6. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with the left vertex $A$ and the right focus $F$, let $P$ be any point on the hyperbola in the first quadrant. If $\angle P F A=2 \angle F A P$ always holds, then the eccentricity of the hyperbola is $\qquad$
6. 2 Detailed Explanation: From the given conditions, we can take a point \( P \) on the hyperbola such that \( P F \perp x \)-axis, obtaining \( P(c, y) \). Then, \(\frac{c^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), hence \( y^{2}=\frac{b^{4}}{a^{2}} \). Since \( y>0 \), we have \( y=\frac{b^{2}}{a}=\frac{c^{2}-a^{2}}{a} \). Given that \( \angle P F A=2 \angle F A P=\frac{\pi}{2} \), we know that \( \triangle A F P \) is an isosceles triangle, with \( A F=P F \), i.e., \( a+c=\frac{c^{2}-a^{2}}{a} \). Simplifying, we get \( c^{2}-a c-2 a^{2}=0 \), or \( e^{2}-e-2=0 \). Since \( e>1 \), solving gives \( e=2 \).