Problem 13 - Algebra for College Students

Question

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ \frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{4}=1 $$

Step-by-Step Solution

Answer

Vertices: (-1, 1) and (5, 1). Minor Axis: (2, -1) and (2, 3). Foci: \((2-\sqrt{5}, 1)\) and \((2+\sqrt{5}, 1)\).

Step 1: Identify the form of the ellipse

The given ellipse equation is \( \frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1 \). This equation is in the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) for horizontal ellipses, where \( (h, k) \) is the center of the ellipse, \( a \) is the semi-major axis, and \( b \) is the semi-minor axis.

Step 2: Find the center of the ellipse

The formulas \( x-h \) and \( y-k \) indicate that the center of the ellipse is at the point \( (h, k) = (2, 1) \).

Step 3: Determine the lengths of the axes

For \( \frac{(x-2)^2}{9} \), \( a^2 = 9 \) which gives \( a = 3 \). For \( \frac{(y-1)^2}{4} \), \( b^2 = 4 \) which gives \( b = 2 \). The major axis is horizontal, and the minor axis is vertical since \( a > b \).

Step 4: Find the vertices of the ellipse

The vertices are determined by moving \( a = 3 \) units left and right from the center. So, the vertices are at \( (2-3, 1) = (-1, 1) \) and \( (2+3, 1) = (5, 1) \).

Step 5: Find the endpoints of the minor axis

The endpoints of the minor axis are obtained by moving \( b = 2 \) units up and down from the center. Therefore, the endpoints are at \( (2, 1-2) = (2, -1) \) and \( (2, 1+2) = (2, 3) \).

Step 6: Calculate the foci of the ellipse

The distance \( c \) from the center to the foci is found using the formula \( c^2 = a^2 - b^2 = 9 - 4 = 5 \). Thus, \( c = \sqrt{5} \). Therefore, the foci are located at \( (2 - \sqrt{5}, 1) \) and \( (2 + \sqrt{5}, 1) \), because the ellipse is horizontal.

Step 7: Sketch the ellipse

Using the center \( (2, 1) \), vertices \( (-1, 1), (5, 1) \), minor axis endpoints \( (2, -1), (2, 3) \), and foci, sketch the ellipse, ensuring the major axis lies horizontally and the minor axis vertically.

Key Concepts

Vertices of Ellipse

When discussing ellipses, the vertices are essential points that are found along the major axis. In any standard-form equation of an ellipse, these vertices lie at the extremes of this axis. For the given ellipse equation \( \frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1 \), the major axis is determined by comparing the denominators under the squared terms. Here, you can tell that 9 is greater than 4, indicating a horizontal major axis.

To find the vertices, we move the length of the semi-major axis, \( a = 3 \), on either side of the center \((2, 1)\). This results in the vertices being located at \((-1, 1)\) and \((5, 1)\). These vertices lie along the x-axis, reinforcing that the major axis is horizontal in this case. Knowing the vertices helps outline the general shape and stretch of the ellipse along its longest dimension.

Foci of Ellipse

The foci of an ellipse are key points that define its shape and are situated on the major axis. The importance of the foci lies in their relationship with the sum of distances from any point on the ellipse, which remains constant.

To determine the foci for our given equation, we use the equation \( c^2 = a^2 - b^2 \), where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively. We established that \( a = 3 \) and \( b = 2 \). Plug these into our formula:

• \(a^2 = 9\),
• \(b^2 = 4\),
• \(c^2 = 9 - 4 = 5\),
• therefore, \(c = \sqrt{5}\).

Thus, the foci are \((2 - \sqrt{5}, 1)\) and \((2 + \sqrt{5}, 1)\), indicating their position on the x-axis, symmetrically placed around the center at \((2, 1)\). These points are closer to the center than the vertices, outlining the ellipse's defining elliptical shape.

Minor Axis Endpoints

The endpoints of the minor axis of an ellipse are equally important and offer information about its height. The minor axis is shorter than the major axis and runs perpendicular to it.

In our example, the minor axis is vertical because the major axis is horizontal (as 9 is greater than 4 from our denominators under the squared terms). The minor axis endpoints are found by moving the length of the semi-minor axis, \( b = 2 \), vertically up and down from the center point \((2, 1)\). So, the endpoints for the minor axis are \((2, -1)\) and \((2, 3)\).

These points help describe the narrow reach of the ellipse, providing a clearer view of its entire shape when graphed. They offer a guide to the ellipse's width across its shorter aspect, which is crucial to plotting an accurate and proportional ellipse.

Graphing Ellipses

Graphing an ellipse requires an understanding of its various parameters: the center, vertices, minor axis endpoints, and sometimes the foci.
Here are the steps to take when graphing the given ellipse equation \( \frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1 \):

• Start by plotting the center of the ellipse, \((2, 1)\).
• Next, mark the vertices \((-1, 1)\) and \((5, 1)\) along the x-axis.
• Place the minor axis endpoints \((2, -1)\) and \((2, 3)\) vertically above and below the center.

After plotting these key points, draw a smooth, oval-shaped curve. The curve should pass through all marked points and mirror the symmetry about both axes. The major axis will appear longer and stretch horizontally, while the minor axis will be clearly shorter and more compact vertically. The visual representation of the ellipse confirms the mathematical properties identified and aids in understanding the spatial relationships of an ellipse.