2024步步高考二輪數(shù)學(xué)新教材講義第4講圓錐曲線的綜合問題[考情分析]1.圓錐曲線的綜合問題是高考考查的重點(diǎn)內(nèi)容，常見的熱點(diǎn)題型有范圍、最值問題，定點(diǎn)、定直線、定值問題及探索性問題.2.以解答題的形式壓軸出現(xiàn)，難度較大．母題突破1范圍、最值問題母題(2023·全國(guó)甲卷)已知直線x－2y＋1＝0與拋物線C：y2＝2px(p>0)交于A，B兩點(diǎn)，|AB|＝4eq\r(15).(1)求p；(2)設(shè)F為C的焦點(diǎn)，M，N為C上兩點(diǎn)，eq\o(FM,\s\up6(→))·eq\o(FN,\s\up6(→))＝0，求△MFN面積的最小值．思路分析?聯(lián)立方程利用弦長(zhǎng)求p?設(shè)直線MN：x＝my＋n和點(diǎn)M，N的坐標(biāo)?利用eq\o(FM,\s\up6(→))·eq\o(FN,\s\up6(→))＝0，得m，n的關(guān)系?寫出S△MFN的面積?利用函數(shù)性質(zhì)求S△MFN面積的最小值________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題1](2023·武漢模擬)已知橢圓C：eq\f(x2,4)＋y2＝1，橢圓C的右頂點(diǎn)為A，若點(diǎn)P，Q在橢圓C上，且滿足直線AP與AQ的斜率之積為eq\f(1,20)，求△APQ面積的最大值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題2](2023·深圳模擬)已知雙曲線C：x2－y2＝1，設(shè)點(diǎn)A為C的左頂點(diǎn)，若過點(diǎn)(3,0)的直線l與C的右支交于P，Q兩點(diǎn)，且直線AP，AQ與圓O：x2＋y2＝1分別交于M，N兩點(diǎn)，記四邊形PQNM的面積為S1，△AMN的面積為S2，求eq\f(S1,S2)的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)律方法求解范圍、最值問題的常見方法(1)利用判別式來(lái)構(gòu)造不等關(guān)系．(2)利用已知參數(shù)的范圍，在兩個(gè)參數(shù)之間建立函數(shù)關(guān)系．(3)利用隱含或已知的不等關(guān)系建立不等式．(4)利用基本不等式．1．(2023·佛山模擬)在平面直角坐標(biāo)系中，點(diǎn)O為坐標(biāo)原點(diǎn)，Meq\b\lc\(\rc\)(\a\vs4\al\co1(－1，0))，N(1,0)，Q為線段MN上異于M，N的一動(dòng)點(diǎn)，點(diǎn)P滿足eq\f(|PM|,|QM|)＝eq\f(|PN|,|QN|)＝2.(1)求點(diǎn)P的軌跡E的方程；(2)點(diǎn)A，C是曲線E上兩點(diǎn)，且在x軸上方，滿足AM∥NC，求四邊形AMNC面積的最大值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2．(2023·溫州模擬)已知拋物線C1：y2＝4x－4與雙曲線C2：eq\f(x2,a2)－eq\f(y2,4－a2)＝1(1<a<2)相交于A，B兩點(diǎn)，F(xiàn)是C2的右焦點(diǎn)，直線AF分別交C1，C2于C，D(不同于A，B點(diǎn))兩點(diǎn)，直線BC，BD分別交x軸于P，Q兩點(diǎn)．(1)設(shè)A(x1，y1)，C(x2，y2)，求證：y1y2是定值；(2)求eq\f(|FQ|,|FP|)的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________母題突破2定點(diǎn)(定直線)問題母題已知雙曲線C：eq\f(x2,a2)－eq\f(y2,b2)＝1(a>0，b>0)的離心率為eq\f(2\r(3),3)，左、右焦點(diǎn)分別為F1，F(xiàn)2，點(diǎn)P的坐標(biāo)為(3,1)，且eq\o(PF1,\s\up6(→))·eq\o(PF2,\s\up6(→))＝6.