圆锥曲线：斜率问题、向量问题、三点共线问题专项训练（解析版）

(2)直线l：y=k(x-1((k>0(与C交于A，B两点，过C上的点P(与A，B不重合且不在坐标轴上)作xBP.，因此sin∠APQ=sin∠BPQ，故直线AP,BP的斜率kAP,kBP互为相反数，则kAP(ii)设P(x0,y0(，x2-4k2x+2k2-8=0，化简得(y1-y0((x2-x0(+(y2-y0((x1-x0(=0，即[k(x1-1(-y0[(x2-x0(+[k(x2-1(-y0[(x1-x0(=2kx1x2-(y0+kx0+k((x1+x2(+2x0(故2k(2k2-8(-4k2(y0+kx0+k(+2x0(y0+k((2k2+1(=0，化简得2y0(x0-1(k2+(x0-8(k+x0y0=0，从而2y0(x0-1(k2+(x0-x-2y(k+x0y0=0，整理得(2y0k-x0([(x0-1(k-y0[=0，又因P(x0,y0(不在直线l:y=k(x-1(上，即(x0-1(k-y0≠0，(2)过点E(2,0(作直线l1交双曲线的右支于A，B两点，连接AO并延长交双曲线左(3)(-1,0(∩(0,1(故右焦点F(c,0(到渐近线的距离为,y2(， (x=my+2x2-=1⇒(3m2-1(y2+12my+9=02××|OE|×(|y1-y2|(=2、(y1+y2(2-4y1y2，-1(y2+6λty+3t2-3=0，方程(3λ2-1(y2+6λty+3t2-3=0的判别式Δ=36λ2t2-4(3λ2-1((3t2-3(=36λ2+12t2-12>0，x3x4=(λy3+t((λy4+t(=λ2y3y4+λt(y3+y4(+t2=，x3x4-(x3+x4(x0+xkSM+kSN=x4y3+x3y4-(x3+x4(y0x3x4-(x3+x4(x0+x所以kSM+kSN=3λ-13λ所以kSM+kSN=3λ-13λ-13λ-1+x所以tx0-1=03x-3-2tx0+t2+x(t-x0(2t-x0，所以6x0y0=2ty03x-3-2tx0+t2+x(t-x0(2t-x0，所以t的范围为(-1,0(∩(0,1(.(2)已知P,Q,R是C上的三个不同点.k3,k4①若P(2,0(，设Q(x2,y2(,R(x3,y3(， -2(2+y=(x3-2(2+y.②根据条件k1,k2,k3,k4均存在知k1,k2,k3,k4均不设点P(x1,y1(,Q(x2,y2(,R(x3,y3(，三角形外心D(x0,y0(，则有-y=1,-y=1,-y=1，则PQ的中垂线为y将k1代入则：y-=-x-，整理得=-+，由①-②得，-=-，则有k1k2k3k4=-.2三等分线段BD.的右顶点，M(2,3(，N(2,-3(在双曲线E上.(2)过点G(-3,0(且斜率为k1的直线l与双曲线E的左支交于A，B两点，△ABD的外接圆的圆心为P，1k2为定值.【详解】(1)因为M(2,3(，N(2,-3(在双曲线E上，(2--2la22x2-=1．得(3-k(x2-6kx-9k-3=0，① 设圆P：(x-s(2+(y-t(2=r2，r>0，由{得(1+k(x2+(6k-2k1t-2s(x+s2+(3k1-t(2-r2=0，②由双曲线的右顶点D在圆上得(1-s(2+t2=r2，t-3s=0③于点F，点M为直线EF与BD的交点，点N为直线BE与OM的交点.证明：直线OM与直线DE的斜率6k(1-3k(x+27k2-18k-9=0，由M为直线EF与BD的交点，则设M(xM,1)，当直线EF的斜率不存在时，xM=9k2-6k- 1-xM- 1-xM-,解得xM=-3k，,解得xM=-3k，=xM-(2)①设M(x1,y1(,N(x2,y2(．.、(x1+x2(2-4x1x2=．则x2-x1=-(x1+x2(2-4x1x2=-． =-(5+26(．2,,8.(24-25高二下·重庆沙坪坝·期末)椭圆E:x2+y2=1(a>b>0(的左、右焦点分别为F12,,F又PF1.PF2=(-x0-c,-y0(.