The area of the region bounded by the parabola $$(y-2)^2 = (x-1)$$, the tangent to it at the point whose ordinate is 3 and the x-axis, is:

We have the parabola $$ (y-2)^2 = x-1 $$.

First we locate the point on the parabola whose ordinate (that is, its $$y$$-coordinate) equals $$3$$. Putting $$y = 3$$ in the equation of the curve gives

$$ (3-2)^2 = x-1 \;\Longrightarrow\; 1 = x-1 \;\Longrightarrow\; x = 2. $$

Thus the point of contact is $$P(2,3).$$

To write the tangent at $$P$$, we need its slope. Differentiating the parabola implicitly with respect to $$x$$ gives

$$ 2(y-2)\frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{2(y-2)}. $$

At $$y = 3$$ we obtain

$$\left.\frac{dy}{dx}\right|_{P} = \frac{1}{2(3-2)} = \frac12. $$

Hence the tangent through $$(2,3)$$ is

$$ y-3 = \frac12\,(x-2)\;\Longrightarrow\; y = \frac{x}{2}+2. $$

Because we shall use horizontal strips (integration with respect to $$y$$), it is convenient to express every bounding curve as $$x$$ in terms of $$y$$. Solving the tangent for $$x$$ we get

$$ y = \frac{x}{2}+2 \;\Longrightarrow\; x = 2y-4. $$

Similarly, from the parabola,

$$ (y-2)^2 = x-1 \;\Longrightarrow\; x = (y-2)^2 + 1. $$

The third boundary is the $$x$$-axis, i.e. $$y = 0.$$

We next determine the $$y$$-range over which all three curves enclose a single region. Along the $$x$$-axis the lowest value of $$y$$ is obviously $$y=0.$$ The tangent and the parabola meet only at the point $$P$$, whose ordinate is $$y=3.$$ Thus every horizontal segment between $$y=0$$ and $$y=3$$ starts on the tangent (left edge) and ends on the parabola (right edge); above $$y=3$$ the two curves diverge and no longer form a closed region with the $$x$$-axis. Hence the required strip-wise integration limits are $$0\le y\le 3.$$

For a fixed $$y$$ in this interval the left boundary is $$x_{\text{tangent}} = 2y-4,$$ while the right boundary is $$x_{\text{parabola}} = (y-2)^2 + 1.$$ The horizontal width of the slice is therefore

$$ x_{\text{parabola}} - x_{\text{tangent}} \;=\; \bigl((y-2)^2 + 1\bigr)\;-\;(2y-4) \;=\; (y^2 - 4y +4) +1 -2y +4 \;=\; y^2 - 6y + 9 \;=\; (y-3)^2. $$

Notice that $$(y-3)^2$$ is always non-negative and becomes $$0$$ precisely at $$y=3$$, exactly as expected for tangency.

The area $$A$$ can now be written as the definite integral of this width from $$y=0$$ up to $$y=3$$:

$$ \begin{aligned} A &= \int_{0}^{3} (y-3)^2 \,dy. \end{aligned} $$

We expand the integrand and integrate term by term:

$$ \begin{aligned} A &= \int_{0}^{3} \bigl(y^2 - 6y + 9\bigr)\,dy \\[4pt] &= \int_{0}^{3} y^2\,dy \;-\; 6\int_{0}^{3} y\,dy \;+\; 9\int_{0}^{3} dy \\[4pt] &= \left[\frac{y^{3}}{3}\right]_{0}^{3} \;-\; 6\left[\frac{y^{2}}{2}\right]_{0}^{3} \;+\; 9\,[y]_{0}^{3}. \end{aligned} $$

Evaluating each bracket gives

$$ \begin{aligned} A &= \Bigl(\tfrac{27}{3} - 0\Bigr) \;-\; 6\Bigl(\tfrac{9}{2} - 0\Bigr) \;+\; 9\,(3 - 0) \\[4pt] &= 9 \;-\; 6\times \tfrac{9}{2} \;+\; 27 \\[4pt] &= 9 \;-\; 27 \;+\; 27 \\[4pt] &= 9. \end{aligned} $$

Thus the area enclosed by the parabola, its tangent at the point of ordinate $$3$$, and the $$x$$-axis equals $$9\text{ square units}.$$

Hence, the correct answer is Option C.