设P(x,y)为椭圆x^2/a^2+y^2/b^2=1上任一点F1(-c,0)F2(c,o)为焦点,∠PF1F2=a,∠PF2F1=b(1).求tan（a/2）tan（b/2）的值

Question

设P(x,y)为椭圆x^2/a^2+y^2/b^2=1上任一点

F1(-c,0) F2(c,o)为焦点,∠PF1F2=a,∠PF2F1=b(1).求tan（a/2）tan（b/2）的值

Answer 1

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)