已知椭圆(x^2)/2+y^2=1,过点A（2,1）的直线l与椭圆相交,求l被截得的弦的中点轨迹方程.

可设直线方程为y-1=k(x-2) (k≠0,否则与椭圆相切),设两交点分别为（x1,y1）,(x2,y2),则x1^2/2+y1^2=1,x2^2/2+y2^2=1,两式相减得
(x1+x2)(x1-x2)/2+(y1+y2)(y1-y2)=0,显然x1≠x2(两点不重合),故（x1+x2）/2+(y1+y2)(y1-y2)/(x1-x2)=0,令中点坐标为（x,y）,则x+2y(y1-y2)/(x1-x2)=0,又（x,y）在直线上,所以(y-1)/(x-2)=k,显然(y1-y2)/(x1-x2)=k,故x+2y*k=x+2y(y-1)/(x-2)=0,即所求轨迹方程为x^2+2y^2-2x-2y=0.
还有一点要注意,即x,y的范围：x^2/2+Y^2