Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 53
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. $$9 x^{2}-4 y^{2}=1$$
Short Answer
Expert verified
Center: (0,0); Domain: \((-\infty, +\infty)\); Range: \((-\infty, -\frac{1}{2})\) or \((\frac{1}{2}, \infty)\).
Step by step solution
01
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.
02
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.
03
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.
04
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.
05
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.
06
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.
07
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 \((-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty)\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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鈥檚 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.
