过椭圆x^2/5+y^2/4=1的左焦点作椭圆的弦,求弦中点的轨迹方程.

a^2=5,b^2=4,c^2=a^2-b^2=1。故左焦点坐标是:F1(-1,0)。设弦AB坐标分别是:A(X1,Y1),B(X2,Y2),中点P坐标是(X,Y)那么有:X1+X2=2X,Y1+Y2=2Y又:x1^2/5+y1^2/4=1x2^2/5+y2^2/4=1二式相减得:(x1+x2)(x1-x2)/5+(y1+y2)(y1-y2)/4=0即AB的斜率k=(y1-y2)/(x1-x2)=-(2x/5)/(2y/4)=-4x/5y。又k=k(PF1)=(y-0)/(x+1)故有:-4x/5y=y/(x+1)即:5y^2=-4x(x+1)化简得:4x^2+5y^2+4x=0。