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