The length of the latus-rectum of the ellipse, whose foci are $$(2, 5)$$ and $$(2, -3)$$ and eccentricity is $$\dfrac{4}{5}$$, is

The two foci are $$(2,5)$$ and $$(2,-3)$$. Their common $$x$$-coordinate is $$2$$, so the major axis is the vertical line $$x = 2$$.

For an ellipse with vertical major axis and centre $$(h,k)$$, the standard form is
$$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$$ with $$a \gt b$$, foci $$(h,\,k \pm c)$$, and relations $$c^2 = a^2 - b^2$$, $$e = \frac{c}{a}$$.

Step 1: Find $$c$$
Distance between the given foci:
$$2c = |5 - (-3)| = 8$$
$$\Rightarrow c = 4$$.

Step 2: Find $$a$$ using eccentricity
Given $$e = \frac{4}{5}$$ and $$e = \frac{c}{a}$$,
$$\frac{4}{5} = \frac{4}{a} \;\Rightarrow\; a = 5$$.

Step 3: Find $$b$$
Using $$c^2 = a^2 - b^2$$:
$$4^2 = 5^2 - b^2$$
$$16 = 25 - b^2$$
$$b^2 = 25 - 16 = 9$$
$$\Rightarrow b = 3$$.

Step 4: Length of the latus-rectum
For any ellipse, length of latus-rectum $$L = \frac{2b^2}{a}$$.
Substituting $$b^2 = 9$$ and $$a = 5$$:
$$L = \frac{2 \times 9}{5} = \frac{18}{5}$$.

Hence, the length of the latus-rectum is $$\frac{18}{5}$$.

Option D is correct.