设P(x,y)为椭圆x^2/a^2+y^2/b^2=1上任一点

2个回答

  • tan(A/2)

    =sin(A/2)/cos(A/2)

    =[2sin(A/2)cos(A/2)]/[2cos(A/2)cos(A/2)]

    =sinA/(cosA+1)

    tan(B/2)

    =sin(B/2)/cos(B/2)

    =[2sin(B/2)sin(B/2)]/[2sin(B/2)cos(B/2)]

    =(1-cosB)/sinB

    设|PF2|=t 则依椭圆定义知|PF1|=2a-t |F1F2|=2c

    由正弦定理

    sinA/sinB=|PF2|/|PF1|=t/(2a-t)

    由余弦定理

    cosA=[(2a-t)^2+4c^2-t^2]/[2*2c*(2a-t)]

    =(c^2+a^2-at)/[c*(2a-t)]

    cosB=[t^2+4c^2-(2a-t)^2]/[2*2c*t]

    =(c^2+at-a^2)/(c*t)

    tan(A/2)*tan(B/2)

    =(sinA/sinB)*[(1-cosB)/(cosA+1)]

    =-(c^2+at-a^2-ct)/(c^2+a^2-at+2ac-ct)

    =-[(c-a)(c+a-t)]/[(c+a)(c+a-t)]

    =(a-c)/(c+a)