高考数学复习第三篇攻坚克难压轴大题多得分第29练直线与圆锥曲线的位置关系文市赛课公开课一等奖省名师优.pptx

第三篇攻坚克难,压轴大题多得分,第,29,练直线与圆锥曲线位置关系,1/67,明考情,直线与圆锥曲线位置关系是高考必考题，难度为中高档，常作为压轴题出现，大致在第,20,题位置,.,知考向,1.,直线与椭圆,.,2.,直线与抛物线,.,2/67,研透考点,关键考点突破练,栏目索引,规范解答,模板答题规范练,3/67,研透考点,关键考点突破练,考点一直线与椭圆,方法技巧,对于直线与圆锥曲线位置关系问题，普通要把圆锥曲线方程与直线方程联立来处理,.,(1),设直线方程，在直线斜率不确定情况下要分斜率存在和不存在两种情况进行讨论，或者将直线方程设成,x,my,b,形式,.,(2),联立直线方程与曲线方程并将其转化成一元二次方程，利用方程根判别式或根与系数关系得到交点横坐标或纵坐标关系,.,4/67,(1),求椭圆方程和抛物线方程；,解答,1,2,3,4,5,5/67,1,2,3,4,解答,5,6/67,解,设直线,AP,方程为,x,my,1(,m,0),，与直线,l,方程,x,1,联立，,1,2,3,4,5,7/67,1,2,3,4,5,8/67,(1),当,t,4,，,|,AM,|,|,AN,|,时，求,AMN,面积；,解答,1,2,3,4,5,9/67,解,设,M,(,x,1,，,y,1,),，则由题意知,y,1,0.,所以直线,AM,方程为,y,x,2.,1,2,3,4,5,10/67,(2),当,2|,AM,|,|,AN,|,时，求,k,取值范围,.,解答,1,2,3,4,5,11/67,1,2,3,4,5,12/67,即,(,k,3,2),t,3,k,(2,k,1),，,1,2,3,4,5,13/67,解答,1,2,3,4,5,14/67,解,由椭圆定义，,设椭圆半焦距为,c,，由已知,PF,1,PF,2,，,1,2,3,4,5,15/67,(2),若,|,PF,1,|,|,PQ,|,，求椭圆离心率,e,.,解,如图，由椭圆定义，,|,PF,1,|,|,PF,2,|,2,a,，,|,QF,1,|,|,QF,2,|,2,a,.,从而由,|,PF,1,|,|,PQ,|,|,PF,2,|,|,QF,2,|,，有,|,QF,1,|,4,a,2|,PF,1,|.,由,PF,1,PF,2,知，,|,PF,1,|,2,|,PF,2,|,2,|,F,1,F,2,|,2,(2,c,),2,，,解答,1,2,3,4,5,16/67,(1),求椭圆,H,离心率,e,；,解,由题意得直线,MN,方程为,x,ay,a,0,，,解答,1,2,3,4,5,17/67,解答,1,2,3,4,5,18/67,当直线,l,斜率为,0,时，其方程为,y,0,，,1,2,3,4,5,19/67,因为点,C,在椭圆上，,所以,x,1,x,2,3,y,1,y,2,0.,1,2,3,4,5,20/67,化简得,m,2,1,0.,所以,m,1.,1,2,3,4,5,21/67,(1),求动点,P,轨迹,C,方程；,解答,1,2,3,4,5,22/67,得,(,x,，,y,),(,x,1,，,y,1,),2(,x,2,，,y,2,),，即,x,x,1,2,x,2,，,y,y,1,2,y,2,.,又因为,x,1,x,2,2,y,1,y,2,0,，所以,x,2,2,y,2,20,，,所以动点,P,轨迹,C,方程为,x,2,2,y,2,20.,1,2,3,4,5,23/67,(2),若直线,l,：,y,x,m,(,m,0),与曲线,C,交于,A,，,B,两点，求,OAB,面积最大值,.,解答,1,2,3,4,5,24/67,消去,y,得,3,x,2,4,mx,2,m,2,20,0.,因为直线,l,与曲线,C,交于,A,，,B,两点，设,A,(,x,3,，,y,3,),，,B,(,x,4,，,y,4,),，,所以,16,m,2,4,3,(2,m,2,20),0.,又,m,0,，所以,0,m,2,30,，,1,2,3,4,5,25/67,而且仅当,m,2,30,m,2,，即,m,2,15,时取等号,.,1,2,3,4,5,26/67,考点二直线与抛物线,方法技巧,(1),判断直线与抛物线位置关系时，能够借助数形结正当，当直线与抛物线轴平行时，直线与抛物线只有一个交点，但并非相切,.,(2),包括中点弦问题，可用,“,点差法,”,求解，但要注意对其存在性检验,.,27/67,(1),求直线,AB,斜率；,解,设,A,(,x,1,，,y,1,),，,B,(,x,2,，,y,2,),，,解答,6,7,8,28/67,(2),设,M,为曲线,C,上一点，,C,在,M,处切线与直线,AB,平行，且,AM,BM,，求直线,AB,方程,.,解答,6,7,8,29/67,设直线,AB,方程为,y,x,m,，,故线段,AB,中点为,N,(2,，,2,m,),，,|,MN,|,|,m,1|.,当,16(,m,1)0,，,6,7,8,30/67,解得,m,7.,所以直线,AB,方程为,y,x,7.,6,7,8,31/67,(1),求曲线,E,方程；,故点,C,轨迹,E,方程为,y,2,x,.,解答,6,7,8,32/67,解答,6,7,8,33/67,消去,x,后，整理得,ky,2,y,k,0.,设,A,(,x,1,，,y,1,),，,B,(,x,2,，,y,2,),，,设直线,l,与,x,轴交于点,N,，则,N,(,1,，,0).,6,7,8,