在平面直角坐标系中A(-2,0),B(2,0)点P为动点,且直线AP与直线BP的斜率之积为-3/4 1.求动点P的轨迹C

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  • L:y=k(x-1)

    x^2/4+y^2/3=1

    3x^2+4y^2=12

    3x^2+4*[k(x-1)]^2=12

    (3+4k^2)x^2-8k^2*x+4k^2-12=0

    yM+yN=8k^2/(3+4k^2),yM*yN=(4k^2-12)/(3+4k^2)

    (yM-yN)^2=(yM+yN)^2-4yM*yN=

    (y1-y2)^2=(y1+y2)^2-4y1*y2=

    |y1-y2|=√[(y1+y2)^2-4y1*y2]

    S△MON

    =S△MOD+S△NOD

    =(1/2)底*|OD|*高|yM| +(1/2)底*|OD|*高|yN|

    =(1/2)*|OD|*|yM|+(1/2)*|OD|*|yN|

    =(1/2)*|OD|*(|yM|+|yN|)

    =(1/2)*1*|yM-yN|

    =(1/2)*|yM-yN|

    =(1/2)*|y1-y2|

    =(1/2)*√[(y1+y2)^2-4y1*y2]