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

Step-by-Step Solution

Answer

Ellipse equation: \( \frac{(x-2)^2}{81} + \frac{(y+1)^2}{16} = 1 \). Major axis endpoints: \((-7, -1), (11, -1)\). Minor axis endpoints: \((2, -5), (2, 3)\). Foci: \((2-\sqrt{65}, -1), (2+\sqrt{65}, -1)\).

Step 1: Identify the Standard Form

The equation of an ellipse in standard form is given by \( \frac{{(x-h)^2}}{a^2} + \frac{{(y-k)^2}}{b^2} = 1 \) for a horizontal major axis, or \( \frac{{(x-h)^2}}{b^2} + \frac{{(y-k)^2}}{a^2} = 1 \) for a vertical major axis. The given equation \( \frac{(x-2)^{2}}{81} + \frac{(y+1)^{2}}{16} = 1 \) matches the horizontal major axis form with \( a^2 = 81 \) and \( b^2 = 16 \).

Step 2: Determine the Center of the Ellipse

The center \((h, k)\) of the ellipse is found from the values \( h = 2 \) and \( k = -1 \) in the given equation. Thus, the center is \((2, -1)\).

Step 3: Set the Values of a and b

Since \( a^2 = 81 \) and \( b^2 = 16 \), we find \( a = \sqrt{81} = 9 \) and \( b = \sqrt{16} = 4 \).

Step 4: Find the End Points of the Major Axis

For a horizontal major axis, the end points are determined by \((h-a, k)\) and \((h+a, k)\). Substituting the values, we get the end points as \((2-9, -1) = (-7, -1)\) and \((2+9, -1) = (11, -1)\).

Step 5: Find the End Points of the Minor Axis

For the minor axis, the end points are \((h, k-b)\) and \((h, k+b)\). Substituting in the values, the end points are \((2, -1-4) = (2, -5)\) and \((2, -1+4) = (2, 3)\).

Step 6: Calculate the Foci of the Ellipse

The foci are calculated using the formula \( c = \sqrt{a^2 - b^2} \). Here, \( c = \sqrt{81 - 16} = \sqrt{65} \). For a horizontal ellipse, the foci are at \((h-c, k)\) and \((h+c, k)\), resulting in \((2 - \sqrt{65}, -1)\) and \((2 + \sqrt{65}, -1)\).

Key Concepts

Standard Form of an Ellipse

The standard form of the equation of an ellipse is a crucial concept in understanding ellipses. It helps us to identify key features like the major and minor axes and the foci. An ellipse equation in standard form can look like this:

• For a horizontal major axis: \( \frac{{(x-h)^2}}{a^2} + \frac{{(y-k)^2}}{b^2} = 1 \)
• For a vertical major axis: \( \frac{{(x-h)^2}}{b^2} + \frac{{(y-k)^2}}{a^2} = 1 \)

In both cases, \((h, k)\) is the center of the ellipse, and \(x\) and \(y\) are the variables. The terms \(a^2\) and \(b^2\) represent the squares of the lengths of the semi-major and semi-minor axes, respectively. The choice between horizontal and vertical orientation depends on which denominator—\(a^2\) or \(b^2\)—is larger.
In our example, for \( \frac{(x-2)^{2}}{81} + \frac{(y+1)^{2}}{16} = 1 \), the major axis is horizontal since \(a^2 = 81\) (larger than \(b^2 = 16\)), making \(a = 9\) and \(b = 4\). The center of this ellipse is located at \((2, -1)\). This is how we identify the equation in its standard form and understand its geometry.

Major Axis of the Ellipse

The major axis of an ellipse is the longest diameter that passes through its center. It dictates the overall extent of the ellipse in one direction. Depending on whether an ellipse is horizontally or vertically oriented, the major axis length varies.
In the standard ellipse equation, the major axis is determined by the larger denominator \(a^2\) or \(b^2\). When \(a^2\) is greater, the major axis is horizontal, and when \(b^2\) is greater, it's vertical.
For a horizontal ellipse, the endpoints of the major axis can be calculated using the formula:

• Endpoints: \((h-a, k)\) and \((h+a, k)\)

In our case, since \(a^2 = 81\) and \(b^2 = 16\), the major axis is horizontal with \(a = 9\). Thus, the endpoints of the major axis are \((-7, -1)\) and \((11, -1)\). These points showcase the full length of the ellipse along the horizontal axis.

Minor Axis of the Ellipse

The minor axis of an ellipse is the shorter diameter through its center, perpendicular to the major axis. It defines the ellipse's width in the direction opposite to the major axis. This axis is crucial for understanding the balanced shape of the ellipse.

• For horizontal ellipses, the minor axis is vertical.
• For vertical ellipses, the minor axis is horizontal.

When writing equations in the standard form, the minor axis length is determined by the smaller denominator \(a^2\) or \(b^2\). For the ellipse in our exercise, since \(b^2 = 16\), the minor axis is determined by \(b = 4\).
The endpoints of the minor axis for a horizontal ellipse are calculated as:

• Endpoints: \((h, k-b)\) and \((h, k+b)\)

For our ellipse centered at \((2, -1)\), this results in endpoints at \((2, -5)\) and \((2, 3)\). These points define the vertical reach of the ellipse.

Foci of the Ellipse

The foci of an ellipse are two special points located along the major axis of the ellipse. These points are essential because they define the shape and nature of the ellipse. The sum of the distances from any point on the ellipse to the two foci is constant.
To find the foci, we use the equation:

• \(c = \sqrt{a^2 - b^2} \)

From this, \(c\) tells us how far each focus is from the center, along the major axis.
In our example, using \(a^2 = 81\) and \(b^2 = 16\), we calculate \(c = \sqrt{81 - 16} = \sqrt{65}\). Since the major axis is horizontal, the foci are located at \((h-c, k)\) and \((h+c, k)\).
Therefore, the foci of our ellipse are at \((2 - \sqrt{65}, -1)\) and \((2 + \sqrt{65}, -1)\), strategically placed along the major axis.