2024步步高考二轮数学新教材讲义专题六　第4讲　母题突破1　范围、最值问题1

2024步步高考二轮数学新教材讲义第4讲圆锥曲线的综合问题[考情分析]1.圆锥曲线的综合问题是高考考查的重点内容，常见的热点题型有范围、最值问题，定点、定直线、定值问题及探索性问题.2.以解答题的形式压轴出现，难度较大．母题突破1范围、最值问题母题(2023·全国甲卷)已知直线x－2y＋1＝0与抛物线C：y2＝2px(p>0)交于A，B两点，|AB|＝4eq\r(15).(1)求p；(2)设F为C的焦点，M，N为C上两点，eq\o(FM,\s\up6(→))·eq\o(FN,\s\up6(→))＝0，求△MFN面积的最小值．思路分析❶联立方程利用弦长求p❷设直线MN：x＝my＋n和点M，N的坐标❸利用eq\o(FM,\s\up6(→))·eq\o(FN,\s\up6(→))＝0，得m，n的关系❹写出S△MFN的面积❺利用函数性质求S△MFN面积的最小值________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题1](2023·武汉模拟)已知椭圆C：eq\f(x2,4)＋y2＝1，椭圆C的右顶点为A，若点P，Q在椭圆C上，且满足直线AP与AQ的斜率之积为eq\f(1,20)，求△APQ面积的最大值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题2](2023·深圳模拟)已知双曲线C：x2－y2＝1，设点A为C的左顶点，若过点(3,0)的直线l与C的右支交于P，Q两点，且直线AP，AQ与圆O：x2＋y2＝1分别交于M，N两点，记四边形PQNM的面积为S1，△AMN的面积为S2，求eq\f(S1,S2)的取值范围．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法求解范围、最值问题的常见方法(1)利用判别式来构造不等关系．(2)利用已知参数的范围，在两个参数之间建立函数关系．(3)利用隐含或已知的不等关系建立不等式．(4)利用基本不等式．1．(2023·佛山模拟)在平面直角坐标系中，点O为坐标原点，Meq\b\lc\(\rc\)(\a\vs4\al\co1(－1，0))，N(1,0)，Q为线段MN上异于M，N的一动点，点P满足eq\f(|PM|,|QM|)＝eq\f(|PN|,|QN|)＝2.(1)求点P的轨迹E的方程；(2)点A，C是曲线E上两点，且在x轴上方，满足AM∥NC，求四边形AMNC面积的最大值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2．(2023·温州模拟)已知抛物线C1：y2＝4x－4与双曲线C2：eq\f(x2,a2)－eq\f(y2,4－a2)＝1(1<a<2)相交于A，B两点，F是C2的右焦点，直线AF分别交C1，C2于C，D(不同于A，B点)两点，直线BC，BD分别交x轴于P，Q两点．(1)设A(x1，y1)，C(x2，y2)，求证：y1y2是定值；(2)求eq\f(|FQ|,|FP|)的取值范围．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________母题突破2定点(定直线)问题母题已知双曲线C：eq\f(x2,a2)－eq\f(y2,b2)＝1(a>0，b>0)的离心率为eq\f(2\r(3),3)，左、右焦点分别为F1，F2，点P的坐标为(3,1)，且eq\o(PF1,\s\up6(→))·eq\o(PF2,\s\up6(→))＝6.(1)求双曲线C的方程；(2)过点P的动直线l与C的左、右两支分别交于A，B两点，若点M在线段AB上，满足eq\f(|AP|,|AM|)＝eq\f(|BP|,|BM|)，证明：点M在定直线上．思路分析❶利用离心率和eq\o(PF1,\s\up6(→))·eq\o(PF2,\s\up6(→))＝6求方程❷设直线方程y－1＝kx－3并联立❸利用比例关系eq\f(|AP|,|AM|)＝eq\f(|BP|,|BM|)列式❹将根与系数的关系代入化简❺消去参数得点在定直线上________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题1](2023·信阳模拟)已知椭圆C：eq\f(x2,4)＋eq\f(y2,2)＝1的左、右顶点分别为A1，A2，过点D(1,0)的直线l与椭圆C交于异于A1，A2的M，N两点．若直线A1M与直线A2N交于点P，证明：点P在定直线上，并求出该定直线的方程．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题2](2023·岳阳模拟)已知双曲线C：x2－eq\f(y2,3)＝1，P为双曲线的右顶点，设直线l不经过P点且与C相交于A，B两点，若直线PA与直线PB的斜率之和为－1.证明：直线l恒过定点．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法动线过定点问题的两大类型及解法(1)动直线l过定点问题，解法：设动直线方程(斜率存在)为y＝kx＋t，由题设条件将t用k表示为t＝mk，得y＝k(x＋m)，故动直线过定点(－m,0)．(2)动曲线C过定点问题，解法：引入参变量建立曲线C的方程，再根据其对参变量恒成立，令其系数等于零，得出定点．1.(2023·襄阳模拟)过抛物线x2＝2py(p>0)内部一点P(m，n)作任意两条直线AB，CD，如图所示，连接AC，BD并延长交于点Q，当P为焦点并且AB⊥CD时，四边形ACBD面积的最小值为32.(1)求抛物线的方程；(2)若点P(1,1)，证明：点Q在定直线上运动，并求出定直线方程．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2．(2023·全国乙卷)已知椭圆C：eq\f(y2,a2)＋eq\f(x2,b2)＝1(a>b>0)的离心率是eq\f(\r(5),3)，点A(－2,0)在C上．(1)求C的方程；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)过点(－2,3)的直线交C于P，Q两点，直线AP，AQ与y轴的交点分别为M，N，证明：线段MN的中点为定点．