1. Write the parabola in a convenient form

Given: $(y-2)^2 = x-1$ So, $x = (y-2)^2 + 1$ This is a right-opening parabola with vertex at $(1,2)$.

2. Find the point on the parabola whose ordinate is 3

Ordinate means the $y$-coordinate. So put $y=3$: $x = (3-2)^2 + 1 = 1+1 = 2$ Hence the point is: $(2,3)$

3. Find the tangent at $(2,3)$

Differentiate implicitly: $(y-2)^2 = x-1$ $2(y-2)\frac{dy}{dx} = 1$ So, $\frac{dy}{dx} = \frac{1}{2(y-2)}$ At $y=3$, $\frac{dy}{dx} = \frac{1}{2}$ Thus tangent equation at $(2,3)$ is: $y-3 = \frac{1}{2}(x-2)$ $2y-6 = x-2$ $x = 2y-4$

4. Understand the bounded region

The region is bounded by:

• the parabola: $x=(y-2)^2+1$
• the tangent: $x=2y-4$
• the $x$-axis: $y=0$

We should find where the tangent and parabola touch: that is at $y=3$. So the enclosed region runs from $y=0$ to $y=3$.

Now compare left/right curves in this interval.

For example at $y=1$:

• parabola: $x=(1-2)^2+1=2$
• tangent: $x=2(1)-4=-2$

So parabola is on the right and tangent is on the left.

Therefore area is: $A = \int_0^3 \left[\big((y-2)^2+1\big) - (2y-4)\right]dy$

5. Simplify the integrand

$(y-2)^2+1-(2y-4) = (y^2-4y+4)+1-2y+4$ $= y^2 - 6y + 9$ $= (y-3)^2$

So, $A = \int_0^3 (y-3)^2 \, dy$

6. Evaluate the integral

$\int (y-3)^2 dy = \frac{(y-3)^3}{3}$ Thus, $A = \left[\frac{(y-3)^3}{3}\right]_0^3$ $= \frac{0^3}{3} - \frac{(-3)^3}{3}$ $= 0 - \left(\frac{-27}{3}\right)$ $= 9$

7. Final answer

The required area is: $\boxed{9}$

So the correct option is A.