(-x0+c,-y0(=x+y-c则Δ=8(m2+1(>0,y1+y2=-,y1y2=-，T(2,0(的直线l与椭圆C交于不同的两点M,N．(2)λ=-或-(x=ty+2且y1+y2=-y1y，-2，若t=-1，同理可得λ=-或λ=-，综上，λ=-或λ=-. (2)过点F(1,0(的直线l(斜率存在且不为0)与C-1,0(和(1,0(为两焦点的椭圆，则直线PM的方程为y-yx-x1(， -18m+-6m则直线PM的方程为yx-4(，(ⅱ)M=(4-x1,y1(,N=(4-x2,y2(，M.N=(4-x1((4-x2(+y1y2=(3-my1((3-my2(+y1y2=(m2+1(y1y2-3m(y1+y2(+9-，:设A(x1,2x1(，B(x2,-2x2(，P(x,y(，((:动点P的轨迹方程为x2+=1.:P.P=(P+C(.(P+C(=(P+C(.(P-C(=P2-C2，:P.P=(x+2)2+y2-1=-15x2+4x+19=-15(x-2+，“-1≤x≤1，:当x=时，P.P取得最大值．5-4=1,-5k2(x2-10kx5-4=1, 所以O.O=(x1,kx1+1(.(x2,kx2+1(=-，A2(所以双曲线C的方程为x2-=1.因为直线l过点M(-2,0)且斜率不为0，所以设直线l的方程为x=my-2(m>(其中m为设P(x1,y1(,Q(x2,y2(，因为直线OQ交双曲线C于点R，所以R(-x2,-y2(，所以A1R=(-x2+1,-y2(，A2P=(x1-1-1(y2-12my+9=0，所以A1R·A2P=(-x2+1((x所以A1R·A2P=(-x2+1((x1-1(-y1y2=(-my2+2+1((my1-2-1(-y1y2=-(m2+1(y1y2+3m(y1+y2(-9=-(m2+1(+3m-9=0，所以A1R.A2M(2,0(的直线l交双曲线Γ于P,Q两点.(2)P(-2,22(2-1=3.-=1，其中M(2,0(,A1(-1,0(， 其方程为(x-2(2+y2=9；则有解得P(-2,22(.(3)由题知A1(-1,0(,A2(1,0(，l≠0，设点P(x1,y1(,Q(x2,y2(，根据OQ延长线交双曲线Γ于点R，根据双曲线对称性知R(-x2,-y2(，可得A1R=(-x2+1,-可得A1R=(-x2+1,-y2(,A2P=(x1-1,y1(，则A1R.A2P=(-x2+1((x1-1(-y1y2=1，因为P(x1,y1(,Q(x2,y2(在直线l上，2-2=0，的两点M，N.-m2(x2+2m2x-2m2=0，因为直线l与双曲线C的右支相交于M,N不同的两点，因为P=P，所以(x1,y1-1(=(x2,y2-1(，(3)过点(0,-的动直线与椭圆E有两个交点P，Q．在y轴上是否存在点T使得T.T≤0恒成 所以A(-2c,0)，B(0,-3c)，C(0,-，[-3,3]设P(x1,y1(,Q(x2,y2(,T(0,t(，y=kx-，可得(3+4k2(x2-12kx-27=设P(x1,y1(,Q(x2,y2(,T(0,t(，y=kx-，可得(3+4k2(x2-12kx-27=0，而TP=(x1,y1-t(,TQ=(x2,y2-t(，故T.T=x1x2+(y1-t((y2-t(=x1x2+(kx1--t((kx2--t(=(1+k2(x1x2-k+t((x1+x2(++t(2=-27k2-27-18k2-12k2t++t(2+(3+2t(2k2=[(3+2t(2-12t-45[k2++t(2-27t(-3≤t≤.P的轨迹为E．H，线段AB与CD的中点分别为M,N．∴点P到(0,1(的距离与到直线y=-1的距离相等，则直线AB的斜率为kABxM，∴直线AB的方程为y-yx-x1(，即(x2+x1(x-4y-x1x2=0，∴直线CD的方程为y-yx-x3(，即(x4+x3(x-4y-x4x3=0，又直线AC的斜率为kAC∴直线AC的方程为y-yx-x3(，即(x1+x3(x-4y-x1x3=0，直线BD的斜率为kBD∴直线BD的方程为y-yx-x2(，即(x2+x4(x-4y-x2x4=0，所以直线AC与MN的交点和直线BD与MN的交点重合，即为点H．