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________母题突破3定值问题母题(2023·黄山模拟)已知椭圆E：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)过点Meq\b\lc\(\rc\)(\a\vs4\al\co1(\r(3)，\f(1,2)))，点A为下顶点，且AM的斜率为eq\f(\r(3),2).(1)求椭圆E的方程；(2)如图，过点B(0,4)作一条与y轴不重合的直线，该直线交椭圆E于C，D两点，直线AD，AC分别交x轴于H，G两点，O为坐标原点．证明：|OH||OG|为定值，并求出该定值．思路分析❶结合点的坐标和AM的斜率列方程组❷设直线BC的方程并与椭圆的方程联立❸得到x1＋x2，x1x2❹写出直线AD，AC的方程并求出H，G的横坐标❺化简运算|OH||OG|________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题1](2023·西安模拟)如图，在平面直角坐标系中，椭圆E：eq\f(x2,8)＋eq\f(y2,4)＝1，A，B分别为椭圆E的左、右顶点．已知图中四边形ABCD是矩形，且|BC|＝4，点M，N分别在边BC，CD上，AM与BN相交于第一象限内的点P.若点P在椭圆E上，证明：eq\f(|BM|,|CN|)为定值，并求出该定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题2](2023·衡水质检)已知E(2,2)是抛物线C：y2＝2x上一点，经过点(2,0)的直线l与抛物线C交于A，B两点(不同于点E)，直线EA，EB分别交直线x＝－2于点M，N.已知O为原点，求证：∠MON为定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法求解定值问题的两大途径(1)由特例得出一个值(此值一般就是定值)→证明定值：将问题转化为证明待证式与参数(某些变量)无关．(2)先将式子用动点坐标或动线中的参数表示，再利用其满足的约束条件使其绝对值相等的正负项抵消或分子、分母约分得定值．1．已知抛物线C：y2＝2px(p>0)，F为其焦点，若圆E：(x－1)2＋y2＝16与抛物线C交于A，B两点，且|AB|＝4eq\r(3).(1)求抛物线C的方程；(2)若点P为圆E上任意一点，且过点P可以作抛物线C的两条切线PM，PN，切点分别为M，N.求证：|MF|·|NF|恒为定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2．(2023·沧州模拟)已知双曲线C：eq\f(x2,a2)－eq\f(y2,b2)＝1(a>0，b>0)经过点eq\b\lc\(\rc\)(\a\vs4\al\co1(3，\f(\r(6),2)))，右焦点为F(c,0)，且c2，a2，b2成等差数列．(1)求C的方程；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)过F的直线与C的右支交于P，Q两点(P在Q的上方)，PQ的中点为M，M在直线l：x＝2上的射影为N，O为坐标原点，设△POQ的面积为S，直线PN，QN的斜率分别为k1，k2，证明：eq\f(k1－k2,S)是定值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________母题突破4探究性问题母题(2023·廊坊质检)已知椭圆C：eq\f(x2,a2)＋eq\f(y2,b2)＝1(a>b>0)经过点A(－2,0)，且两个焦点及短轴两顶点围成四边形的面积为4.(1)求椭圆C的方程和离心率；(2)设P，Q为椭圆C上两个不同的点，直线AP与y轴交于点E，直线AQ与y轴交于点F，且P，O，Q三点共线．其中O为坐标原点．问：在x轴上是否存在点M，使得∠AME＝∠EFM？若存在，求点M的坐标；若不存在，请说明理由．思路分析❶代入点，结合面积求方程和离心率❷设点P，Q，表示出直线AP，AQ的方程❸求出E，F的坐标❹由∠AME＝∠EFM得eq\o(ME,\s\up6(→))·eq\o(MF,\s\up6(→))＝0❺利用向量运算求点M的坐标________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题1](2023·西安模拟)已知椭圆C：eq\f(x2,4)＋eq\f(y2,3)＝1，过点Teq\b\lc\(\rc\)(\a\vs4\al\co1(\r(3)，0))的直线交该椭圆于P，Q两点，若直线PQ与x轴不垂直，在x轴上是否存在定点S(s,0)，使得∠PST＝∠QST恒成立？若存在，求出s的值；若不存在，请说明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________[子题2]已知双曲线C：eq\f(y2,a2)－eq\f(x2,b2)＝1(a>0，b>0)，直线l在x轴上方与x轴平行，交双曲线C于A，B两点，直线l交y轴于点D.当l经过C的焦点时，点A的坐标为(6,4)．(1)求C的方程；(2)设OD的中点为M，是否存在定直线l，使得经过M的直线与C交于P，Q两点，与线段AB交于点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，请说明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法探索性问题的求解策略(1)若给出问题的一些特殊关系，要探索一般规律，并能证明所得规律的正确性，通常要对已知关系进行观察、比较、分析，然后概括一般规律．(2)若只给出条件，求“不存在”“是否存在”等语句表述问题时，一般先对结论给出肯定的假设，然后由假设出发，结合已知条件进行推理，从而得出结论．1．已知抛物线C：x2＝2py(p>0)，点P(2,8)在抛物线上，直线y＝kx＋2交C于A，B两点，M是线段AB的中点，过M作x轴的垂线交C于点N.(1)求点P到抛物线焦点的距离；(2)是否存在实数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，请说明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________2.