(2)直線l：y=k(x-1((k>0(與C交于A，B兩點(diǎn)，過C上的點(diǎn)P(與A，B不重合且不在坐標(biāo)軸上)作xBP.，因此sin∠APQ=sin∠BPQ，故直線AP,BP的斜率kAP,kBP互為相反數(shù)，則kAP(ii)設(shè)P(x0,y0(，x2-4k2x+2k2-8=0，化簡得(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，化簡得2y0(x0-1(k2+(x0-8(k+x0y0=0，從而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(不在直線l:y=k(x-1(上，即(x0-1(k-y0≠0，(2)過點(diǎn)E(2,0(作直線l1交雙曲線的右支于A，B兩點(diǎn)，連接AO并延長交雙曲線左(3)(-1,0(∩(0,1(故右焦點(diǎn)F(c,0(到漸近線的距離為,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的判別式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的范圍為(-1,0(∩(0,1(.(2)已知P,Q,R是C上的三個(gè)不同點(diǎn).k3,k4①若P(2,0(，設(shè)Q(x2,y2(,R(x3,y3(， -2(2+y=(x3-2(2+y.②根據(jù)條件k1,k2,k3,k4均存在知k1,k2,k3,k4均不設(shè)點(diǎn)P(x1,y1(,Q(x2,y2(,R(x3,y3(，三角形外心D(x0,y0(，則有-y=1,-y=1,-y=1，則PQ的中垂線為y將k1代入則：y-=-x-，整理得=-+，由①-②得，-=-，則有k1k2k3k4=-.2三等分線段BD.的右頂點(diǎn)，M(2,3(，N(2,-3(在雙曲線E上.(2)過點(diǎn)G(-3,0(且斜率為k1的直線l與雙曲線E的左支交于A，B兩點(diǎn)，△ABD的外接圓的圓心為P，1k2為定值.【詳解】(1)因?yàn)镸(2,3(，N(2,-3(在雙曲線E上，(2--2la22x2-=1．得(3-k(x2-6kx-9k-3=0，① 設(shè)圓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由雙曲線的右頂點(diǎn)D在圓上得(1-s(2+t2=r2，t-3s=0③于點(diǎn)F，點(diǎn)M為直線EF與BD的交點(diǎn)，點(diǎn)N為直線BE與OM的交點(diǎn).證明：直線OM與直線DE的斜率6k(1-3k(x+27k2-18k-9=0，由M為直線EF與BD的交點(diǎn)，則設(shè)M(xM,1)，當(dāng)直線EF的斜率不存在時(shí)，xM=9k2-6k- 1-xM- 1-xM-,解得xM=-3k，,解得xM=-3k，=xM-(2)①設(shè)M(x1,y1(,N(x2,y2(．.、(x1+x2(2-4x1x2=．則x2-x1=-(x1+x2(2-4x1x2=-． =-(5+26(．2,,8.(24-25高二下·重慶沙坪壩·期末)橢圓E:x2+y2=1(a>b>0(的左、右焦點(diǎn)分別為F12,,F又PF1.PF2=(-x0-c,-y0(.(-x0+c,-y0(=x+y-c則Δ=8(m2+1(>0,y1+y2=-,y1y2=-，T(2,0(的直線l與橢圓C交于不同的兩點(diǎn)M,N．(2)λ=-或-(x=ty+2且y1+y2=-y1y，-2，若t=-1，同理可得λ=-或λ=-，綜上，λ=-或λ=-. (2)過點(diǎn)F(1,0(的直線l(斜率存在且不為0)與C-1,0(和(1,0(為兩焦點(diǎn)的橢圓，則直線PM的方程為y-yx-x1(， -18m+-6m則直線PM的方程為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-，:設(shè)A(x1,2x1(，B(x2,-2x2(，P(x,y(，((:動(dòng)點(diǎn)P的軌跡方程為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，:當(dāng)x=時(shí)，P.P取得最大值．5-4=1,-5k2(x2-10kx5-4=1, 所以O(shè).O=(x1,kx1+1(.(x2,kx2+1(=-，A2(所以雙曲線C的方程為x2-=1.因?yàn)橹本€l過點(diǎn)M(-2,0)且斜率不為0，所以設(shè)直線l的方程為x=my-2(m>(其中m為設(shè)P(x1,y1(,Q(x2,y2(，因?yàn)橹本€OQ交雙曲線C于點(diǎn)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(的直線l交雙曲線Γ于P,Q兩點(diǎn).(2)P(-2,22(2-1=3.-=1，其中M(2,0(,A1(-1,0(， 其方程為(x-2(2+y2=9；則有解得P(-2,22(.(3)由題知A1(-1,0(,A2(1,0(，l≠0，設(shè)點(diǎn)P(x1,y1(,Q(x2,y2(，根據(jù)OQ延長線交雙曲線Γ于點(diǎn)R，根據(jù)雙曲線對稱性知R(-x2,-y2(，可得A1R=(-x2+1,-可得A1R=(-x2+1,-y2(,A2P=(x1-1,y1(，則A1R.A2P=(-x2+1((x1-1(-y1y2=1，因?yàn)镻(x1,y1(,Q(x2,y2(在直線l上，2-2=0，的兩點(diǎn)M，N.