2. Given points
$$
A\left(x-\frac{3-\sqrt{5}}{2} \cdot y, 1\right) 、 B\left(x-\frac{3+\sqrt{5}}{2} \cdot y,-1\right)
$$

satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$. Then the range of $S=x^{2}-x y+y^{2}-1$ is $\qquad$ .
2. $\left[-\frac{2}{5},+\infty\right)$.

From the given, we have
$$
\begin{array}{l}
\left(x-\frac{3-\sqrt{5}}{2} \cdot y\right)\left(x-\frac{3+\sqrt{5}}{2} \cdot y\right)-1=0 \\
\Rightarrow x^{2}-3 x y+y^{2}-1=0 \\
\Rightarrow 1=(x+y)^{2}-5 x y \geqslant-5 x y \\
\Rightarrow x y \geqslant-\frac{1}{5} .
\end{array}
$$

When $x=-y= \pm \frac{\sqrt{5}}{5}$, the equality in (1) and (2) holds.
Thus, $S=x^{2}-x y+y^{2}-1$
$$
\begin{array}{l}
=\left(x^{2}-3 x y+y^{2}-1\right)+2 x y \\
=2 x y \text { (substitute (1)) } \\
\geqslant-\frac{2}{5} \text { (substitute (2)). }
\end{array}
$$

Therefore, the range of $S$ is $\left[-\frac{2}{5},+\infty\right)$.