Chapter 10: Problem 23
Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ x^{2}-y^{2}=4 $$
Short Answer
Expert verified
Vertices: \((-2,0)\), \((2,0)\); Foci: \((-2\sqrt{2},0)\), \((2\sqrt{2},0)\); Asymptotes: \(y=x\), \(y=-x\).
Step by step solution
01
Recognize the Form of the Equation
The equation given is already in the form for a standard hyperbola: \( x^2 - y^2 = 4 \). This can be rearranged to \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \), showing that it is a hyperbola centered at the origin \((0,0)\) with a horizontal transverse axis.
02
Identify the Values of a and b
From the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we identify that \( a^2 = 4 \) and \( b^2 = 4 \). Therefore, \( a = 2 \) and \( b = 2 \).
03
Calculate the Distance to the Foci
For a hyperbola in this form, the foci are located at \(( \pm c, 0)\), where \( c = \sqrt{a^2 + b^2} \). Substituting \( a^2 = 4 \) and \( b^2 = 4 \), we have \( c = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \). So the foci are \(( \pm 2\sqrt{2}, 0)\).
04
Determine the Vertices
The vertices of the hyperbola are at \(( \pm a, 0) \). Since \( a = 2 \), the vertices are at \(( -2, 0) \) and \( (2, 0) \).
05
Write the Equations of the Asymptotes
The slopes of the asymptotes for this hyperbola are \( \pm \frac{b}{a} \). Since \( a = 2 \) and \( b = 2 \), the slopes are \( \pm 1 \). Therefore, the equations of the asymptotes are \( y = x \) and \( y = -x \).
06
Graph the Hyperbola
To sketch the hyperbola, plot the vertices \((-2, 0)\) and \((2, 0)\), and draw the asymptotes \(y=x\) and \(y=-x\), which intersect at the center (0,0). The hyperbola will approach but never intersect the asymptotes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Asymptotes
Asymptotes are crucial lines that provide guidance on the behavior of a hyperbola. They act as imaginary boundaries that the hyperbola gets infinitely close to but never actually touches or crosses. For the hyperbola given by the equation, \[ x^2 - y^2 = 4 \] after writing it in standard form \[ \frac{x^2}{4} - \frac{y^2}{4} = 1, \] you can deduce the equations for the asymptotes. Since the asymptotes of a hyperbola of this type have slopes determined by the formula \( \pm \frac{b}{a} \), where both \( a \) and \( b \) are equal to 2, the slopes are \( \pm 1 \). Therefore, the asymptote equations are
- \( y = x \)
- \( y = -x \)
Vertices
Vertices are key points that define the stretch and direction of a hyperbola. For a hyperbola centered on the origin, like \( x^2 - y^2 = 4 \), the vertices are located directly along the transverse axis. In standard hyperbolic equations of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices will be at \( (\pm a, 0) \) for horizontally oriented hyperbolas. Here, since \( a = 2 \), the vertices are at
- \((-2, 0)\)
- \((2, 0)\)
Foci
Foci, or focal points, are special points inside each branch of the hyperbola. They are critical in defining the shape of a hyperbola. Unlike ellipses, hyperbolas have foci located outside the vertices. For our hyperbola \[ x^2 - y^2 = 4, \] the foci are calculated using the formula \( c = \sqrt{a^2 + b^2} \), where both \( a^2 \) and \( b^2 \) equal 4. Thus, \( c = \sqrt{8} = 2\sqrt{2} \). Therefore, the coordinates of the foci are:
- \( (2\sqrt{2}, 0) \)
- \( (-2\sqrt{2}, 0) \)
Equations
The equation of a hyperbola is the foundation for understanding its geometry. Starting with \[ x^2 - y^2 = 4, \] it was rearranged to fit the standard hyperbola form \[ \frac{x^2}{4} - \frac{y^2}{4} = 1. \] This rearrangement confirms the hyperbola's center is at (0,0) and that it has a horizontal transverse axis. This standard form makes it possible to identify elements such as
- \( a^2 = 4 \), implying \( a = 2 \)
- \( b^2 = 4 \), implying \( b = 2 \)