Problem 53 - Precalculus: Functions and Graphs

Question

Exer. \(51-58:\) Determine whether the graph of the equation is the upper, lower, left, or right half of an ellipse, and find an equation for the ellipse. $$x=-\frac{1}{3} \sqrt{9-y^{2}}$$

Step-by-Step Solution

Answer

The graph is the left half of the ellipse with equation \( \frac{x^2}{1} + \frac{y^2}{9} = 1 \).

Step 1: Identify the given equation form

The equation given is \( x = -\frac{1}{3} \sqrt{9 - y^2} \). This equation represents an altered version of the form \( x = a \sqrt{b - y^2} \), where it is clear that the variable \( y \) is squared, suggesting a relation similar to that of an ellipse.

Step 2: Rewrite the equation to isolate the radical term

Move the radical to one side and everything else to the other to focus the equation form: \( x = -\frac{1}{3} \sqrt{9 - y^2} \). Square both sides to eliminate the square root, leading to \( x^2 = \frac{1}{9}(9 - y^2) \).

Step 3: Eliminate fractions by multiplying through

Multiply all terms by 9 to clear the fraction on the right side: \( 9x^2 = 9 - y^2 \), which simplifies to \( y^2 = 9 - 9x^2 \). Rearrange as \( 9x^2 + y^2 = 9 \), adjusting to resemble the standard ellipse equation form.

Step 4: Determine the complete ellipse equation

Recognize that the standard form of an ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Given \( 9x^2 + y^2 = 9 \), divide through by 9 to yield \( \frac{x^2}{1} + \frac{y^2}{9} = 1 \), identifying it as an ellipse centered at the origin.

Step 5: Identify which half of the ellipse is given

The original equation \( x = -\frac{1}{3}\sqrt{9 - y^2} \) results in negative \( x \) values. This means it only represents the left half of the ellipse because the square root's negative sign results in \( x \) being negative.

Key Concepts

Half of an Ellipse

When dealing with ellipses, you might sometimes encounter equations that only represent a portion of the entire shape. These portions are often referred to as halves of an ellipse. Based on the orientation and location of variables, these halves could be the upper, lower, left, or right portions. In the given problem, the original equation is not symmetrical around the origin or axes; hence, it does not represent a full ellipse.

• Recognizing the half of the ellipse involves determining whether the equation describes only half of its shape, based on its structure.
• For instance, when the equation is of the form \( x = -\frac{1}{3} \sqrt{9-y^2} \), it suggests a partial representation, likely because of the square root.
• The negative sign preceding this square root indicates the left half, where the x-values are negative.

Understanding how to discern these halves is crucial for determining which part of the coordinate plane the ellipse resides on. This skill helps in laying out the right graphical representation when you solve geometry-related problems.

Standard Form of Ellipse

The standard form of an ellipse's equation provides a clear mathematical structure of its dimensions and orientation. An ellipse in its standard form is expressed as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where:

• \(a\) is the semi-major axis length
• \(b\) is the semi-minor axis length

Keeping these dimensions in mind helps to transform modified or complex equations into ones easier to visualize and plot.
In the problem, we transform the equation \(9x^2 + y^2 = 9\) to its standard form. Dividing each term by 9, we get:

• \( \frac{x^2}{1} + \frac{y^2}{9} = 1 \) – indicating a vertical ellipse centered at the origin \((0,0)\).

The process of bringing equations to this form is essential. This makes it more straightforward to identify the ellipse's size and orientation, distinguishing it from other conic sections like circles or hyperbolas.

Squaring Equations

Working with equations that contain square roots often involves squaring both sides to simplify them. This process can turn complex-looking problems into linear forms that are easier to handle. In mathematical terms, squaring an equation helps eliminate square roots to reveal a more streamlined algebraic form.
For instance, when you start with the equation: \( x = -\frac{1}{3} \sqrt{9 - y^2} \), squaring both sides removes the square root:

• \( x^2 = \left(-\frac{1}{3} \right)^2 (9 - y^2) \)
• This process simplifies to \( x^2 = \frac{1}{9} (9 - y^2) \)

Through this systematic approach, you can expose the underlying mathematical framework, making it easier to progress steps further towards solving for x and y. Always practice caution, as squaring can introduce extra solutions that might not apply to the original context.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebra to understand geometric principles. It provides a means to employ coordinates and equations to describe geometric figures, precisely outlining attributes like location, size, and shape.
For ellipses, coordinate geometry crucially helps in:

• Defining positions and properties using equations such as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
• Facilitating transformations of equations involving ellipses between different forms, helping to locate shapes within a coordinate system efficiently.
• Providing insight into how these equations behave under constraints such as sections of ellipses or specific orientations.

By harnessing these concepts, you can effectively dissect and understand the relationship between algebraic equations and their geometric implications on coordinate planes. This skill is invaluable for problems that ask you to interpret or visualize shapes based on given data.