2. Given real numbers $x, y$ satisfy $x^{2}-x y+2 y^{2}=1$. Then the sum of the maximum and minimum values of $x^{2}+$ $2 y^{2}$ is equal to ( ).
(A) $\frac{8}{7}$
(B) $\frac{16}{7}$
(C) $\frac{8-2 \sqrt{2}}{7}$
(D) $\frac{8+2 \sqrt{2}}{7}$
2. (B).

From the given, we have $x^{2}+2 y^{2}=x y+1$.
$$
\begin{array}{l}
\text { Also, } \frac{x^{2}+2 y^{2}}{2 \sqrt{2}} \geqslant x y \geqslant-\frac{x^{2}+2 y^{2}}{2 \sqrt{2}}, \\
\therefore \frac{x^{2}+2 y^{2}}{2 \sqrt{2}}+1 \geqslant x^{2}+2 y^{2} \geqslant-\frac{x^{2}+2 y^{2}}{2 \sqrt{2}}+1 .
\end{array}
$$

Solving, we get $\frac{2 \sqrt{2}}{2 \sqrt{2}-1} \geqslant x^{2}+2 y^{2} \geqslant \frac{2 \sqrt{2}}{2 \sqrt{2}+1}$.
That is, $\frac{8+2 \sqrt{2}}{7} \geqslant x^{2}+2 y^{2} \geqslant \frac{8-2 \sqrt{2}}{7}$.
When and only when $x=\sqrt{2} y$ or $x=-\sqrt{2} y$, the two equalities in equation (1) are respectively satisfied. Therefore, the sum of the maximum and minimum values of $x^{2}+2 y^{2}$ is $\frac{16}{7}$.