5. Given an ellipse $\frac{x^{2}}{2}+k y^{2}=1$ with its foci on the $x$-axis, points $A$ and $B$ are the two intersection points of a line passing through the origin with the ellipse. If the number $k$ allows for another point $C$ on the ellipse such that $\triangle A B C$ is an equilateral triangle, then for all such $k$, the maximum area of $\triangle A B C$ is $\qquad$ _.
5. $\frac{2 \sqrt{3}}{3}$.

Let $|O A|=r_{1},|O C|=r_{2}$. Then, since $\triangle A B C$ is an equilateral triangle, we know that $O C \perp A B$, and $r_{2}=\sqrt{3} r_{1}$.

Let $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right)$. Then it is easy to know that the coordinates of point $C$ are $\left(r_{2} \cos \left(\theta \pm \frac{\pi}{2}\right), r_{2} \sin \left(\theta \pm \frac{\pi}{2}\right)\right)$. Substituting these into the ellipse equation, we get
$$
\begin{array}{l}
\frac{\left(r_{1} \cos \theta\right)^{2}}{2}+k\left(r_{1} \sin \theta\right)^{2}=1, \\
\frac{\left[r_{2} \cos \left(\theta \pm \frac{\pi}{2}\right)\right]^{2}}{2}+k\left[r_{2} \sin \left(\theta \pm \frac{\pi}{2}\right)\right]^{2}=1 . \\
\text { Therefore, } \frac{1}{r_{1}^{2}}=\frac{\cos ^{2} \theta}{2}+k \sin ^{2} \theta, \frac{1}{r_{2}^{2}}=\frac{\sin ^{2} \theta}{2}+k \cos ^{2} \theta .
\end{array}
$$

Then, from $r_{2}=\sqrt{3} r_{1}$, we get
$$
k=\frac{2 \sin ^{2} \theta-\frac{1}{2}}{4 \sin ^{2} \theta-3}=\frac{1}{2}+\frac{1}{4 \sin ^{2} \theta-3} \text {. }
$$
$\therefore k \geqslant \frac{3}{2} .\left(k \leqslant \frac{1}{6}\right.$ is not valid, discard $)$
Thus, $\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}=\frac{4}{3 r_{1}^{2}}=\frac{1}{2}+k \geqslant 2$.
Therefore, $r_{1}^{2} \leqslant \frac{2}{3}, S_{\triangle A B C}=\sqrt{3} r_{1}^{2} \leqslant \frac{2 \sqrt{3}}{3}$.