Problem 8 - Precalculus Mathematics for Calculus

Question

Find the center, foci, and vertices of the ellipse, and determine the lengths of the major and minor axes. Then sketch the graph. $$\frac{(x+2)^{2}}{4}+y^{2}=1$$

Step-by-Step Solution

Answer

Center: (-2,0); Vertices: (-4,0), (0,0); Foci: (-3.73,0), (-0.27,0); Major axis: 4; Minor axis: 2.

Step 1: Identify the Form of the Ellipse

The given equation \( \frac{(x+2)^{2}}{4} + y^{2} = 1 \) is in the standard form of an ellipse, \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), where \( 0 < b < a \). Here, the ellipse is aligned with the x and y axes, and not rotated.

Step 2: Find the Center of the Ellipse

Identify \( (h, k) \) by comparing the equation to the standard form. The transformed equation is centered at \((h, k) = (-2, 0)\). Thus, the center of the ellipse is \((-2, 0)\).

Step 3: Determine Semi-Axes Lengths

From the standard form, \( a^2 = 4 \) and \( b^2 = 1 \), giving \( a = 2 \) and \( b = 1 \). The major axis, aligned with the x-axis, has length \(2a = 4\), and the minor axis, aligned with the y-axis, has length \(2b = 2\).

Step 4: Find the Vertices

The vertices are located along the major axis, calculated as \((h \pm a, k) = (-2 \pm 2, 0)\). Thus, the vertices are \((-4, 0)\) and \((0, 0)\).

Step 5: Calculate the Foci

Using the formula for the distance to the foci, \( c^2 = a^2 - b^2 \), we find \( c = \sqrt{4-1} = \sqrt{3} \approx 1.73 \). Hence, the foci are at \((h \pm c, k) = (-2 \pm \sqrt{3}, 0)\), approximately \((-2-1.73, 0)\) and \((-2+1.73, 0)\).

Step 6: Sketch the Graph

Draw an ellipse centered at \((-2,0)\) with a horizontal major axis of length 4 from \((-4, 0)\) to \((0, 0)\) and a vertical minor axis of length 2 from \((-2, -1)\) to \((-2, 1)\). Mark the foci at \((-3.73, 0)\) and \((-0.27, 0)\).

Key Concepts

Ellipse Standard Form

Understanding the standard form of an ellipse is crucial when solving problems related to ellipses. An ellipse is a type of conic section that is defined mathematically. In its standard form, the equation of an ellipse is given as:

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

Here,

• \((h, k)\) represents the center of the ellipse.
• \(a\) is the semi-major axis length while \(b\) is the semi-minor axis length.
• If \(a>b\), the major axis is horizontal; if \(b>a\), it is vertical.

The form allows for easy identification of key properties of the ellipse such as its center, orientation, and axis lengths. By rewriting the equation of an ellipse in standard form, we can easily extract this information. In the exercise, we transformed the equation \(\frac{(x+2)^2}{4} + y^2 = 1\) into the standard form, helping us to identify its center, axes, and other features.

Semi-Major Axis

The semi-major axis of an ellipse is the longest radius that extends from the center to the edge of the ellipse along the major axis. It is denoted by \(a\) in the standard form of an ellipse. For the given problem, we found

• \(a^2=4\)
• \(a=2\)

This means the total length of the major axis is twice the semi-major axis, giving it a length of \(4\).
The semi-major axis defines the overall "stretch" of the ellipse. The length and orientation of the semi-major axis help determine the shape and appearance of the ellipse graphically, and understanding this can assist in sketching the accurate representation of an ellipse.

Vertices of an Ellipse

Vertices of an ellipse are the points where the ellipse is widest or longest along its major axis. They are located on the major axis at a distance of \(a\) from the center. For an ellipse centered at \((h, k)\), the vertices are determined using the formula:

• \((h\pm a, k)\) for a horizontal major axis
• \((h, k\pm a)\) for a vertical major axis

In our problem, the vertices were calculated as

• \((-2 \pm 2, 0)\)

which gives the points

• \((-4, 0)\)
• \((0, 0)\)

These points are crucial for sketching the position and shape of the ellipse correctly because they give the furthest extents of the ellipse along the major axis.

Foci of an Ellipse

The foci (or focus points) of an ellipse are two important points along the major axis that help define the shape of the ellipse. The property of an ellipse is that the sum of the distances from any point on the ellipse to each of the foci is constant. To find the distance from the center to each focus, denoted as \(c\), we use the relationship:

• \(c^2 = a^2 - b^2\)

In the solution, this is calculated as follows:

• \(c^2 = 4 - 1\)
• \(c = \sqrt{3} \approx 1.73\)

Thus, the foci are located at

• \((-2 \pm \sqrt{3}, 0)\)

Approximately, this results in the points

• \((-3.73, 0)\)
• \((-0.27, 0)\)

The placement of these foci is central to understanding the behavior of the ellipse, especially when considering geometric properties and real-world applications of elliptical shapes.