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Chapter 11: Problem 43

Find an equation for the ellipse that satisfies the given conditions. Eccentricity: \(0.8,\) foci: \((\pm 1.5,0)\)

Short Answer

Expert verified
The equation of the ellipse is \( \frac{x^2}{3.515625} + \frac{y^2}{1.265625} = 1 \).

Step by step solution

01

Understanding Eccentricity and Foci

The eccentricity of an ellipse is given 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-major axis. In this problem, \( e = 0.8 \) and \( c = 1.5 \), since the foci are at \( (\pm 1.5, 0) \).
02

Calculating the Semi-Major Axis Length

Using the equation for eccentricity \( e = \frac{c}{a} \), we can solve for \( a \):\[ 0.8 = \frac{1.5}{a} \]. This gives us \( a = \frac{1.5}{0.8} = 1.875 \). Hence, the length of the semi-major axis is 1.875.
03

Calculating the Semi-Minor Axis Length

For ellipses, the relationship \( b^2 = a^2 - c^2 \) holds, where \( b \) is the semi-minor axis. We already found \( a = 1.875 \) and \( c = 1.5 \). Substituting these values gives \[ b^2 = (1.875)^2 - (1.5)^2 = 3.515625 - 2.25 = 1.265625 \]. Calculating \( b \), we find \( b = \sqrt{1.265625} \approx 1.125 \).
04

Writing the Equation of the Ellipse

With \( a = 1.875 \) and \( b \approx 1.125 \), the standard form of the ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substituting \( a^2 = 3.515625 \) and \( b^2 = 1.265625 \), the equation becomes \[ \frac{x^2}{3.515625} + \frac{y^2}{1.265625} = 1 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Eccentricity
Eccentricity is a measure of how much an ellipse deviates from being a circle. A circle has an eccentricity of 0, while an ellipse has an eccentricity greater than 0 but less than 1.

In the context of an ellipse, the eccentricity (\( e \)) is calculated with the formula:
  • \( e = \frac{c}{a} \)
Here, \( c \) is the distance from the center of the ellipse to each focus, and \( a \) is the length of the semi-major axis.

For our example, the eccentricity is given as 0.8, and the foci are located at \((\pm 1.5, 0)\), meaning \( c = 1.5 \). Through the eccentricity formula, we can solve for \( a \) and further explore the ellipse's properties.
Semi-Major Axis
In an ellipse, the semi-major axis is the longest radius, running from the center to the furthest edge. The semi-major axis length is crucial because it helps determine the shape and size of the ellipse.

To find the semi-major axis in our exercise, we use the given eccentricity equation:
  • \( e = \frac{c}{a} \)
Given \( e = 0.8 \) and \( c = 1.5 \), we can rearrange this to find \( a \):
  • \( 0.8 = \frac{1.5}{a} \)
  • \( a = \frac{1.5}{0.8} = 1.875 \)
So, the semi-major axis of the ellipse is 1.875 units long.
Semi-Minor Axis
The semi-minor axis of an ellipse is the shortest radius, extending from the center to the edge along the axis perpendicular to the semi-major axis. The semi-minor axis determines the vertical span of the ellipse when it is oriented with its semi-major axis along the x-axis.

We determine the length of the semi-minor axis using the equation:
  • \( b^2 = a^2 - c^2 \)
Given \( a = 1.875 \) and \( c = 1.5 \), the calculation becomes:
  • \( b^2 = (1.875)^2 - (1.5)^2 = 3.515625 - 2.25 \)
  • \( b^2 = 1.265625 \)
  • \( b = \sqrt{1.265625} \approx 1.125 \)
Hence, the semi-minor axis measures approximately 1.125 units.
Ellipse Formula
The standard equation of an ellipse centered at the origin \( (0, 0) \) is derived from the lengths of its semi-major and semi-minor axes.

For an ellipse aligned with its semi-major axis along the x-axis, the equation is expressed as:
  • \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
In our specific problem, with \( a = 1.875 \) and \( b \approx 1.125 \), substitute these into the equations:
  • \( a^2 = 3.515625 \)
  • \( b^2 = 1.265625 \)
This gives the ellipse's equation:
  • \( \frac{x^2}{3.515625} + \frac{y^2}{1.265625} = 1 \)
This equation represents the ellipse precisely, sketching a path through all the points satisfying this relationship.

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