已知过点(2,(√6)/3)的椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左焦点F(-2,0)直线l过点M(-

1个回答

  • 椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左焦点F(-2,0),

    ∴a^=b^+4,①

    椭圆过点(2,√6/3),

    ∴4/a^+(2/3)/b^=1.②

    把①代入②*b^(b^+4),4b^+(2/3)(b^+4)=b^(b^+4),

    b^4-(2/3)b^-8/3=0,

    (b^-2)(b^+4/3)=0,b^=2,

    代入①,a^=6,

    ∴椭圆方程是x^/6+y^/2=1.③

    设l:x=my-3,④代入③*6,m^y^-6my+9+3y^=6,

    (m^+3)y^-6my+3=0,

    y1,2=[3m土√(6m^-9)]/(m^+3),

    代入④,x1,2=[-9土m√(6m^-9)]/(m^+3),

    看图,B(x2,y2),C(x1,-y1),

    BF的斜率k1=y2/(x2+2)=[3m-√(6m^-9)]/[2m^-3-m√(6m^-9)],

    CF的斜率k2=-y1/(x1+2)=[-3m-√(6m^-9)]/[2m^-3+m√(6m^-9)],

    [3m-√(6m^-9)][2m^-3+m√(6m^-9)]-[-3m-√(6m^-9)][2m^-3-m√(6m^-9)]

    =6m(2m^-3)+√(6m^-9)*[-2m√(6m^-9)]

    =0,

    ∴k1=k2,

    ∴B、F、C三点共线.