Question

过点P(-1,1)作直线与椭圆x^2/4+y^2/2=1交于AB两点,若线段AB的中点为P,则弦长|AB|=

求过程

Answer 1

xA+xB=2xP=-2,
yA+yB=2yP=2
k(AB)=(yA-yB)/(xA-xB)
x^2/4+y^2/2=1
x^2+2y^2=4
[(xA)^2+2(yA)^2]-[(xB)^2+2(yB)^2]=4-4=0
(xA+xB)*(xA-xB)+2(yA+yB)*(yA-yB)=0
(xA+xB)+2(yA+yB)*(yA-yB)/(xA-xB)=0
-2+2*2k(AB)=0
k(AB)=0.5
直线的方程:x-2y-3=0
x=2y+3
x^2+2y^2=4
(2y+3)^2+2y^2=4
6y^2+12y+5=0
yA+yB=-2,
yA*yB=5/6
(yA-yB)^2=(yA+yB)^2-4yA*yB=4-4*5/6=2/3
(xA-xB)^2=4(yA-yB)^2
AB^2=(xA-xB)^2+(yA-yB)^2=5(yA-yB)^2=10/3
|AB|=√(10/3)