Question

椭圆ax^2+by^2=1,斜率为-1的平行弦中点轨迹方程为线段CD,CD斜率1/2,CD的长为√30/3,求a,b

Answer 1

平行弦AB中点P(x,y)

xA+xB=2x,yA+yB=2y,(yA-yB)/(xA-xB)=-1

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

a(xA+xB)*(xA-xB)+b(yA+yB)*(yA-yB)=0

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

ax-by=0

直线CD:y=(a/b)x,k(CD)=1/2

a/b=1/2

b=2a

CD:y=(1/2)x,x=2y

ax^2+2ay^2=1

a(2y)^2+2ay^2=1

6ay^2=1

yC-yD=2/(6a)^0.5

xC-xD=2(yC-yD)=4/(6a)^0.5

(xC-xD)^2+(yC-yD)^2=CD^2=30/9=10/3

16/(6a)+4/(6a)=10/3

a=1,b=2