MHB Solving for $2012x^2-2010y^2+2011x-2011y-2011$ with $x,y\in R$

[Albert1]

$x,y\in R$, satisfying $(x-\sqrt{x^2-2011})(y-\sqrt{y^2-2011})=2011$
find $2012x^2-2010y^2+2011x-2011y-2011=?$

 

[anemone]

$x,y\in R$, satisfying $(x-\sqrt{x^2-2011})(y-\sqrt{y^2-2011})=2011$
find $2012x^2-2010y^2+2011x-2011y-2011=?$

My solution:

Spoiler

Note that we could rewrite the given equality as

$(x-\sqrt{x^2-2011})(y-\sqrt{y^2-2011})=2011$

$x-\sqrt{x^2-2011}=\dfrac{2011}{y-\sqrt{y^2-2011}}$

$x-\sqrt{x^2-2011}=\dfrac{2011(y+\sqrt{y^2-2011})}{(y-\sqrt{y^2-2011})(y+\sqrt{y^2-2011})}=y+\sqrt{y^2-2011}$

Note that for $x\ge \sqrt{2011}$, the function on the right is an increasing function and its starting point is at $(\sqrt{2011},\, \sqrt{2011})$ while the function on the left is a decreasing function with its starting point $(\sqrt{2011},\, \sqrt{2011})$, (whereas for the region where $x\le \sqrt{2011}$, the function on the right is an decreasing function and its starting point is at $(-\sqrt{2011},\, -\sqrt{2011})$ while the function on the left is an increasing function with its starting point $(-\sqrt{2011},\, -\sqrt{2011})$) therefore we can conclude the only solution the system has is when $x^2=2011=y^2$, or $x=y=\pm\sqrt{2011}$

Therefore,

$\begin{align*}2012x^2-2010y^2+2011x-2011y-2011&=(2012-2010)(2011)-2011\\&=2011(2-1)\\&=2011\end{align*}$