直线y=mx+1(m>0)与椭圆2x²+y²=2交于A、B两点，|AB|=6√2/5,求M
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{y=mx+1;2x²+y²=2
==>2x²+(mx+1)²=2
==>(2+m²)x²+2mx-1=0
Δ=4m²+4(m²+2)>0恒成立
设A(x1,y1),B(x2,y2)
∴x1+x2=-2m/(m²+2),x1x2=-1/(m²+2)
∴|AB|=√(1+m²)*√[(x1+x2)²-4x1x2]
=√(1+m²)*√[4m²/(m²+2)²+4/(m²+2)]
=√(1+m²)*√[[8(m²+1)/(m²+2)²]
=2√2*(1+m²)/(m²+2)=6√2/5
∴(1+m²)/(m²+2)=3/5
∴m²=1/2 ,m=±√2/2
∵m>0 ∴m=√2/2
==>2x²+(mx+1)²=2
==>(2+m²)x²+2mx-1=0
Δ=4m²+4(m²+2)>0恒成立
设A(x1,y1),B(x2,y2)
∴x1+x2=-2m/(m²+2),x1x2=-1/(m²+2)
∴|AB|=√(1+m²)*√[(x1+x2)²-4x1x2]
=√(1+m²)*√[4m²/(m²+2)²+4/(m²+2)]
=√(1+m²)*√[[8(m²+1)/(m²+2)²]
=2√2*(1+m²)/(m²+2)=6√2/5
∴(1+m²)/(m²+2)=3/5
∴m²=1/2 ,m=±√2/2
∵m>0 ∴m=√2/2
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