所以M,H,N三点共线； “ABⅡCD，:=>1，得yN>yH>yM，:|HM|=yH-yM==1，上面两式相减得(x4-x1(2+(x2-x3(2=8，1-x4=x3-x2，:|x2-x3|=2，过点B作BEⅡMN交CD于点E，“四边形BMNE是平行四边形，:NE=MB，“M,N分别是AB,CD的中点，:NC=2m，MB=m，:NE=m，:EC=m，设△ABH的边AB上的高为h1，△BCE的边EC上的高为h2，则=，“====，:S△CDH=4S△ABH，S△BCH=2S△ABH，S△ADH=2S△ABH，:SABCD=S△ABH+S△AHD+S△CHD+S△BHC=9S△ABH=9×2=18．三点共线.(2)x-y+1=0或x+y-1=0；,y1),Q(x2,y2)，-4kx-4=0，x1+x2=4k,x1x2=-4，所以直线l的方程为x-y+1=0或x+y-1=0.(3)依题意，设M(t,-1)，A(x3,y3),B(x4,y4)，由y=x2，求导得y/=x，因此抛物线C在点A处切线方程为y-y3=x3(x-x3)，所以A,B,F三点共线.方程为y=.(1)求二次函数y=x2-x+1的焦点坐标和准线方程；(3)设过A(4,1(的直线与抛物线y=x2-x+1的另一个交点为B，直线AB与直线y=x-4交于点P，过点P作x轴的垂线交抛物线y=x2-x+1于点N.是否存在定点G，使得B,N,G三点共线?若【详解】(1)二次函数y=x2-x+1=(x2所以二次函数y=x2-x+1的焦点坐标为F(2,1)，准线方程为y=-1． 准线方程为y(3)由(1)知抛物线y=x2-x+1=(x2-4x+4)=(x-2)2可以由抛物线y=x2沿向量=设B/yx-xx-x1(，化简得y=+(x-x1)=(x-2)+--+2；(2)设A为C的左顶点，过F且斜率存在的直线交C的右支于M,N两点，直线AM,AN分别交圆O的另一点于P,Q.所以双曲线C的方程为x(x=my+2x2-=1∴y1+yy1y其中-<m<，∴-4my1y2=3(y1+y2(，∴P,O,Q三点共线.(ii)不妨设直线AM:x=m1y-1,AN:x=m2y-1，其中m由(i)可知m1m2=-1，∴直线OP:x=-my， :点D在定直线x=1上.(2)过点(2,0(且斜率为k(k≠0(的直线与椭圆E交于A,B两点，点C与点B关于x轴对称．在设直线AB的方程为y=k(x-2(，点A(x1,y1(，点B(x2,y2(．x2-8k2x+8k2-6=0．若在x轴上存在定点D(m,0(，使A,D,C三点共线，则kAD-kCD=0．k(x1-2((x2-m(+k(x2-2((x1-x1x2-m(x1+x2(+m2k[2x1x2-(m+2((x1+x2(+4m[=x1x2-m(x1+x2(+m2由kAD-kCD=0，得k[2x1x2-(m+2((x1+x2(+4m[=0．x2-(m+2((x1+x2(+4m=0．x2-(m+2((x1+x2(+4m=2×-(m+2(+4m22-12-8mk2-16k2+4m+8mk2=2 故在x轴上存在定点D(3,0(，使A,C,D三点共线．(2)设点M(4,0(，斜率不为0的直线AM与C交于另一点B.设直线lAB:x=my+4,m设直线lAB:x=my+4,m≠0,A(x1,y1),B(x2,y2)，则D(x1,-y1(，(i)因为弦AB中点的纵坐标为-，解得m=3或(舍)，所以直线AB的斜率为.(ii)由题得F=(x2-1,y2(,F=(x1-1,-y1(，因为(x2-1((-y1(-y2(x1-1(=(my2+4-1(.(-y1(-y2(my1+4-1(=-2my1y2-3(y1+y2(=-=0，所以FBⅡFD，又FB