，滿足SKIPIF1<0.(1)求SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0作直線SKIPIF1<0（與SKIPIF1<0軸不重合）交SKIPIF1<0于SKIPIF1<0兩點(diǎn)，設(shè)直線SKIPIF1<0的斜率分別為SKIPIF1<0，求證：SKIPIF1<0為定值.【解析】（1）因?yàn)镾KIPIF1<0，故可設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.又SKIPIF1<0在橢圓SKIPIF1<0上，故SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0.又SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0.故SKIPIF1<0的方程為SKIPIF1<0.（2）因?yàn)闄E圓方程為SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0斜率為0時(shí)SKIPIF1<0或SKIPIF1<0重合，不滿足題意，故可設(shè)SKIPIF1<0：SKIPIF1<0.聯(lián)立SKIPIF1<0可得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0.故SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故定值為SKIPIF1<0(五)與向量有關(guān)的定值問題與向量有關(guān)的定值問題常見類型一是求數(shù)量積有關(guān)的定值問題，二是根據(jù)向量共線,寫出向量系數(shù)的表達(dá)式,再通過計(jì)算得出與向量系數(shù)有關(guān)的定值結(jié)論.【例6】（2023屆湖南省部分校高三上學(xué)期9月月考）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.【例7】（2022屆上海市金山區(qū)高三上學(xué)期一模）已知SKIPIF1<0為橢圓C：SKIPIF1<0內(nèi)一定點(diǎn),Q為直線l：SKIPIF1<0上一動(dòng)點(diǎn),直線PQ與橢圓C交于A?B兩點(diǎn)（點(diǎn)B位于P?Q兩點(diǎn)之間）,O為坐標(biāo)原點(diǎn).（1）當(dāng)直線PQ的傾斜角為SKIPIF1<0時(shí),求直線OQ的斜率；（2）當(dāng)SKIPIF1<0AOB的面積為SKIPIF1<0時(shí),求點(diǎn)Q的橫坐標(biāo)；（3）設(shè)SKIPIF1<0,SKIPIF1<0,試問SKIPIF1<0是否為定值？若是,請求出該定值；若不是,請說明理由.【解析】（1）因?yàn)橹本€PQ的傾斜角為SKIPIF1<0,且SKIPIF1<0,所以直線PQ的方程為：SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,所以直線OQ的斜率是SKIPIF1<0；（2）易知直線PQ的斜率存在,設(shè)直線PQ的方程為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0,所以直線PQ的方程為SKIPIF1<0或SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0；由SKIPIF1<0,得SKIPIF1<0；（3）易知直線PQ的斜率存在,設(shè)直線PQ的方程為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0.(六)與代數(shù)式有關(guān)的定值問題與代數(shù)式有關(guān)的定值問題．一般是依題意設(shè)條件,得出與代數(shù)式參數(shù)有關(guān)的等式,代入代數(shù)式、化簡即可得出定值【例8】在平面直角坐標(biāo)系SKIPIF1<0中，橢圓SKIPIF1<0的右準(zhǔn)線為直線SKIPIF1<0，動(dòng)直線SKIPIF1<0交橢圓于SKIPIF1<0兩點(diǎn)，線段SKIPIF1<0的中點(diǎn)為SKIPIF1<0，射線SKIPIF1<0分別交橢圓及直線SKIPIF1<0于點(diǎn)SKIPIF1<0，如圖，當(dāng)SKIPIF1<0兩點(diǎn)分別是橢圓SKIPIF1<0的右頂點(diǎn)及上頂點(diǎn)時(shí)，點(diǎn)SKIPIF1<0的縱坐標(biāo)為SKIPIF1<0（其中SKIPIF1<0為橢圓的離心率），且SKIPIF1<0．