6. The parabola $C_{1}: x^{2}=2 p y$ has a focus at $F$, and the hyperbola $C_{2}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ has foci at $F_{1}$ and $F_{2}$. Point $P$ is a common point of the two curves in the first quadrant. If $P F F_{1}$ are collinear, and $C_{1}$ and $C_{2}$ have a common tangent line at $P$, then the eccentricity of $C_{2}$ is
Answer: $e=\sqrt{2}$
Let the common tangent be $l$. Then, by the optical property of the hyperbola, $l$ is the internal (external) angle bisector of the two angles marked in the figure, where $PG$ is a ray parallel to the $y$-axis:
So $F_{2}PG$ are collinear, i.e., $PF_{2} \parallel y$-axis, hence $P\left(c, \frac{b^{2}}{a}\right)$
Let $l$ intersect the $x$-axis at $Q$, by $x_{Q}=\frac{x_{P}}{2}=\frac{c}{2}$, we have $\frac{\left|F_{1} P\right|}{\left|F_{2} P\right|}=\frac{\left|F_{1} Q\right|}{\left|F_{2} Q\right|}=3,\left|F_{P} P\right|=\frac{3 b^{2}}{a}$
$$
2 a=\left|F_{1} P-F_{2} P\right|=\frac{2 b^{2}}{a}
$$

Therefore, $e=\sqrt{2}$