2=、(x1+x2)2-4x1x1|kx0-y0+m|11y2-x22y2-x22EQ\*jc3\*hps21\o\al(\s\up6(y),2)-EQ\*jc3\*hps21\o\al(\s\up5(x),2)2=1EQ\*jc3\*hps21\o\al(\s\up5(c),a)EQ\*jc3\*hps21\o\al(\s\up5(2),a)2-a2=2EQ\*jc3\*hps21\o\al(\s\up6(y),2)EQ\*jc3\*hps21\o\al(\s\up5(x),2)所以m2≠1且Δ=16m2-24(m2-1(=2|m2-1| 、、3-m22|m2-1| 、、3-m2|m2-1||m2-1|1+m2所以直線l的方程為x=2y+2或x=-2y+2.EQ\*jc3\*hps22\o\al(\s\up9(—→),A)EQ\*jc3\*hps22\o\al(\s\up9(—→),B)22EQ\*jc3\*hps21\o\al(\s\up9(—→),A)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up7(—→),A)EQ\*jc3\*hps21\o\al(\s\up7(—→),B)P(t,0)，所以PEQ\*jc3\*hps21\o\al(\s\up9(—→),A).PEQ\*jc3\*hps21\o\al(\s\up9(—→),B)=(x1-t)(x2-t)+y1y2=(my1+2-t)(my2+2-t)+y1y2y2+m(2-t)(y1+y2)+(2-t)2=(m2+1)×-m(2-t)×+(2-t)2=-+(2-t)2．0)滿足PEQ\*jc3\*hps21\o\al(\s\up8(→),A).PEQ\*jc3\*hps21\o\al(\s\up8(—→),B)=-1．EQ\*jc3\*hps21\o\al(\s\up9(—→),A)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)PEQ\*jc3\*hps21\o\al(\s\up8(—→),A).PEQ\*jc3\*hps21\o\al(\s\up8(—→),B)=-1．EQ\*jc3\*hps21\o\al(\s\up9(—→),A)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)33所以C的方程為x2-=1；(2)由方程得T(-1,0(,F(2,0(,設直線AB:x=my+2,A(x1,y1(,B(x2,y2(，(x=my+2,x2-=1,可得(3m2-1(y2+12my+9=0，-36(3m2-1(>044點M的坐標；(2),或(-,-則圓心F1(-1,0(到直線3x-y=0的距離為，y1+y23x1-x2，作差可得x1+x2y1+y23x1-x2，EQ\*jc3\*hps15\o\al(\s\up3(2),0)所以點M的坐標為,或(-,-.(y=kx+455 所以S△ODE=×4×|x3-x4|=2|x3-x4|=2、(x3+x4(2-4x3x4 (2)若不過點A(2,0)的直線l交曲線M于P，Q兩點；EQ\*jc3\*hps21\o\al(\s\up9(—→),P)EQ\*jc3\*hps21\o\al(\s\up9(—→),Q)662mky+m2-4=0，①由題意AEQ\*jc3\*hps21\o\al(\s\up9(—→),P).AEQ\*jc3\*hps21\o\al(\s\up9(—→),Q)=(x1-2,y1).(x2-2,y2)=x1x2-2(x1+x2)+4+y1y2y2+k(m-2)(y1+y2)+m2-4m+422近線方程為x-3y=0.EQ\*jc3\*hps16\o\al(\s\up5(a),b)EQ\*jc3\*hps12\o\al(\s\up8(2),77EQ\*jc3\*hps16\o\al(\s\up15(2),b)EQ\*jc3\*hps12\o\al(\s\up8(2),3)EQ\*jc3\*hps21\o\al(\s\up6(a),b)x23x23消去y并整理，得(1-3k2(x2-6kmx-3m2-3=0，1-3k2((3m2+3(=12(m2-3k2+1(>0，EQ\*jc3\*hps21\o\al(\s\up6(m),3k)EQ\*jc3\*hps21\o\al(\s\up6(3),1)EQ\*jc3\*hps21\o\al(\s\up6(+),3k)EQ\*jc3\*hps21\o\al(\s\up6(3),2),0(，=-(kx1+m((x0-x2(-(kx2+m((x0-x1(=(x0-x1((x0-x2(=2kx1x2-(kx0-m((x1+x2(-2mx0=xEQ\*jc3\*hps12\o\al(\s\up3(2),0)-(