The minimum area of a triangle formed by any tangent to the ellipse $$\frac{x^2}{16} + \frac{y^2}{81} = 1$$ and the co-ordinate axes is:

We are given the ellipse equation: $$\frac{x^2}{16} + \frac{y^2}{81} = 1$$. Here, $$a^2 = 16$$ so $$a = 4$$, and $$b^2 = 81$$ so $$b = 9$$. We need to find the minimum area of the triangle formed by any tangent to this ellipse and the coordinate axes.

First, recall the equation of a tangent to the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ in terms of slope $$m$$. It is given by: $$y = mx \pm \sqrt{a^2 m^2 + b^2}$$ Substituting $$a^2 = 16$$ and $$b^2 = 81$$, we get: $$y = mx \pm \sqrt{16m^2 + 81}$$

This tangent intersects the coordinate axes. To find the x-intercept, set $$y = 0$$: $$0 = mx \pm \sqrt{16m^2 + 81}$$ Solving for $$x$$: $$mx = \mp \sqrt{16m^2 + 81}$$ $$x = \frac{\mp \sqrt{16m^2 + 81}}{m}$$ The absolute value of the x-intercept is: $$\left| x \right| = \frac{\sqrt{16m^2 + 81}}{\left| m \right|}$$

To find the y-intercept, set $$x = 0$$: $$y = m(0) \pm \sqrt{16m^2 + 81} = \pm \sqrt{16m^2 + 81}$$ The absolute value of the y-intercept is: $$\left| y \right| = \sqrt{16m^2 + 81}$$

The area $$A$$ of the triangle formed by the tangent and the axes is half the product of the intercepts: $$A = \frac{1}{2} \times \left| x \right| \times \left| y \right| = \frac{1}{2} \times \frac{\sqrt{16m^2 + 81}}{\left| m \right|} \times \sqrt{16m^2 + 81}$$ Simplifying: $$A = \frac{1}{2} \times \frac{16m^2 + 81}{\left| m \right|}$$ Since the area is positive and the expression depends on $$m^2$$, we can assume $$m > 0$$ without loss of generality. Thus: $$A = \frac{1}{2} \times \frac{16m^2 + 81}{m} = \frac{1}{2} \left( 16m + \frac{81}{m} \right)$$

To minimize $$A$$, we take the derivative with respect to $$m$$ and set it to zero. Let: $$A(m) = \frac{1}{2} \left( 16m + \frac{81}{m} \right)$$ The derivative is: $$A'(m) = \frac{1}{2} \left( 16 - \frac{81}{m^2} \right)$$ Set $$A'(m) = 0$$: $$\frac{1}{2} \left( 16 - \frac{81}{m^2} \right) = 0$$ $$16 - \frac{81}{m^2} = 0$$ $$16 = \frac{81}{m^2}$$ $$m^2 = \frac{81}{16}$$ $$m = \frac{9}{4} \quad \text{(since $$m > 0$$)}$$

To confirm this is a minimum, check the second derivative: $$A''(m) = \frac{1}{2} \left( \frac{162}{m^3} \right) = \frac{81}{m^3}$$ For $$m = \frac{9}{4} > 0$$, $$A''(m) > 0$$, so it is a minimum.

Now substitute $$m = \frac{9}{4}$$ into the area formula: $$A = \frac{1}{2} \left( 16 \times \frac{9}{4} + \frac{81}{\frac{9}{4}} \right)$$ Simplify inside the parentheses: $$16 \times \frac{9}{4} = 36$$ $$\frac{81}{\frac{9}{4}} = 81 \times \frac{4}{9} = 36$$ So: $$A = \frac{1}{2} (36 + 36) = \frac{1}{2} \times 72 = 36$$

Therefore, the minimum area is 36. Comparing with the options: A. 12, B. 18, C. 26, D. 36. Hence, the correct answer is Option D.