The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola $$\frac{x^2}{4} - \frac{y^2}{5} = 1$$, meets the x-axis and y-axis at A and B, respectively. Then $$OA^2 - OB^2$$, where O is the origin, equals:

The given hyperbola is $$\frac{x^2}{4} - \frac{y^2}{5} = 1$$. Comparing this with the standard form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, we find $$a^2 = 4$$ and $$b^2 = 5$$, so $$a = 2$$ and $$b = \sqrt{5}$$.

The foci of the hyperbola are at $$(\pm c, 0)$$, where $$c = \sqrt{a^2 + b^2} = \sqrt{4 + 5} = \sqrt{9} = 3$$. Thus, the foci are $$(3, 0)$$ and $$(-3, 0)$$.

We need the extremity of the latus rectum in the first quadrant. The latus rectum through the focus $$(3, 0)$$ is perpendicular to the x-axis, so its equation is $$x = 3$$. Substituting $$x = 3$$ into the hyperbola equation:

$$\frac{(3)^2}{4} - \frac{y^2}{5} = 1 \implies \frac{9}{4} - \frac{y^2}{5} = 1.$$

Rearranging terms:

$$\frac{9}{4} - 1 = \frac{y^2}{5} \implies \frac{5}{4} = \frac{y^2}{5}.$$

Solving for $$y^2$$:

$$y^2 = \frac{5}{4} \times 5 = \frac{25}{4},$$

so $$y = \pm \frac{5}{2}$$. The extremities are $$(3, \frac{5}{2})$$ and $$(3, -\frac{5}{2})$$. The first quadrant point is $$(3, \frac{5}{2})$$.

The tangent to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ at a point $$(x_1, y_1)$$ is given by $$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$$. Substituting $$a^2 = 4$$, $$b^2 = 5$$, $$x_1 = 3$$, and $$y_1 = \frac{5}{2}$$:

$$\frac{x \cdot 3}{4} - \frac{y \cdot \frac{5}{2}}{5} = 1.$$

Simplifying the second term:

$$\frac{y \cdot \frac{5}{2}}{5} = \frac{5y}{2} \times \frac{1}{5} = \frac{y}{2},$$

so the equation becomes:

$$\frac{3x}{4} - \frac{y}{2} = 1.$$

Multiplying through by 4 to clear denominators:

$$4 \times \frac{3x}{4} - 4 \times \frac{y}{2} = 4 \times 1 \implies 3x - 2y = 4.$$

Thus, the tangent equation is $$3x - 2y = 4$$.

This tangent meets the x-axis at A (where $$y = 0$$):

$$3x - 2(0) = 4 \implies 3x = 4 \implies x = \frac{4}{3},$$

so A is $$\left(\frac{4}{3}, 0\right)$$.

It meets the y-axis at B (where $$x = 0$$):

$$3(0) - 2y = 4 \implies -2y = 4 \implies y = -2,$$

so B is $$(0, -2)$$.

O is the origin $$(0, 0)$$. Now, compute $$OA^2$$ and $$OB^2$$:

OA is the distance from O to A $$\left(\frac{4}{3}, 0\right)$$, so $$OA = \frac{4}{3}$$ and $$OA^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$.

OB is the distance from O to B $$(0, -2)$$, so $$OB = 2$$ and $$OB^2 = 2^2 = 4$$.

Therefore,

$$OA^2 - OB^2 = \frac{16}{9} - 4 = \frac{16}{9} - \frac{36}{9} = -\frac{20}{9}.$$

Hence, the correct answer is Option A.