Problem 18 - 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. $$ \frac{(x-7)^{2}}{49}+\frac{(y-7)^{2}}{49}=1 $$

Step-by-Step Solution

Answer

The ellipse is a circle centered at (7, 7), with endpoints (14, 7), (0, 7), (7, 14), (7, 0), and the foci at (7, 7).

Step 1: Recognize the Equation Form

The given equation \( \frac{(x-7)^2}{49} + \frac{(y-7)^2}{49} = 1 \) is the equation of an ellipse in the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) where \( h = 7 \) and \( k = 7 \), and \( a^2 = 49 \) and \( b^2 = 49 \).

Step 2: Identify the Center of the Ellipse

In the equation \( \frac{(x-7)^2}{49} + \frac{(y-7)^2}{49} = 1 \), the center of the ellipse is \( (h, k) = (7, 7) \).

Step 3: Determine the Lengths of the Axes

Since \( a^2 = b^2 = 49 \), it means \( a = 7 \) and \( b = 7 \). The ellipse is a circle because both axes are equal. Hence, both the major and minor axes have lengths of 14.

Step 4: Identify Endpoints of Major and Minor Axes

For this circle, which is a special case of an ellipse, the endpoints of both axes simply circle around the center. The endpoints are the same for both axes: - On the horizontal axis: \((7+7, 7)=(14, 7)\) and \((7-7, 7)=(0, 7)\)- On the vertical axis: \((7, 7+7)=(7, 14)\) and \((7, 7-7)=(7, 0)\)

Step 5: Determine the Foci

For an ellipse, the distance to the foci \( c \) is given by \( c^2 = a^2 - b^2 \). Here, \( c^2 = 49 - 49 = 0 \). Hence, \( c = 0 \), indicating that the foci coincide at the center (since it is a circle). Therefore, the foci are at \((7, 7)\).

Key Concepts

Ellipse Standard Form

The equation in the form \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]represents an ellipse. The values \(h\) and \(k\) are the coordinates of the center, while \(a^2\) and \(b^2\) are the squares of the lengths of the semi-major and semi-minor axes, respectively.
The standard form allows us to easily identify key features of the ellipse:

• If \(a = b\), the ellipse becomes a circle.
• If \(a eq b\), the ellipse stretches longer along one axis.

In the exercise, our given equation \[ \frac{(x-7)^2}{49} + \frac{(y-7)^2}{49} = 1 \]fits this form. Both \(a^2\) and \(b^2\) are 49, meaning \(a = b\) and the ellipse is perfectly circular. So for practical purposes, it's a circle.

Major and Minor Axes

In an ellipse, the major and minor axes are the longest and shortest diameters, respectively. They intersect at the center of the ellipse.

• The major axis runs through the longest part of the ellipse.
• The minor axis is perpendicular to the major axis and runs through the shortest part.

For the given special case of an ellipse being a circle (where \(a = b\)), both axes are equal in length. Here, since \(a = 7\) and \(b = 7\), both axes measure 14 units in total.Let's break it down:

• Endpoints on the horizontal axis: - Right: \((7+7, 7) = (14, 7)\) - Left: \((7-7, 7) = (0, 7)\)
• Endpoints on the vertical axis: - Top: \((7, 7+7) = (7, 14)\) - Bottom: \((7, 7-7) = (7, 0)\)

With this configuration, both axes coincide perfectly at each point because they form a circle.

Center of Ellipse

The center of an ellipse is represented by the coordinates \((h, k)\) in the standard form equation.
It is the midpoint of both the major and minor axes, providing symmetry to the ellipse. This is particularly helpful when sketching or analyzing the ellipse, as it serves as the "balance point" of the shape.

• In our exercise, the center is at \((7, 7)\).
• This is derived from the terms \((x-7)^2\) and \((y-7)^2\), indicating shifts of 7 units from the origin in both the x and y directions.

This center is key in not only determining the rest of the important points of the ellipse but also working out problems tied to positioning and spatial relations.

Foci of Ellipse

The foci (plural for "focus") are two special points located along the major axis of an ellipse, and they help define its shape. The distance to each focus is determined using the equation \(c^2 = a^2 - b^2\), where \(c\) is the distance from the center of the ellipse to each focus.

• If \(a > b\), the foci are farther apart along the major axis.
• If \(a = b\) (as in the case of a circle), the foci collapse into a single point at the center.

In this exercise, \(a^2 = b^2 = 49\), so \(c = 0\). This means there is no distance between the center and the foci. Hence, the foci are at the same point as the center: \((7, 7)\). This aligns with the fact that the given ellipse is a circle, where the foci and center overlap.