Question

已知椭圆C的方程为x^2/9+y^2=1,直线y=x+2交椭圆C于A,B两点,求线段AB的长及线段AB的中点

Answer 1

y=x+2,x=y-2

x^2/9+y^2=1

x^2+9y^2=9

(y-2)^2+9y^2=9

10y^2-4y-5=0

yA+yB=4/5=2/5,yA*yB=-1/2

(yA-yB)^2=(xA-xB)^2=(yA+yB)^2-4*yA*yB=66/25

AB^2=(xA-xB)^2+(yA-yB)^2=2*66/25=132/25

|AB|=2√33/5

(yA+yB)/2=(4/10)/2=1/5

(xA+xB)/2=(1/5)-2=-9/5

AB的中点(1/5,-9/5)