Problem 40 - Algebra and Trigonometry

Question

Find an equation for the ellipse that satisfies the given conditions. Endpoints of minor axis: \((0, \pm 3),\) distance between foc: 8

Step-by-Step Solution

Answer

The equation is \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

Step 1: Understanding the Problem

To find the equation of the ellipse, note that the given endpoints of the minor axis are \((0, -3)\) and \((0, 3)\), which implies that the ellipse is centered at the origin \((0, 0)\) and has its minor axis along the y-axis.

Step 2: Calculate Semi-Minor Axis Length

The semi-minor axis length is half the distance between the endpoints of the minor axis. The length of the minor axis is \(2 imes 3 = 6\), so the semi-minor axis \(b\) is 3.

Step 3: Express Relationship with Foci and Semi-Major Axis

The distance between the foci is given as 8, thus \(2c = 8\), which gives \(c = 4\). In an ellipse, we have the relationship \(c^2 = a^2 - b^2\), where \(a\) is the length of the semi-major axis.

Step 4: Calculate Semi-Major Axis Length

Substitute the known values into the equation: \(c^2 = a^2 - b^2\), thus \(4^2 = a^2 - 3^2\) or \(16 = a^2 - 9\). Solving for \(a^2\) gives \(a^2 = 25\), so \(a = 5\).

Step 5: Write the Equation of the Ellipse

The standard form of an ellipse with a semi-major axis along the x-axis and semi-minor axis along the y-axis is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Substituting \(a = 5\) and \(b = 3\) gives the equation \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

Key Concepts

Semi-Major Axis

The semi-major axis of an ellipse is the longest radius, stretching from the center to the farthest edge of the ellipse. In many ways, it resembles the larger sibling of the semi-minor axis. If we imagine the ellipse as a "stretched circle," the semi-major axis gives us the maximum stretch. This axis is important because it dictates the width of the ellipse. When you place an ellipse with its center at the origin, the semi-major axis can lie along the x-axis or y-axis depending on its orientation.
In the given problem, after computing the necessary components, we found the semi-major axis to be 5. This means that on a coordinate grid, the ellipse extends 5 units outward from the center in the direction of the semi-major axis, which in this problem, is aligned with the x-axis.

• The formula to find the semi-major axis, when you know the foci and semi-minor axis, is: \(a^2 = b^2 + c^2\) where \(c\) is the distance from the center to each focus.
• For our ellipse: \(a = 5\).

Semi-Minor Axis

The semi-minor axis is essentially the "shorter" radius of an ellipse. While the semi-major axis decides the ellipse's width, the semi-minor axis determines its height when oriented vertically. In simple terms, it measures how "tall" the ellipse will be.
In the context of our ellipse, the semi-minor axis is 3 units, which comes directly from half the distance between the endpoints of the minor axis, given as (0, -3) and (0, 3). This means, from the center of the ellipse to the edge along the y-axis, it measures 3 units.

• The semi-minor axis is always perpendicular to the semi-major axis.
• The length of the semi-minor axis can be easily deduced once you know the endpoints: \(b = \frac{distance}{2}\).
• For this problem: \(b = 3\).

Foci

Foci are two fixed points on the interior of an ellipse used in the formal mathematical description of the shape. The interesting property of an ellipse is that its shape results from the sum of the distances from these two foci points to any point on the ellipse line being constant.
In our step-by-step solution, we had a distance of 8 between the two foci, giving us \(c = 4\). Foci in an ellipse play a role in defining the level of "eccentricity"—or deviation from being a perfect circle. The farther apart the foci, the more elongated the ellipse appears.

• The distance \(2c\) is equal to the length between the foci. This can be split equally, giving \(c\) as the half-distance from the center to a focus.
• In this problem, the foci lie along the major axis, precisely 4 units away from the ellipse center on either side.

Standard Form of Ellipse

The standard form of the equation for an ellipse allows mathematicians and students alike to determine the ellipse's orientation, center, and axes lengths easily by examining the algebraic expression. The essence of the standard form is to provide a uniform method to represent ellipses on a coordinate system.
If an ellipse is centered at the origin, with its semi-major axis on the x-axis and semi-minor axis on the y-axis, the equation can be written as \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). The numbers in the denominators—\(a^2\) and \(b^2\)—directly inform us of the lengths of the semi-axes.

• Substituting our calculated \(a = 5\) and \(b = 3\) into this form gives us \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).
• This formula succinctly describes the ellipse, allowing users to quickly pick out the semi-major and semi-minor axes and confirm the ellipse's orientation and major properties.