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

Answer: (B)

9

Answer 2

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, $A = -6 + 3 + 12 = 9$