Problem 22 - A Graphical Approach to College Algebra

Question

Graph each ellipse by hand. Give the domain and range. Give the foci and identify the center. Do not use a calculator. $$\frac{(x-4)^{2}}{9}+\frac{(y+2)^{2}}{4}=1$$

Step-by-Step Solution

Answer

Center: (4, -2); Foci: (4±√5, -2); Domain: [1, 7]; Range: [-4, 0].

Step 1: Identify the Standard Form

The given equation \( \frac{(x-4)^2}{9} + \frac{(y+2)^2}{4} = 1 \) is in the standard form of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), with center \((h, k)\). Here, we identify \(h = 4\), \(k = -2\), \(a^2 = 9\), and \(b^2 = 4\).

Step 2: Determine the Center

The center of the ellipse is given by \((h, k)\). For our equation, this means the center is \((4, -2)\).

Step 3: Calculate the Semi-Major and Semi-Minor Axes

The semi-major axis length \(a\) is \( \sqrt{9} = 3 \) and the semi-minor axis length \(b\) is \( \sqrt{4} = 2 \). Since \(a > b\), the ellipse is horizontally oriented.

Step 4: Find the Foci

For horizontal ellipses, the foci are located at \((h \pm c, k)\), where \(c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5}\). Therefore, the foci are \((4 \pm \sqrt{5}, -2)\).

Step 5: Determine the Domain

The domain of the ellipse is determined by the horizontal span inclusive of the semi-major axis: \([4 - 3, 4 + 3] = [1, 7]\).

Step 6: Determine the Range

The range is determined by the vertical span inclusive of the semi-minor axis: \([-2 - 2, -2 + 2] = [-4, 0]\).

Step 7: Summarize and Sketch the Ellipse

With a center at \((4, -2)\), semi-major axis \(3\), and semi-minor axis \(2\), the foci at \((4 \pm \sqrt{5}, -2)\), sketch the ellipse centered at \((4, -2)\), extending \(3\) units left and right and \(2\) units up and down.

Key Concepts

Domain and Range

The domain and range give us the set of possible values for the x and y coordinates on the graph of an ellipse. This tells us the extent of the ellipse horizontally and vertically. For our ellipse, we need to first identify the lengths of the semi-major and semi-minor axes:
- The semi-major axis length is 3.- The semi-minor axis length is 2.
Given that the center of the ellipse is at (4, -2), we can determine the domain by considering how far the ellipse extends in the horizontal direction. Since it stretches 3 units in both the left and right directions from the center, the domain is \[ [4 - 3, 4 + 3] = [1, 7] \].
Similarly, the range is determined by the vertical span of the ellipse. It extends 2 units up and 2 units down from the center’s y-coordinate, giving us a range of \[ [-2 - 2, -2 + 2] = [-4, 0] \].
Therefore, when graphing the ellipse, ensure that it spans the full extent within these domain and range values.

Foci of an Ellipse

The foci of an ellipse are two special points located inside the ellipse. They are on the major axis (the longer axis if not equal) and play a crucial role in defining the shape of the ellipse. For an ellipse centered at \( (h, k) \) with the equation in standard form, the foci for a horizontal ellipse are positioned at \( (h \pm c, k) \), where \( c \) is found using the formula \( c = \sqrt{a^2 - b^2} \). Here, \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.
In our example, the calculations show that:

• \( a^2 = 9 \) and \( b^2 = 4 \)
• So, \( c = \sqrt{9 - 4} = \sqrt{5} \)

This means the foci are located at \( (4 - \sqrt{5}, -2) \) and \( (4 + \sqrt{5}, -2) \). These points indicate where the sum of the distances from any point on the ellipse to the two foci remains constant.

Ellipse Center

The center of an ellipse in standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) is given by the coordinates \( (h, k) \). It is the point at which the ellipse is perfectly balanced. For our equation, the center is calculated as follows:

• The horizontal shift is represented by \( (x-4) \), which means \( h = 4 \).
• The vertical shift is shown by \( (y+2) \), meaning \( k = -2 \).

Thus, the center of our ellipse is located at the point \( (4, -2) \). Understanding the center is vital for sketching the graph correctly, as it is the reference point from which the axis lengths radiate outward.

Standard Form of an Ellipse

The standard form of an ellipse is crucial as it provides all necessary information to graph it correctly. The form is expressed as \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where:

• \( (h, k) \) is the center of the ellipse.
• \( a^2 \) and \( b^2 \) represent the squares of the lengths of the semi-major and semi-minor axes, respectively.

In this specific example, we have \( h = 4 \), \( k = -2 \), \( a^2 = 9 \), and \( b^2 = 4 \). These values tell us that:

• The center is at \( (4, -2) \).
• The length of the semi-major axis is \( a = 3 \).
• The length of the semi-minor axis is \( b = 2 \).

This information ensures that graphing the ellipse is straightforward, as the center and extents of the axes are clearly defined, allowing the ellipse to be drawn accurately.