20:已知:三角形ABC为直角三角形,角C为直角,A(0,-8),顶点C在x轴上运动,M在y轴上,向量AM=1/2(向量

1个回答

  • (1)

    A(0,-8) ,C(c,0),M(0,m),∠C=90°,B(x,y)

    AM = (1/2)(AB+AC)

    OM-OA =(1/2)(OB-OA +OC -OA)

    (0,m+8) = (1/2)( (x,y) + (c,16))

    = (1/2)(x+c,y+16)

    => (1/2)(x+c)=0

    c = -x

    ∠C=90°

    => AC.CB =0

    (c,8).(x-c,y) =0

    c(x-c) - 8y = 0

    -x(2x) -8y =0

    x^2 + 4y =0

    E的方程:x^2 + 4y =0 (1)

    (2)

    let Q(x1,y1),N(x2,y2)

    QP=PN

    OP-OQ = ON-OP

    (2-x1,4-y1)= (x2-2,y2-4)

    => 2-x1= x2-2 and 4-y1 = y2-4

    => x1+x2 = 4 and y1+y2 = 8

    l:y= mx + c

    过点P(2,4)

    => 4 = 2m +c

    c = 4-2m

    l:y = mx+ 4-2m (2)

    sub (2) into (1)

    x^2 + 4y =0

    x^2 + 4(mx+ 4-2m) =0

    x^2 + 4mx + 16-8m =0

    x1+x2 = -4m = 4

    => m = -1

    l:y = mx+ 4-2m

    y = -x + 6