4. Real numbers $x, y$ satisfy
$$
x^{2}+12 x y+52 y^{2}-8 y+1=0 \text {. }
$$

Then $x^{2}-y^{2}=$ $\qquad$
4. $\frac{35}{16}$.

From the problem, we get
$$
\begin{array}{l}
(x+6 y)^{2}+(4 y-1)^{2}=0 \\
\Rightarrow x+6 y=4 y-1=0 \\
\Rightarrow y=\frac{1}{4}, x=-\frac{3}{2} \\
\Rightarrow x^{2}-y^{2}=\frac{35}{16} .
\end{array}
$$