正方形ABCD的边长为2,动点P到该正方形两组对边距离的积相等,求动

正方形ABCD的边长为2,动点P到该正方形两组对边距离的积相等,求动点P的轨迹方程.

答案

解:以正方形的中心为原点O,两坐标轴分别平行于正方形的两边建立直角坐标系,取ABCD的坐标分别为A(1,1)、B(-1,1)、C(-1,-1)、D(1,-1).设动点P的坐标为(xy),则P到两组对边的距离分别为|x-1|、|x+1|、|y-1|、|y+1|.

由题意,得|x-1||x+1|=|y-1||y+1|.

x2-1=±(y2-1),即x±y=0或x2y2=2为所求.

点评:此题也可以正方形两邻边所在的直线为坐标轴建立直角坐标系进行求解.

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