题目

已知椭圆x^2/2+y^2=1,求(1)斜率为2的平行弦中点的轨迹方程;(2)过A(2,1)引椭圆割线,求截得弦中点的轨迹方程;(3)过点P(1/2,1/2)且被P平分的弦中点的直线方程

提问时间：2020-07-25

答案

求(1)斜率为2的平行弦AB中点P的轨迹方程
中点P(x,y)
(yA-yB)/(xA-xB)=2
xA+xB=2x,yA+yB=2y
x^2/2+y^2=1
x^2+2y^2=2
(xA)^2+2(yA)^2=2.(1)
(xB)^2+2(yB)^2=2.(2)
(1)-(2):
(xA+xB)*(xA-xB)+2(yA+yB)*(yA-yB)=0
(xA+xB)+2(yA+yB)*(yA-yB)/(xA-xB)=0
2x+2*2y*2=0
弦中点的轨迹方程:x+4y=0
(2)过A(2,1)引椭圆割线BC,求截得弦中点P的轨迹方程
P(x,y)
xB+xC=2x,yB+yC=2y
k(BC)=(yB-yC)/(xB-xC)=(yA-yP)/(xA-xP)=(1-y)/(2-x)
(xB+xC)+2(yB+yC)*(yB-yC)/(xB-xC)=0
2x+2*2y*(1-y)/(2-x)=0
(x-1)^2+2*(y-0.5)^2=1.5
弦中点的轨迹方程椭圆:(x-1)^2/1.5+(y-0.5)/0.75=1
(3)过点P(1/2,1/2)且被P平分的弦AB的直线方程
xA+xB=2*1/2=1
yA+yB=1
(yA-yB)/(xA-xB=(y-1/2)/(x-1/2)
(xA+xB)+2(yA+yB)*(yA-yB)/(xA-xB)=0
1+2*1*(y-1/2)/(x-1/2)=0
2x-4y-3=0