Question

一道椭圆问题过椭圆x^2/25+y^2/9=1的右焦点F作直线l,交椭圆于A,B两点,求线段AB终点P的轨迹方程

Answer 1

椭圆x^2/25+y^2/9=1

a^2=25,b^2=9,c=4

右焦点F(4,0)

设P(x,y),则

k(L)=(yA-yB)/(xA-xB)=y/(x-4)

xA+xB=2x,yA+yB=2y

x^2/25+y^2/9=1

9x^2+25y^2=225

9xA^2+25yA^2=225.(1)

9xB^2+25yB^2=225.(2)

(1)-(2),得

9(xA+xB)*(xA-xB)+25(yA+yB)*(yA-yB)=0

9(xA+xB)+25(yA+yB)*(yA-yB)/(xA-xB)=0

9*2x+25*2y*y/(x-4)=0

9x(x-4)+25y^2=0

9(x^2-4x)+25y^2=0

9(x-2)^2+25y^2=36

线段AB终点P的轨迹方程是椭圆:

(x-2)^2/4+y^2/(36/25)=1