Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. \(\frac{x^{2}}{2}+\frac{(y+1)^{2}}{5}=1\)

Step-by-Step Solution

Answer

The center is \((0, -1)\); vertices are \((0, -1\pm\sqrt{5})\), foci are \((0, -1\pm\sqrt{3})\).

Step 1: Identify the Standard Form

First, identify the standard form of the ellipse. The given equation is \(\frac{x^2}{2} + \frac{(y+1)^2}{5} = 1\), which matches the general form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), indicating a vertical ellipse because \(b^2 > a^2\). Here, \(h=0\), \(k=-1\), \(a^2 = 2\), and \(b^2 = 5\).

Step 2: Find the Center

The center of the ellipse \((h, k)\) is obtained directly from the standard form, so the center is \((0, -1)\).

Step 3: Determine the Vertices

Since it's a vertical ellipse, the vertices are located \(b\) units above and below the center along the y-axis. Here, \(b = \sqrt{5}\), so the vertices are \((0, -1 + \sqrt{5})\) and \((0, -1 - \sqrt{5})\).

Step 4: Determine the Foci

The foci are calculated using \(c^2 = b^2 - a^2\). For this ellipse, \(c^2 = 5 - 2 = 3\), so \(c = \sqrt{3}\). The foci are \(c\) units up and down from the center along the y-axis, giving \((0, -1 + \sqrt{3})\) and \((0, -1 - \sqrt{3})\).

Step 5: Graph the Ellipse

To graph the ellipse, plot the center at \((0, -1)\). Mark the vertices at \((0, -1 + \sqrt{5})\) and \((0, -1 - \sqrt{5})\) and the foci at \((0, -1 + \sqrt{3})\) and \((0, -1 - \sqrt{3})\). Draw an ellipse that fits these points, elongated vertically.

Key Concepts

Standard Form of Ellipse

The standard form of an ellipse is essential for understanding how it is aligned and stretched on a graph. An ellipse has a standard equation form that can be written as: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). This equation provides valuable information about the ellipse. Here:

• \(h\) and \(k\) denote the coordinates of the ellipse's center.
• \(a\) and \(b\) are the semi-major and semi-minor axes. They determine the length and direction.

For the given equation \(\frac{x^2}{2} + \frac{(y+1)^2}{5} = 1\), it matches the form of a vertical ellipse. This is indicated by \(b^2\) being greater than \(a^2\). Understanding the standard form is critical because it shows the vertices' alignment and the ellipse's orientation on a graph.

Ellipse Center

The ellipse's center, represented by the coordinates \((h, k)\), is like a home base for the rest of the ellipse. In our equation \(\frac{x^2}{2} + \frac{(y+1)^2}{5} = 1\), you can find the center by identifying the values of \(h\) and \(k\):

• \(h = 0\)
• \(k = -1\)

Hence, the center is at \((0, -1)\). This point is crucial as it helps establish the location of the ellipse on the coordinate plane. From this center, you can measure distances to find the vertices and foci, aiding in the complete understanding of the ellipse's layout.

Ellipse Vertices

Vertices are key features of an ellipse, marking the most extended edge points. The vertices determine how far the ellipse stretches along its major axis. For a vertical ellipse like ours, vertices lie along the y-axis. Here's how you find them:

• The semi-major axis length is \(b\), which is \(\sqrt{5}\) for our example.
• Measure this distance up and down from the center \((0, -1)\).

So, the vertices are located at \((0, -1 + \sqrt{5})\) and \((0, -1 - \sqrt{5})\). Remember, identifying vertices helps graph the ellipse accurately, giving a visual sense of its direction and size based on the major axis.

Ellipse Foci

The foci (singular: focus) of an ellipse are two special points. They play a role in its unique property of constant sum of distances to any point on the ellipse. Finding these points involves a simple calculation:

• Use the formula \(c^2 = b^2 - a^2\) to find \(c\).
• In our equation, \(a^2 = 2\) and \(b^2 = 5\), so \(c^2 = 3\) and \(c = \sqrt{3}\).

The foci are positioned \(c\) units from the center along the y-axis. Therefore, they are \((0, -1 + \sqrt{3})\) and \((0, -1 - \sqrt{3})\). By plotting these foci, along with vertices and center, you can sketch the ellipse with its correct shape and orientation.