7. 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
$\qquad$ -
7. $\sqrt{2}$.

From $10 x-2 x y-2 y+1=0$, we get
$$
\begin{array}{l}
x^{2}+6 x+y^{2}-6 y-2 x y+9=x^{2}-4 x+4+y^{2}-4 y+4 \\
\Rightarrow \sqrt{(x-2)^{2}+(y-2)^{2}}=|x-y+3| \\
\Rightarrow \frac{\sqrt{(x-2)^{2}+(y-2)^{2}}}{\frac{|x-y+3|}{\sqrt{2}}}=\sqrt{2} .
\end{array}
$$

Therefore, the eccentricity of the conic section is $\sqrt{2}$.