(1)求雙曲線C的方程；(2)過點(diǎn)P的動(dòng)直線l與C的左、右兩支分別交于A，B兩點(diǎn)，若點(diǎn)M在線段AB上，滿足eq\f(|AP|,|AM|)＝eq\f(|BP|,|BM|)，證明：點(diǎn)M在定直線上．思路分析?利用離心率和eq\o(PF1,\s\up6(→))·eq\o(PF2,\s\up6(→))＝6求方程?設(shè)直線方程y－1＝kx－3并聯(lián)立?利用比例關(guān)系eq\f(|AP|,|AM|)＝eq\f(|BP|,|BM|)列式?將根與系數(shù)的關(guān)系代入化簡(jiǎn)?消去參數(shù)得點(diǎn)在定直線上________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題1](2023·信陽(yáng)模擬)已知橢圓C：eq\f(x2,4)＋eq\f(y2,2)＝1的左、右頂點(diǎn)分別為A1，A2，過點(diǎn)D(1,0)的直線l與橢圓C交于異于A1，A2的M，N兩點(diǎn)．若直線A1M與直線A2N交于點(diǎn)P，證明：點(diǎn)P在定直線上，并求出該定直線的方程．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題2](2023·岳陽(yáng)模擬)已知雙曲線C：x2－eq\f(y2,3)＝1，P為雙曲線的右頂點(diǎn)，設(shè)直線l不經(jīng)過P點(diǎn)且與C相交于A，B兩點(diǎn)，若直線PA與直線PB的斜率之和為－1.證明：直線l恒過定點(diǎn)．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)律方法動(dòng)線過定點(diǎn)問題的兩大類型及解法(1)動(dòng)直線l過定點(diǎn)問題，解法：設(shè)動(dòng)直線方程(斜率存在)為y＝kx＋t，由題設(shè)條件將t用k表示為t＝mk，得y＝k(x＋m)，故動(dòng)直線過定點(diǎn)(－m,0)．(2)動(dòng)曲線C過定點(diǎn)問題，解法：引入?yún)⒆兞拷⑶€C的方程，再根據(jù)其對(duì)參變量恒成立，令其系數(shù)等于零，得出定點(diǎn)．1.(2023·襄陽(yáng)模擬)過拋物線x2＝2py(p>0)內(nèi)部一點(diǎn)P(m，n)作任意兩條直線AB，CD，如圖所示，連接AC，BD并延長(zhǎng)交于點(diǎn)Q，當(dāng)P為焦點(diǎn)并且AB⊥CD時(shí)，四邊形ACBD面積的最小值為32.(1)求拋物線的方程；(2)若點(diǎn)P(1,1)，證明：點(diǎn)Q在定直線上運(yùn)動(dòng)，并求出定直線方程．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2．(2023·全國(guó)乙卷)已知橢圓C：eq\f(y2,a2)＋eq\f(x2,b2)＝1(a>b>0)的離心率是eq\f(\r(5),3)，點(diǎn)A(－2,0)在C上．(1)求C的方程；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)過點(diǎn)(－2,3)的直線交C于P，Q兩點(diǎn)，直線AP，AQ與y軸的交點(diǎn)分別為M，N，證明：線段MN的中點(diǎn)為定點(diǎn)．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________母題突破3定值問題母題(2023·黃山模擬)已知橢圓E：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)過點(diǎn)Meq\b\lc\(\rc\)(\a\vs4\al\co1(\r(3)，\f(1,2)))，點(diǎn)A為下頂點(diǎn)，且AM的斜率為eq\f(\r(3),2).