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\((\qquad)\)
\(\mathrm{A}. \left(1,\dfrac{\sqrt{6}}{2}\right)\) \(\qquad\mathrm{B}.\left(\dfrac{\sqrt{6}}{2},3\right)\) \(\qquad\mathrm{C}.\left(1,\sqrt{5}\right)\) \(\qquad\mathrm{D}.\left(\sqrt{5},+\infty\right)\)
解析
根据题意,由于\[
|PA|^2+|PB|^2=|AB|^2.\]
所以\(PA \perp PB\),若设\(P\left(x,y\right)\),则\(x\in\left(a,2a\right)\).由于\(P\)点坐标满足\[
\left(x-\dfrac{3}{2}a\right)^2+y^2=\dfrac{1}{4}a^2,\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.\]联立两式消去\(y\)可得关于\(x\)的一元二次方程\[
c^2x^2-3a^3x+3a^4-a^2c^2=0.\]其中\(c\)是双曲线半焦距,因式分解可得\[
(x-a)\cdot \left(c^2x+ac^2-3a^3\right)=0.\]所以\(x=\dfrac{3a^3-ac^2}{c^2}\),从而可得\[
a<\dfrac{3a^3-ac^2}{c^2}<2a.\]解得双曲线的离心率的取值范围为\(\left(1,\dfrac{\sqrt{6}}{2}\right)\).
原文：https://www.cnblogs.com/Math521/p/11959689.html