How to show minimum distance between two branches of the curve $xy=1$ is $2\sqrt 2$ 
This is the curve of $xy=1$.
can anyone please tell me how to prove that the minimum distance between two points one of which is on the brach of the first quadrant and other one on the branch of the third quadrant is $\sqrt 8$?
I know the closest two points will be $(1 , 1) , (-1 , -1)$. I can prove it byy using $AM-GM$ inequality. But how to prove it geometrically ?
 A: Any line segment joining the two branches can be split into two sub-segments, each of which goes from one of the branches to some point on the line $y = -x.$  The shortest distance (for example) from any point on the 1st quadrant branch to the line $y = -x$ is represented by the line going from $(1,1)$ to $(0,0)$.
This may be geometrically verified by rotating the graph $(45^\circ)$ so that the line $y = -x$ becomes horizontal.  Then, in the rotated graph, find the point on the 1st quadrant branch that has the lowest height.
A: We know that rectangular hyperbolas are obtained by rotating a standard hyperbola ($a=b$) by $45^\circ$. And in a standard hyperbola, we can clearly see that the minimum distance between two points lies on the major axis i.e, points of intersection of major axis with hyperbola (vertex). Similarly, in a rectangular hyperbola, the minimum distance will lie on its major axis that is now $y=x$. Earlier the major axis was $y=0$ and asymptotes were $y=\pm x$ and now the asymptotes have become the major axis. Note that we are only rotating the curve and not shifting anything so the concept of shortest distance which was earlier the vertex, still remains the same, the only change is major axis, ($y=o \space \text{to} \space y=\pm x$). So, the point of intersection is clearly $(1,1),(-1,-1)$ and hence, $d=2\sqrt2$.

Source: Wolfram MathWorld
