已知直线l：y=kx+b与椭圆C：3x^2+y^2=3相交于A、B两点,M是的线段AB的中点,O是坐标原点

1）
l和直线x+y=0平行
k=-1
l:y=-x+b
3x^2+(b-x)^2=3
4x^2-2bx=3
x1+x2=2b/4=b/2
Mx=(x1+x2)/2=b/4
My=-Mx+b=3b/4
OM 斜率k'=My/Mx=3
2
3x^2+(kx+b)^2=3
(3+k^2)x^2+2bkx=3-b^2
x1+x2=-2bk/((3+k^2)
Mx=(x1+x2)/2=-bk/(3+k^2)
My=kMx+b=-bk^2/(3+k^2)+b
|OM|^2=Mx^2+My^2
|OM|=1
(bk)^2/(3+k^2)^2 +[-bk^2/(3+k^2)+b]^2=1
b^2k^2+b^2k^4 -2b^2k^4(3+k^2)+b^2(3+k^2)^2=(3+k^2)^2
b^2=(k^2+3)^2/(k^2+9)
3
求线段AD?长

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