Let $$f(x) = x^2 + 9$$, $$g(x) = \frac{x}{x-9}$$ and $$a = f \circ g(10)$$, $$b = g \circ f(3)$$. If $$e$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\frac{x^2}{a} + \frac{y^2}{b} = 1$$, then $$8e^2 + l^2$$ is equal to :

$$f(x) = x^2 + 9$$, $$g(x) = \frac{x}{x-9}$$.

$$a = f(g(10)) = f\left(\frac{10}{1}\right) = f(10) = 100 + 9 = 109$$.

$$b = g(f(3)) = g(9+9) = g(18) = \frac{18}{18-9} = \frac{18}{9} = 2$$.

Ellipse: $$\frac{x^2}{109} + \frac{y^2}{2} = 1$$. Here $$a^2 = 109 > b^2 = 2$$.

Eccentricity: $$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{109} = \frac{107}{109}$$.

Latus rectum: $$l = \frac{2b^2}{a} = \frac{4}{\sqrt{109}}$$. So $$l^2 = \frac{16}{109}$$.

$$8e^2 + l^2 = 8 \cdot \frac{107}{109} + \frac{16}{109} = \frac{856 + 16}{109} = \frac{872}{109} = 8$$.

The correct answer is Option 1: 8.