jmc

The graph of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is an ellipse centered at the origin, with semi-axes of length $a$ and $b.$ Because the foci of the ellipse lie along the $y-$axis, the major axis of the ellipse must be the vertical axis. The distance between each focus of the ellipse and the center is $4,$ so we have $$b^2-a^2=4^2=16.$$The graph of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is an hyperbola centered at the origin. The distance between each focus of the hyperbola and the center is $6,$ so we have $$a^2+b^2=6^2=36.$$Therefore, we have the system of equations $$\begin{array}{rl}{b}^2-a^2& =16,\\ a^2+b^2& =36.\end{array}$$To solve this system, we add the two equations, giving $2b^2=52,$ so $b^2=26,$ and $b=\pm \sqrt{26}.$ Then, $26-a^2=16,$ so $a^2=10,$ and $a=\pm \sqrt{10}.$ Thus, $$ab=(\pm \sqrt{10})(\pm \sqrt{26})=\pm 2\sqrt{65},$$so $|ab|=\boxed{2\sqrt{65}}.$