Let $$E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,a > b$$ and $$H: \frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$$.Let the distance between the foci of E and the foci of H be $$2sqrt{3}$$. If a-A=2, and the ratio of the eccentricities of E and H is $$\frac{1}{3}$$, then the sum of the lengths of their latus rectums is equal to:

We need to find the sum of lengths of latus rectums of ellipse $$E$$ and hyperbola $$H$$.

The ellipse is given by $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b$$ and the hyperbola by $$H: \frac{x^2}{A^2}-\frac{y^2}{B^2}=1$$. The distance between the foci of $$E$$ is $$2ae$$, while that of $$H$$ is $$2Ae_H$$, and it is given that $$2ae = 2Ae_H = 2\sqrt{3}$$, so that $$ae = Ae_H = \sqrt{3}$$. Additionally, $$a - A = 2$$ and $$\frac{e}{e_H} = \frac{1}{3}$$, i.e., $$e_H = 3e$$.

From $$ae = \sqrt{3}$$ it follows that $$e = \frac{\sqrt{3}}{a}$$. Since $$e_H = 3e = \frac{3\sqrt{3}}{a}$$ and $$Ae_H = \sqrt{3}$$, we have $$A \cdot \frac{3\sqrt{3}}{a} = \sqrt{3} \Rightarrow A = \frac{a}{3}$$. Substituting into $$a - A = 2$$ yields $$a - \frac{a}{3} = 2 \Rightarrow \frac{2a}{3} = 2 \Rightarrow a = 3$$, hence $$A = 1$$, $$e = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$, and $$e_H = \sqrt{3}$$.

For the ellipse, $$b^2 = a^2(1-e^2) = 9(1-\frac{1}{3}) = 6$$, and for the hyperbola, $$B^2 = A^2(e_H^2-1) = 1(3-1) = 2$$.

The length of the latus rectum of the ellipse is $$\frac{2b^2}{a} = \frac{12}{3} = 4$$, and that of the hyperbola is $$\frac{2B^2}{A} = \frac{4}{1} = 4$$, so that the sum is $$4 + 4 = 8$$.

The sum of lengths of latus rectums is 8, which matches Option C. Therefore, the answer is Option C.