(1)求橢圓E的方程；(2)如圖，過點(diǎn)B(0,4)作一條與y軸不重合的直線，該直線交橢圓E于C，D兩點(diǎn)，直線AD，AC分別交x軸于H，G兩點(diǎn)，O為坐標(biāo)原點(diǎn)．證明：|OH||OG|為定值，并求出該定值．思路分析?結(jié)合點(diǎn)的坐標(biāo)和AM的斜率列方程組?設(shè)直線BC的方程并與橢圓的方程聯(lián)立?得到x1＋x2，x1x2?寫出直線AD，AC的方程并求出H，G的橫坐標(biāo)?化簡(jiǎn)運(yùn)算|OH||OG|________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題1](2023·西安模擬)如圖，在平面直角坐標(biāo)系中，橢圓E：eq\f(x2,8)＋eq\f(y2,4)＝1，A，B分別為橢圓E的左、右頂點(diǎn)．已知圖中四邊形ABCD是矩形，且|BC|＝4，點(diǎn)M，N分別在邊BC，CD上，AM與BN相交于第一象限內(nèi)的點(diǎn)P.若點(diǎn)P在橢圓E上，證明：eq\f(|BM|,|CN|)為定值，并求出該定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題2](2023·衡水質(zhì)檢)已知E(2,2)是拋物線C：y2＝2x上一點(diǎn)，經(jīng)過點(diǎn)(2,0)的直線l與拋物線C交于A，B兩點(diǎn)(不同于點(diǎn)E)，直線EA，EB分別交直線x＝－2于點(diǎn)M，N.已知O為原點(diǎn)，求證：∠MON為定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)律方法求解定值問題的兩大途徑(1)由特例得出一個(gè)值(此值一般就是定值)→證明定值：將問題轉(zhuǎn)化為證明待證式與參數(shù)(某些變量)無(wú)關(guān)．(2)先將式子用動(dòng)點(diǎn)坐標(biāo)或動(dòng)線中的參數(shù)表示，再利用其滿足的約束條件使其絕對(duì)值相等的正負(fù)項(xiàng)抵消或分子、分母約分得定值．1．已知拋物線C：y2＝2px(p>0)，F(xiàn)為其焦點(diǎn)，若圓E：(x－1)2＋y2＝16與拋物線C交于A，B兩點(diǎn)，且|AB|＝4eq\r(3).(1)求拋物線C的方程；(2)若點(diǎn)P為圓E上任意一點(diǎn)，且過點(diǎn)P可以作拋物線C的兩條切線PM，PN，切點(diǎn)分別為M，N.求證：|MF|·|NF|恒為定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2．(2023·滄州模擬)已知雙曲線C：eq\f(x2,a2)－eq\f(y2,b2)＝1(a>0，b>0)經(jīng)過點(diǎn)eq\b\lc\(\rc\)(\a\vs4\al\co1(3，\f(\r(6),2)))，右焦點(diǎn)為F(c,0)，且c2，a2，b2成等差數(shù)列．(1)求C的方程；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)過F的直線與C的右支交于P，Q兩點(diǎn)(P在Q的上方)，PQ的中點(diǎn)為M，M在直線l：x＝2上的射影為N，O為坐標(biāo)原點(diǎn)，設(shè)△POQ的面積為S，直線PN，QN的斜率分別為k1，k2，證明：eq\f(k1－k2,S)是定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________母題突破4探究性問題母題(2023·廊坊質(zhì)檢)已知橢圓C：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)經(jīng)過點(diǎn)A(－2,0)，且兩個(gè)焦點(diǎn)及短軸兩頂點(diǎn)圍成四邊形的面積為4.