x1+x2(x0+x1x2=-2k.EQ\*jc3\*hps16\o\al(\s\up4(3),1)m-2EQ\*jc3\*hps16\o\al(\s\up4(+),3k)EQ\*jc3\*hps16\o\al(\s\up4(3),2)-(kx0-m(.16-kEQ\*jc3\*hps16\o\al(\s\up4(m),3k)2-2mx0EQ\*jc3\*hps12\o\al(\s\up3(2),0)EQ\*jc3\*hps16\o\al(\s\up5(6),1)EQ\*jc3\*hps12\o\al(\s\up4(x0),k2)EQ\*jc3\*hps16\o\al(\s\up4(3),1)EQ\*jc3\*hps16\o\al(\s\up4(+),3k)EQ\*jc3\*hps16\o\al(\s\up4(3),2)-6k-2mx0=-3xEQ\*jc3\*hps12\o\al(\s\up3(2),0)k2-6mx0k-3m2-3+xEQ\*jc3\*hps12\o\al(\s\up3(2),0)=-6k-2(1-3k(x0=-3xEQ\*jc3\*hps12\o\al(\s\up3(2),0)k2-6(1-3k(x0k-3(1-3k)2-3+xEQ\*jc3\*hps12\o\al(\s\up3(2),0)(23x0-6(k-2x-3(x0-3(2k2+(63-6x0(k-6+xEQ\*jc3\*hps12\o\al(\s\up3(2),0)，88定直線上.(2)N(-1,0)或N(-1,-4)(2)設l(2)設l，A(x1,y1),B(x2,y2)，N(-1,n)，99由y1+y2=4t,y1y2=-4，代入可得2+4n)=0，解得n=0或n=-4，則直線直線解得x=-1，即直線AP與BQ的交點在一條定直線x=-1上從而簡化運算.,((EQ\*jc3\*hps21\o\al(\s\up17(a),c)2-c2=1，所以P，Q均在直線x+my-1=0上，即lPQ:x+my-1=0，EQ\*jc3\*hps21\o\al(\s\up6(x),y)EQ\*jc3\*hps21\o\al(\s\up6(1),0)EQ\*jc3\*hps21\o\al(\s\up6(x),y)EQ\*jc3\*hps21\o\al(\s\up6(1),0)EQ\*jc3\*hps21\o\al(\s\up5(1),m)12×|y2-yN|11EQ\*jc3\*hps16\o\al(\s\up5(1),k)12×|y2-yN|11EQ\*jc3\*hps16\o\al(\s\up5(1),k)2×|y1-yN|+=1+(k1EQ\*jc3\*hps16\o\al(\s\up5(1),1EQ\*jc3\*hps16\o\al(\s\up5(1),k)2EQ\*jc3\*hps16\o\al(\s\up4(1),k)2×|y1-yN|y1y2|x1EQ\*jc3\*hps12\o\al(\s\up3(2),1)EQ\*jc3\*hps12\o\al(\s\up3(2),2) EQ\*jc3\*hps21\o\al(\s\up3(1),2)EQ\*jc3\*hps15\o\al(\s\up7(b2),c2)EQ\*jc3\*hps21\o\al(\s\up6((2)過橢圓C的右焦點F2作直線交橢圓于M，N兩點.直線設A(x1,y1),B(x2,y2),不妨設A在第一象限，B在第三象限，-9=0，顯然Δ>0，設點M(x3,y3),N(x4,y4)，于是直線NH的方程為由圖形的對稱性(交換點M和點N的位置)可知直線NH必經過x軸上的一個定點， 5即直線NH經過定點E,0(.故直線NH經過定點E,0(.解證明.的右支于E,F兩點，其中點E在x軸上方，直線EA與FB交得出結論.F所以直線F1Q的方程為x=-5；設E(x1,y1(,F(x2,y2(， 則直線EA:y=(x+1(，直線FBmy1y2+(y1+y2=-1 +y2=-1所以點P到直線F1Q的距離為-(-5(=+5，所以點P到直線F1Q的距離為定值.PEQ\*jc3\*hps22\o\al(\s\up9(—→),F)EQ\*jc3\*hps22\o\al(\s\up9(—→),F)EQ\*jc3\*hps21\o\al(\s\up8(—→),F).PEQ\*jc3\*hps21\o\al(\s\up8(—→),F)EQ\*jc3\*hps21\o\al(\s\up8(—→),F)EQ\*jc3\*hps21\o\al(\s\up8(—→),F)2EQ\*jc3\*hps21\o\al(\s\up8(—→),O)EQ\*jc3\*hps21\o\al(\s\up8(—→),F)2-c2=-2求出c2,a2即可；|QT|表達出即可.