34/67,6,7,8,35/67,8.,已知抛物线,C,：,y,2,2,px,(,p,0),过点,A,(1,，,2).,(1),求抛物线,C,方程，并求其准线方程；,解,将,(1,，,2),代入,y,2,2,px,，,得,(,2),2,2,p,1,，,所以,p,2.,故所求抛物线,C,方程为,y,2,4,x,，其准线方程为,x,1.,解答,6,7,8,36/67,解答,6,7,8,37/67,因为直线,l,与抛物线,C,有公共点，,解,假设存在符合题意直线,l,，其方程为,y,2,x,t,.,所以符合题意直线,l,存在，其方程为,2,x,y,1,0.,6,7,8,38/67,规范解答,模板答题规范练,(1),求椭圆,C,方程；,模,板体验,求,ABQ,面积最大值,.,39/67,审题路线图,40/67,规范解答,评分标准,41/67,设,A,(,x,1,，,y,1,),，,B,(,x,2,，,y,2,).,将,y,kx,m,代入椭圆,E,方程，,可得,(1,4,k,2,),x,2,8,kmx,4,m,2,16,0,，,由,0,，可得,m,2,4,16,k,2,，,(*),因为直线,y,kx,m,与,y,轴交点坐标为,(0,，,m,),，,42/67,可得,(1,4,k,2,),x,2,8,kmx,4,m,2,4,0,，,由,0,，可得,m,2,1,4,k,2,.(*),43/67,构建答题模板,第一步,求曲线方程,：,依据基本量法确定圆锥曲线方程,.,第二步,联立消元,：,将直线方程和圆锥曲线方程联立，得到方程,Ax,2,Bx,C,0,，然后研究判别式，利用根与系数关系,.,第三步,找关系,：,从题设中寻求变量等量或不等关系,.,第四步,建函数,：,对范围最值类问题，要建立关于目标变量函数关系,.,第五步,得范围,：,经过求解函数值域或解不等式得目标变量范围或最值，要注意变量条件制约，检验最值取得条件,.,44/67,(1),求椭圆,E,离心率；,1,2,3,4,5,解答,规范演练,45/67,解,由椭圆定义知,|,AF,2,|,|,BF,2,|,|,AB,|,4,a,，,消去,y,，化简得,(,a,2,b,2,),x,2,2,a,2,cx,a,2,(,c,2,b,2,),0,，,1,2,3,4,5,46/67,因为直线,AB,斜率为,1,，,1,2,3,4,5,47/67,(2),设点,P,(0,，,1),满足,|,PA,|,|,PB,|,，求椭圆,E,方程,.,解,设,AB,中点为,N,(,x,0,，,y,0,),，由,(1),知，,1,2,3,4,5,解答,48/67,(1),求椭圆,E,离心率；,解,过点,(,c,，,0),，,(0,，,b,),直线方程为,bx,cy,bc,0,，,1,2,3,4,5,解答,49/67,1,2,3,4,5,解答,50/67,解,方法一,由,(1),知，椭圆,E,方程为,x,2,4,y,2,4,b,2,.,易知，,AB,与,x,轴不垂直，设其方程为,y,k,(,x,2),1,，,代入,得,(1,4,k,2,),x,2,8,k,(2,k,1),x,4(2,k,1),2,4,b,2,0,，,设,A,(,x,1,，,y,1,),，,B,(,x,2,，,y,2,),，,从而,x,1,x,2,8,2,b,2,.,1,2,3,4,5,51/67,方法二,由,(1),知，椭圆,E,方程为,x,2,4,y,2,4,b,2,，,两式相减并结合,x,1,x,2,4,，,y,1,y,2,2,，,得,4(,x,1,x,2,),8(,y,1,y,2,),0,，,1,2,3,4,5,52/67,易知,AB,与,x,轴不垂直，则,x,1,x,2,，,代入,得,x,2,4,x,8,2,b,2,0,，,所以,x,1,x,2,4,，,x,1,x,2,8,2,b,2,，,1,2,3,4,5,53/67,(1),求椭圆方程；,1,2,3,4,5,解答,54/67,1,2,3,4,5,解答,55/67,解,由,(1),可知,F,(,1,，,0),，则直线,CD,方程为,y,k,(,x,1).,设,C,(,x,1,，,y,1,),，,D,(,x,2,，,y,2,),，,1,2,3,4,5,56/67,1,2,3,4,5,57/67,(1),求抛物线,C,方程，并求其焦点坐标和准线方程；,所以抛物线,C,方程为,y,2,x,，,1,2,3,4,5,解答,58/67,(2),求证：,A,为线段,BM,中点,.,1,2,3,4,5,证实,59/67,l,与抛物线,C,交点为,M,(,x,1,，,y,1,),，,N,(,x,2,，,y,2,).,因为点,P,坐标为,(1,，,1),，,所以直线,OP,方程为,y,x,，点,A,坐标为,(,x,1,，,x,1,).,1,2,3,4,5,60/67,故,A,为线段,BM,中点,.,1,2,3,4,5,61/67,(1),求椭圆,C,方程；,1,2,3,4,5,解答,62/67,1,2,3,4,5,解答,63/67,解,设直线,l,方程为,y,kx,t,(,k,0),，,A,(,x,1,，,y,1,),，,B,(,x,2,，,y,2,).,1,2,3,4,5,64/67,由,(8,kt,),2,4(1,4,k,2,)(4,t,2,4),16(4,k,2,1,t,2,),0,，,设,A,，,B,中点为,D,(,m,，,n,),，,因为直线,PD,与直线,l,垂直，,1,2,3,4,5,65/67,1,2,3,4,5,66/67,本课结束,67/67,