12. Given a moving point $P(x, y)$ satisfies the quadratic equation $10 x-2 x y-2 y+1=0$, then the eccentricity of this quadratic curve is $\qquad$.
12. $\sqrt{2}$. Hint: From $10 x-2 x y-2 y+1=0$, we have $x^{2}+6 x+y^{2}-6 y-2 x y+9=x^{2}-4 x+4+y^{2}-4 y+4$. That is, $\sqrt{(x-2)^{2}+(y-2)^{2}}=|x-y+3|$, which means $\frac{\sqrt{(x-2)^{2}+(y-2)^{2}}}{\frac{|x-y+3|}{\sqrt{2}}}=\sqrt{2}$, hence $e=\sqrt{2}$.