-1PEQ\*jc3\*hps21\o\al(\s\up8(→),F1).PEQ\*jc3\*hps21\o\al(\s\up8(—→),F)2EQ\*jc3\*hps21\o\al(\s\up8(—→),O)EQ\*jc3\*hps21\o\al(\s\up8(—→),F)(.(PEQ\*jc3\*hps21\o\al(\s\up8(—→),O)+OEQ\*jc3\*hps21\o\al(\s\up8(—→),F)2(=PEQ\*jc3\*hps21\o\al(\s\up8(—→),O)2-OEQ\*jc3\*hps21\o\al(\s\up8(—→),F)12=PEQ\*jc3\*hps21\o\al(\s\up8(—→),O)2-c2≥b2-c2，(y=kx+m((y=kx+m 2m+1)x1+(1-m2)x2(1-m2)x1+(m2-2m+1)x2，記y=2與y軸的交點為T(0,2)，線l過點P．可求解.4x的焦點坐標為F(1,0)，準線方程為x=-1，則P(EQ\*jc3\*hps21\o\al(\s\up6(y),x)EQ\*jc3\*hps21\o\al(\s\up6(4),m)EQ\*jc3\*hps21\o\al(\s\up6(x),y)-4my+4=0．此時l的方程為x=y-1或x=-y-1．EQ\*jc3\*hps21\o\al(\s\up11(—→),A)EQ\*jc3\*hps21\o\al(\s\up11(—→),B)EQ\*jc3\*hps21\o\al(\s\up11(—→),A)所以△FAB的面積S△FAB=S△FAP-S△FPB1+y2)2-4y1y2=16m2-16=4.和上頂點B．(2)與直線AB平行的直線l交C于M,N兩點(M,N均不與C的頂點重合)，設直線AM，BN的斜率-b22-b2=5由(1)知直線AB的斜率為-，設直線l的方程為y=-x+t，M(x1,y1(，N(x2,y2(，EQ\*jc3\*hps21\o\al(\s\up6(3),2)t．y1y2x1-3x2x1-3x2x1x2-3x2因為y1y2-2y1=(-EQ\*jc3\*hps21\o\al(\s\up6(2),3)x1+t((-EQ\*jc3\*hps21\o\al(\s\up6(2),3)x2+t(-2(-EQ\*jc3\*hps21\o\al(\s\up6(2),3)x1+t(=t2-EQ\*jc3\*hps21\o\al(\s\up6(2),3)t(x1+x2(+EQ\*jc3\*hps21\o\al(\s\up6(4),9)x1x2-2t+EQ\*jc3\*hps21\o\al(\s\up6(4),3)x1，且t=EQ\*jc3\*hps21\o\al(\s\up5(2),3)(x1+x2(，所以y1y2-2y1=EQ\*jc3\*hps21\o\al(\s\up5(4),9)(x1x2EQ\*jc3\*hps21\o\al(\s\up5(4),9)3.(2024高三·全國·專題練習)已知雙曲線C:EQ\*jc3\*hps16\o\al(\s\up7(x2),a2)-EQ\*jc3\*hps22\o\al(\s\up7(y),b)EQ\*jc3\*hps16\o\al(\s\up11(2),2)=1(a>0,b>0(過點M(2,3(，且M到直線l:x=-a的距離為5EQ\*jc3\*hps21\o\al(\s\up7(y),3)EQ\*jc3\*hps21\o\al(\s\up5(3),1)EQ\*jc3\*hps21\o\al(\s\up5(6),3)aa2aa2=222y22=y2故雙曲線C的標準方程為x2-=-2,0(,F2(2,0(,l:x=-EQ\*jc3\*hps21\o\al(\s\up5(1),2)．(x=my-2EQ\*jc3\*hps16\o\al(\s\up2147483647(y),3)-1(y2-12my+9=0，(3m2-1≠093m2-13m2-1,y1y2=(3m2-1≠093m2-13m2-1,y1y2=x-x2(．令y=0，得到直線BP與x軸的交點橫坐標x=x2-== 5(my2-2(y1-(my1-2(y2+2y24my1y2-10(y1+y 2(my1-2(y2-4y2+5y12my1y2+5(y1+y2(-13y236m-120m+14(3m2-1(y2=-84m+14(3m2-1(y2=-1418m+60m-13(3m2-1(y278m-13(3m2-1(y213，所以直線BP過定點(-,0(．故△ABD的面積的最小值為．4.