每日一题_191130

已知双曲线(dfrac{x^2}{a^2}-dfrac{y^2}{b^2}=1) ((a>0,b>0)),(A(a,0)),(B(2a,0)),(P)为(C)上异于点(A)的一点,且满足(overrightarrow{PA}^2+overrightarrow{PB}^2=a^2),则(C)的离心率(e)的取值范围是((qquad))
(mathrm{A}. left(1,dfrac{sqrt{6}}{2} ight)) (qquadmathrm{B}.left(dfrac{sqrt{6}}{2},3 ight)) (qquadmathrm{C}.left(1,sqrt{5} ight)) (qquadmathrm{D}.left(sqrt{5},+infty ight))
解析
根据题意,由于$$
|PA|^2+|PB|2=|AB|^2.$$
所以(PA perp PB),若设(Pleft(x,y ight)),则(xinleft(a,2a ight)).由于(P)点坐标满足$$
left(x-dfrac{3}{2}a ight)^2+y2=dfrac{1}{4}a^2,dfrac{x2}{a^2}-dfrac{y2}{b^2}=1.$$联立两式消去(y)可得关于(x)的一元二次方程$$
c^2x2-3a^3x+3a4-a^2c2=0.$$其中(c)是双曲线半焦距,因式分解可得$$
(x-a)cdot left(c^2x+ac2-3a^3 ight)=0.$$所以(x=dfrac{3a^3-ac^2}{c^2}),从而可得$$
a<dfrac{3a^3-ac2}{c^2}<2a.$$解得双曲线的离心率的取值范围为(left(1,dfrac{sqrt{6}}{2} ight)).