I don't know even what a type of surface will be. And what equation will be? The equation of hyperbola - $ xy = l. $ Now, let's
$ x = x'cos(\varphi ) - y'sin(\varphi ), y = x'sin(\varphi ) + y'cos(\varphi ) \Rightarrow \frac{1}{2}sin(2 \varphi )x'^{2} - \frac{1}{2}sin(2 \varphi )y'^{2} + x'y'cos(2 \varphi) = l. $
So
$ cos( 2 \varphi ) = 0 \Rightarrow \varphi = \frac{\pi}{4} \Rightarrow \frac{x'^{2}}{2l} - \frac{y'^{2}}{2l} = 1. $