-m2(x2+2m2x-2m2=0，因?yàn)橹本€l與雙曲線C的右支相交于M,N不同的兩點(diǎn)，因?yàn)镻=P，所以(x1,y1-1(=(x2,y2-1(，(3)過點(diǎn)(0,-的動(dòng)直線與橢圓E有兩個(gè)交點(diǎn)P，Q．在y軸上是否存在點(diǎn)T使得T.T≤0恒成 所以A(-2c,0)，B(0,-3c)，C(0,-，[-3,3]設(shè)P(x1,y1(,Q(x2,y2(,T(0,t(，y=kx-，可得(3+4k2(x2-12kx-27=設(shè)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的軌跡為E．H，線段AB與CD的中點(diǎn)分別為M,N．∴點(diǎn)P到(0,1(的距離與到直線y=-1的距離相等，則直線AB的斜率為kABxM，∴直線AB的方程為y-yx-x1(，即(x2+x1(x-4y-x1x2=0，∴直線CD的方程為y-yx-x3(，即(x4+x3(x-4y-x4x3=0，又直線AC的斜率為kAC∴直線AC的方程為y-yx-x3(，即(x1+x3(x-4y-x1x3=0，直線BD的斜率為kBD∴直線BD的方程為y-yx-x2(，即(x2+x4(x-4y-x2x4=0，所以直線AC與MN的交點(diǎn)和直線BD與MN的交點(diǎn)重合，即為點(diǎn)H．所以M,H,N三點(diǎn)共線； “ABⅡCD，:=>1，得yN>yH>yM，:|HM|=yH-yM==1，上面兩式相減得(x4-x1(2+(x2-x3(2=8，1-x4=x3-x2，:|x2-x3|=2，過點(diǎn)B作BEⅡMN交CD于點(diǎn)E，“四邊形BMNE是平行四邊形，:NE=MB，“M,N分別是AB,CD的中點(diǎn)，:NC=2m，MB=m，:NE=m，:EC=m，設(shè)△ABH的邊AB上的高為h1，△BCE的邊EC上的高為h2，則=，“====，: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=18．三點(diǎn)共線.(2)x-y+1=0或x+y-1=0；,y1),Q(x2,y2)，-4kx-4=0，x1+x2=4k,x1x2=-4，所以直線l的方程為x-y+1=0或x+y-1=0.(3)依題意，設(shè)M(t,-1)，A(x3,y3),B(x4,y4)，由y=x2，求導(dǎo)得y/=x，因此拋物線C在點(diǎn)A處切線方程為y-y3=x3(x-x3)，所以A,B,F三點(diǎn)共線.方程為y=.(1)求二次函數(shù)y=x2-x+1的焦點(diǎn)坐標(biāo)和準(zhǔn)線方程；(3)設(shè)過A(4,1(的直線與拋物線y=x2-x+1的另一個(gè)交點(diǎn)為B，直線AB與直線y=x-4交于點(diǎn)P，過點(diǎn)P作x軸的垂線交拋物線y=x2-x+1于點(diǎn)N.是否存在定點(diǎn)G，使得B,N,G三點(diǎn)共線?若【詳解】(1)二次函數(shù)y=x2-x+1=(x2所以二次函數(shù)y=x2-x+1的焦點(diǎn)坐標(biāo)為F(2,1)，準(zhǔn)線方程為y=-1． 準(zhǔn)線方程為y(3)由(1)知拋物線y=x2-x+1=(x2-4x+4)=(x-2)2可以由拋物線y=x2沿向量=設(shè)B/yx-xx-x1(，化簡得y=+(x-x1)=(x-2)+--+2；(2)設(shè)A為C的左頂點(diǎn)，過F且斜率存在的直線交C的右支于M,N兩點(diǎn)，直線AM,AN分別交圓O的另一點(diǎn)于P,Q.所以雙曲線C的方程為x(x=my+2x2-=1∴y1+yy1y其中-<m<，∴-4my1y2=3(y1+y2(，∴P,O,Q三點(diǎn)共線.(ii)不妨設(shè)直線AM:x=m1y-1,AN:x=m2y-1，其中m由(i)可知m1m2=-1，∴直線OP:x=-my， :點(diǎn)D在定直線x=1上.(2)過點(diǎn)(2,0(且斜率為k(k≠0(的直線與橢圓E交于A,B兩點(diǎn)，點(diǎn)C與點(diǎn)B關(guān)于x軸對稱．在設(shè)直線AB的方程為y=k(x-2(，點(diǎn)A(x1,y1(，點(diǎn)B(x2,y2(．x2-8k2x+8k2-6=0．若在x軸上存在定點(diǎn)D(m,0(，使A,D,C三點(diǎn)共線，則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軸上存在定點(diǎn)D(3,0(，使A,C,D三點(diǎn)共線．(2)設(shè)點(diǎn)M(4,0(，斜率不為0的直線AM與C交于另一點(diǎn)B.設(shè)直線lAB:x=my+4,m設(shè)直線lAB:x=my+4,m≠0,A(x1,y1),B(x2,y2)，則D(x1,-y1(，(i)因?yàn)橄褹B中點(diǎn)的縱坐標(biāo)為-，解得m=3或(舍)，所以直線AB的斜率為.(ii)由題得F=(x2-1,y2(,F=(x1-1,-y1(，因?yàn)?x2-1((-y1(-y2(x1-1(=(my2+4-1(.(-y1(-y2(my1+4-1(=-2my1y2-3(y1+y2(=-=0，所以FBⅡFD，又FB