Given parabola

$(y-2)^2 = x-1$

Rewrite in terms of $x$:

$x = (y-2)^2 + 1$

Find the tangent at $(2,3)$

First verify the point lies on the parabola:

$(3-2)^2 = 2-1 = 1$

so $(2,3)$ is on the curve.

Differentiate implicitly:

$(y-2)^2 = x-1$ $2(y-2)\frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2(y-2)}$

At $(2,3)$, we have $y-2=1$, hence

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

Equation of tangent through $(2,3)$:

$y-3 = \frac12(x-2)$

$2y-6 = x-2$ $x = 2y-4$

Find intersections with the $x$-axis

The $x$-axis is $y=0$.

• With the parabola: $(0-2)^2 = x-1$ $4 = x-1$ $x=5$ So point is $(5,0)$.

• With the tangent: $0-3 = \frac12(x-2)$ $-3 = \frac12(x-2)$ $-6 = x-2$ $x=-4$ So point is $(-4,0)$.

Understand the bounded region

The closed region is bounded by:

• the tangent from $(-4,0)$ to $(2,3)$,
• the parabola from $(2,3)$ to $(5,0)$,
• the $x$-axis from $(5,0)$ back to $(-4,0)$.

Since both curves are conveniently written as $x$ in terms of $y$, integrate with respect to $y$ from $y=0$ to $y=3$.

For a horizontal strip:

• right boundary: parabola $x = (y-2)^2+1 = y^2-4y+5$
• left boundary: tangent $x = 2y-4$

So area is

$A = \int_0^3 \Big[(y^2-4y+5) - (2y-4)\Big] \, dy$

$A = \int_0^3 (y^2 - 6y + 9)\,dy$

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

Evaluate the integral

$A = \left[\frac{y^3}{3} - 3y^2 + 9y\right]_0^3$

At $y=3$: $\frac{27}{3} - 3(9) + 27 = 9 - 27 + 27 = 9$

At $y=0$: $0$

Therefore,

$A = 9$

Check options

The correct option is:

$\boxed{\text{B: }9}$