(1)求橢圓C的方程和離心率；(2)設(shè)P，Q為橢圓C上兩個(gè)不同的點(diǎn)，直線AP與y軸交于點(diǎn)E，直線AQ與y軸交于點(diǎn)F，且P，O，Q三點(diǎn)共線．其中O為坐標(biāo)原點(diǎn)．問：在x軸上是否存在點(diǎn)M，使得∠AME＝∠EFM？若存在，求點(diǎn)M的坐標(biāo)；若不存在，請(qǐng)說(shuō)明理由．思路分析?代入點(diǎn)，結(jié)合面積求方程和離心率?設(shè)點(diǎn)P，Q，表示出直線AP，AQ的方程?求出E，F(xiàn)的坐標(biāo)?由∠AME＝∠EFM得eq\o(ME,\s\up6(→))·eq\o(MF,\s\up6(→))＝0?利用向量運(yùn)算求點(diǎn)M的坐標(biāo)________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題1](2023·西安模擬)已知橢圓C：eq\f(x2,4)＋eq\f(y2,3)＝1，過點(diǎn)Teq\b\lc\(\rc\)(\a\vs4\al\co1(\r(3)，0))的直線交該橢圓于P，Q兩點(diǎn)，若直線PQ與x軸不垂直，在x軸上是否存在定點(diǎn)S(s,0)，使得∠PST＝∠QST恒成立？若存在，求出s的值；若不存在，請(qǐng)說(shuō)明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子題2]已知雙曲線C：eq\f(y2,a2)－eq\f(x2,b2)＝1(a>0，b>0)，直線l在x軸上方與x軸平行，交雙曲線C于A，B兩點(diǎn)，直線l交y軸于點(diǎn)D.當(dāng)l經(jīng)過C的焦點(diǎn)時(shí)，點(diǎn)A的坐標(biāo)為(6,4)．(1)求C的方程；(2)設(shè)OD的中點(diǎn)為M，是否存在定直線l，使得經(jīng)過M的直線與C交于P，Q兩點(diǎn)，與線段AB交于點(diǎn)N(N，D不重合)，eq\o(PM,\s\up6(→))＝λeq\o(PN,\s\up6(→))，eq\o(MQ,\s\up6(→))＝λeq\o(QN,\s\up6(→))均成立？若存在，求出l的方程；若不存在，請(qǐng)說(shuō)明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)律方法探索性問題的求解策略(1)若給出問題的一些特殊關(guān)系，要探索一般規(guī)律，并能證明所得規(guī)律的正確性，通常要對(duì)已知關(guān)系進(jìn)行觀察、比較、分析，然后概括一般規(guī)律．(2)若只給出條件，求“不存在”“是否存在”等語(yǔ)句表述問題時(shí)，一般先對(duì)結(jié)論給出肯定的假設(shè)，然后由假設(shè)出發(fā)，結(jié)合已知條件進(jìn)行推理，從而得出結(jié)論．1．已知拋物線C：x2＝2py(p>0)，點(diǎn)P(2,8)在拋物線上，直線y＝kx＋2交C于A，B兩點(diǎn)，M是線段AB的中點(diǎn)，過M作x軸的垂線交C于點(diǎn)N.(1)求點(diǎn)P到拋物線焦點(diǎn)的距離；(2)是否存在實(shí)數(shù)k使得eq\o(NA,\s\up6(→))·eq\o(NB,\s\up6(→))＝0，若存在，求k的值；若不存在，請(qǐng)說(shuō)明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2.(2023·池州模擬)如圖，點(diǎn)A為橢圓E：eq\f(x2,4)＋y2＝1的上頂點(diǎn)，圓C：x2＋y2＝1，過坐標(biāo)原點(diǎn)O的直線l交橢圓E于M，N兩點(diǎn)．(1)求直線AM，AN的斜率之積；(2)設(shè)直線AM：y＝kx＋1(k≠0)，AN與圓C分別交于點(diǎn)P，Q，記直線MN，PQ的斜率分別為k1，k2，探究是否存在實(shí)數(shù)λ，使得k1＝λk2？