If a curve passes through the point $$(2, \frac{7}{2})$$ and has slope $$\left(1 - \frac{1}{x^2}\right)$$ at any point $$(x, y)$$ on it, then the ordinate of the point on the curve whose abscissa is -2 is :

The slope of the curve at any point $$(x, y)$$ is given as $$\frac{dy}{dx} = 1 - \frac{1}{x^2}$$. This is a differential equation that we can solve by separating the variables.

Rewrite the equation as $$dy = \left(1 - \frac{1}{x^2}\right) dx$$. Now, integrate both sides:

$$\int dy = \int \left(1 - x^{-2}\right) dx$$

The left side integrates to $$y$$. For the right side, integrate term by term:

$$\int 1 dx = x$$

$$\int -x^{-2} dx = -\left(\frac{x^{-1}}{-1}\right) = -\left(-\frac{1}{x}\right) = \frac{1}{x}$$

So, the right side becomes $$x + \frac{1}{x} + C$$, where $$C$$ is the constant of integration. Therefore, the equation of the curve is:

$$y = x + \frac{1}{x} + C$$

The curve passes through the point $$(2, \frac{7}{2})$$. Substitute $$x = 2$$ and $$y = \frac{7}{2}$$ to find $$C$$:

$$\frac{7}{2} = 2 + \frac{1}{2} + C$$

Simplify the right side:

$$2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$$

So,

$$\frac{7}{2} = \frac{5}{2} + C$$

Subtract $$\frac{5}{2}$$ from both sides:

$$\frac{7}{2} - \frac{5}{2} = C$$

$$\frac{2}{2} = C$$

$$C = 1$$

Thus, the equation of the curve is:

$$y = x + \frac{1}{x} + 1$$

We need to find the ordinate (y-coordinate) when the abscissa (x-coordinate) is $$-2$$. Substitute $$x = -2$$:

$$y = (-2) + \frac{1}{(-2)} + 1$$

Simplify each term:

$$-2 + \left(-\frac{1}{2}\right) + 1$$

Combine the integers first:

$$-2 + 1 = -1$$

Then add the fraction:

$$-1 - \frac{1}{2} = -\left(1 + \frac{1}{2}\right) = -\left(\frac{2}{2} + \frac{1}{2}\right) = -\frac{3}{2}$$

So, the ordinate is $$-\frac{3}{2}$$.

Comparing with the options, $$-\frac{3}{2}$$ corresponds to option A.

Hence, the correct answer is Option A.