Key Concepts and Formulas

• Area between curves (integration with respect to x): If a region is bounded by $y = f(x)$, $y = g(x)$, $x = a$, and $x = b$, where $f(x) \ge g(x)$ for $x \in [a,b]$, the area is given by $A = \int_a^b [f(x) - g(x)] \, dx$.
• Equation of Tangent: The equation of the tangent to the curve $y = f(x)$ at the point $(x_0, y_0)$ is given by $y - y_0 = f'(x_0)(x - x_0)$. Implicit differentiation might be required to find the derivative.
• Finding Intersection Points: To find the points where two curves intersect, we set their equations equal to each other and solve for the variables.

Step-by-Step Solution

Step 1: Find the equation of the tangent to the parabola at (2, 3)

We are given the parabola $(y - 2)^2 = x - 1$. We need to find the equation of the tangent at the point (2, 3). To do this, we will use implicit differentiation to find $\frac{dy}{dx}$.

Differentiating both sides of the equation with respect to $x$, we get: $2(y - 2) \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2(y - 2)}$

Now, we evaluate the derivative at the point (2, 3): $\frac{dy}{dx}\Big|_{(2, 3)} = \frac{1}{2(3 - 2)} = \frac{1}{2}$

So, the slope of the tangent at (2, 3) is $\frac{1}{2}$. Using the point-slope form of a line, the equation of the tangent is: $y - 3 = \frac{1}{2}(x - 2)$ $y = \frac{1}{2}x - 1 + 3$ $y = \frac{1}{2}x + 2$

Step 2: Find the intersection point of the tangent and the x-axis

To find where the tangent intersects the x-axis, we set $y = 0$: $0 = \frac{1}{2}x + 2$ $\frac{1}{2}x = -2$ $x = -4$

So, the tangent intersects the x-axis at the point (-4, 0).

Step 3: Find the intersection point of the parabola and the x-axis

To find where the parabola intersects the x-axis, we set $y = 0$: $(0 - 2)^2 = x - 1$ $4 = x - 1$ $x = 5$

So, the parabola intersects the x-axis at the point (5, 0).

Step 4: Express x in terms of y for both the parabola and the tangent

We already have the tangent equation as $y = \frac{1}{2}x + 2$. Solving for $x$, we get: $x = 2y - 4$

For the parabola, we have $(y - 2)^2 = x - 1$. Solving for $x$, we get: $x = (y - 2)^2 + 1$

Step 5: Set up the integral to find the area

The region is bounded by the parabola $x = (y - 2)^2 + 1$, the tangent $x = 2y - 4$, and the x-axis ($y = 0$). The tangent and parabola intersect at (2, 3), so $y=3$ at their intersection. We will integrate with respect to y from $y = 0$ to $y = 3$. The area is given by:

$A = \int_0^3 [(2y - 4) - ((y - 2)^2 + 1)] \, dy$ $A = \int_0^3 [2y - 4 - (y^2 - 4y + 4) - 1] \, dy$ $A = \int_0^3 [2y - 4 - y^2 + 4y - 4 - 1] \, dy$ $A = \int_0^3 [-y^2 + 6y - 9] \, dy$ $A = \int_0^3 -(y - 3)^2 \, dy$ $A = -\int_0^3 (y^2 - 6y + 9) \, dy$ $A = -\left[\frac{y^3}{3} - 3y^2 + 9y\right]_0^3$ $A = -\left[\frac{3^3}{3} - 3(3^2) + 9(3) - 0\right]$ $A = -\left[\frac{27}{3} - 27 + 27\right]$ $A = -[9 - 27 + 27]$ $A = -9$

Since area must be positive, we take the absolute value: $A = |-9| = 9$

Step 6: Re-evaluate integral to find error in previous calculation

Upon reviewing the previous step, there appears to be an error in the setup of the integral. We need to find the area BETWEEN the curves. The equation should be $A = \int_0^3 [((y - 2)^2 + 1) - (2y - 4)] \, dy$ $A = \int_0^3 [(y^2 - 4y + 4 + 1) - (2y - 4)] \, dy$ $A = \int_0^3 [y^2 - 4y + 5 - 2y + 4] \, dy$ $A = \int_0^3 [y^2 - 6y + 9] \, dy$ $A = \int_0^3 (y-3)^2 \, dy$ $A = \left[ \frac{(y-3)^3}{3} \right]_0^3$ $A = \frac{(3-3)^3}{3} - \frac{(0-3)^3}{3}$ $A = 0 - \frac{(-3)^3}{3}$ $A = 0 - \frac{-27}{3}$ $A = 9$

Step 7: Re-evaluate integral to find error in previous calculation

There are conflicting answers from the previous calculation. Let's try calculating the integral with the correct order of subtraction: $A = \int_{0}^{3} [(y-2)^2 + 1 - (2y - 4)] dy$ $A = \int_{0}^{3} [y^2 - 4y + 4 + 1 - 2y + 4] dy$ $A = \int_{0}^{3} [y^2 - 6y + 9] dy$ $A = \int_{0}^{3} (y-3)^2 dy$ $A = \left[ \frac{1}{3} (y-3)^3 \right]_{0}^{3}$ $A = \frac{1}{3} [(3-3)^3 - (0-3)^3]$ $A = \frac{1}{3} [0 - (-27)] = \frac{1}{3} [27] = 9$

Step 8: Verify the integration limits

The parabola and the line intersect when $(y-2)^2 + 1 = 2y - 4$ $y^2 - 4y + 4 + 1 = 2y - 4$ $y^2 - 6y + 9 = 0$ $(y-3)^2 = 0$ $y = 3$ The tangent intersects the x-axis at $y = 0$. The parabola intersects the x-axis when $y=0$, at $x = 5$. Thus the integration limits are from $y = 0$ to $y = 3$.

Step 9: Final Check

The area is given by the integral of the difference between the parabola and the tangent, integrated from y = 0 to y = 3. The setup is: $\int_{0}^{3} [((y-2)^2 + 1) - (2y - 4)] dy = 9$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when expanding and simplifying the integrand.
• Correct Order of Subtraction: Ensure you are subtracting the correct functions to get a positive area. If you get a negative area, take the absolute value.
• Integration Limits: Double-check the integration limits to ensure they accurately represent the region's boundaries.

Summary

To find the area of the region, we first found the equation of the tangent to the parabola at the given point using implicit differentiation. We then expressed both the parabola and the tangent as functions of y and integrated the difference between these functions with respect to y from 0 to 3. The area of the region is 9.

Final Answer

The final answer is \boxed{9}, which corresponds to option (B).