已知(F_1,F_2)分别是双曲线(C:dfrac{x^2}{a^2}-dfrac{y^2}{b^2}=1) ((a>0,b>0))的左,右焦点,过(F_1(-2,0))的直线与双曲线(C)的左支交于(A,B)两点,若(Dleft(-dfrac{1}{2},dfrac{5}{8}
ight))为( riangle ABF_2)的内切圆的圆心,则该内切圆的半径为((qquad))
(mathrm{A}.dfrac{sqrt{41}}{8}) (mathrm{B}.qquaddfrac{3}{2}) (mathrm{C}.qquaddfrac{13}{8}) (mathrm{D}.qquaddfrac{5sqrt{17}}{8})
解析: 过(D)作(DHperp AB),垂足为(H),则$$
|AF_2|-|BF_2|=|AH|-|BH|.$$又$$
egin{cases}
& |AF_2|=|AF_1|+2a,
& |BF_2|=|BF_1|+2a,
end{cases}
[所以
]
|AF_2|-|BF_2|=|AF_1|-|BF_1|,$$于是$$|AH|-|BH|=|AF_1|-|BF_1|.$$因此(H)与(F_1)重合,因此所求内切圆半径即$$|DF_1|=sqrt{
left(-dfrac{1}{2}+2
ight)2+left(dfrac{5}{8}-0
ight)2}=dfrac{13}{8}.$$