Problem 88 - College Algebra with Modeling and Visualization

Question

Shade the region in the xy-plane that satisfies the given inequality. Find the area of this region if units are in feet. $$ \frac{(x+3)^{2}}{4}+\frac{(y-2)^{2}}{8} \leq 1 $$

Step-by-Step Solution

Answer

The area of the shaded region is \(4\pi\sqrt{2}\) square feet.

Step 1: Understanding the Inequality

The inequality \( \frac{(x+3)^2}{4} + \frac{(y-2)^2}{8} \leq 1 \) represents an elliptical region centered at \((-3, 2)\). The form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) is that of a standard ellipse, where \( (h, k) \) is the center and \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.

Step 2: Determine the Center and Axes

From the inequality, the center of the ellipse is \((-3, 2)\). The term \(\frac{(x+3)^2}{4}\) indicates the horizontal semi-axis (\(a = 2\)) and \(\frac{(y-2)^2}{8}\) indicates the vertical semi-axis (\(b = \sqrt{8} = 2\sqrt{2}\)).

Step 3: Graph the Ellipse

Plot an ellipse centered at \((-3, 2)\) with a horizontal radius of 2 and a vertical radius of \(2\sqrt{2}\). The inequality \(\leq\) indicates that the region to shade is inside this ellipse.

Step 4: Calculate the Area of the Elliptical Region

The area of an ellipse is calculated using the formula \( \pi \times a \times b \). Substitute \(a = 2\) and \(b = 2\sqrt{2}\) to get the area: \[ \text{Area} = \pi \times 2 \times 2\sqrt{2} = 4\pi\sqrt{2}\text{ square feet} \]

Key Concepts

Inequalities Representing Ellipses

Ellipses can be represented through inequalities of the form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \leq 1 \). This inequality defines the region inside or on the boundary of an ellipse centered at \((h, k)\). To understand this better, consider if the comparison symbol was \(=\) instead of \(\leq\). This would define the boundary of the ellipse itself.
When the inequality uses \(\leq\), it includes the entire area within the ellipse. The inequality splits the xy-plane into regions that either satisfy or do not satisfy the condition. In this specific problem, the inequality \(\frac{(x+3)^2}{4} + \frac{(y-2)^2}{8} \leq 1 \) indicates that all points \((x, y)\) that lie within a particular ellipse, as well as on its boundary, are part of the solution set. This ellipse is centered at \((-3, 2)\), with semi-axes determined from the denominators \(4\) and \(8\). This denotes a horizontal semi-axis of \(2\) units and a vertical semi-axis of \(2\sqrt{2}\) units, inside which all points satisfy the inequality.

Area Calculation of Ellipses

Finding the area of an ellipse can be straightforward once you identify its dimensions. The formula used is \( \pi \times a \times b \), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes. For the inequality \( \frac{(x+3)^2}{4} + \frac{(y-2)^2}{8} \leq 1 \), we have already found

• Horizontal semi-axis \(a = 2\)
• Vertical semi-axis \(b = 2\sqrt{2}\)

To determine the area of the region within this ellipse, substitute these values into the formula: \[\text{Area} = \pi \times 2 \times 2\sqrt{2} = 4\pi\sqrt{2} \text{ square feet}\]This provides the total area within the elliptical boundary that you need to shade in a graph, considering units are in feet.

Graphing Ellipses from Inequalities

Graphing ellipses can reinforce comprehension of their characteristics and areas. For graphing the ellipse defined by the inequality \(\frac{(x+3)^2}{4} + \frac{(y-2)^2}{8} \leq 1\), start by determining the center and the semi-axes.
Here’s a simple method to graph this ellipse:

• Find the center: The center \((h, k)\) is extracted directly from the expression, which is \((-3, 2)\) for this problem.
• Identify the axes:
  The denominator of the fraction under \(x\) terms gives \(a^2\), where \(a\) is the semi-major or minor axis. Here, \(a = 2\) for the x-direction. Similarly, the denominator for \(y\) terms reveals \(b = 2\sqrt{2}\), describing the vertical reach.
• Plot the ellipse:
  Draw the ellipse using these measurements as a guide. The horizontal radius extends 2 units left and right from the center, while the vertical radius spans \(2\sqrt{2}\) units up and down.
• Shade the region: The inequality \(\leq\) indicates shading inside the ellipse, depicting all solutions visually.

This visual plotting of the ellipse rooted in its inequality allows for clearer interpretation of the region specified by your inequality constraint.