Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the given central conic. Ellipse with a focus at \((0,-5)\) and eccentricity \(\frac{1}{3}\)

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{x^2}{200} + \frac{y^2}{225} = 1 \).

Step 1: Understanding Eccentricity of Ellipse

For an ellipse, the eccentricity \( e \) is the ratio of the distance from the center to a focus (\( c \)) to the distance from the center to a vertex (\( a \)). Given \( e = \frac{1}{3} \), it follows that \( c = ae \).

Step 2: Understanding the Placement of Focus

The focus of the ellipse is given at \((0, -5)\). Since the y-coordinate of the focus is -5, it suggests that the ellipse is oriented vertically with its center, say at \((0, 0)\).

Step 3: Calculating the Distance c

Use the eccentricity formula \( c = ae \). Substituting \( e = \frac{1}{3} \) gives \( c = a \times \frac{1}{3} \). Since \( c \) is the distance from the center to the focus, \( c = 5 \) (because distance from \((0, 0)\) to \((0, -5)\) is 5). Now solve for \( a \): \( 5 = a \times \frac{1}{3} \Rightarrow a = 15 \).

Step 4: Finding b using Ellipse Formula

The relationship \( b^2 = a^2 - c^2 \) holds for ellipses. Using this relationship, calculate \( b^2 \): \( a = 15 \, \text{and} \, c = 5 \), so \( a^2 = 225 \) and \( c^2 = 25 \). Substituting gives \( b^2 = 225 - 25 = 200 \).

Step 5: Writing the Ellipse Equation

Since the conic section is a vertically oriented ellipse centered at the origin, its geometric equation is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Substituting values for \( a^2 \) and \( b^2 \) yields: \( \frac{x^2}{200} + \frac{y^2}{225} = 1 \).

Key Concepts

Ellipse Equation

An ellipse can be defined mathematically by its equation. The general form of the equation for an ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) when centered at the origin. Here, \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. These parameters determine the shape and size of the ellipse.
The values for \(a^2\) and \(b^2\) are crucial to writing the equation correctly:

• If \(a \gt b\), the major axis is along the y-direction, resulting in a vertically oriented ellipse.
• If \(b \gt a\), the major axis is along the x-direction, resulting in a horizontally oriented ellipse.

In the context of our solution, given \(a^2 = 225\) and \(b^2 = 200\), the ellipse is elongated along the y-axis, leading to a vertically oriented ellipse. Hence, the specific equation for this ellipse is \( \frac{x^2}{200} + \frac{y^2}{225} = 1 \). This equation characterizes all the points \((x, y)\) that form the ellipse.

Eccentricity of Ellipse

Eccentricity (denoted by \(e\)) is a measure of how much an ellipse deviates from being circular. For ellipses, the eccentricity is always between 0 and 1, given as \( e = \frac{c}{a} \).
Here, \(c\) is the distance from the center to a focus, and \(a\) is the distance from the center to a vertex (semi-major axis).
To understand this:

• An ellipse with \(e = 0\) would be a perfect circle.
• As \(e\) approaches 1, the ellipse becomes more elongated.

In the exercise, the eccentricity was given as \(\frac{1}{3}\). This tells us that the ellipse is not extremely elongated, having a modest deviation from being a circle. Knowing \(e\) is essential for determining both the focus and vertex, especially in deriving the equation \(c = ae\). With \(e = \frac{1}{3}\), the focus lies closer to the center relative to the vertex, helping us solve for other parameters.

Focus and Vertex of Ellipse

In an ellipse, the foci are two fixed points located along the major axis. The focus is fundamental in defining the shape of the ellipse. For our ellipse, one given focus is at \((0, -5)\).
From the center \((0,0)\) to this focus, the distance \(c = 5\). Combining this with the provided eccentricity \(e = \frac{1}{3}\), we calculate that \(a = 15\) using the relationship \(c = ae\).
These calculated parameters help sense the placement and spread of the ellipse:

• The vertices, being \(a\) distance away from the center, lie at \((0, 15)\) and \((0, -15)\).
• The foci, utilizing \(c\) distance, further define the horizontal elongation.

Positioning the foci and vertices is a core step in visualizing and constructing the ellipse accurately in physical or theoretical models.

Orientation of Ellipse

The orientation of an ellipse determines the direction along which its major axis is aligned. It influences the arrangement of its axes:

• Vertical Orientation: Major axis runs along the y-axis, causing the ellipse to extend vertically.
• Horizontal Orientation: Major axis runs along the x-axis, causing the ellipse to extend horizontally.

In this exercise, the focus at \((0, -5)\) and the distance calculations through eccentricity suggested a vertical orientation. This means that the major axis is parallel to the y-axis, resulting in vertex points along that line.
Aligning the ellipse vertically, with \(a = 15\) and \(b = 200\), our equation becomes \( \frac{x^2}{200} + \frac{y^2}{225} = 1 \). This orientation is crucial because it affects how the ellipse interacts with other geometrical constructs or when solving problems involving systems with ellipses.