Problem 38 - Algebra and Trigonometry

Question

Find an equation for the ellipse that satisfies the given conditions. Eccentricity \(\frac{1}{9},\) foci \((0, \pm 2)\)

Step-by-Step Solution

Answer

The equation is \(\frac{x^2}{320} + \frac{y^2}{324} = 1\).

Step 1: Understand the Ellipse Parameters

Ellipses are defined by several parameters, including their foci, eccentricity, and semi-major axis (usually denoted by \(a\) for horizontal ellipses or by \(b\) for vertical ellipses). We know the eccentricity \(e\) is given by \(\frac{1}{9}\) and the foci are at \((0, \pm 2)\). This gives us an understanding of the orientation and position of the ellipse.

Step 2: Determine the Orientation of the Ellipse

The foci are at \((0, \pm 2)\), indicating that the major axis is vertical (aligned with the \(y\)-axis). This means our ellipse is centered at the origin \((0, 0)\). So we are dealing with a vertical ellipse.

Step 3: Use the Eccentricity to Find \(a\)

The eccentricity \(e\) of an ellipse is given by \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus. From the foci information, \(c = 2\). Since \(e = \frac{1}{9}\), set it in the equation \(\frac{2}{a} = \frac{1}{9}\). Solve for \(a\) to find \(a = 18\).

Step 4: Calculate \(b^2\) Using \(a\) and \(c\)

For a vertical ellipse, the equation relating these parameters is \(c^2 = a^2 - b^2\). Plugging \(a = 18\) and \(c = 2\) into the equation gives \(2^2 = 18^2 - b^2\). Simplifying, \(4 = 324 - b^2\). Therefore, \(b^2 = 320\).

Step 5: Write the Equation of the Ellipse

For a vertical ellipse with center at the origin, the equation is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\). Substitute the known values: \(b^2 = 320\) and \(a^2 = 324\). Thus, the equation is \(\frac{x^2}{320} + \frac{y^2}{324} = 1\).

Key Concepts

Eccentricity

Every ellipse has a unique property called eccentricity, denoted as \(e\), which measures how stretched or elongated the ellipse is. If the eccentricity is close to 0, the ellipse looks more like a circle. If \(e\) is closer to 1, it appears more elongated. Eccentricity is a key parameter in determining an ellipse's shape. It is calculated using the formula:

• \(e = \frac{c}{a}\)

where \(c\) is the distance from the center to one of the foci and \(a\) is the length of the semi-major axis. In our problem, the eccentricity is given as \(\frac{1}{9}\), indicating a relatively circular shape. Understanding eccentricity helps in visualizing how much the ellipse deviates from being a perfect circle.

Foci

In an ellipse, the foci are two specially located points that play a central role in its geometry. For any point on the ellipse, the sum of distances to these two foci remains constant. Knowing the location of the foci helps in finding other parameters of the ellipse. For this exercise, the foci are at \((0, \pm 2)\). This implies that both foci are located on the vertical \(y\)-axis, symmetrically above and below the center, which is at the origin \((0,0)\). The value \(c = 2\) is derived from the distance from the center to either focus. This value of \(c\) is crucial for using the eccentricity to find other properties of the ellipse.

Semi-Major Axis

The semi-major axis, denoted by \(a\), is the longest radius of the ellipse, extending from the center to the furthest point on the edge of the ellipse. It essentially defines the "length" of the ellipse. In our vertical ellipse problem, the semi-major axis lies along the vertical \(y\)-axis. The eccentricity formula \(e = \frac{c}{a}\) allows us to solve for \(a\). With \(e = \frac{1}{9}\) and \(c = 2\), solving \(\frac{2}{a} = \frac{1}{9}\) gives us \(a = 18\). Thus, the semi-major axis extends 18 units from the center in the vertical direction.

Vertical Ellipse

A vertical ellipse is characterized by a major axis that runs parallel to the \(y\)-axis. This implies that the ellipse stretches more vertically than horizontally. For a vertical ellipse centered at the origin, the general equation is:

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

Here, \(a\) is the semi-major axis and \(b\) is the semi-minor axis. Based on the given foci of \((0, \pm 2)\), we know the ellipse is vertical. Knowing whether an ellipse is vertical or horizontal is vital in forming and understanding its equation, as it influences the components derived from the foci and axes.

Ellipse Parameters

The parameters of an ellipse provide vital information about its shape and orientation. Key parameters include:

• Center: In this problem, the center is \((0,0)\).
• Semi-major axis \(a\): Length along the major direction, here \(a = 18\).
• Semi-minor axis \(b\): Length along the minor direction.
• Foci distance \(c\): Given as 2, which helps calculate \(b\^2\).

The equation for this vertical ellipse is constructed using these parameters: \(\frac{x^2}{320} + \frac{y^2}{324} = 1\), where \(a^2 = 324\) and \(b^2 = 320\). Understanding how these parameters are derived and used helps you paint a clear picture of any ellipse you encounter.