Problem 12 - Thomas Calculus

Question

Exercises \(9-12\) give the foci or vertices and the eccentricities of ellipses centered at the origin of the \(x y\) -plane. In each case, find the ellipse's standard-form equation in Cartesian coordinates. $$\begin{array}{l}{\text { Vertices: }( \pm 10,0)} \\ {\text { Eccentricity: } 0.24}\end{array}$$

Step-by-Step Solution

Answer

The ellipse equation is \(\frac{x^2}{100} + \frac{y^2}{94.24} = 1\).

Step 1: Identify the Type of Ellipse

Given that the vertices are \((\pm 10,0)\), the major axis is along the x-axis, implying a horizontal ellipse. This indicates that the form of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

Step 2: Determine the Semi-Major Axis

The vertices \((\pm 10,0)\) indicate that the length of the major axis is 20. Thus, \(a = 10\) because the semi-major axis \(a\) is half of the length of the major axis.

Step 3: Use Eccentricity to Calculate \(c\)

The eccentricity \(e\) is given as 0.24. The eccentricity relation for an ellipse is \(e = \frac{c}{a}\). By substituting the given values and solving for \(c\), we have \(c = e \times a = 0.24 \times 10 = 2.4\).

Step 4: Calculate the Semi-Minor Axis

Using the relationship \(b^2 = a^2 - c^2\), we can find \(b\). Substituting the values: \(b^2 = 10^2 - 2.4^2 = 100 - 5.76 = 94.24\). Thus, \(b = \sqrt{94.24}\).

Step 5: Write the Equation of the Ellipse

Now that we have \(a = 10\) and \(b = \sqrt{94.24}\), we can write the equation of the ellipse in standard form: \(\frac{x^2}{100} + \frac{y^2}{94.24} = 1\).

Key Concepts

Eccentricity

Eccentricity (denoted as \(e\)) is a parameter that determines how elongated an ellipse is. It varies between \(0\) to \(1\). A circle, which is a special case of an ellipse, has an eccentricity of \(0\), because it is perfectly round. As eccentricity approaches \(1\), the ellipse becomes more stretched out.
In this particular problem, the eccentricity is \(0.24\), indicating that the ellipse isn't particularly elongated and is fairly close to a circle in shape.
The value of eccentricity relates the foci and the semi-major axis using 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. With these relationships, you can understand and derive other properties of the ellipse, such as its overall shape and spread.

Vertices

Vertices are key points on an ellipse that lie along the major axis. They represent the furthest points from the center on this axis.
For our given exercise, the vertices are \((\pm 10,0)\). This means they lie on the x-axis, indicating a horizontal orientation of the ellipse.
The distance between these vertices gives twice the length of the major axis, which is always denoted as \(2a\). Therefore, the calculation is:

• \(2a = 20\)
• \(a = 10\)

Knowing the vertices help in sketching the ellipse accurately, as they determine its bounding dimension. Along with identifying the orientation (horizontal or vertical), they also provide an anchor to connect the dots of other calculated values, like foci, to visualize the entire shape.

Semi-Major Axis

The semi-major axis of an ellipse is half of the longest diameter, extending from the center to a vertex. It plays a crucial role because it dictates the spread of the ellipse along its major direction.
In the problem, the semi-major axis length \(a\) was found to be \(10\) as it is half the distance between the vertices \((\pm 10,0)\).

• It aligns with the x-axis because the vertices lie horizontally.
• Helps to establish the major shape and size of the ellipse.

This axis is also used to find the semi-minor axis, which complements the major axis in describing the ellipse completely. By utilizing the relationship \(b^2 = a^2 - c^2\), where \(b\) is the semi-minor axis and \(c\) is the distance to the foci, you can uncover all the dimensions necessary to write the standard equation of the ellipse.