解:(1)由已知e2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853305216.gif)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853320250.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853305216.gif)
,故a2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853305216.gif)
c2, ①
从而b2=c2-a2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853539212.gif)
c2, ②
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853554227.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853570335.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853585209.gif)
,设∠AOF=∠BOF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853601193.gif)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853617361.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853585209.gif)
.
故tan∠AOB=tan2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853601193.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853679545.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853695214.gif)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853710355.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853695214.gif)
.
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853024260.gif)
=3m(m>0) ,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853024257.gif)
=4m,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853055267.gif)
=5m,满足
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853024260.gif)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853055267.gif)
=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853024257.gif)
,
所以,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853024260.gif)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853024257.gif)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853055267.gif)
依次成等差数列.
(2)由已知c2=5,代入①,②得a2=4, b2=1,
于是双曲线的方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853882428.gif)
.
设直线AB的斜率为k,则k=tan∠BFx=tan∠AFO=cot
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853601193.gif)
=2.
于是直线AB的议程为 y=2(x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853133213.gif)
).…………………………………………9分
联立
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181854163739.gif)
消y得15x2-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181854178262.gif)
x+84=0.
故弦CD的长度 | CD |=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181854194518.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853133213.gif)
×
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181854225725.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823181853695214.gif)
…13分