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, \[ A = -6 + 3 + 12 = 9 \]

Answer 2

The problem asks for the area of the region bounded by the parabola \(y^2 = 4(x - 2)\) and the line \(y = 2x - 8\).

Concept Used:

To find the area between two curves, it is often convenient to integrate with respect to \(y\). If we have two curves expressed as \(x = f(y)\) (the "right" curve) and \(x = g(y)\) (the "left" curve) that intersect at \(y = c\) and \(y = d\), the area of the region enclosed by them is given by the definite integral:

\[ A = \int_{c}^{d} [f(y) - g(y)] \, dy \]

This method requires finding the points of intersection to determine the limits of integration (\(c\) and \(d\)) and expressing both equations in the form \(x\) as a function of \(y\).

Step-by-Step Solution:

Step 1: Find the points of intersection of the parabola and the line.

We have the two equations:

1. Parabola: \(y^2 = 4(x - 2)\)
2. Line: \(y = 2x - 8\)

From the line equation, we can express \(x\) in terms of \(y\):

\[ 2x = y + 8 \implies x = \frac{y + 8}{2} \]

Substitute this expression for \(x\) into the parabola's equation:

\[ y^2 = 4\left(\frac{y + 8}{2} - 2\right) \] \[ y^2 = 4\left(\frac{y + 8 - 4}{2}\right) \] \[ y^2 = 2(y + 4) \] \[ y^2 = 2y + 8 \]

Rearrange this into a quadratic equation in \(y\):

\[ y^2 - 2y - 8 = 0 \]

Factor the quadratic equation:

\[ (y - 4)(y + 2) = 0 \]

The y-coordinates of the points of intersection are \(y = 4\) and \(y = -2\). These will be our limits of integration, so \(c = -2\) and \(d = 4\).

Step 2: Express both curves as functions of \(y\).

For the parabola:

\[ y^2 = 4x - 8 \implies 4x = y^2 + 8 \implies x_P = \frac{y^2}{4} + 2 \]

For the line:

\[ y = 2x - 8 \implies 2x = y + 8 \implies x_L = \frac{y}{2} + 4 \]

Step 3: Set up the definite integral for the area.

To determine which curve is on the right (\(x_{right}\)) and which is on the left (\(x_{left}\)) in the interval \([-2, 4]\), we can test a point within the interval, for example, \(y = 0\).

At \(y=0\), for the parabola: \(x_P = \frac{0^2}{4} + 2 = 2\).

At \(y=0\), for the line: \(x_L = \frac{0}{2} + 4 = 4\).

Since \(x_L > x_P\), the line is the right curve and the parabola is the left curve in the region. Thus, \(f(y) = x_L\) and \(g(y) = x_P\).

The area \(A\) is given by the integral:

\[ A = \int_{-2}^{4} \left[ \left(\frac{y}{2} + 4\right) - \left(\frac{y^2}{4} + 2\right) \right] \, dy \]

Simplify the integrand:

\[ A = \int_{-2}^{4} \left( -\frac{y^2}{4} + \frac{y}{2} + 2 \right) \, dy \]

Final Computation & Result:

Step 4: Evaluate the definite integral.

First, find the antiderivative of the integrand:

\[ \int \left( -\frac{y^2}{4} + \frac{y}{2} + 2 \right) \, dy = -\frac{1}{4}\frac{y^3}{3} + \frac{1}{2}\frac{y^2}{2} + 2y = -\frac{y^3}{12} + \frac{y^2}{4} + 2y \]

Now, apply the limits of integration:

\[ A = \left[ -\frac{y^3}{12} + \frac{y^2}{4} + 2y \right]_{-2}^{4} \]

Evaluate at the upper limit (\(y = 4\)):

\[ \left( -\frac{4^3}{12} + \frac{4^2}{4} + 2(4) \right) = \left( -\frac{64}{12} + \frac{16}{4} + 8 \right) = \left( -\frac{16}{3} + 4 + 8 \right) = \left( -\frac{16}{3} + 12 \right) = \frac{-16 + 36}{3} = \frac{20}{3} \]

Evaluate at the lower limit (\(y = -2\)):

\[ \left( -\frac{(-2)^3}{12} + \frac{(-2)^2}{4} + 2(-2) \right) = \left( -\frac{-8}{12} + \frac{4}{4} - 4 \right) = \left( \frac{2}{3} + 1 - 4 \right) = \left( \frac{2}{3} - 3 \right) = \frac{2 - 9}{3} = -\frac{7}{3} \]

Subtract the lower limit value from the upper limit value:

\[ A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9 \]

The area of the bounded region is 9 square units.