Let one focus of the hyperbola H: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be at $$(\sqrt{10}, 0)$$ and the corresponding directrix be $$x = \frac{9}{\sqrt{10}}$$. If $$e$$ and $$l$$ respectively are the eccentricity and the length of the latus rectum of H, then $$9(e^2 + l)$$ is equal to:

We are given the hyperbola $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with one focus at $$(\sqrt{10}, 0)$$ and the corresponding directrix $$x = \frac{9}{\sqrt{10}}$$.

For a hyperbola, the focus is at $$(ae, 0)$$ and the directrix is $$x = \frac{a}{e}$$. So we have:

$$ae = \sqrt{10} \quad \text{...(1)}$$

$$\frac{a}{e} = \frac{9}{\sqrt{10}} \quad \text{...(2)}$$

Multiplying equations (1) and (2): $$ae \cdot \frac{a}{e} = \sqrt{10} \cdot \frac{9}{\sqrt{10}}$$, which gives $$a^2 = 9$$, so $$a = 3$$.

From equation (1): $$3e = \sqrt{10}$$, so $$e = \frac{\sqrt{10}}{3}$$.

Therefore $$e^2 = \frac{10}{9}$$.

Now, $$b^2 = a^2(e^2 - 1) = 9\left(\frac{10}{9} - 1\right) = 9 \cdot \frac{1}{9} = 1$$.

The length of the latus rectum is $$l = \frac{2b^2}{a} = \frac{2 \cdot 1}{3} = \frac{2}{3}$$.

Finally, $$9(e^2 + l) = 9\left(\frac{10}{9} + \frac{2}{3}\right) = 9\left(\frac{10}{9} + \frac{6}{9}\right) = 9 \cdot \frac{16}{9} = 16$$.

Hence, the correct answer is Option C.