设动点M的坐标为(x,y),B(x,-3),OA=(0,-1);则
MA= (-x,-1-y);AB=(x,-2);MB=(0,-3-y);BA= (-x,2);
∵MA•AB=MB•BA,故有等式:-x²-2(-1-y)=2(-3-y),即有y=(1/4)x²-2,就是动点M的轨迹C的方程.
设动点M的坐标为(x,y),B(x,-3),OA=(0,-1);则
MA= (-x,-1-y);AB=(x,-2);MB=(0,-3-y);BA= (-x,2);
∵MA•AB=MB•BA,故有等式:-x²-2(-1-y)=2(-3-y),即有y=(1/4)x²-2,就是动点M的轨迹C的方程.