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,0)为焦点,∠PF1F2=α,∠PF2F1=β求证离心率e=sin(α+β)/(sinα+sinβ),分析：设|PF1|=r1,|PF2|=r2由正弦定理得r1/sinα=r2/sinβ=2c/sin(α+β)得(r1+r2）/(sinα+sinβ)=2c/si

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,0)为焦点,∠PF1F2=α,∠PF2F1=β
求证离心率e =sin(α+β)/(sinα+sinβ),
分析：设|PF1|=r1,|PF2|=r2由正弦定理得r1/sinα=r2/sinβ=2c/sin(α+β)
得(r1+r2）/(sinα+sinβ)=2c/sin(α+β),e=c/a=sin(α+β)/(sinα+sinβ)
最后一步是怎么得出来的啊为什么知道(r1+r2）/(sinα+sinβ)=2c/sin(α+β),

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)
以上回答你满意么?