2. Real numbers $x, y$ satisfy $4 x^{2}-5 x y+4 y^{2}=5$, let $S=x^{2}+y^{2}$, then $\frac{1}{S_{\max }}+$ $\frac{1}{S_{\text {min }}}$ is . $\qquad$
2. It is known that $s=x^{2}+y^{2}>0$, let
$$
\left\{\begin{array}{l}
x=\sqrt{s} \cos \theta, \\
y=\sqrt{s} \sin \theta .
\end{array}\right.
$$

Substitute into
$$
4 x^{2}-5 x y+4 y^{2}=5 \text {, }
$$

we get
$$
\sin 2 \theta=\frac{8 s-10}{5 s},
$$

thus
$$
\left|\frac{8 s-10}{5 s}\right| \leqslant 1,
$$

which gives
$$
\begin{array}{cc} 
& \frac{10}{13} \leqslant s \leqslant \frac{10}{3}, \\
\therefore & S_{\max }=\frac{10}{3}, S_{\min }=\frac{10}{13}, \\
\therefore \quad & \frac{1}{S_{\max }}+\frac{1}{\mathrm{~S}_{\min }}=\frac{3}{10}+\frac{13}{10}=\frac{8}{5} .
\end{array}
$$