若存在，求出λ的值；若不存在，請(qǐng)說(shuō)明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________規(guī)范答題6解析幾何(12分)(2023·新高考全國(guó)Ⅰ)在直角坐標(biāo)系Oxy中，點(diǎn)P到x軸的距離等于點(diǎn)P到點(diǎn)eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(1,2)))的距離，記動(dòng)點(diǎn)P的軌跡為W.(1)求W的方程；[切入點(diǎn)：直接法求軌跡方程](2)已知矩形ABCD有三個(gè)頂點(diǎn)在W上，證明：矩形ABCD的周長(zhǎng)大于3eq\r(3).[關(guān)鍵點(diǎn)：對(duì)周長(zhǎng)放縮](1)解設(shè)P(x，y)，則|y|＝eq\r(x2＋\b\lc\(\rc\)(\a\vs4\al\co1(y－\f(1,2)))2)，?兩邊同時(shí)平方化簡(jiǎn)得y＝x2＋eq\f(1,4)，故W：y＝x2＋eq\f(1,4).(2分)(2)證明設(shè)矩形的三個(gè)頂點(diǎn)Aeq\b\lc\(\rc\)(\a\vs4\al\co1(a，a2＋\f(1,4)))，Beq\b\lc\(\rc\)(\a\vs4\al\co1(b，b2＋\f(1,4)))，Ceq\b\lc\(\rc\)(\a\vs4\al\co1(c，c2＋\f(1,4)))在W上，且a<b<c，易知矩形四條邊所在直線的斜率均存在，且不為0，則kAB·kBC＝－1，a＋b<b＋c，(4分)令kAB＝eq\f(b2＋\f(1,4)－\b\lc\(\rc\)(\a\vs4\al\co1(a2＋\f(1,4))),b－a)＝a＋b＝m<0，同理令kBC＝b＋c＝n>0，且mn＝－1，則m＝－eq\f(1,n)，?(6分)設(shè)矩形周長(zhǎng)為C，由對(duì)稱性不妨設(shè)|m|≥|n|，kBC－kAB＝c－a＝n－m＝n＋eq\f(1,n)，則eq\f(1,2)C＝|AB|＋|BC|＝(b－a)eq\r(1＋m2)＋(c－b)eq\r(1＋n2)?≥(c－a)eq\r(1＋n2)＝eq\b\lc\(\rc\)(\a\vs4\al\co1(n＋\f(1,n)))eq\r(1＋n2)，n>0，?(8分)易知eq\b\lc\(\rc\)(\a\vs4\al\co1(n＋\f(1,n)))eq\r(1＋n2)>0.則令f(x)＝eq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(1,x)))2(1＋x2)，x>0，?f′(x)＝2eq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(1,x)))2eq\b\lc\(\rc\)(\a\vs4\al\co1(2x－\f(1,x)))，令f′(x)＝0，解得x＝eq\f(\r(2),2)，當(dāng)x∈eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(\r(2),2)))時(shí)，f′(x)<0，此時(shí)f(x)單調(diào)遞減，當(dāng)x∈eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)，＋∞))時(shí)，f′(x)>0，此時(shí)f(x)單調(diào)遞增，則f(x)min＝f

eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)))＝eq\f(27,4)，?(10分)故eq\f(1,2)C≥eq\r(\f(27,4))＝eq\f(3\r(3),2)，即C≥3eq\r(3).當(dāng)C＝3eq\r(3)時(shí)，n＝eq\f(\r(2),2)，m＝－eq\r(2)，與當(dāng)(b－a)eq\r(1＋m2)＝(b－a)eq\r(1＋n2)，即m＝n時(shí)等號(hào)成立，矛盾，?故C>3eq\r(3)，得證．(12分)①處直接法求軌跡方程②處得到m，n之間的關(guān)系③處用弦長(zhǎng)公式表示周長(zhǎng)④處進(jìn)行周長(zhǎng)放縮⑤處建立函數(shù)⑥處利用導(dǎo)數(shù)求最值⑦處排除邊界值母題突破1范圍、最值問題母題解(1)設(shè)A(xA，yA)，B(xB，yB)，由eq\b\lc\{\rc\(\a\vs4\al\co1(x－2y＋1＝0，,y2＝2px，))可得y2－4py＋2p＝0，所以yA＋yB＝4p，yAyB＝2p，所以|AB|＝eq\r(5)×eq\r(yA＋yB2－4yAyB)＝4eq\r(15)，即2p2－p－6＝0，解得p＝2(負(fù)值舍去)．(2)由(1)知y2＝4x，所以焦點(diǎn)F(1,0)，顯然直線MN的斜率不可能為零，設(shè)直線MN：x＝my＋n，M(x1，y1)，N(x2，y2)，由eq\b\lc\{\rc\(\a\vs4\al\co1(y2＝4x，,x＝my＋n，))可得y2－4my－4n＝0，所以y1＋y2＝4m，y1y2＝－4n，Δ＝16m2＋16n>0?m2＋n>0，因?yàn)閑q\o(FM,\s\up6(→))·eq\o(FN,\s\up6(→))＝0，eq\o(FM,\s\up6(→))＝(x1－1，y1)，eq\o(FN,\s\up6(→))＝(x2－1，y2)，所以(x1－1)(x2－1)＋y1y2＝0，即(my1＋n－1)(my2＋n－1)＋y1y2＝0，即(m2＋1)y1y2＋m(n－1)(y1＋y2)＋(n－1)2＝0，將y1＋y2＝4m，y1y2＝－4n代入得，4m2＝n2－6n＋1，所以4(m2＋n)＝(n－1)2>0，所以n≠1，且n2－6n＋1≥0，解得n≥3＋2eq\r(2)或n≤3－2eq\r(2).設(shè)點(diǎn)F到直線MN的距離為d，所以d＝eq\f(|n－1|,\r(1＋m2))，|MN|＝eq\r(1＋m2)eq\r(y1＋y22－4y1y2)＝eq\r(1＋m2)eq\r(16m2＋16n)＝eq\r(1＋m2)eq\r(4n2－6n＋1＋16n)＝2eq\r(1＋m2)|n－1|，所以△MFN的面積S＝eq\f(1,2)×|MN|×d＝eq\f(1,2)×2eq\r(1＋m2)·|n－1|×eq\f(|n－1|,\r(1＋m2))＝(n－1)2，而n≥3＋2eq\r(2)或n≤3－2eq\r(2)，所以當(dāng)n＝3－2eq\r(2)時(shí)，△MFN的面積最小，為Smin＝(2－2eq\r(2))2＝12－8eq\r(2)＝4(3－2eq\r(2))．[子題1]解易知直線AP與AQ的斜率同號(hào)，所以直線PQ不垂直于x軸，故可設(shè)直線PQ：y＝kx＋m，P(x1，y1)，Q(x2，y2)，由eq\b\lc\{\rc\(\a\vs4\al\co1(\f(x2,4)＋y2＝1，,y＝kx＋m，))可得(1＋4k2)x2＋8mkx＋4m2－4＝0，所以x1＋x2＝eq\f(－8mk,1＋4k2)，x1x2＝eq\f(4m2－4,1＋4k2)，Δ＝16(4k2＋1－m2)>0，即4k2＋1>m2，而kAPkAQ＝eq\f(1,20)，A(2,0)，所以eq\f(y1,x1－2)·eq\f(y2,x2－2)＝eq\f(1,20)，化簡(jiǎn)可得20(kx1＋m)(kx2＋m)＝(x1－2)(x2－2)，即20k2x1x2＋20km(x1＋x2)＋20m2＝x1x2－2(x1＋x2)＋4，20k2·eq\f(4m2－4,1＋4k2)＋20km·eq\f(－8mk,1＋4k2)＋20m2＝eq\f(4m2－4,1＋4k2)－2×eq\f(－8mk,1＋4k2)＋4，整理得6k2＋mk－m2＝0，所以m＝－2k或m＝3k，所以直線PQ：y＝k(x－2)或y＝k(x＋3)，因?yàn)橹本€PQ不經(jīng)過點(diǎn)A(2,0)，所以直線PQ經(jīng)過定點(diǎn)(－3,0)，即m＝3k.