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Chapter 13: Problem 16

Graph hyperbola. Label all vertices and sketch all asymptotes. \(x^{2}-y^{2}=4\)

Short Answer

Expert verified
The vertices are at (2, 0) and (-2, 0) and the asymptotes are y = x and y = -x.

Step by step solution

01

Identify the standard form of a hyperbola

The given equation is \(x^{2} - y^{2} = 4\). Rewrite it in the standard form for hyperbolas. Divide the equation by 4: \(\frac{x^2}{4} - \frac{y^2}{4} = 1)\). This is the standard form of the hyperbola, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1)\), with \(a^2 = 4\) and \(b^2 = 4\).
02

Determine the values of a and b

From the standard form \( \frac{x^2}{4} - \frac{y^2}{4} = 1\), we have \(a^2 = 4\) and \(b^2 = 4\). Thus, \(a = 2\) and \(b = 2\).
03

Identify the vertices

In the standard form of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are located at \( (\text{卤}a, 0)\). Given that \(a = 2\), the vertices are at \( (2, 0) \) and \( (-2, 0)\).
04

Determine the equations of the asymptotes

The equations of the asymptotes for the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by \( y = \text{卤}\frac{b}{a}x \). Substitute \( a \) and \( b \) with 2 to get the asymptote equations: \( y = \text{卤}x \).
05

Sketch the hyperbola and asymptotes

Draw the hyperbola opening left and right because \( x^2 \) comes first in the equation. Plot the vertices at (2, 0) and (-2, 0). Next, draw the asymptotes \( y = x \) and \( y = -x \), which are straight lines passing through the origin (0, 0), as guidelines.

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Key Concepts

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

vertices in hyperbola
Vertices in hyperbolas are key points where the hyperbola intersects its transverse axis. For the standard form of a hyperbola, \(\frac{\left( x \right)^{2}}{a^{2}} - \frac{\left( y \right)^{2}}{b^{2}} = 1\), the vertices can be easily identified.
Here, \(a^{2} = 4\), so \(a = 2\). The transverse axis is the horizontal line since \(x^{2}\) comes first.
  • Because \(a = 2\), the vertices are at \(\left( 2, 0 \right)\) and \(\left( -2, 0 \right).\)
These points are crucial because the hyperbola opens left and right along this axis. The vertices help in sketching the core shape of the hyperbola.
asymptotes of hyperbola
Asymptotes are lines that the hyperbola approaches but never touches. For a hyperbola in the standard form \(\frac{\left( x \right)^{2}}{a^{2}} - \frac{\left( y \right)^{2}}{b^{2}} = 1\), the asymptotes have the equations:
\(\text{\pm}\frac{b}{a}x\).
  • Given \(a = 2\) and \(b = 2\), the equations for the asymptotes become \(y = \text{\pm}x.\)
These lines pass through the origin and serve as a guide for sketching the hyperbola.
They help ensure the arms of the hyperbola get wider but remain close to these lines.
standard form of hyperbola
The standard form of a hyperbola is critical for graphing. The given equation is \(x^{2} - y^{2} = 4\). To convert to standard form, divide by 4:
  • \(\frac{\left( x \right)^{2}}{4} - \frac{\left( y \right)^{2}}{4} = 1\).
Here, \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\) with \(a^{2} = 4\) and \(b^{2} = 4\).
Thus, \(a = 2\) and \(b = 2\).
This standard form helps identify vertices and asymptotes:
  • Vertices at \(2\) and \(-2\)
  • Asymptotes \(y = \text{\pm}x\)
This setup provides a clear way to draw and understand the hyperbola.

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Most popular questions from this chapter

Match the equation with the center or vertex of its graph, listed in the column on the right. a) Vertex: \((-2,5)\) b) Vertex: \((5,-2)\) c) Vertex: \((2,-5)\) d) Vertex: \((-5,2)\) e) Center: \((-2,5)\) f) Center: \((2,-5)\) g) Center: \((5,-2)\) h) Center: \((-5,2)\) $$(x-5)^{2}+(y+2)^{2}=9$$

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