Problem - 4574
Consider the parabola $y=ax^2 + 2019x + 2019$. There exists exactly one circle which is centered on the $x$-axis and is tangent to the parabola at exactly two points. It turns out that one of these tangent points is $(0, 2019)$. Find $a$.
Because the $x$-coordinate of the parabola's vertex is $-\frac{2019}{2a}$. Therefore, the center of the circle must be $\left(-\frac{2019}{2a}, 0\right)$.
Meanwhile, the slope of the tangent to the parabola equals $2ax + 2019$. Hence, at the point $(0,2019)$, the slope is $2019$. Accordingly, the slope of the line from the circle center to the tangent point must be $-\frac{1}{2019}$. This means
$$-\frac{0-2019}{-\frac{2019}{2a}-0}=-\frac{1}{2019}\implies a=\boxed{-\frac{1}{4038}}$$