Problem 16 - College Algebra with Corequisite Support

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. \(\frac{(x-2)^{2}}{81}+\frac{(y+1)^{2}}{16}=1\)

Step-by-Step Solution

Answer

The ellipse is centered at (2, -1) with major axis endpoints at (-7, -1) and (11, -1), minor axis at (2, 3) and (2, -5), and foci at \((-6.06, -1)\) and \((10.06, -1)\).

Step 1: Identify the Standard Form

The given equation \(\frac{(x-2)^{2}}{81}+\frac{(y+1)^{2}}{16}=1\) is already expressed in the standard form of an ellipse equation: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \(a^2 = 81\) and \(b^2 = 16\). Therefore, \(a=9\) and \(b=4\), and the center of the ellipse is at \((h, k) = (2, -1)\).

Step 2: Determine the Orientation of the Ellipse

The ellipse is horizontal because \(a^2 = 81\) is greater than \(b^2 = 16\). Hence, the major axis is along the \(x\)-axis.

Step 3: Find the Major and Minor Axis Endpoints

For a horizontal ellipse centered at \((h, k) = (2, -1)\), the endpoints of the major axis (horizontal direction) are at \((h \pm a, k) = (2 \pm 9, -1) = (-7, -1)\) and \((11, -1)\). The endpoints of the minor axis (vertical direction) are at \((h, k \pm b) = (2, -1 \pm 4) = (2, 3)\) and \((2, -5)\).

Step 4: Calculate the Foci of the Ellipse

To find the foci, use the formula \(c^2 = a^2 - b^2\). Thus, \(c^2 = 81 - 16 = 65\), so \(c = \sqrt{65}\). The foci for a horizontal ellipse are at \((h \pm c, k) = (2 \pm \sqrt{65}, -1)\), which are approximately \((-6.06, -1)\) and \((10.06, -1)\).

Step 5: Recap and Verify

The equation is in standard form with the major axis along the \(x\)-axis. Endpoints of the major axis are \((-7, -1)\) and \((11, -1)\); minor axis endpoints are \((2, 3)\) and \((2, -5)\). Foci are approximately \((-6.06, -1)\) and \((10.06, -1)\).

Key Concepts

Major and Minor Axes

In ellipse equations, the major and minor axes are key characteristics that allow us to understand its shape and orientation. For our given ellipse \(\frac{(x-2)^{2}}{81}+\frac{(y+1)^{2}}{16}=1\), these axes help us determine how the ellipse stretches across the coordinate plane.
* The **major axis** is the longest diameter of the ellipse and in this instance, aligns with the x-axis because \(a^2 = 81\) is greater than \(b^2 = 16\). This means the horizontal direction has the larger spread.
* The endpoints of the major axis are calculated using the formula \((h \pm a, k)\), which computes to points \((-7, -1)\) and \((11, -1)\).

• These points signify where the ellipse extends furthest horizontally from the center.

* The **minor axis** is the shortest diameter and runs perpendicular to the major axis. Its endpoints use \((h, k \pm b)\), giving us the locations \((2, 3)\) and \((2, -5)\).

• These illustrate the closest points vertically from the center to the edge of the ellipse.

Foci of Ellipse

The foci (plural of focus) of an ellipse are two key points located symmetrically along the major axis from the center. In an ellipse, the total distance from one focus to any point on the ellipse and back to the other focus remains constant.
To locate the foci of the ellipse equation \(\frac{(x-2)^{2}}{81}+\frac{(y+1)^{2}}{16}=1\), we utilize the relationship between the semi-axes lengths.
* Apply the formula \(c^2 = a^2 - b^2\) to find \(c\), the distance from the center to each focus.

• In our example, \(c^2 = 81 - 16 = 65\), which means \(c = \sqrt{65}\).

* For a horizontally oriented ellipse, the foci are positioned at \((h \pm c, k)\), making them approximately \((-6.06, -1)\) and \((10.06, -1)\).

• This demonstrates that these points lie on the major axis, equidistant from the center, and contribute to the unique property of an ellipse."

Standard Form

The standard form of an ellipse equation provides a structured way to understand its dimensions and positioning in the coordinate plane. It typically appears as \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) or vice versa depending on orientation.
* The equation \(\frac{(x-2)^{2}}{81}+\frac{(y+1)^{2}}{16}=1\) is in standard form. The segments beneath each squared component, \(a^2\) and \(b^2\), define the extent along the x and y axes respectively.
* The terms \((h, k)\) identify the center at \((2, -1)\).

• Determining \(a^2\) (81) and \(b^2\) (16) allows us to assess the size and orientation, with \(a = 9\) and \(b = 4\).

* If \(a^2 > b^2\), the ellipse is wider than it is tall, revealing its orientation—as is the case here with a horizontal layout.

• Comprehension of this form is crucial for identifying all key features, such as axes and foci, succinctly from just the equation.