An ellipse passes through the foci of the hyperbola, $$9x^2 - 4y^2 = 36$$ and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is $$\frac{1}{2}$$, then which of the following points does not lie on the ellipse?

The given equation is $$9x^2 - 4y^2 = 36$$. 

$$\frac{x^2}{4} - \frac{y^2}{9} = 1$$

Comparing with the standard form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, $$a^2 = 4 \implies a = 2$$, $$b^2 = 9 \implies b = 3$$

$$e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$$

The coordinates of the foci are $$(\pm ae_H, 0)$$:  $$\text{Foci} = \left(\pm 2 \cdot \frac{\sqrt{13}}{2}, 0\right) = (\pm \sqrt{13}, 0)$$

Let the ellipse be $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$.

The ellipse passes through the hyperbola's foci $$(\pm \sqrt{13}, 0)$$. 

Therefore, the semi-major axis is $$A = \sqrt{13} \implies A^2 = 13$$.

$$e_E \cdot e_H = \frac{1}{2} \implies e_E \cdot \frac{\sqrt{13}}{2} = \frac{1}{2} \implies e_E = \frac{1}{\sqrt{13}}$$

For an ellipse, $$B^2 = A^2(1 - e_E^2)$$. $$B^2 = 13 \left( 1 - \frac{1}{13} \right) = 13 \left( \frac{12}{13} \right) = 12$$

Thus, the equation of the ellipse is $$\frac{x^2}{13} + \frac{y^2}{12} = 1$$

The point that does not lie on the ellipse from the given options is $$\left( \frac{\sqrt{13}}{2}, \frac{\sqrt{3}}{2} \right)$$