Question

直线y=mx+1(m>0)与椭圆2x2+y2=2交与AB两点弦长为6根号2/5,求m

Answer 1

解：直线代入椭圆：（2+m²)x²+2mx-1=0
设A(x1,y1)、B(x2,y2)。则 x1+x2=-2m/(m²+2) x1x2=-1/(m²+2)
y1+y2=m(x1+x2)+2=4/(m²+2) y1y2=m²x1x2+m(x1+x2)+1=(-2m²+2)/(m²+2)
∵AB两点弦长为6根号2/5
∴(x1-x2)²+(y1-y2)²=72/25
(x1+x2)²-4x1x2+(y1+y2)²-4y1y2=72/25
(8m²+8)/(m²+2)²+(8m^4+8m²)/(m²+2)²=72/25
16m^4+14m²-11=0
∴m²=1/2(-11/8舍去）
∴m＝±√2／2

Answer 2

先把直线与椭圆方程联立（m2+2)x2+2mx-1=0
有椭圆的距离公式d = √(1+k^2)|x1-x2| = √(1+k^2)[(x1+x2)^2 - 4x1x2],=6根号2/5,,,,,,k=m
由韦达定理x1+x2=-b/a,,x1x2=c/a 得x1+x2=-2m/(m2+2),,,x1x2=-1/(m2+2)
代入距离公式得到方程化简为(m2+1)/(m2+2)=3/5,,,m>0,,,,所以,m=根号2/2