一道关于椭圆的题！在线等！！

x^2/25+y^/16=1
a=5,b=4,c=3
F1(-3,0)
设P(x1,y1),Q(x2,y2),
把直线PQ方程：y=k(x+3)代人x^2/25+y^/16=1得：
x^2/25+k^2(x+3)^2/16=1
(16+25k^2)x^2+150k^2x+225k^2-400=0
x1+x2=-150k^2/(16+25k^2)
中点横坐标=(x1+x2)/2=-75k^2/(16+25k^2)
由y=k(x+3)知：k=y/(x+3)
代人：x=-75k^2/(16+25k^2)得：
x=-(75y^2/(x+3)^2) / (16+25y^2/(x+3)^2)
=-75y^2/(16(x+3)^2+25y^2)
x(16(x+3)^2+25y^2)+75y^2=0
这就是PQ中点的轨迹方程