The area of the region bounded by the curve \( (x-2)^{2}=y-1 \) and its tangent at the point \( (3,2) \) and the \( y \)-axis equals..... area units. \( \begin{array}{llll}\text { (a) } 4 & \text { (b) } 6 & \text { (c) } 9 & \text { (d) } 12\end{array} \)

To find the area of the region bounded by the curve \( (x-2)^{2} = y-1 \) and its tangent at the point \( (3,2) \) and the \( y \)-axis, we first need to determine the equation of the tangent line at the given point. 1. **Rearranging the curve equation:** The equation can be rewritten as \( y = (x-2)^{2} + 1 \). 2. **Finding the derivative:** The slope of the tangent line can be found using the derivative: \[ \frac{dy}{dx} = 2(x-2). \] At the point \( (3,2) \): \[ \frac{dy}{dx}\bigg|_{x=3} = 2(3-2) = 2. \] 3. **Equation of the tangent line:** Using the point-slope form, the equation of the tangent line at \( (3,2) \) is: \[ y - 2 = 2(x - 3) \implies y = 2x - 4. \] 4. **Finding the intersection with the \( y \)-axis:** The \( y \)-intercept occurs when \( x = 0 \): \[ y = 2(0) - 4 = -4. \] 5. **Finding the area bounded by the curve, tangent, and the \( y \)-axis:** We need to calculate the area between the curve and the tangent line from \( (0,-4) \) to \( (3,2) \). a. Find the points of intersection between the curve and the tangent line: \[ (x - 2)^{2} + 1 = 2x - 4. \] Rearranging gives: \[ (x - 2)^{2} - 2x + 5 = 0 \implies x^2 - 4x + 4 - 2x + 5 = 0 \implies x^2 - 6x + 9 = 0 \implies (x - 3)^{2} = 0. \] Thus, the point of intersection is \( x = 3 \). b. Set up the integral to find the area: The area \( A \) can be calculated as: \[ A = \int_{0}^{3} [(2x - 4) - ((x - 2)^{2} + 1)] dx. \] Simplifying the integrand: \[ A = \int_{0}^{3} [2x - 4 - (x^2 - 4x + 4)] dx = \int_{0}^{3} [-x^2 + 6x - 8] dx. \] Evaluating the integral: \[ A = \left[-\frac{x^3}{3} + 3x^2 - 8x \right]_{0}^{3} = \left[-\frac{27}{3} + 27 - 24 \right] - 0 = [-9 + 27 - 24] = -6. \] Since area cannot be negative, take the absolute value: \[ A = 6. \] Thus, the area of the region bounded by the curve, its tangent line, and the \( y \)-axis is \( \boxed{6} \) area units.