8. In the Cartesian coordinate system $x O y$, there are 2018 circles: $\odot A_{1}, \odot A_{2}, \cdots, \odot A_{2018}$, where the center of $\odot A_{k}$ is $A_{k}\left(a_{k}, \frac{1}{4} a_{k}^{2}\right)$, and the radius is $\frac{1}{4} a_{k}^{2}(k=1, \cdots, 2018)$, here $a_{1}>\cdots>a_{2018}=\frac{1}{2018}$, and $\odot A_{k+1}$ is externally tangent to $\odot A_{k}$ $(k=1,2, \cdots, 2017)$, then $a_{1}=$ $\qquad$
Answer $\frac{2}{2019}$.
Analysis Since they are externally tangent, we have
$$
\left(\frac{1}{4} a_{k+1}^{2}+\frac{1}{4} a_{k}^{2}\right)^{2}=\left(a_{k}-a_{k+1}\right)^{2}+\left(\frac{1}{4} a_{k}^{2}-\frac{1}{4} a_{k+1}^{2}\right)^{2}, \quad a_{k} a_{k+1}=2\left(a_{k}-a_{k+1}\right), \quad \frac{1}{a_{k+1}}=\frac{1}{a_{k}}+\frac{1}{2}
$$

Therefore, $\frac{1}{a_{2018}}=\frac{1}{a_{1}}+\frac{2017}{2}$, so $a_{1}=\frac{2}{2019}$.