(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)如果SKIPIF1<0是SKIPIF1<0的等比中項(xiàng)，那么SKIPIF1<0是否為常數(shù)？若是，求出該常數(shù)；若不是，請說明理由．【解析】（1）橢圓SKIPIF1<0的右準(zhǔn)線為直線SKIPIF1<0，動(dòng)直線SKIPIF1<0交橢圓于SKIPIF1<0兩點(diǎn)，當(dāng)SKIPIF1<0零點(diǎn)分別是橢圓SKIPIF1<0的有頂點(diǎn)和上頂點(diǎn)時(shí)，則SKIPIF1<0，因?yàn)榫€段SKIPIF1<0的中點(diǎn)為SKIPIF1<0，射線SKIPIF1<0分別角橢圓及直線SKIPIF1<0與SKIPIF1<0兩點(diǎn)，所以SKIPIF1<0，由SKIPIF1<0三點(diǎn)共線，可得SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，又由SKIPIF1<0，解得SKIPIF1<0，所以橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0.（2）解：把SKIPIF1<0代入橢圓SKIPIF1<0，可得SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0,所以直線SKIPIF1<0的方程為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，因?yàn)镾KIPIF1<0是SKIPIF1<0的等比中項(xiàng)，所以SKIPIF1<0，可得SKIPIF1<0，又由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，此時(shí)滿足SKIPIF1<0，所以SKIPIF1<0為常數(shù)SKIPIF1<0.(六)與定值有關(guān)的結(jié)論1.若點(diǎn)A,B是橢圓C：SKIPIF1<0上關(guān)于原點(diǎn)對稱的兩點(diǎn),點(diǎn)P是橢圓C上與A,B不重合的點(diǎn),則SKIPIF1<0；2.若點(diǎn)A,B是雙曲線C：SKIPIF1<0上關(guān)于原點(diǎn)對稱的兩點(diǎn),點(diǎn)P是雙曲線C上與A,B不重合的點(diǎn),則SKIPIF1<0.3.設(shè)點(diǎn)是橢圓C：上一定點(diǎn),點(diǎn)A,B是橢圓C上不同于P的兩點(diǎn),若SKIPIF1<0,則直線AB斜率為定值；4.設(shè)點(diǎn)是雙曲線C：一定點(diǎn),點(diǎn)A,B是雙曲線C上不同于P的兩點(diǎn),若SKIPIF1<0,直線AB斜率為定值；5.設(shè)點(diǎn)是拋物線C：一定點(diǎn),點(diǎn)A,B是拋物線C上不同于P的兩點(diǎn),若SKIPIF1<0,直線AB斜率為定值.6.設(shè)SKIPIF1<0是橢圓上不同3點(diǎn),B,C關(guān)于x軸對稱,直線AC,BC與x軸分別交于點(diǎn)SKIPIF1<0,則SKIPIF1<0.7.點(diǎn)A,B是橢圓C：SKIPIF1<0上動(dòng)點(diǎn),O為坐標(biāo)原點(diǎn),若SKIPIF1<0,則SKIPIF1<0=SKIPIF1<0（即點(diǎn)O到直線AB為定值）8.經(jīng)過橢圓SKIPIF1<0（a＞b＞0）的長軸的兩端點(diǎn)A1和A2的切線,與橢圓上任一點(diǎn)的切線相交于P1和P2,則SKIPIF1<0.9.過橢圓SKIPIF1<0（a＞b＞0）的右焦點(diǎn)F作直線交該橢圓右支于M,N兩點(diǎn),弦MN的垂直平分線交x軸于P,則SKIPIF1<0.10.點(diǎn)SKIPIF1<0為橢圓SKIPIF1<0(包括圓在內(nèi))在第一象限的弧上任意一點(diǎn),過SKIPIF1<0引SKIPIF1<0軸、SKIPIF1<0軸的平行線,交SKIPIF1<0軸、SKIPIF1<0軸于SKIPIF1<0,交直線SKIPIF1<0于SKIPIF1<0,記SKIPIF1<0與SKIPIF1<0的面積為SKIPIF1<0,則：SKIPIF1<0.【例9】（2022屆上海市黃浦區(qū)高三一模）設(shè)常數(shù)SKIPIF1<0且SKIPIF1<0,橢圓SKIPIF1<0：SKIPIF1<0,點(diǎn)SKIPIF1<0是SKIPIF1<0上的動(dòng)點(diǎn)．