Question

已知椭圆x^2/a^2+y^2/b^2=1双曲线x^2/m^2-y^2/n^2=1有相同的焦点,P是它们的公共点,设角F

Answer 1

cosα=[1-tan^2(α/2)]/[1+tan^2(α/2)]

要证 tana=n/b ==== 只要证 cos2a = (b^2 - n^2)/(b^2 + n^2)

F1P + PF2 = 2a F1P - PF2 = 2m

a^2 =c^2 +b^2 m^2 = c^2 -n^2

F1P^2 + PF2^2 = 2a^2 + 2m^2 = 4c^2 + 2b^2 - 2n^2

PF1*PF2 = a^2 - m^2 = b^2 + n^2

cos2a = (F1P^2 + F2P^2 - F1F2^2)/ 2PF1*PF2 = (b^2 - n^2)/(b^2 + n^2)

得证：