已知双曲线C:
(I)求a,b; (II)设过F2的直线l与C的左、右两支分别相交于A、B两点,且|AF1|=|BF1|,证明:|AF2|、|AB|、|BF2|成等比数列. |
(I)由题设知
=3,即c a
=9,故b2=8a2b2+a2 a2
所以C的方程为8x2-y2=8a2
将y=2代入上式,并求得x=±
,a2+ 1 2
由题设知,2
=a2+ 1 2
,解得a2=16
所以a=1,b=22
(II)由(I)知,F1(-3,0),F2(3,0),C的方程为8x2-y2=8 ①
由题意,可设l的方程为y=k(x-3),|k|<2
代入①并化简得(k2-8)x2-6k2x+9k2+8=02
设A(x1,y1),B(x2,y2),
则x1≤-1,x2≥1,x1+x2=
,x1x2=6k2 k2-8
,于是9k2+8 k2-8
|AF1|=
=(x1+3)2+y1 2
=-(3x1+1),(x1+3)2+8x1 2-8
|BF1|=
=(x2+3)2+y2 2
=3x2+1,(x2+3)2+8x2 2-8
|AF1|=|BF1|得-(3x1+1)=3x2+1,即x1+x2=-2 3
故
=-6k2 k2-8
,解得k2=2 3
,从而x1x2=4 5
=-9k2+8 k2-8 19 9
由于|AF2|=
=(x1-3)2+y1 2
=1-3x1,(x1+3)2+8x1 2-8
|BF2|=
=(x2-3)2+y2 2
=3x2-1,(x2+3)2+8x2 2-8
故|AB|=|AF2|-|BF2|=2-3(x1+x2)=4,|AF2||BF2|=3(x1+x2)-9x1x2-1=16
因而|AF2||BF2|=|AB|2,所以|AF2|、|AB|、|BF2|成等比数列