Question

已知椭圆x^2/2+y^2=1,（1）过椭圆的左焦点F引椭圆的割线,求截得的弦的中点P的轨迹方程（2）求斜率为2的

Answer 1

(1)c=1,F(-1,0)

弦的AB中点P(x,y)

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

k(AB)=k(PF)

(yA-yB)/(xA-xB)=y/(x+1)

[(xA)^2/2+(yA)^2]-[(xB)^2/2+(yB)^2=1-1=0

(xA+xB)*(xA-xB)/2+(yA+yB)*(yA-yB)=0

2x/2+2y*(yA-yB)/(xA-xB)=0

0.5x+y*y/(x+1)=0

(x+0.5)^2+y^2/0.125=1

(2)

0.5x+y*2=0

x+4y=0