Key Concepts and Formulas

• Area between curves: The area between two curves $x = f(y)$ and $x = g(y)$ from $y=a$ to $y=b$, where $f(y) \ge g(y)$, is given by $A = \int_{a}^{b} (f(y) - g(y)) \, dy$.
• Equation of a tangent line: Given a curve and a point $(x_0, y_0)$ on the curve, the equation of the tangent line is $y - y_0 = m(x - x_0)$, where $m$ is the derivative of the curve evaluated at $x_0$.
• Finding the point of tangency: If the ordinate is given, substitute the value into the equation of the curve to find the corresponding abscissa.

Step-by-Step Solution

Step 1: Find the point of tangency

The equation of the parabola is $(y-2)^2 = x-1$. The ordinate of the point of tangency is $y=3$. Substituting $y=3$ into the equation of the parabola: $(3-2)^2 = x - 1$ $1^2 = x - 1$ $1 = x - 1$ $x = 2$ So, the point of tangency is $(2, 3)$.

Step 2: Find the equation of the tangent line

We can express $x$ in terms of $y$ for the parabola: $x = (y-2)^2 + 1$. To find the slope of the tangent, we differentiate $x$ with respect to $y$: $\frac{dx}{dy} = 2(y-2)$ The slope of the tangent is given by $\frac{dy}{dx}$, which is the reciprocal of $\frac{dx}{dy}$: $\frac{dy}{dx} = \frac{1}{2(y-2)}$ At the point $(2, 3)$, the slope of the tangent is: $m = \frac{1}{2(3-2)} = \frac{1}{2}$ Using the point-slope form of a line, the equation of the tangent is: $y - 3 = \frac{1}{2}(x - 2)$ $2(y - 3) = x - 2$ $2y - 6 = x - 2$ $x = 2y - 4$

Step 3: Find the limits of integration

The region is bounded by the parabola $x = (y-2)^2 + 1$, the tangent line $x = 2y - 4$, and the x-axis $y=0$. We need to find the intersection points of these curves to determine the limits of integration.

The intersection of the parabola and the x-axis ($y=0$) is: $x = (0-2)^2 + 1 = 4 + 1 = 5$ The intersection of the tangent and the x-axis ($y=0$) is: $x = 2(0) - 4 = -4$ The intersection of the parabola and the tangent is the point of tangency $(2,3)$, so $y = 3$. The region is bounded by $y=0$ and $y=3$.

Step 4: Set up and evaluate the definite integral

The area of the region is given by the integral of the difference between the $x$-values of the parabola and the tangent line with respect to $y$, from $y=0$ to $y=3$: $A = \int_{0}^{3} \left[((y-2)^2 + 1) - (2y - 4)\right] dy$ $A = \int_{0}^{3} (y^2 - 4y + 4 + 1 - 2y + 4) dy$ $A = \int_{0}^{3} (y^2 - 6y + 9) dy$ $A = \int_{0}^{3} (y-3)^2 dy$ $A = \left[\frac{(y-3)^3}{3}\right]_{0}^{3}$ $A = \frac{(3-3)^3}{3} - \frac{(0-3)^3}{3}$ $A = 0 - \frac{(-27)}{3}$ $A = 9$

Common Mistakes & Tips

• Incorrectly identifying the limits of integration: Always sketch the curves to visualize the region and determine the correct limits.
• Forgetting to find the equation of the tangent: This is a crucial step. Make sure to differentiate correctly and use the point-slope form.
• Choosing the wrong variable for integration: Integrating with respect to $y$ is easier in this case because we have $x$ as a function of $y$ for both curves.

Summary

We found the area of the region bounded by the parabola, the tangent line, and the x-axis by first finding the point of tangency, then finding the equation of the tangent line. We then set up a definite integral with respect to $y$, using the equations of the parabola and the tangent line, and the limits of integration from $y=0$ to $y=3$. Evaluating the integral gave us the area of the region, which is 9.

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