Question

The area (in square units) of the region bounded by the parabola \(y^2 = 4(x - 2)\) and the line \(y = 2x - 8\) is:

• A. 8
• B. 9
• C. 6
• D. 7

Answer 1

The Correct Option is B

The problem provides two equations: \( y^2 = 4(x - 2) \) (Equation 1) and \( y = 2x - 8 \) (Equation 2). The objective is to determine the area enclosed by these curves. Step 1: Rearrange the Parabola Equation. The parabola \( y^2 = 4(x - 2) \) is rewritten as \( x = \frac{y^2}{4} + 2 \). Step 2: Determine Intersection Points. To find where the line \( y = 2x - 8 \) and the parabola \( y^2 = 4(x - 2) \) intersect, substitute \( y = 2x - 8 \) into the parabola's equation: \( (2x - 8)^2 = 4(x - 2) \). Simplifying the equation yields \( 4x^2 - 36x + 72 = 0 \), which factors to \( (x - 6)(x - 3) = 0 \). This gives \( x = 6 \) and \( x = 3 \). Substitute these x-values back into \( y = 2x - 8 \) to find the corresponding y-values: For \( x = 6 \), \( y = 2 \times 6 - 8 = 4 \) (Equation 3). For \( x = 3 \), \( y = 2 \times 3 - 8 = -2 \) (Equation 4). The intersection points are \( (6, 4) \) and \( (3, -2) \). Step 3: Formulate the Integral. The area \( A \) of the region between the parabola and the line, from \( y = -2 \) to \( y = 4 \), is calculated using the integral: \( A = \int_{-2}^{4} (x_{\text{line}} - x_{\text{parabola}}) \, dy \). The expressions for x are: \( x_{\text{line}} = \frac{y + 8}{2} \) (Equation 5) and \( x_{\text{parabola}} = \frac{y^2}{4} + 2 \) (Equation 6). The integral becomes: \( A = \int_{-2}^{4} \left( \frac{y + 8}{2} - \left( \frac{y^2}{4} + 2 \right) \right) \, dy \). Step 4: Simplify the Integral. The integrand is simplified to: \( A = \int_{-2}^{4} \left( -\frac{y^2}{4} + \frac{y}{2} + 2 \right) \, dy \). Step 5: Evaluate the Integral. The integral is evaluated term by term: \( A = \int_{-2}^{4} -\frac{y^2}{4} \, dy + \int_{-2}^{4} \frac{y}{2} \, dy + \int_{-2}^{4} 2 \, dy \). The individual integrals are: \( \int_{-2}^{4} -\frac{y^2}{4} \, dy = -6 \) (Equation 7), \( \int_{-2}^{4} \frac{y}{2} \, dy = 3 \) (Equation 8), and \( \int_{-2}^{4} 2 \, dy = 12 \) (Equation 9). The total area is \( A = -6 + 3 + 12 = 9 \).