Let the common tangents to the curves $$4(x^2 + y^2) = 9$$ and $$y^2 = 4x$$ intersect at the point $$Q$$. Let an ellipse, centered at the origin $$O$$, has lengths of semi-minor and semi-major axes equal to $$OQ$$ and $$6$$, respectively. If $$e$$ and $$l$$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $$\frac{l}{e^2}$$ is equal to ______

We need common tangents to $$4(x^2 + y^2) = 9$$ (circle $$x^2 + y^2 = 9/4$$, radius $$3/2$$) and $$y^2 = 4x$$ (parabola).

Since any tangent to the parabola $$y^2 = 4x$$ can be written as $$y = mx + \frac{1}{m}$$, we require that its distance from the origin equals the circle’s radius. The distance from the origin to the line $$y = mx + \frac{1}{m}$$ is $$\frac{|1/m|}{\sqrt{1 + m^2}} = \frac{3}{2}$$.

Squaring both sides gives $$\frac{1}{m^2(1 + m^2)} = \frac{9}{4}$$, which leads to $$4 = 9m^2(1 + m^2) = 9m^2 + 9m^4$$. This yields the equation $$9m^4 + 9m^2 - 4 = 0$$, factorable as $$(3m^2 + 4)(3m^2 - 1) = 0$$. From this we find $$m^2 = \frac{1}{3}$$ and hence $$m = \pm\frac{1}{\sqrt{3}}$$.

Accordingly the two common tangents are $$y = \frac{x}{\sqrt{3}} + \sqrt{3}$$ and $$y = -\frac{x}{\sqrt{3}} - \sqrt{3}$$. Adding these equations gives $$2y = 0$$, so $$y = 0$$. Substituting back into the first tangent yields $$0 = \frac{x}{\sqrt{3}} + \sqrt{3}$$, which implies $$x = -3$$. Hence the intersection point is $$Q = (-3, 0)$$.

Since the distance $$OQ$$ is $$3$$, we take the semi-minor axis of the ellipse as $$b = 3$$ and the semi-major axis as $$a = 6$$. Therefore the eccentricity satisfies $$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{9}{36} = \frac{27}{36} = \frac{3}{4}$$. The length of the latus rectum is then $$l = \frac{2b^2}{a} = \frac{2 \times 9}{6} = 3$$.

Finally, the required ratio is $$\frac{l}{e^2} = \frac{3}{3/4} = 4$$.

The answer is $$\boxed{4}$$.