Let $$P(3, 3)$$ be a point on the hyperbola, $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$. If the normal to it at P intersects the $$x$$-axis at (9, 0) and $$e$$ is its eccentricity, then the ordered pair $$(a^2, e^2)$$ is equal to:

The hyperbola is given by $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ and the point $$P(3,3)$$ lies on it. Substituting the coordinates of P, we have

$$\dfrac{3^{2}}{a^{2}}-\dfrac{3^{2}}{b^{2}}=1 \;\;\Longrightarrow\;\; \dfrac{9}{a^{2}}-\dfrac{9}{b^{2}}=1.$$

To find the normal at P, we first need the slope of the tangent. Differentiating the hyperbola implicitly with respect to x, we obtain

$$\dfrac{2x}{a^{2}}-\dfrac{2y}{b^{2}}\dfrac{dy}{dx}=0.$$ Solving for $$\dfrac{dy}{dx}$$ gives

$$\dfrac{dy}{dx}=\dfrac{b^{2}x}{a^{2}y}.$$

At the point $$P(3,3)$$, the slope of the tangent is therefore

$$m_{t}=\left.\dfrac{dy}{dx}\right|_{(3,3)}=\dfrac{b^{2}\cdot3}{a^{2}\cdot3}=\dfrac{b^{2}}{a^{2}}.$$

The slope of the normal is the negative reciprocal of the slope of the tangent, so

$$m_{n}=-\,\dfrac{a^{2}}{b^{2}}.$$

Using the point-slope form, the equation of the normal through $$P(3,3)$$ is

$$y-3=-\dfrac{a^{2}}{b^{2}}\,(x-3).$$

The normal meets the $$x$$-axis where $$y=0$$. Setting $$y=0$$ and solving for $$x$$, we get

$$0-3=-\dfrac{a^{2}}{b^{2}}\,(x-3) \;\;\Longrightarrow\;\; 3=\dfrac{a^{2}}{b^{2}}\,(x-3) \;\;\Longrightarrow\;\; x-3=\dfrac{3b^{2}}{a^{2}} \;\;\Longrightarrow\;\; x=3+\dfrac{3b^{2}}{a^{2}}.$$

But the normal is known to cut the $$x$$-axis at $$(9,0)$$, so

$$9=3+\dfrac{3b^{2}}{a^{2}} \;\;\Longrightarrow\;\; 6=\dfrac{3b^{2}}{a^{2}} \;\;\Longrightarrow\;\; 2=\dfrac{b^{2}}{a^{2}} \;\;\Longrightarrow\;\; b^{2}=2a^{2}.$$

Substituting $$b^{2}=2a^{2}$$ in the earlier relation $$\dfrac{9}{a^{2}}-\dfrac{9}{b^{2}}=1$$, we obtain

$$\dfrac{9}{a^{2}}-\dfrac{9}{2a^{2}}=1 \;\;\Longrightarrow\;\; \dfrac{18-9}{2a^{2}}=1 \;\;\Longrightarrow\;\; \dfrac{9}{2a^{2}}=1 \;\;\Longrightarrow\;\; 2a^{2}=9 \;\;\Longrightarrow\;\; a^{2}=\dfrac{9}{2}.$$

Now, the eccentricity $$e$$ of the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ satisfies the standard formula $$e^{2}=1+\dfrac{b^{2}}{a^{2}}.$$ Since $$\dfrac{b^{2}}{a^{2}}=2$$, we get

$$e^{2}=1+2=3.$$

Thus the ordered pair is $$\left(a^{2},\,e^{2}\right)=\left(\dfrac{9}{2},\,3\right).$$

Hence, the correct answer is Option A.