If the tangent at a point on the ellipse $$\frac{x^2}{27} + \frac{y^2}{3} = 1$$ meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is

The equation of the ellipse is $$\frac{x^2}{27} + \frac{y^2}{3} = 1$$. Here, $$a^2 = 27$$ and $$b^2 = 3$$, so $$a = 3\sqrt{3}$$ and $$b = \sqrt{3}$$. Consider a point $$(x_1, y_1)$$ on the ellipse. The equation of the tangent at this point is given by $$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$$, which simplifies to $$\frac{x x_1}{27} + \frac{y y_1}{3} = 1$$.

This tangent intersects the coordinate axes. To find the x-intercept (point A), set $$y = 0$$:

$$\frac{x x_1}{27} + \frac{0 \cdot y_1}{3} = 1 \implies \frac{x x_1}{27} = 1 \implies x = \frac{27}{x_1}$$

So, point A is at $$\left( \frac{27}{x_1}, 0 \right)$$.

To find the y-intercept (point B), set $$x = 0$$:

$$\frac{0 \cdot x_1}{27} + \frac{y y_1}{3} = 1 \implies \frac{y y_1}{3} = 1 \implies y = \frac{3}{y_1}$$

So, point B is at $$\left( 0, \frac{3}{y_1} \right)$$.

Triangle OAB has vertices at O(0,0), A$$\left( \frac{27}{x_1}, 0 \right)$$, and B$$\left( 0, \frac{3}{y_1} \right)$$. The area of this triangle is given by $$\frac{1}{2} \times \text{base} \times \text{height}$$, where the base is the x-intercept and the height is the y-intercept. Thus, the area is:

$$\text{Area} = \frac{1}{2} \times \left| \frac{27}{x_1} \right| \times \left| \frac{3}{y_1} \right|$$

Since the ellipse is symmetric and we are minimizing the area, we can consider $$x_1 > 0$$ and $$y_1 > 0$$ (first quadrant) without loss of generality, as the area depends on the absolute values. Therefore,

$$\text{Area} = \frac{1}{2} \times \frac{27}{x_1} \times \frac{3}{y_1} = \frac{1}{2} \times \frac{81}{x_1 y_1} = \frac{81}{2} \cdot \frac{1}{x_1 y_1}$$

The point $$(x_1, y_1)$$ lies on the ellipse, so it satisfies $$\frac{x_1^2}{27} + \frac{y_1^2}{3} = 1$$. Solving for $$y_1^2$$:

$$\frac{x_1^2}{27} + \frac{y_1^2}{3} = 1 \implies \frac{y_1^2}{3} = 1 - \frac{x_1^2}{27} \implies y_1^2 = 3 \left(1 - \frac{x_1^2}{27}\right) = 3 - \frac{x_1^2}{9}$$

Thus, $$y_1 = \sqrt{3 - \frac{x_1^2}{9}} = \frac{\sqrt{27 - x_1^2}}{3}$$, since $$y_1 > 0$$.

Now, $$x_1 y_1 = x_1 \cdot \frac{\sqrt{27 - x_1^2}}{3} = \frac{1}{3} x_1 \sqrt{27 - x_1^2}$$. Let $$g(x_1) = x_1 \sqrt{27 - x_1^2}$$, so $$x_1 y_1 = \frac{1}{3} g(x_1)$$. To minimize the area, we need to maximize $$x_1 y_1$$ (since area is inversely proportional to $$x_1 y_1$$).

Maximize $$g(x_1) = x_1 \sqrt{27 - x_1^2}$$ for $$0 < x_1 < 3\sqrt{3}$$. To simplify, maximize $$h(x_1) = [g(x_1)]^2 = (x_1 \sqrt{27 - x_1^2})^2 = x_1^2 (27 - x_1^2) = 27x_1^2 - x_1^4$$.

Set $$u = x_1^2$$, so $$h(u) = 27u - u^2$$ for $$0 < u < 27$$. This is a quadratic function in $$u$$, and since the coefficient of $$u^2$$ is negative, it opens downwards. The maximum occurs at $$u = -\frac{b}{2a} = -\frac{27}{2(-1)} = \frac{27}{2} = 13.5$$.

Since $$13.5$$ is in $$(0, 27)$$, substitute $$u = 13.5$$:

$$h(u) = 27(13.5) - (13.5)^2 = 364.5 - 182.25 = 182.25 = (13.5)^2$$

Thus, the maximum of $$h(x_1)$$ is $$(13.5)^2$$, so the maximum of $$g(x_1)$$ is $$13.5$$ (since $$g(x_1) > 0$$).

Therefore, the maximum of $$x_1 y_1 = \frac{1}{3} \times 13.5 = \frac{13.5}{3} = 4.5$$.

Now, the minimum area is:

$$\text{Area} = \frac{81}{2} \cdot \frac{1}{x_1 y_1} = \frac{81}{2} \cdot \frac{1}{4.5}$$

Since $$4.5 = \frac{9}{2}$$,

$$\frac{1}{4.5} = \frac{1}{\frac{9}{2}} = \frac{2}{9}$$

So,

$$\text{Area} = \frac{81}{2} \times \frac{2}{9} = \frac{81}{9} = 9$$

This minimum area occurs when $$x_1^2 = 13.5 = \frac{27}{2}$$, so $$x_1 = \sqrt{\frac{27}{2}} = \frac{3\sqrt{6}}{2}$$, and $$y_1 = \sqrt{3 - \frac{13.5}{9}} = \sqrt{3 - 1.5} = \sqrt{1.5} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$$. Then $$x_1 y_1 = \frac{3\sqrt{6}}{2} \times \frac{\sqrt{6}}{2} = \frac{3 \times 6}{4} = \frac{18}{4} = 4.5$$, confirming the calculation.

The intercepts are A$$\left( \frac{27}{\frac{3\sqrt{6}}{2}}, 0 \right) = \left( \frac{27 \times 2}{3\sqrt{6}}, 0 \right) = \left( \frac{54}{3\sqrt{6}}, 0 \right) = \left( \frac{18}{\sqrt{6}}, 0 \right) = \left( 3\sqrt{6}, 0 \right)$$ and B$$\left( 0, \frac{3}{\frac{\sqrt{6}}{2}} \right) = \left( 0, \frac{3 \times 2}{\sqrt{6}} \right) = \left( 0, \frac{6}{\sqrt{6}} \right) = \left( 0, \sqrt{6} \right)$$. The area is $$\frac{1}{2} \times 3\sqrt{6} \times \sqrt{6} = \frac{1}{2} \times 3 \times 6 = \frac{1}{2} \times 18 = 9$$.

At the endpoints, when $$x_1 = 3\sqrt{3}$$ or $$y_1 = \sqrt{3}$$, the area becomes infinite, so 9 is indeed the minimum. Comparing with the options, 9 corresponds to option C.

Hence, the correct answer is Option C.