(4) A chord $AB$ with an inclination angle of $45^{\circ}$ is drawn through the right focus $F_{2}$ of the ellipse $\frac{x^{2}}{2}+y^{2}=1$. Then $|AB|$ is ( ).
(A) $\frac{2 \sqrt{6}}{3}$
(B) $\frac{4 \sqrt{6}}{3}$
(C) $\frac{4 \sqrt{2}}{3}$
(D) $\frac{4 \sqrt{3}}{3}$
(4) $\mathrm{C}$ Hint: The right focus of the ellipse is $(1,0)$, then the chord $A B$ is $y=x-1$, substituting into the ellipse equation yields
$$
\begin{aligned}
& 3 x^{2}-4 x=0 \Rightarrow x_{1}=0, x_{2}=\frac{4}{3} \\
\Rightarrow & |A B|=\sqrt{2\left(x_{1}-x_{2}\right)^{2}}=\frac{4 \sqrt{2}}{3} .
\end{aligned}
$$