Question

已知椭圆x^2/16+y^2/9=1.斜率为1/2的直线与椭圆相交.求直线截椭圆所得弦的中点轨迹方程.

Answer 1

弦AB的中点M(x,y),k(AB)=1/2

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

k(AB)=(yA-yB)/(xA-xB)=1/2

x^2/16+y^2/9=1

9x^2+16y^2=144

[9(xA)^2+16(yA)^2]-[9(xB)^2+16(yB)^2]=144-144=0

9(xA+xB)*(xA-xB)+16(yA+yB)*(yA-yB)=0

9*2x+16*2y*(yA-yB)/(xA-xB)=0

9x+16y*(1/2)=0

直线截椭圆所得弦的中点轨迹方程:

9x+8y=0