一道与椭圆有关的问题,已知椭圆x^2/a^2 +y^2/b^2=1(a>b>0)的离心率为根号3/2,过右焦点F且斜率为

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  • B.根号2

    已知椭圆x^2/a^2 +y^2/b^2=1(a>b>0)的离心率为根号3/2,设椭圆方程x^2/4+y^2=1

    设直线方程为x=my+根号3 代入椭圆方程得

    (m^2+4)y^2+2根号3y-1=0

    若AF=3FB

    所以|yA|=3|yB|

    解方程(m^2+4)y^2+2根号3y-1=0

    y=[-2根号3m±√(12m^2+4(m^2+4))] /[2(m^2+4)]

    m>0 yA=[-2根号3m-√(12m^2+4(m^2+4))] /[2(m^2+4)]

    yB=[-2根号3m+√(12m^2+4(m^2+4))] /[2(m^2+4)]

    |yA|=3|yB|

    |[-2根号3m-√(12m^2+4(m^2+4))] /[2(m^2+4)]|=3|[-2根号3m+√(12m^2+4(m^2+4))] /[2(m^2+4)]

    2根号3m+√(12m^2+4(m^2+4))=3 √[12m^2+4(m^2+4))] -6根号3m

    8根号3m=2√(16m^2+16) 64*3*m^2=64(m^2+1) 2m^2=1

    m=根号2/2 k=1/m=根号2