jmc

By the distance formula, the distance from $(2a,a-4)$ to $(4,-1)$ is $\sqrt{(2a-4{)}^2+((a-4)-(-1){)}^2}$. Setting this equal to $2\sqrt{10}$, we find $$\begin{array}{rl}(2a-4{)}^2+(a-3{)}^2& ={\sqrt{40}}^2\\ 4a^2-16a+16+a^2-6a+9& =40\\ 5a^2-22a-15& =0\\ (a-5)(5a+3)& =0\end{array}$$The possible values for $a$ are $5$ and $-\frac{3}{5}$. Thus, the answer is $5\times -\frac{3}{5}=\boxed{-3}$.