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Chapter 13: Problem 10

Graph hyperbola. Label all vertices and sketch all asymptotes. \(\frac{x^{2}}{9}-\frac{y^{2}}{9}=1\)

Short Answer

Expert verified
Vertices: (3,0), (-3,0). Asymptotes: y=x and y=-x.

Step by step solution

01

Identifying the Standard Form

Recognize that the given equation \(\frac{x^{2}}{9}-\frac{y^{2}}{9}=1\) is in the standard form of a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\).
02

Determine Values of a and b

Identify \(a^{2}=9\) and \(b^{2}=9\), thus \(a=3\) and \(b=3\).
03

Find Vertices

Use the value of \(a\) to determine the vertices. Since the hyperbola is horizontal, vertices are at \((\text{+ or -}a,0)= (3,0) \) and \((-3,0)\).
04

Sketching the Asymptotes

For horizontal hyperbolas, asymptotes are given by the lines \(y=\text{+ or -}\frac{b}{a}x\). Substitute the values of \(a\) and \(b\): \(y=\frac{3}{3}x\) which simplifies to \(y=x\) and \(-y=x\).
05

Drawing the Graph

Sketch the hyperbola opening left and right from the center at (0,0) with vertices at (3,0) and (-3,0). Draw the asymptotes, which are straight lines passing through the origin with slopes of 1 and -1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbola Equation
To start, let鈥檚 understand the given hyperbola equation \(\frac{x^2}{9}-\frac{y^2}{9}=1\). This equation represents a standard form of a hyperbola. A hyperbola in standard form has one of two structures: horizontal or vertical. For a hyperbola centered at the origin, the equations look like this:
  • Horizontal: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\
  • Vertical: \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\
In the given equation, since the x-term is positive and appears first, it denotes a horizontal hyperbola.
Finding Vertices
The vertices of a hyperbola are key points where the hyperbola intersects its transverse axis. Here鈥檚 how you find them. First, look at the value of \(a^2\) and \(b^2\). In our equation, we identify:
  • \(a^2=9\), \(a=3\)
  • \(b^2=9\), \(b=3\)
For a horizontal hyperbola, the vertices are at \(\pm a, 0\). So, the vertices for this hyperbola are at:
  • (3,0)
  • (-3,0)
These points are where the hyperbola is closest to or furthest from the center, which is at (0,0).
Sketching Asymptotes
Asymptotes are lines that the hyperbola approaches but never touches. They help define the shape and direction in which the hyperbola opens. For a horizontal hyperbola, the asymptotes are given by the equations \y=\frac{b}{a}x\ and \y=-\frac{b}{a}x\. In our case:
  • \(b=3\) and \(a=3\)
Substituting these into the asymptote formulas gives:
  • \(y=\frac{3}{3}x\) simplifies to \y = x\
  • \(y=-\frac{3}{3}x\) simplifies to \y = -x\
These are the equations of the asymptotes, which are straight lines passing through the origin with slopes of 1 and -1.
Standard Form of a Hyperbola
It's essential to recognize the standard form of the hyperbola given in the equation \ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\. This form provides valuable information, such as the orientation (horizontal or vertical), and the values for \a\ and \b\, which indicate the distances to the vertices and co-vertices. For a hyperbola centered at the origin, the standard form looks like:
  • \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for a horizontal hyperbola
  • \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for a vertical hyperbola
In many problems, recognizing this standard form is the first step in sketching the graph, finding vertices, and determining asymptotes.

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Most popular questions from this chapter

A rectangle has an area of 20 in \(^{2}\) and a perimeter of 18 in. Find its dimensions.

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