5. Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its two foci and one endpoint of the minor axis forming the vertices of an equilateral triangle. If the shortest distance from a focus to a point on the ellipse is $\sqrt{3}$, then $(a, b)=$ $\qquad$
5. $(2 \sqrt{3}, 3)$.

As shown in Figure 3, let the foci be $F_{1}(-c, 0), F_{2}(c, 0)$, where $c=\sqrt{a^{2}-b^{2}}$, one endpoint of the major axis be $A(a, 0)$, and one endpoint of the minor axis be $B(0, b)$.
Then, $B F_{1}=B F_{2}=F_{1} F_{2}$, and $B F_{1}+B F_{2}=2 a$.
Thus, $a=2 c$.
Let $P$ be any point on the ellipse, then $P F_{1}+P F_{2}=2 a=(a+c)+(a-c)$.
Since $P F_{1} \leqslant F_{1} O+O P \leqslant c+a$, it follows that $P F_{2} \geqslant a-c$.
Therefore, the shortest distance from a focus to a point on the ellipse is $a-c$, which occurs when point $P$ coincides with the endpoint of the major axis, i.e.,
$a-c=\sqrt{3}$.
Thus, $c=\sqrt{3}, a=2 \sqrt{3}, b=3$.
Hence, $(a, b)=(2 \sqrt{3}, 3)$.