Problem 10 - 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. $$\begin{array}{l}{\text { Foci: }( \pm 8,0)} \\ {\text { Eccentricity: } 0.2}\end{array}$$

Step-by-Step Solution

Answer

The ellipse's equation is \(\frac{x^2}{1600} + \frac{y^2}{1536} = 1\).

Step 1: Understand the Relationship between Parameters

The standard form equation for an ellipse centered at the origin with the major axis along the x-axis is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) the semi-minor axis. The foci \((\pm c, 0)\) and the eccentricity \(e\) are related by \(c = ae\).

Step 2: Solve for c

Given that the foci are \((\pm 8, 0)\), we know \(c = 8\).

Step 3: Solve for a using Eccentricity

The eccentricity is given as \(e = 0.2\). Using the relationship \(c = ae\), substitute \(c = 8\) and \(e = 0.2\):\[ae = 8 \Rightarrow a(0.2) = 8\Rightarrow a = \frac{8}{0.2} = 40\]

Step 4: Calculate b using Relationship in Ellipse

Using the relationship \(b^2 = a^2 - c^2\):\[b^2 = 40^2 - 8^2 = 1600 - 64 = 1536\]Thus, \(b = \sqrt{1536}\).

Step 5: Write the Standard Form Equation of the Ellipse

The formula for the standard form of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). We have already calculated that \(a = 40\) and \(b^2 = 1536\), so:\[\frac{x^2}{1600} + \frac{y^2}{1536} = 1\]

Key Concepts

Ellipse Foci

When studying ellipses, the foci are key components. An ellipse has two fixed points, called foci, located symmetrically along the major axis. In any ellipse, every point on its perimeter has the unique property that the sum of the distances to these two foci remains constant. This is what defines the elliptical shape. In the given exercise, the foci are located at the points \(\pm 8, 0\). This means the distance from the center to each focus, known as \(c\), is 8.

• Foci help determine the shape and orientation of an ellipse.
• They are always along the major axis of the ellipse.
• For an ellipse centered at the origin, foci coordinates are often expressed as \(\pm c, 0\) or \(0, \pm c\) depending on the axis orientation.

Understanding the exact position of the foci helps in defining the ellipse's geometry and orientation in the Cartesian plane.

Eccentricity

Eccentricity is a crucial aspect of ellipses, as it quantifies their shape, indicating how much they deviate from being circular. The eccentricity, denoted by \(e\), is a number between 0 and 1 for ellipses:

• If \(e = 0\), the shape is a perfect circle, where the lengths of the semi-major and semi-minor axes are equal.
• If \(0 < e < 1\), the figure is an ellipse, with greater values of \(e\) yielding more elongated shapes.

In our particular problem, the ellipse has an eccentricity of \(e = 0.2\). This suggests a slightly elongated shape. To solve for the semi-major axis \(a\), we use the relation \(c = ae\), where \(c\) is the distance between the foci and the center. Here, \(c = 8\) and \(e = 0.2\), leading to the calculation of \(a\) as 40.

Semi-major and Semi-minor Axes

The semi-major and semi-minor axes of an ellipse are fundamental geometric properties:

• The semi-major axis, denoted as \(a\), is the longest radius extending from the center to the perimeter along the major axis. For our ellipse, \(a = 40\).
• The semi-minor axis, denoted as \(b\), is the shortest radius extending from the center and perpendicular to the major axis. The length of \(b\) is found using the equation \(b^2 = a^2 - c^2\).

In the calculation, \(a = 40\) and \(c = 8\), hence:\[
b^2 = 40^2 - 8^2 = 1600 - 64 = 1536
\]Therefore, \(b = \sqrt{1536}\).These axes help to determine the precise dimensions of an ellipse and how it stretches across the plane. The lengths of \(a\) and \(b\) directly influence the appearance and extent of the ellipse in both the horizontal and vertical directions.