Question

帮我解道题，谢了第4题

Answer 1

C：√30/3

AB的中点M(1,1)
xA+xB=2xM=2*1=2
设直线AB的方程为:y-1=k(x-1),
y=kx+1-k，代入x^2+2y^2=4得
x^2+2(kx+1-k)^2=4
(1+2k^2)x^2+4k(1-k)x-2-4k+2k^2=0
xA+xB=-4k(1-k)/(1+2k^2)=2
k=-0.5
xA*xB=(-2-4k+2k^2)/(1+2k^2)=1/3
(xA-xB)^2
=(xA+xB)^2-4xA*xB
=2^2-4*1/3
=8/3
(yA-yB)^2
=k^2*(xA-xB)^2
=（0.5）^2*8/3
=2/3

AB^2=(xA-xB)^2+(yA-yB)^2
=8/3+2/3
=10/3
AB=√(10/3)
AB=√30/3