Problem 37 - A Graphical Approach to College Algebra

Question

Find an equation for each ellipse. Vertices ( \(4,9\) ) and (4, 1); minor axis of length 6

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{(x-4)^2}{9} + \frac{(y-5)^2}{16} = 1 \).

Step 1: Identify the Orientation

Given the vertices (4, 9) and (4, 1), we identify that they have the same x-coordinate, which implies a vertical major axis for the ellipse. Thus, the standard form of the ellipse equation is \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \).

Step 2: Locate the Center

The center \((h, k)\) is the midpoint of the vertices. Calculate it as follows: \( h = 4 \) (since the x-coordinate is the same), and \( k = \frac{9 + 1}{2} = 5 \). So, the center is \( (4, 5) \).

Step 3: Determine the Length of the Major Axis

The distance between the vertices \( (4,9) \) and \( (4, 1) \) gives us the length of the major axis, which is the distance in the y-direction: \( 9 - 1 = 8 \). The semi-major axis length \( a \) is half of this: \( a = 8/2 = 4 \).

Step 4: Determine the Length of the Minor Axis

Given that the minor axis length is 6, the semi-minor axis length \( b \) is half of this: \( b = 6/2 = 3 \).

Step 5: Write the Equation of the Ellipse

Using the standard form for a vertical ellipse and substituting the values of \( (h, k) = (4, 5)\), \( a = 4 \), and \( b = 3 \), the equation becomes \( \frac{(x-4)^2}{3^2} + \frac{(y-5)^2}{4^2} = 1 \) or \( \frac{(x-4)^2}{9} + \frac{(y-5)^2}{16} = 1 \).

Key Concepts

Major Axis

In the context of an ellipse, the major axis is the longest diameter that passes through the center and both foci of the ellipse. This line essentially bisects the ellipse into two equal halves and determines its orientation and shape.
The major axis can be oriented horizontally or vertically. If the major axis is vertical, like in this particular exercise, it means the ellipse is taller rather than wider.
The vertices given in the problem are \( (4, 9) \) and \( (4, 1) \). These points have the same x-coordinate, indicating a vertical orientation. Therefore, the distance between these two points, \( 9 - 1 = 8 \), represents the length of the entire major axis.
This length is crucial as it determines the position and size of the ellipse. The semi-major axis, which is half the length of the major axis, is used in the equation for the ellipse to establish its towering dimension.

Minor Axis

The minor axis of an ellipse is the shorter diameter, perpendicular to the major axis, and passes through the center of the ellipse. It is an equally important element in defining the shape of the ellipse.
In this particular problem, the length of the minor axis is provided as 6. To find the semi-minor axis, which we need for the standard ellipse equation, we simply divide this length by 2:

• Semi-minor axis: \( b = \frac{6}{2} = 3 \)

This semi-minor length becomes a crucial part of the equation of the ellipse alongside the semi-major axis. It directly influences the width of the ellipse, indicating how wide the ellipse will appear once plotted.

Midpoint Calculation

The midpoint, often referred to as the center of the ellipse, is essential in forming the standard equation of the ellipse. It is determined as the average of the coordinates of the vertices for ellipses with a vertical major axis.
For the given vertices, \( (4, 9) \) and \( (4, 1) \), the center is found by averaging the y-coordinates because the x-coordinates remain constant:

• Horizontal center \( h = 4 \)
• Vertical center \( k = \frac{9+1}{2} = 5 \)

Hence, the midpoint or the center of the ellipse is \( (4, 5) \). This point is used to translate the ellipse within its coordinate plane and is pivotal for constructing the ellipse equation, anchoring its position and symmetry.