(24-25高三上·上海浦東新·期中)已知A(EQ\*jc3\*hps22\o\al(\s\up9(—→),M)EQ\*jc3\*hps22\o\al(\s\up9(—→),N)(3)-,3EQ\*jc3\*hps21\o\al(\s\up9(—→),M)EQ\*jc3\*hps21\o\al(\s\up9(—→),N)依題解得設P(x1,y1(，B(x2,y2(EQ\*jc3\*hps21\o\al(\s\up20(3),1)EQ\*jc3\*hps20\o\al(\s\up24(3),2)消去y整理可得，(4k2+3(x2-(24k2-12k(x+36k2-36k-27=0則直線l的方程為y=x-3或y=x-3；設M(x1,y1(，N(x2,y2(，T(0,t(，r3x設M(x1,y1(，N(x2,y2(，T(0,t(，r3x而TEQ\*jc3\*hps21\o\al(\s\up9(—→),M)=(x1,y1-t(，TEQ\*jc3\*hps21\o\al(\s\up9(—→),N)=(x2,y2-t(，故TEQ\*jc3\*hps21\o\al(\s\up6(—→),M).TEQ\*jc3\*hps21\o\al(\s\up6(—→),N)=x1x2+(y1-t((y2-t(=x1x2+(kx1+-t((kx2+-t(=(1+k2(x1x2+k-t((x1+x2(+-t(2=(1+k2(×-t(×+-t(2=(-36+4t2(k2--9t+3t2EQ\*jc3\*hps21\o\al(\s\up8(—→),M)EQ\*jc3\*hps21\o\al(\s\up8(—→),N)若過點D(0,的動直線的斜率不存在，則M(-3,-N(3,，此時需-3≤t≤3，兩者結合可得-≤t≤3故這個點T縱坐標的取值范圍為-,3.2-a2=9-4=5，,y1(EQ\*jc3\*hps21\o\al(\s\up7(x=),5x2)EQ\*jc3\*hps21\o\al(\s\up7(y),4)-y1(my2+6(-3y2(my1+2(-4my1y2-6(y1+y23(x1-2((x2+2(3(x1-2((x2+2(=5m-45m-=5m-45m-4=03(x1-2((x2+2(聯立可得(m2+4(y2+2mx0y+EQ\*jc3\*hps15\o\al(\s\up3(2),0)則且BC與直線x=-2交于點E，F．證明：以4x(x≥0(或y=0(x<0)EQ\*jc3\*hps21\o\al(\s\up9(—→),T)EQ\*jc3\*hps21\o\al(\s\up9(—→),T)所以動點P的軌跡方程為y2=4x(x≥0(或y=0(x<0)．直線AC的方程為∴E(-2,，同理可得F(-2,，以線段EF為直徑的圓與x軸交點設為T(a,0(，則且∴a=22-2或a=-22-2，∴以EF為直徑的圓在x軸上截得的弦長為定值|(22-2(-(-22-2(|=42．點P.NE交直線MF2于點Q2-c2=2-n2+9=0①,△MPR=2S△OPM，故設直線DE的方程為x=ty+9,D(x1,y1(,E(x2,y2(，所以4(y1+y2(=-ty1y2，所以4y1=-ty1y2-4y2，故DQ丄y軸.方法二：設D(x1,y1(,E(x2,y2(,DP=EQ\*jc3\*hps21\o\al(\s\up7(λ),λ)EQ\*jc3\*hps21\o\al(\s\up7(x),y)EQ\*jc3\*hps15\o\al(\s\up5(2),2)EQ\*jc3\*hps21\o\al(\s\up7(+),y)EQ\*jc3\*hps21\o\al(\s\up9(8),8)x(EQ\*jc3\*hps21\o\al(\s\up9(+),x)2EQ\*jc3\*hps21\o\al(\s\up9(=),9)EQ\*jc3\*hps21\o\al(\s\up9(2),y)2+λx2((x1-λx2(+9(y1+λy2((y1-λy2(=72(1+λ((1-λ(②,又P(9,0(,F2(1,0(,N(5,0(,E(x2,y2(，則直線NE的方程為y-0=-(x-5(,(x2≠5(9.(24-25高三上·安徽銅陵·期末)圓E上.(2)若過點A(2,1(的直線l與橢圓E交(ii)當P為AC中點時，求直線AP的方程.(2)(i)證明見解析；(ii)x-y-1=0或x-5y+3=0.-2=0， APPC APPC|BC|即證(x0-2)(x2-x1)=2(x1-2)(x2-x0)，整理得(x0EQ\*jc3\*hps21\o\al(\s\up7(AB),BC)EQ\*jc3\*hps21\o\al(\s\up7(AP),PC)(2)-EQ\*jc3\*hps15\o\al(\s\up3(2),0)EQ\*jc3\*hps15\o\al(\s\up3(2),0)EQ\*jc3\*hps15\o\al(\s\up3(2),0)=-．(x=my+n+y2=1得(m2+4(y2+2mny+n2-4=0，Δ=4m2n2-4(m2+4((n2-4(=16m2-16n2+64(x=my+n解得n=-1或n=-2(舍)．2y-23=-15y1EQ\*jc3\*hps21\o\al(\s\up6(a),b)2y-2，