For which of the following curves, the line $$x + \sqrt{3}y = 2\sqrt{3}$$ is the tangent at the point $$\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$$?

We need to find which curve has the line $$x + \sqrt{3}y = 2\sqrt{3}$$ as a tangent at the point $$\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$$.

Consider the ellipse $$x^2 + 9y^2 = 9$$. First, we verify that the point lies on this curve: $$\left(\frac{3\sqrt{3}}{2}\right)^2 + 9\left(\frac{1}{2}\right)^2 = \frac{27}{4} + \frac{9}{4} = \frac{36}{4} = 9$$. The point is indeed on the ellipse.

The equation of the tangent to the ellipse $$\frac{x^2}{9} + \frac{y^2}{1} = 1$$ at the point $$(x_1, y_1)$$ is $$\frac{xx_1}{9} + yy_1 = 1$$.

Substituting $$(x_1, y_1) = \left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$$, the tangent is $$\frac{x \cdot \frac{3\sqrt{3}}{2}}{9} + y \cdot \frac{1}{2} = 1$$, which simplifies to $$\frac{\sqrt{3}x}{6} + \frac{y}{2} = 1$$.

Multiplying through by 6 gives $$\sqrt{3}x + 3y = 6$$. Dividing by $$\sqrt{3}$$, we get $$x + \sqrt{3}y = \frac{6}{\sqrt{3}} = 2\sqrt{3}$$.

This matches the given tangent line $$x + \sqrt{3}y = 2\sqrt{3}$$.

Therefore, the curve is $$x^2 + 9y^2 = 9$$.