6、$F_{1}, F_{2}$ are the two foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $P$ is a point on the ellipse. If the area of $\Delta P F_{1} F_{2}$ is $1$, $\tan \angle P F_{1} F_{2}=\frac{1}{2}$, and $\tan \angle P F_{2} F_{1}=-2$. Then $a=\frac{\sqrt{15}}{2}$.
Suppose $F_{1}=F_{1}(-c, 0), F_{2}=F_{2}(c, 0),(c>0)$, and $P=P\left(x_{0}, y_{0}\right)$. According to our conditions, the slope of the line connecting $F_{1}, P$ is $k_{1}=\tan \angle P F_{1} F_{2}=\frac{1}{2}$, and the slope of the line connecting $F_{2}, P$ is $k_{2}=-\tan \angle P F_{2} F_{1}=2$. Therefore, we have
$$
\left\{\begin{array}{l}
y_{0}=\frac{1}{2}\left(x_{0}+c\right) \\
y_{0}=2\left(x_{0}-c\right)
\end{array} .\right.
$$

From these two equations, we can solve for $x_{0}=\frac{5}{3} c, y_{0}=\frac{4}{3} c$.
Since $1=S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right| y_{0}=\frac{4}{3} c^{2}$, we get $c=\frac{\sqrt{3}}{2}$. Therefore, the coordinates of point $P$ are $\left(\frac{5 \sqrt{3}}{6}, \frac{2 \sqrt{3}}{3}\right)$. Thus,
$$
2 a=\left|P F_{1}\right|+\left|P F_{2}\right|=\sqrt{\left(x_{0}+c\right)^{2}+y_{0}^{2}}+\sqrt{\left(x_{0}-c\right)^{2}+y_{0}^{2}}=\frac{2 \sqrt{15}}{3}+\frac{\sqrt{15}}{3}=\sqrt{15} .
$$

That is, $a=\frac{\sqrt{15}}{2}$.