Problem 54 - Precalculus: Mathematics for Calculus

Question

Finding the Equation of an Ellipse Find an equation for the ellipse that satisfies the given conditions. Eccentricity: \(0.75,\) foci: \((\pm 1.5,0)\)

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{4} + \frac{y^2}{1.75} = 1 \).

Step 1: Understand the Relationship between Eccentricity and Axes

The eccentricity of an ellipse, denoted as \( e \), is given by the formula \( e = \frac{c}{a} \), where \( c \) is the distance from the center to each focus and \( a \) is the semi-major axis length. Given \( e = 0.75 \) and \( c = 1.5 \) (from the foci \( (\pm 1.5, 0) \)), we can calculate \( a \).

Step 2: Calculate the Semi-Major Axis (a)

Using the formula \( e = \frac{c}{a} \), substituting the given values, we get \( 0.75 = \frac{1.5}{a} \). Solving for \( a \), we have \( a = \frac{1.5}{0.75} = 2 \).

Step 3: Find the Semi-Minor Axis (b)

From the relationship \( b^2 = a^2 - c^2 \), substitute \( a = 2 \) and \( c = 1.5 \) to find \( b^2 \). This gives \( b^2 = 2^2 - 1.5^2 = 4 - 2.25 = 1.75 \). Therefore, \( b = \sqrt{1.75} \).

Step 4: Write the Equation of the Ellipse

The standard form of an ellipse centered at the origin with its major axis along the x-axis is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substitute \( a = 2 \) and \( b^2 = 1.75 \) into the equation, yielding \( \frac{x^2}{4} + \frac{y^2}{1.75} = 1 \).

Key Concepts

Eccentricity of an Ellipse

The concept of eccentricity in ellipses helps describe how "stretched" they are. An ellipse ranges from a circle to a more elongated shape, and eccentricity quantifies this aspect.
Eccentricity, denoted by the symbol \( e \), is calculated using the formula:

• \( e = \frac{c}{a} \)

where \( c \) is the distance from the center of the ellipse to one of its foci, and \( a \) is the length of the semi-major axis.
For circles, \( e = 0 \) because \( c = 0 \), as the center and foci coincide. As eccentricity increases, the ellipse becomes more elongated. For the given exercise, the eccentricity of 0.75 indicates a moderately stretched ellipse.
Understanding this ratio helps when analyzing or constructing an ellipse from specific characteristics like foci or axis lengths.

Semi-Major Axis

The semi-major axis of an ellipse is a crucial element in determining its overall size. It runs along the longest diameter of the ellipse, extending from the center to one endpoint.
In mathematical terms, it is denoted by \( a \). Knowledge of the semi-major axis is essential when calculating the eccentricity or standard form of an ellipse.
In the exercise, given \( c = 1.5 \) and \( e = 0.75 \), the semi-major axis \( a \) is determined using:

• \( a = \frac{c}{e} = \frac{1.5}{0.75} = 2 \)

This calculation reveals the length of \( a \), which is fundamental for figuring out other properties of the ellipse.

Semi-Minor Axis

The semi-minor axis of an ellipse is the shorter diameter, crossing perpendicular to the semi-major axis at the center. It is represented as \( b \), and helps determine the dimensions of an ellipse.
Using the relationship between the axes and foci described by:

• \( b^2 = a^2 - c^2 \)

we can calculate \( b \). In our example with \( a = 2 \) and \( c = 1.5 \):

• \( b^2 = 2^2 - 1.5^2 = 4 - 2.25 = 1.75 \)
• \( b = \sqrt{1.75} \)

This calculation tells us the semi-minor axis length, enriching our understanding of the ellipse's shape and proportions.

Standard Form of an Ellipse

The standard form of an ellipse provides a way to represent any ellipse mathematically, especially when centered at the origin.
The equation is typically given as:

• \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively.
For the specific ellipse in this exercise with \( a = 2 \) and \( b^2 = 1.75 \), the standard form is:

• \( \frac{x^2}{4} + \frac{y^2}{1.75} = 1 \)

This equation helps visualize and interpret the ellipse on a coordinate plane, ensuring a complete understanding of its mathematical properties.