（1）若點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,求SKIPIF1<0的焦點(diǎn)坐標(biāo)；（2）設(shè)SKIPIF1<0,若定點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,求SKIPIF1<0的最大值與最小值；（3）設(shè)SKIPIF1<0,若SKIPIF1<0上的另一動(dòng)點(diǎn)SKIPIF1<0滿足SKIPIF1<0（SKIPIF1<0為坐標(biāo)原點(diǎn)）,求證：SKIPIF1<0到直線PQ的距離是定值．【解析】（1）∵橢圓SKIPIF1<0：SKIPIF1<0,點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0的焦點(diǎn)坐標(biāo)為SKIPIF1<0；（2）設(shè)SKIPIF1<0,又SKIPIF1<0,由題知SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,又SKIPIF1<0,∴當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最大值為25；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值為SKIPIF1<0；∴SKIPIF1<0的最大值為5,最小值為SKIPIF1<0.（3）當(dāng)SKIPIF1<0時(shí),橢圓SKIPIF1<0：SKIPIF1<0,設(shè)SKIPIF1<0,當(dāng)直線PQ斜率存在時(shí)設(shè)其方程為SKIPIF1<0,則由SKIPIF1<0,得SKIPIF1<0,∴SKIPIF1<0,由SKIPIF1<0可知SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,可得SKIPIF1<0,滿足SKIPIF1<0,∴SKIPIF1<0到直線PQ的距離為SKIPIF1<0為定值；當(dāng)直線PQ斜率不存在時(shí),SKIPIF1<0,可得直線方程為SKIPIF1<0,SKIPIF1<0到直線PQ的距離為SKIPIF1<0.綜上,SKIPIF1<0到直線PQ的距離是定值．三、跟蹤檢測1．（2023屆江蘇省南通市海安市高三上學(xué)期質(zhì)量監(jiān)測）已知橢圓SKIPIF1<0：SKIPIF1<0的離心率為SKIPIF1<0，短軸長為2.(1)求SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0且斜率不為0的直線SKIPIF1<0與SKIPIF1<0自左向右依次交于點(diǎn)SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在線段SKIPIF1<0上，且SKIPIF1<0，SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，記直線SKIPIF1<0，SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，求證：SKIPIF1<0為定值.2．（2023屆湖北省“宜荊荊恩”高三上學(xué)期考試）已知雙曲線SKIPIF1<0與雙曲線SKIPIF1<0有相同的漸近線，且過點(diǎn)SKIPIF1<0.(1)求雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)已知SKIPIF1<0是雙曲線SKIPIF1<0上不同于SKIPIF1<0的兩點(diǎn)，且SKIPIF1<0于SKIPIF1<0，證明：存在定點(diǎn)SKIPIF1<0，使SKIPIF1<0為定值.3．（2023屆江蘇省南京市高三上學(xué)期9月學(xué)情調(diào)研）已知拋物線C：SKIPIF1<0的焦點(diǎn)為F，過點(diǎn)P(0，2)的動(dòng)直線l與拋物線相交于A，B兩點(diǎn)．當(dāng)l經(jīng)過點(diǎn)F時(shí)，點(diǎn)A恰好為線段PF中點(diǎn)．(1)求p的值；(2)是否存在定點(diǎn)T，使得SKIPIF1<0為常數(shù)？若存在，求出點(diǎn)T的坐標(biāo)及該常數(shù)；若不存在，說明理由．4．（2023屆重慶市2023屆高三上學(xué)期質(zhì)量檢測）已知拋物線SKIPIF1<0的焦點(diǎn)為F，斜率不為0的直線l與拋物線C相切，切點(diǎn)為A，當(dāng)l的斜率為2時(shí)，SKIPIF1<0.(1)求p的值；(2)平行于l的直線交拋物線C于B，D兩點(diǎn)，且SKIPIF1<0，點(diǎn)F到直線BD與到直線l的距離之比是否為定值？若是，求出此定值；否則，請說明理由.5．