Problem 28 - College Algebra with Modeling and Visualization

Question

Find an equation of the ellipse, centered at the origin, satisfying the conditions. Eccentricity \(\frac{3}{4},\) vertices \((0, \pm 8)\)

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{25x^2}{1024} + \frac{y^2}{64} = 1\).

Step 1: Understand the problem

An ellipse centered at the origin has the standard equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a\) is the semi-major axis length and \(b\) is the semi-minor axis length. We are given the eccentricity as \(\frac{3}{4}\) and the vertices at \((0,\pm 8)\). The vertices indicate that the major axis is vertical.

Step 2: Identify the length of the semi-major axis

The vertices are located at \((0,\pm 8)\). This implies that the length of the semi-major axis, \(b\), is \(8\). Hence, \(b = 8\).

Step 3: Relate eccentricity to semi-major and semi-minor axes

The eccentricity \(e\) is defined by the formula \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus and \(a\) is the length of the semi-minor axis. Given that \(e = \frac{3}{4}\), we can express \(c\) in terms of \(a\): \(c = \frac{3}{4}a\).

Step 4: Use relation between axes and eccentricity

For ellipses, \(c^2 = b^2 - a^2\). Substitute \(c = \frac{3}{4}a\) into this equation. From the previous steps, \(b = 8\), so \(b^2 = 64\). Thus, \(\left(\frac{3}{4}a\right)^2 = 64 - a^2\).

Step 5: Solve for semi-minor axis

Expand and solve \(\left(\frac{3}{4}a\right)^2 = \frac{9}{16}a^2 = 64 - a^2\). Hence, \(\frac{25}{16}a^2 = 64\). Multiply both sides by 16: \(25a^2 = 1024\). Divide by 25: \(a^2 = \frac{1024}{25}\). Hence, \(a = \frac{32}{5}\).

Step 6: Write the equation of the ellipse

Substitute \(a^2 = \frac{1024}{25}\) and \(b^2 = 64\) into the standard form equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). This leads to \(\frac{x^2}{\frac{1024}{25}} + \frac{y^2}{64} = 1\), which simplifies to \(\frac{25x^2}{1024} + \frac{y^2}{64} = 1\).

Key Concepts

Eccentricity

The eccentricity of an ellipse is a measure of how much an ellipse deviates from being a perfect circle. It is a numerical value between 0 and 1, where an ellipse with eccentricity 0 is a perfect circle. Higher eccentricity values indicate a more elongated shape.
For any ellipse, the eccentricity, denoted by the letter \( e \), is calculated using the formula: \[ e = \frac{c}{a} \] where \( c \) is the distance from the center to one of the foci, and \( a \) is the length of the semi-major axis.
In contexts like exercises, knowing this relationship helps in understanding the geometry of the ellipse and is crucial for solving equations related to it. In this exercise, the eccentricity is given as \( \frac{3}{4} \), meaning the ellipse is moderately elongated. It frames how the semi-major and semi-minor axes relate to each other, guiding the formulation of the ellipse's equation.

Semi-major axis

The semi-major axis of an ellipse is the longest radius extending from the center to the perimeter. In simple terms, it's half of the longest diameter. This axis plays a key role in defining the ellipse's size and shape. For this specific problem, the vertices provided are \((0, \pm 8)\), indicating the semi-major axis is vertical. Therefore, the length of the semi-major axis, represented by \( b \) in the equation, is 8 units.
Recognizing which axis is the semi-major allows you to correctly apply the ellipse's standard formula. With a vertical major axis, parameters need to be calculated carefully to maintain the proper geometric representation of the ellipse.

Vertices

Vertices are crucial points in understanding an ellipse's dimensions. They are the endpoints of the ellipse along the major axis. These points tell us not only about the length and orientation of the major axis but also help find other critical parameters of the ellipse.
Given vertices \((0, \pm 8)\) specify that these points are located directly above and below the center, indicating a vertical orientation. This typically means we use the y-coordinates to identify the semi-major axis. From these vertices, we can deduce that the distance to these extreme points is the measure of the semi-major axis, which is 8. Therefore, the vertices inform that the length of the vertical semi-major axis is 16, with 8 being half this distance on either side of the center.

Equation of an ellipse

The standard form equation of an ellipse centered at the origin depends on whether the major axis is horizontal or vertical. For a vertically oriented ellipse like in this exercise, the equation is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where \( a \) is the semi-minor axis length and \( b \) is the semi-major axis length.
Starting with the known eccentricity \( e = \frac{3}{4} \) and the vertical semi-major axis \( b = 8 \), we derived the semi-minor axis \( a \) from the relation \( b^2 = c^2 + a^2 \). Solving these equations, we adjust the standard form parameters specifically for the given ellipse conditions to: \[ \frac{25x^2}{1024} + \frac{y^2}{64} = 1 \] This showcases how the interplay of eccentricity, axis lengths, and ellipse positioning aids in forming its equation. Understanding this balance is essential for interpreting or constructing the ellipse equation correctly.