10. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0), A_{1}, A_{2}$ are the vertices of the real axis, $F$ is the right focus, $B(0, b)$ is the endpoint of the imaginary axis. If there exist two different points $P_{i}(i=1,2)$ on the line segment $B F$ (excluding the endpoints), such that $\triangle P_{i} A_{1} A_{2}(i=1,2)$ form right triangles with $A_{1} A_{2}$ as the hypotenuse, then the range of the eccentricity $e$ of the hyperbola is $\qquad$.
Answer
As shown in the figure, the circle with diameter $A_{1} A_{2}$ intersects the line segment $B F$ at two points.
Given $B(0, b), F(c, 0) \Rightarrow k_{B F}=-\frac{b}{c}$,
the line $B F: b x+c y-b c=0$. Therefore, the distance $d\sqrt{2}$ from $O$ to the line $B F$,
so the range of the eccentricity $e$ of the hyperbola is $\left(\sqrt{2}, \frac{\sqrt{5}+1}{2}\right)$.