Problem 23 - College Algebra

Question

For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. $$ 4 x^{2}+40 x+25 y^{2}-100 y+100=0 $$

Step-by-Step Solution

Answer

The ellipse is \(\frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1\), the major axis endpoints are \((-10, 2)\) and \((0, 2)\), the minor axis endpoints are \((-5, 0)\) and \((-5, 4)\), and the foci are approximately \((-5 \pm \sqrt{21}, 2)\).

Step 1: Rearrange and Group Terms

Start by rearranging the given equation to group the terms involving \(x\) and \(y\):\[ 4x^2 + 40x + 25y^2 - 100y = -100 \]

Step 2: Complete the Square for x-terms

Factor the coefficient of \(x^2\) from the \(x\) terms:\[ 4(x^2 + 10x) + 25y^2 - 100y = -100 \]To complete the square for \(x\), add and subtract \((\frac{10}{2})^2 = 25\) inside the parenthesis:\[ 4(x^2 + 10x + 25 - 25) + 25y^2 - 100y = -100 \]\[ 4((x + 5)^2 - 25) + 25y^2 - 100y = -100 \]\[ 4(x + 5)^2 - 100 + 25y^2 - 100y = -100 \]

Step 3: Complete the Square for y-terms

Now, focus on the \(y\) terms:\[ 4(x+5)^2 + 25(y^2 - 4y) = 0 \]Complete the square for \(y\), by adding and subtracting \((\frac{4}{2})^2=4\):\[ 4(x+5)^2 + 25((y-2)^2 - 4) = 0 \]\[ 4(x+5)^2 + 25(y-2)^2 - 100 = 0 \]

Step 4: Simplify and Write in Standard Form

Isolate the ellipse term:\[ 4(x+5)^2 + 25(y-2)^2 = 100 \]Divide through by 100 to get the standard form:\[ \frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1 \]

Step 5: Identify Parameters

Identify the parameters from the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \(a^2 = 25\) and \(b^2 = 4\), so \(a = 5\) and \(b = 2\). The center is \((-5, 2)\).

Step 6: Identify the Major and Minor Axes

Since \(a > b\), the major axis is horizontal. The endpoints for the major axis are spread along the x-axis, centered at \(-5\):\(x = -5 \pm 5\), resulting in endpoints \((-10, 2)\) and \((0, 2)\).The minor axis endpoints are spread along the y-axis\(y = 2 \pm 2\), resulting in \((-5, 0)\) and \((-5, 4)\).

Step 7: Locate the Foci

The distance to the foci from the center is \(c\), where \(c^2 = a^2 - b^2 = 25 - 4 = 21\), so \(c = \sqrt{21}\). The foci are located horizontally from the center:\((-5 \pm \sqrt{21}, 2)\), approximately at \((-5 + \sqrt{21}, 2)\) and \((-5 - \sqrt{21}, 2)\).

Key Concepts

Understanding the Standard Form of an Ellipse Equation

The standard form of an ellipse equation is crucial for identifying its properties. It takes the form: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here,

• \((h, k)\) represents the center of the ellipse,
• \(a\) is the distance from the center to a vertex along the major axis,
• \(b\) is the distance from the center to a vertex along the minor axis.

For our specific problem, the equation \[ \frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1 \]indicates that the ellipse is centered at \((-5, 2)\). Rewriting an ellipse equation in this form allows us to easily identify and utilize its features.

Locating the Major Axis

The major axis of an ellipse is the longest diameter that passes through the center. For an equation in standard form, the length of the major axis is determined by the larger of the denominators, \(a^2\) or \(b^2\). In our example, since \(a^2 = 25\) and \(b^2 = 4\), \(a = 5\), indicating that the major axis is horizontal because \(a > b\). The endpoints of the major axis are found by moving \(a\) units to the left and right of the center along the x-axis:

• Center: \((-5, 2)\)
• Endpoints: \((-10, 2)\) and \((0, 2)\)

These points represent the widest part of the ellipse.

Defining the Minor Axis

The minor axis is the shortest diameter, perpendicular to the major axis and also passing through the center. For our ellipse, since \(b^2 = 4\), we have \(b = 2\). This means the minor axis is vertical.The endpoints of the minor axis are determined by moving \(b\) units up and down from the center along the y-axis:

• Center: \((-5, 2)\)
• Endpoints: \((-5, 0)\) and \((-5, 4)\)

These points illustrate the shortest span of the ellipse.

Identifying the Foci

A unique feature of ellipses is the foci (plural of focus). These are two fixed points inside the ellipse such that the sum of the distances from any point on the ellipse to the foci is constant. In the standard equation, the distance from the center to each focus, \(c\), is calculated using the formula: \[ c^2 = a^2 - b^2 \]In this scenario:

• \(a^2 = 25\)
• \(b^2 = 4\)
• Therefore, \(c^2 = 21\), making \(c = \sqrt{21}\)

Given \(c\), the foci are located on the major axis, \(c\) units away from the center:

• Center: \((-5, 2)\)
• Foci: \((-5 + \sqrt{21}, 2)\) and \((-5 - \sqrt{21}, 2)\)

These points signify the unique symmetry and geometry of the ellipse.