所以直線PQ的方程為y＝k(x＋3)，易知k≠0，設(shè)定點(diǎn)B(－3,0)，S△APQ＝eq\b\lc\|\rc\|(\a\vs4\al\co1(S△ABP－S△ABQ))＝eq\f(1,2)|AB||y1－y2|＝eq\f(5,2)|k|eq\b\lc\|\rc\|(\a\vs4\al\co1(x1－x2))＝eq\f(5,2)|k|eq\r(x1＋x22－4x1x2)＝eq\f(5,2)|k|eq\r(\b\lc\(\rc\)(\a\vs4\al\co1(\f(－8mk,1＋4k2)))2－4×\f(4m2－4,1＋4k2))＝eq\f(5,2)|k|·eq\f(\r(164k2＋1－m2),1＋4k2)＝eq\f(10\r(\b\lc\(\rc\)(\a\vs4\al\co1(1－5k2))k2),1＋4k2)，因?yàn)棣?gt;0，且m＝3k，所以1－5k2>0，所以0<k2<eq\f(1,5)，設(shè)t＝4k2＋1∈eq\b\lc\(\rc\)(\a\vs4\al\co1(1，\f(9,5)))，所以S△APQ＝eq\f(5,2)eq\r(\f(－5t2＋14t－9,t2))＝eq\f(5,2)eq\r(－9\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,t)－\f(7,9)))2＋\f(4,9))≤eq\f(5,3)，當(dāng)且僅當(dāng)t＝eq\f(9,7)，即k2＝eq\f(1,14)時(shí)取等號(hào)，即△APQ面積的最大值為eq\f(5,3).[子題2]解如圖所示，設(shè)直線lPQ的方程為x＝ty＋3，P(x1，y1)，Q(x2，y2)，由eq\b\lc\{\rc\(\a\vs4\al\co1(x＝ty＋3，,x2－y2＝1，))得(t2－1)y2＋6ty＋8＝0，因?yàn)橹本€l與雙曲線C的右支交于兩點(diǎn)，所以eq\b\lc\{\rc\(\a\vs4\al\co1(t2－1≠0，,Δ＝6t2－32t2－1>0，,y1y2＝\f(8,t2－1)<0，))解得－1<t<1，y1＋y2＝eq\f(－6t,t2－1)，y1y2＝eq\f(8,t2－1)，所以kAP·kAQ＝eq\f(y1,x1＋1)·eq\f(y2,x2＋1)＝eq\f(y1y2,\b\lc\(\rc\)(\a\vs4\al\co1(ty1＋4))\b\lc\(\rc\)(\a\vs4\al\co1(ty2＋4)))＝eq\f(y1y2,t2y1y2＋4ty1＋y2＋16)＝eq\f(\f(8,t2－1),t2·\f(8,t2－1)＋4t·\f(－6t,t2－1)＋16)＝eq\f(8,8t2－24t2＋16t2－16)＝－eq\f(1,2)，設(shè)AP：x＝m1y－1，AQ：x＝m2y－1，且|m1|>1，|m2|>1，所以eq\f(1,m1)·eq\f(1,m2)＝－eq\f(1,2)，即m1·m2＝－2，所以|m1|·|m2|＝2，又因?yàn)閨m2|＝eq\f(2,|m1|)>1，所以1<|m1|<2，由eq\b\lc\{\rc\(\a\vs4\al\co1(x＝m1y－1，,x2－y2＝1，))得(meq\o\al(2,1)－1)y2－2m1y＝0，所以yP＝eq\f(2m1,m\o\al(2,1)－1)，同理可得yQ＝eq\f(2m2,m\o\al(2,2)－1)，由eq\b\lc\{\rc\(\a\vs4\al\co1(x＝m1y－1，,x2＋y2＝1，))得(meq\o\al(2,1)＋1)y2－2m1y＝0，所以yM＝eq\f(2m1,m\o\al(2,1)＋1)，同理可得yN＝eq\f(2m2,m\o\al(2,2)＋1)，所以eq\f(S△APQ,S△AMN)＝eq\f(\f(1,2)|AP||AQ|sin∠PAQ,\f(1,2)|AM||AN|sin∠MAN)＝eq\f(|AP|,|AM|)·eq\f(|AQ|,|AN|)＝eq\f(yP,yM)·eq\f(yQ,yN)＝eq\f(\f(2m1,m\o\al(2,1)－1),\f(2m1,m\o\a