（2023屆江蘇省百校聯(lián)考高三上學(xué)期考試）設(shè)SKIPIF1<0為橢圓SKIPIF1<0：SKIPIF1<0的右焦點(diǎn)，過點(diǎn)SKIPIF1<0且與SKIPIF1<0軸不重合的直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0，SKIPIF1<0兩點(diǎn).(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0；(2)在SKIPIF1<0軸上是否存在異于SKIPIF1<0的定點(diǎn)SKIPIF1<0，使SKIPIF1<0為定值（其中SKIPIF1<0，SKIPIF1<0分別為直線SKIPIF1<0，SKIPIF1<0的斜率）？若存在，求出SKIPIF1<0的坐標(biāo)；若不存在，請說明理由.6．（2022屆湖南省長沙市寧鄉(xiāng)市高三下學(xué)期5月模擬）已知拋物線SKIPIF1<0SKIPIF1<0的焦點(diǎn)與橢圓SKIPIF1<0SKIPIF1<0的右焦點(diǎn)SKIPIF1<0重合，橢圓SKIPIF1<0的長軸長為SKIPIF1<0.(1)求橢圓SKIPIF1<0的方程；(2)過點(diǎn)SKIPIF1<0且斜率為SKIPIF1<0的直線SKIPIF1<0交橢圓SKIPIF1<0于SKIPIF1<0兩點(diǎn)，交拋物線SKIPIF1<0于SKIPIF1<0兩點(diǎn)，請問是否存在實(shí)常數(shù)SKIPIF1<0，使SKIPIF1<0為定值？若存在，求出SKIPIF1<0的值；若不存在，說明理由.7．（2023屆江蘇省南京市高三上學(xué)期數(shù)學(xué)大練）已知點(diǎn)B是圓C:SKIPIF1<0上的任意一點(diǎn)，點(diǎn)F（SKIPIF1<0，0），線段BF的垂直平分線交BC于點(diǎn)P.(1)求動(dòng)點(diǎn)Р的軌跡E的方程;(2)設(shè)曲線E與x軸的兩個(gè)交點(diǎn)分別為A1，A2，Q為直線x=4上的動(dòng)點(diǎn)，且Q不在x軸上，QA1與E的另一個(gè)交點(diǎn)為M，QA2與E的另一個(gè)交點(diǎn)為N，證明:△FMN的周長為定值.8．（2023屆安徽省皖南八校高三上學(xué)期考試）已知橢圓SKIPIF1<0的左?右焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，且左焦點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0為橢圓上的一個(gè)動(dòng)點(diǎn)，SKIPIF1<0的最大值為SKIPIF1<0.(1)求橢圓SKIPIF1<0的標(biāo)準(zhǔn)方程；(2)若過點(diǎn)SKIPIF1<0的直線SKIPIF1<0與橢圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，點(diǎn)SKIPIF1<0，記直線SKIPIF1<0的斜率為SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，證明：SKIPIF1<0.9．（2023屆北京市房山區(qū)高三上學(xué)期考試）已知橢圓SKIPIF1<0的長軸的兩個(gè)端點(diǎn)分別為SKIPIF1<0離心率為SKIPIF1<0．(1)求橢圓C的標(biāo)準(zhǔn)方程；(2)M為橢圓C上除A，B外任意一點(diǎn)，直線SKIPIF1<0交直線SKIPIF1<0于點(diǎn)N，點(diǎn)O為坐標(biāo)原點(diǎn)，過點(diǎn)O且與直線SKIPIF1<0垂直的直線記為l，直線SKIPIF1<0交y軸于點(diǎn)P，交直線l于點(diǎn)Q，求證：SKIPIF1<0為定值．10．（2023屆湖南師范大學(xué)附屬中學(xué)高三上學(xué)期月考）已知SKIPIF1<0，直線SKIPIF1<0的斜率之積為SKIPIF1<0，記動(dòng)點(diǎn)SKIPIF1<0的軌跡為曲線SKIPIF1<0.(1)求SKIPIF1<0的方程;(2)直線SKIPIF1<0與曲線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，若直線SKIPIF1<0的斜率之積為SKIPIF1<0，證明:SKIPIF1<0的面積為定值.11．（2023屆貴州省遵義市新高考協(xié)作體高三上學(xué)期質(zhì)量監(jiān)測）已知點(diǎn)SKIPIF1<0是橢圓SKIPIF1<0的左焦點(diǎn)，SKIPIF1<0是橢圓SKIPIF1<0上的任意一點(diǎn)，SKIPIF1<0．(1)求SKIPIF1<0的最大值；(2)過點(diǎn)SKIPIF1<