Problem 38 - College Algebra

Question

Graph each ellipse and give the location of its foci. $$ \frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1 $$

Step-by-Step Solution

Answer

The center of the ellipse is at the point (1, -2), the lengths of the half-axes are a=4 and b=3, and the foci are located at points \((1 + \sqrt{7}, -2)\) and \(1 - \sqrt{7}, -2).\)

Step 1: Identify the center (h, k)

The center of the ellipse is given by (h, k), where h and k are the values in the formula that make the expressions within the brackets equal to zero. Looking at our equation \(\frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1\), we find that setting \(x-1=0\) and \(y+2=0\), we get x=1 and y=-2 respectively. So our center is at (1, -2).

Step 2: Identify a and b

The coefficients \(a^2\) and \(b^2\) in our formula are the squares of the lengths of the half-axes a and b, respectively. In the given formula, these values are underneath the fractions. Hence, \(a^2 = 16\) and \(b^2 = 9\), so \(a = 4\) and \(b = 3\). Since a > b, our major axis is along the x-axis.

Step 3: Find the foci

The distance from the center to the foci can be calculated with \(c = \sqrt{a^2 - b^2}\). Substituting a and b, we get \(c = \sqrt{4^2 - 3^2} = \sqrt{7}\). Since the major axis is along the x-axis, the foci are located at \((h+c,k)\) and \((h-c,k)\), or \((1 + \sqrt{7}, -2)\) and \((1 - \sqrt{7}, -2)\).

Key Concepts

Center of the Ellipse

Understanding the center of an ellipse is a fundamental step in interpreting its equation. The equation of an ellipse typically takes the form:

• Horizontal: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
• Vertical: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)

In both forms, the center of the ellipse is found at \((h, k)\). This means that \(h\) and \(k\) are the values for the x and y coordinates, respectively, that make the expressions inside the brackets equal to zero.

In our specific example, the equation is \( \frac{(x-1)^2}{16} + \frac{(y+2)^2}{9} = 1 \). By setting \(x-1 = 0\), we find \(x = 1\), and by setting \(y+2 = 0\), we determine \(y = -2\). Thus, the center of the ellipse is \((1, -2)\). This point is crucial as it serves as the starting point to assess distances like half-axes and foci from the center.

Major and Minor Axes

Every ellipse has two main lines through its center, known as axes. These are called the major and minor axes, and they run through the widest and narrowest parts of the ellipse, respectively.

In the ellipse equation, the values \(a^2\) and \(b^2\) appear under the terms \((x-h)^2\) and \((y-k)^2\). Here:

• \(a^2 = 16\) gives \(a = 4\)
• \(b^2 = 9\) gives \(b = 3\)

Since \(a > b\), the major axis is along the direction with \(a\), which is the x-axis in this case.

The major axis length is \(2a\) and the minor axis length is \(2b\):

• Major axis length is 8 (\(2 \times 4\))
• Minor axis length is 6 (\(2 \times 3\))

This determination is based on which value (\(a\) or \(b\)) is larger, defining the orientation and span of the ellipse's axes across the Cartesian plane.

Foci of an Ellipse

The foci of an ellipse are special points located along the major axis. They are essential in understanding the properties of ellipses, specifically relating to how they are drawn.

To find the distance \(c\) from the center to each focus, use the formula \(c = \sqrt{a^2 - b^2}\). This calculation stems from the geometric property of ellipses, connecting their axes and spanning their foci.

• In our example, substituting \(a = 4\) and \(b = 3\) gives \(c = \sqrt{16 - 9} = \sqrt{7}\).

Since the ellipse's major axis is along the x-axis, the foci are positioned horizontally from the center, \((h, k)\). Therefore, the coordinates of the foci are at:

• \((1 + \sqrt{7}, -2)\)
• \((1 - \sqrt{7}, -2)\)

This setup highlights the symmetry of ellipses and their unique geometric properties that ensure any point on their perimeter maintains consistent relationships with the foci.