题目

已知椭圆：x^2+2y^2=2,1.求斜率为2的平行弦的中点轨迹方程； 2.求过点P（1/2,1/2）.且
被点P平分的弦的弦长.

提问时间：2020-10-31

答案

1,斜率为2的平行弦AB的中点M(x,y)
xA+xB=2xM=2x,yA+yB=2y
k(AB)=(yA-yB)/(xA-xB)=2
[(xA)^2+2(yA)^2]-[(xB)^2+2(yB)^2]=0
(xA+xB)+2(yA+yB)*(yA-yB)/(xA-xB)=0
2x+2y*2=0
x+y=0
2,xA+xB=1,yA+yB=1
(xA+xB)+2(yA+yB)*(yA-yB)/(xA-xB)=0
k(AB)=(yA-yB)/(xA-xB)=-(xA+xB)/[2(yA+yB)]=-1/2
AB:y-1/2=(-1/2)*(x-1/2)
x=1.5-2y
x^2+2y^2=2
(1.5-2y)^2+2y^2=2
yA*yB=1/24
(yA-yB)^2=(yA+yB)^2-4yA*yB=1-4/24=5/6
(xA-xB)^2=4(yA-yB)^2=20/6
AB^2=25/6
|AB|=5√6/6