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

We are given: \[ y^2 = 4(x - 2) \tag{1} \] \[ y = 2x - 8 \tag{2} \] We need to find the area of the region bounded by these two curves. Step 1: Rewrite the Equation of the Parabola} Rewrite the parabola \( y^2 = 4(x - 2) \) in terms of \( x \): \[ x = \frac{y^2}{4} + 2 \] {Step 2: Find Points of Intersection} To find the points of intersection of the line \( y = 2x - 8 \) and the parabola \( y^2 = 4(x - 2) \), substitute \( y = 2x - 8 \) into the parabola equation: \[ (2x - 8)^2 = 4(x - 2) \] Expanding and simplifying: \[ 4x^2 - 36x + 72 = 0 \] \[ (x - 6)(x - 3) = 0 \] So, \( x = 6 \) and \( x = 3 \). Substitute these values of \( x \) back into \( y = 2x - 8 \) to find the corresponding \( y \)-values: \[ \text{For } x = 6 : \quad y = 2 \times 6 - 8 = 4 \tag{3} \] \[ \text{For } x = 3 : \quad y = 2 \times 3 - 8 = -2 \tag{4} \] Thus, the points of intersection are \( (6, 4) \) and \( (3, -2) \). {Step 3: Set Up the Integral} The area \( A \) of the region bounded by the parabola and the line from \( y = -2 \) to \( y = 4 \) is given by: \[ A = \int_{-2}^{4} (x_{\text{line}} - x_{\text{parabola}}) \, dy \] where: \[ x_{\text{line}} = \frac{y + 8}{2} \tag{5} \] \[ x_{\text{parabola}} = \frac{y^2}{4} + 2 \tag{6} \] So 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} Simplify the integrand: \[ A = \int_{-2}^{4} \left( -\frac{y^2}{4} + \frac{y}{2} + 2 \right) \, dy \] {Step 5: Evaluate the Integral} Now, integrate term by term: \[ A = \int_{-2}^{4} -\frac{y^2}{4} \, dy + \int_{-2}^{4} \frac{y}{2} \, dy + \int_{-2}^{4} 2 \, dy \] Calculate each integral: \[ \int_{-2}^{4} -\frac{y^2}{4} \, dy = -6 \tag{7} \] \[ \int_{-2}^{4} \frac{y}{2} \, dy = 3 \tag{8} \] \[ \int_{-2}^{4} 2 \, dy = 12 \tag{9} \] So, \[ A = -6 + 3 + 12 = 9 \]