Let an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a < b$$, pass through the point $$(4, 3)$$ and have eccentricity $$\frac{\sqrt{5}}{3}$$. Then the length of its latus rectum is :

Since $$a \lt b$$, the major axis is along the $$y$$-axis. The eccentricity formula for this case is $$e^2 = 1 - \frac{a^2}{b^2}$$.

Given $$e = \frac{\sqrt{5}}{3}$$, we get $$\frac{5}{9} = 1 - \frac{a^2}{b^2}$$, so $$\frac{a^2}{b^2} = \frac{4}{9}$$, which gives $$a^2 = \frac{4b^2}{9}$$.

The ellipse passes through $$(4, 3)$$:

$$\frac{16}{a^2} + \frac{9}{b^2} = 1$$

Substituting $$a^2 = \frac{4b^2}{9}$$:

$$\frac{16}{\frac{4b^2}{9}} + \frac{9}{b^2} = 1$$

$$\frac{36}{b^2} + \frac{9}{b^2} = 1$$

$$\frac{45}{b^2} = 1 \implies b^2 = 45$$

So $$a^2 = \frac{4 \times 45}{9} = 20$$. We verify $$a = \sqrt{20} = 2\sqrt{5} \lt b = \sqrt{45} = 3\sqrt{5}$$.

The length of the latus rectum (with major axis along the $$y$$-axis) is $$\frac{2a^2}{b}$$:

$$\frac{2 \times 20}{3\sqrt{5}} = \frac{40}{3\sqrt{5}} = \frac{40\sqrt{5}}{15} = \frac{8\sqrt{5}}{3}$$

Hence, the correct answer is Option 4.