jmc

From the graph of $xy=1,$ we can tell that the foci will be at the points $(t,t)$ and $(-t,-t)$ for some positive real number $t.$



Thus, if $P=(x,y)$ is a point on the hyperbola, then one branch of the hyperbola is defined by $$\sqrt{(x+t{)}^2+(y+t{)}^2}-\sqrt{(x-t{)}^2+(y-t{)}^2}=d$$for some positive real number $d.$ Then $$\sqrt{(x+t{)}^2+(y+t{)}^2}=\sqrt{(x-t{)}^2+(y-t{)}^2}+d.$$Squaring both sides, we get $$(x+t{)}^2+(y+t{)}^2=(x-t{)}^2+(y-t{)}^2+2d\sqrt{(x-t{)}^2+(y-t{)}^2}+d^2.$$This simplifies to $$4tx+4ty-d^2=2d\sqrt{(x-t{)}^2+(y-t{)}^2}.$$Squaring both sides, we get $$\begin{array}{rl}& 16t^2{x}^2+16t^2{y}^2+d^4+32t^{2}xy-8d^{2}tx-8d^{2}ty\\ & =4d^2{x}^2-8d^{2}tx+4d^2{y}^2-8d^{2}ty+8d^2{t}^2.\end{array}$$We can cancel some terms, to get $$16t^2{x}^2+16t^2{y}^2+d^4+32t^{2}xy=4d^2{x}^2+4d^2{y}^2+8d^2{t}^2.$$We want this equation to simplify to $xy=1.$ For this to occur, the coefficients of $x^2$ and $y^2$ on both sides must be equal, so $$16t^2=4d^2.$$Then $d^2=4t^2,$ so $d=2t.$ The equation above becomes $$16t^4+32t^{2}xy=32t^4.$$Then $32t^{2}xy=16t^4,$ so $xy=\frac{t^2}{2}.$ Thus, $t=\sqrt{2},$ so the distance between the foci $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ is $\boxed{4}.$