一道高二数学题(求点的轨迹方程)

1个回答

  • 直线L过定点A(0,3)

    曲线C:y^2= 4x

    Let P(x,y),P(x1,y1),P2(x2,y2)

    P is mid point of P1P2,=>

    x1+x2 = 2x,y1+y2=2y

    P1,P2 is on C,then

    y1^2=4x1 and y2^2=4x2

    => 4(x2-x1)= y2^2-y1^2

    => (y2-y1)/(x2-x1) = 4/(y2+y1)

    => (y2-y1)/(x2-x1) = 4/2y

    slope of AP = (y-3)/x

    slope of P1P2 = (y2-y1)/(x2-x1)

    AP 垂至于P1P2 => slope of AP x slope of P!P2 = -1 =>

    ((y-3)/x) ( y2-y1)/(x2-x1) = -1

    ((y-3)/x) (4/2y) = -1

    2(y-3) = -xy

    xy-2y-6 = 0

    求直线L与动弦p1p2的交点M的轨迹方程:

    xy-2y-6 = 0