(2023·池州模拟)如图，点A为椭圆E：eq\f(x2,4)＋y2＝1的上顶点，圆C：x2＋y2＝1，过坐标原点O的直线l交椭圆E于M，N两点．(1)求直线AM，AN的斜率之积；(2)设直线AM：y＝kx＋1(k≠0)，AN与圆C分别交于点P，Q，记直线MN，PQ的斜率分别为k1，k2，探究是否存在实数λ，使得k1＝λk2？若存在，求出λ的值；若不存在，请说明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规范答题6解析几何(12分)(2023·新高考全国Ⅰ)在直角坐标系Oxy中，点P到x轴的距离等于点P到点eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(1,2)))的距离，记动点P的轨迹为W.(1)求W的方程；[切入点：直接法求轨迹方程](2)已知矩形ABCD有三个顶点在W上，证明：矩形ABCD的周长大于3eq\r(3).[关键点：对周长放缩](1)解设P(x，y)，则|y|＝eq\r(x2＋\b\lc\(\rc\)(\a\vs4\al\co1(y－\f(1,2)))2)，❶两边同时平方化简得y＝x2＋eq\f(1,4)，故W：y＝x2＋eq\f(1,4).(2分)(2)证明设矩形的三个顶点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，易知矩形四条边所在直线的斜率均存在，且不为0，则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，则m＝－eq\f(1,n)，❷(6分)设矩形周长为C，由对称性不妨设|m|≥|n|，kBC－kAB＝c－a＝n－m＝n＋eq\f(1,n)，则eq\f(1,2)C＝|AB|＋|BC|＝(b－a)eq\r(1＋m2)＋(c－b)eq\r(1＋n2)❸≥(c－a)eq\r(1＋n2)＝eq\b\lc\(\rc\)(\a\vs4\al\co1(n＋\f(1,n)))eq\r(1＋n2)，n>0，❹(8分)易知eq\b\lc\(\rc\)(\a\vs4\al\co1(n＋\f(1,n)))eq\r(1＋n2)>0.则令f(x)＝eq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(1,x)))2(1＋x2)，x>0，❺f′(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)，当x∈eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(\r(2),2)))时，f′(x)<0，此时f(x)单调递减，当x∈eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)，＋∞))时，f′(x)>0，此时f(x)单调递增，则f(x)min＝f

eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)))＝eq\f(27,4)，❻(10分)故eq\f(1,2)C≥eq\r(\f(27,4))＝eq\f(3\r(3),2)，即C≥3eq\r(3).当C＝3eq\r(3)时，n＝eq\f(\r(2),2)，m＝－eq\r(2)，与当(b－a)eq\r(1＋m2)＝(b－a)eq\r(1＋n2)，即m＝n时等号成立，矛盾，❼故C>3eq\r(3)，得证．(12分)①处直接法求轨迹方程②处得到m，n之间的关系③处用弦长公式表示周长④处进行周长放缩⑤处建立函数⑥处利用导数求最值⑦处排除边界值母题突破1范围、最值问题母题解(1)设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(负值舍去)．(2)由(1)知y2＝4x，所以焦点F(1,0)，显然直线MN的斜率不可能为零，设直线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，因为eq\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，将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).设点F到直线MN的距离为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的面积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)，所以当n＝3－2eq\r(2)时，△MFN的面积最小，为Smin＝(2－2eq\r(2))2＝12－8eq\r(2)＝4(3－2eq\r(2))．[子题1]解易知直线AP与AQ的斜率同号，所以直线PQ不垂直于x轴，故可设直线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)，化简可得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，所以直线PQ：y＝k(x－2)或y＝k(x＋3)，因为直线PQ不经过点A(2,0)，所以直线PQ经过定点(－3,0)，即m＝3k.所以直线PQ的方程为y＝k(x＋3)，易知k≠0，设定点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)，因为Δ>0，且m＝3k，所以1－5k2>0，所以0<k2<eq\f(1,5)，设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)，当且仅当t＝eq\f(9,7)，即k2＝eq\f(1,14)时取等号，即△APQ面积的最大值为eq\f(5,3).[子题2]解如图所示，设直线lPQ的方程为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，因为直线l与双曲线C的右支交于两点，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)，设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，又因为|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