Problem 53 - A Graphical Approach to Precalculus with Limits

Question

CHECKING ANALYTIC SKILLS Graph each hyberbola by hand. Give the domain and range. Give the center in Exercises \(55-61 .\) Do not use a calculator. $$9 x^{2}-4 y^{2}=1$$

Step-by-Step Solution

Answer

Center: (0,0); Domain: \((-\infty, +\infty)\); Range: \((-\infty, -\frac{1}{2})\) or \((\frac{1}{2}, \infty)\).

Step 1: Identify the Standard Form

The equation of the hyperbola is given by \(9x^2 - 4y^2 = 1\). Recognize that this is a hyperbola centered at the origin, of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). We need to rewrite the given equation in this standard form.

Step 2: Rewrite in Standard Form

Divide every term in the equation \(9x^2 - 4y^2 = 1\) by 1 to express it as \(\frac{9x^2}{1} - \frac{4y^2}{1} = 1\). Simplify this to \(\frac{x^2}{\frac{1}{9}} - \frac{y^2}{\frac{1}{4}} = 1\). Now, it matches the standard form of a hyperbola centered at the origin.

Step 3: Identify Hyperbola Characteristics

In the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we identify \(a^2 = \frac{1}{9}\) (so \(a = \frac{1}{3}\)) and \(b^2 = \frac{1}{4}\) (so \(b = \frac{1}{2}\)). The transverse axis is along the x-axis because \(x\) comes first in the equation.

Step 4: Determine the Center

The center of the hyperbola, based on its standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center, is clearly at the origin \((0, 0)\) because the terms do not include linear x or y shifts.

Step 5: Sketch the Hyperbola

Draw the center at the origin \((0,0)\). Since \(a = \frac{1}{3}\), plot points to the right and left of the origin along the x-axis at \(\pm \frac{1}{3}\). Similarly, using \(b = \frac{1}{2}\), plot points above and below the origin along the y-axis at \(\pm \frac{i}{2}\). Draw the rectangle from these points and sketch the asymptotes through its diagonals.

Step 6: Asymptotes Equations

The equations for the asymptotes of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are \(y = \pm \frac{b}{a}x\). Substituting \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\) gives asymptotes: \(y = \pm \frac{3}{2}x\). Sketch these lines through the origin and symmetric about the axes.

Step 7: Domain and Range

The domain of the hyperbola is all real numbers, \((-\infty, +\infty)\), because there are no restrictions on \(x\). The range is all real numbers except those between \(-\frac{1}{2}\) and \(\frac{1}{2}\), written as \((-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty)\).

Key Concepts

Standard Form of a Hyperbola

The standard form of a hyperbola reveals essential information about its orientation and dimensions. For the given equation \(9x^2 - 4y^2 = 1\), we need to rewrite it to match the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). This helps us identify that it is a hyperbola opening left and right, centered at the origin. To convert the equation, divide every term by 1, simplifying to \(\frac{x^2}{\frac{1}{9}} - \frac{y^2}{\frac{1}{4}} = 1\). Now, the equation is in standard form, with \(a^2 = \frac{1}{9}\) and \(b^2 = \frac{1}{4}\), meaning \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\). The transverse axis is along the x-axis, showing that the hyperbola opens horizontally.

Asymptotes

Asymptotes are straight lines that the hyperbola approaches but never touches. In a hyperbola, they provide a guideline for sketching the curve, helping predict its behavior at infinity. For the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes are given by the equations \(y = \pm \frac{b}{a}x\). Inserting the values \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\), we find these equations become \(y = \pm \frac{3}{2}x\). Draw these lines through the origin to guide your hyperbola sketch. They're crucial to capturing the infinite nature of hyperbolas.

Domain and Range

Understanding the domain and range of a hyperbola allows you to recognize the set of possible inputs and outputs. The domain of this hyperbola is all real numbers, \((-\infty, \infty)\), because the expression under the square root is always valid regardless of \(x\). However, the range is limited. For this hyperbola, the \(y\) values cannot fall between \(-\frac{1}{2}\) and \(\frac{1}{2}\), thus the range is \((-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty)\). This means the hyperbola’s tails extend infinitely along the y-axis, avoiding the gap between these intervals.

Graphing by Hand

To graph a hyperbola by hand, start by identifying the center. For \(9x^2 - 4y^2 = 1\), it's at the origin \((0,0)\). Next, use \(a = \frac{1}{3}\) and \(b = \frac{1}{2}\) to mark points \(\pm \frac{1}{3}\) on the x-axis and \(\pm \frac{1}{2}\) on the y-axis, creating a rectangle. Draw the diagonals to represent the asymptotes' paths, \(y = \pm \frac{3}{2}x\). The arms of the hyperbola will approach these asymptotes without ever crossing. Finally, sketch the hyperbola's branches opening horizontally from the center, hugging the asymptotes, emphasizing the curve's open nature.