Let (h, k) lie on the circle $$C: x^{2}+y^{2}=4$$ and the point (2h + l , 3k + 2) lie on an ellipse with eccentricity e. Then the value of $$\frac{5}{e^{2}}$$ is equal to __________.

We need to find $$\frac{5}{e^2}$$ where the point (2h+1, 3k+2) lies on an ellipse, with (h,k) on the circle $$x^2 + y^2 = 4$$.

Since the point lies on the circle, let $$h = 2\cos\theta$$ and $$k = 2\sin\theta$$. Then the corresponding point on the ellipse is $$X = 2h + 1 = 4\cos\theta + 1$$ and $$Y = 3k + 2 = 6\sin\theta + 2$$.

Solving for $$\cos\theta$$ and $$\sin\theta$$ gives $$\cos\theta = \frac{X-1}{4}$$ and $$\sin\theta = \frac{Y-2}{6}$$. Substituting into $$\cos^2\theta + \sin^2\theta = 1$$ yields $$\frac{(X-1)^2}{16} + \frac{(Y-2)^2}{36} = 1$$, which is an ellipse with center (1,2) and semi-axes $$a = 6$$ (along Y) and $$b = 4$$ (along X), where $$a > b$$.

The eccentricity satisfies $$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = 1 - \frac{4}{9} = \frac{5}{9}$$.

Since $$e^2 = \frac{5}{9}$$, it follows that $$\frac{5}{e^2} = \frac{5}{5/9} = 9$$.

Therefore, $$\frac{5}{e^2} = $$ 9.