已知点Q是圆C:(x-4)^2+y^2=9上的动点,A(-2,3),且向量AP=1/2向量PQ

1个回答

  • (1) 设P(x,y)Q(xq,yq)

    AP=(x+2,y-3)

    PQ=(xq-x,yq-y)

    AP=1/2PQ

    x+2=1/2(xq-x)

    y-3=1/2(yq-y)

    xq=3x+4

    yq=3y-6

    Q 在圆C上

    (3x+4-4)^2+(3y-6)^2=9

    即x^2+(y-2)^2=1

    (2)设P点圆轨迹的圆心为D(0,2),半径长为r

    MN中点为E

    OM*ON=(OE+EM)*(OE+EN)=OE^2+OE(EM+EN)+EM*EN=OE^2-EM^2

    =OE^2-(r^2-DE^2)=OE^2+DE^2-1

    设E(x1,y1)M(xm,ym)N(xn,yn)

    x1=(xn+xm)/2

    直线代人圆方程得

    x^2+(x+m-2)^2=1

    2x^2+2(m-2)x+(m-2)^2-1=0

    x1=(xn+xm)/2=(2-m)/2

    代人直线方程得y1=(m+2)/2

    OM*ON=OE^2+DE^2-1=x1^2+y1^2+x1^2+(y1-2)^2-1

    =1/4((m-2)^2+(m+2)^2+(m-2)^2+(m-2)^2)-1

    =1/4(4m^2-8m+16)-1=m^2-2m+3=(m-1)^2+2>=2

    等号当m=1是成立

    m=1时,E 点坐标为(1/2,3/2)

    (1/2)^2+(3/2-2)^2=1/2