Question

如图3-3-3,A₁A₂=2a,F₁F₂=2c,求证: =.

图3-3-3

Answer 1

证明:连结O₁O₂、O₁F₁、O₂F₂、O₁Q₁、O₂Q₂.

∵直线F₁F₂与Dandelin双球相切,

∴O₁F₁⊥F₁F₂,O₂F₂⊥F₁F₂.

过O₁作O₁H⊥O₂F₂于H,

则O₁HF₂F₁是矩形,

∴O₁H=F₁F₂=2c.

又平面π与圆锥的轴的夹角为β,

∴O₁O₂=.                                                ①

由双曲线的定义,知双曲线上任意一点到F₁、F₂的距离之差为定值Q₁Q₂,故取如图位置.

由切线长定理,

A₂Q₁=A₂F₁=A₁F₂.

∴F₁F₂-A₂F₁=F₁F₂-A₁F₂.∴A₂F₂=A₁F₁.

又A₂Q₂=A₂F₂,

∴Q₁Q₂=A₂Q₁-A₂Q₂

=A₂F₁-A₁F₁

=A₁A₂=2a.

∵母线与轴夹角为α,即∠O₂OQ₂=α,

∴O₁O₂=O₁O+OO₂

=

=.                                         ②

由①②,∴.∴==e.