Let each of the two ellipses $$E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a > b)$$ and $$E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1,(A > B)$$ have eccentricity $$\frac{4}{5}$$. Let the lengths of the latus recta of $$E_{1}\text{ and }E_{2}$$ be $$l_{1}\text{ and }l_{2}$$ respectively, such that $$2\ l_{1}^{2}=9\ l_{2}$$. If the distance between the foci of $$E_{1}$$ is 8, then the distance between the foci of $$E_{2}$$ is

Given two ellipses $$E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ ($$a > b$$) and $$E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$ ($$A > B$$), both with eccentricity $$\frac{4}{5}$$. The lengths of their latus recta are $$l_1$$ and $$l_2$$ respectively, satisfying $$2l_1^2 = 9l_2$$. The distance between the foci of $$E_1$$ is 8.

For any ellipse $$\frac{x^2}{p^2} + \frac{y^2}{q^2} = 1$$ ($$p > q$$), the eccentricity $$e$$ is given by $$e = \sqrt{1 - \frac{q^2}{p^2}}$$, the length of the latus rectum is $$\frac{2q^2}{p}$$, and the distance between foci is $$2pe$$.

For $$E_1$$, distance between foci is $$2a e_1 = 8$$. Given $$e_1 = \frac{4}{5}$$:

$$2a \cdot \frac{4}{5} = 8$$

$$\frac{8a}{5} = 8$$

$$8a = 40$$

$$a = 5$$

Using eccentricity for $$E_1$$:

$$e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \frac{4}{5}$$

$$\left(\frac{4}{5}\right)^2 = 1 - \frac{b^2}{a^2}$$

$$\frac{16}{25} = 1 - \frac{b^2}{25} \quad (\text{since } a = 5)$$

$$\frac{b^2}{25} = 1 - \frac{16}{25} = \frac{9}{25}$$

$$b^2 = 9$$

Latus rectum of $$E_1$$:

$$l_1 = \frac{2b^2}{a} = \frac{2 \cdot 9}{5} = \frac{18}{5}$$

Given $$2l_1^2 = 9l_2$$:

$$2 \left(\frac{18}{5}\right)^2 = 9l_2$$

$$2 \cdot \frac{324}{25} = 9l_2$$

$$\frac{648}{25} = 9l_2$$

$$l_2 = \frac{648}{25 \cdot 9} = \frac{648}{225} = \frac{72}{25}$$

For $$E_2$$, eccentricity $$e_2 = \frac{4}{5}$$:

$$e_2 = \sqrt{1 - \frac{B^2}{A^2}} = \frac{4}{5}$$

$$\left(\frac{4}{5}\right)^2 = 1 - \frac{B^2}{A^2}$$

$$\frac{16}{25} = 1 - \frac{B^2}{A^2}$$

$$\frac{B^2}{A^2} = \frac{9}{25}$$

$$B^2 = \frac{9}{25} A^2$$

Latus rectum of $$E_2$$:

$$l_2 = \frac{2B^2}{A} = \frac{2}{A} \cdot \frac{9}{25} A^2 = \frac{18}{25} A$$

Substituting $$l_2 = \frac{72}{25}$$:

$$\frac{18}{25} A = \frac{72}{25}$$

$$18A = 72$$

$$A = \frac{72}{18} = 4$$

Distance between foci of $$E_2$$:

$$2A e_2 = 2 \cdot 4 \cdot \frac{4}{5} = \frac{32}{5}$$

Thus, the distance is $$\frac{32}{5}$$, corresponding to option D.