Question

24、已知过点P(-2,1)的直线被椭圆x^2+2y^2 =8截得的弦AB的中点恰好为点P (1)求直线AB的方程(2)

Answer 1

[(xA)^2+2(yA)^2-[(xB)^2+2(yB)^2=2-2

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

2*(-2)+2*(2*1)*(yA-yB)/(xA-xB)=0

-1+(y-1)/(x+2)=0

AB:y=x+3

x=y-3

(y-3)^2+2y^2=8

3y^2-6y+1=0

yA+yB=2,yA*yB=1/3

(xA-xB)^2=(yA-yB)^2=(yA+yB)^2-4*yA*yB=2^2-4*1/3=8/3

AB^2=16/3

|AB|=4/√3