Chapter 8: Problem 17
Write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(\frac{(x-2)^{2}}{49}-\frac{(y+7)^{2}}{49}=1\)
Short Answer
Expert verified
Vertices: (9, -7) and (-5, -7). Foci: \((2 \pm 7\sqrt{2}, -7)\). Asymptotes: \(y = x - 9\) and \(y = -x - 5\).
Step by step solution
01
Verify Standard Form
The given equation is \( \frac{(x-2)^{2}}{49}-\frac{(y+7)^{2}}{49}=1 \). The standard form for a hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) is satisfied as it compares directly. Therefore, the equation is already in standard form.
02
Identify the Center
In the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center of the hyperbola is given by \((h, k)\). Here, \( h = 2 \) and \( k = -7 \). Thus, the center is \((2, -7)\).
03
Determine Semi-major and Semi-minor Axes
The values \( a^2 = 49 \) and \( b^2 = 49 \) are given. Thus, \( a = 7 \) and \( b = 7 \). Since \( a = b \), the hyperbola is symmetric about the axes and both directions have equal importance. However, the equation form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) suggests a horizontally oriented hyperbola.
04
Identify the Vertices
For a horizontal hyperbola, vertices are at \((h \pm a, k)\). So substituting the values, the vertices are \((2 + 7, -7) = (9, -7)\) and \((2 - 7, -7 ) = (-5, -7)\).
05
Calculate the Foci
For a horizontal hyperbola, the foci are located at \((h \pm c, k)\) where \( c = \sqrt{a^2 + b^2} \). Hence, \( c = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \). The foci are \((2 + 7\sqrt{2}, -7)\) and \((2 - 7\sqrt{2}, -7)\).
06
Write Equations of Asymptotes
The equations of the asymptotes for a horizontal hyperbola are given by \( y = k \pm \frac{b}{a}(x-h) \). Substitution yields the asymptote equations \( y = -7 \pm \frac{7}{7}(x - 2) \), which simplifies to \( y = -7 \pm (x - 2) \). Therefore, the asymptotes are \( y = x - 9 \) and \( y = -x - 5 \).
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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 helps us identify its key characteristics and position in the coordinate plane. Standard form comes in two typical equations:
This recognition allows us to proceed with identifying other properties like the center, vertices, and asymptotes based on the equation in this specific standard form.
- Horizontal hyperbola: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]
- Vertical hyperbola: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]
This recognition allows us to proceed with identifying other properties like the center, vertices, and asymptotes based on the equation in this specific standard form.
Center of a Hyperbola
The center of a hyperbola is a crucial part of understanding its placement on the graph. In a hyperbola's standard form, the center is denoted by the coordinates \((h, k)\).These are derived directly from the terms in the equation, specifically from the forms \((x-h)^2\) and \((y-k)^2\).
For our example \( \frac{(x-2)^{2}}{49} - \frac{(y+7)^{2}}{49} = 1 \), we identify \(h = 2\) and \(k = -7\).
Therefore, the center of this hyperbola is at \((2, -7)\).This point is the midpoint between the vertices and serves as a reference point for describing other elements like the foci and asymptotes.
For our example \( \frac{(x-2)^{2}}{49} - \frac{(y+7)^{2}}{49} = 1 \), we identify \(h = 2\) and \(k = -7\).
Therefore, the center of this hyperbola is at \((2, -7)\).This point is the midpoint between the vertices and serves as a reference point for describing other elements like the foci and asymptotes.
Vertices and Foci of a Hyperbola
The vertices and foci are essential elements that describe the shape and extent of a hyperbola. For a horizontal hyperbola, like in our exercise, the vertices are located on the transverse axis. They can be calculated using the formula \((h \pm a, k)\).
Given \(a = 7\), the vertices are found at \((2 \pm 7, -7)\), which computes to \((9, -7)\) and \((-5, -7)\).
Foci for a horizontal hyperbola stretch beyond the vertices along the transverse axis. Their locations are determined by \((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\).
Plugging in the values \(c = \sqrt{98} = 7\sqrt{2}\), the foci become \((2 \pm 7\sqrt{2}, -7)\).This places them further out than the vertices, illustrating the hyperbola's "opening" along the x-direction.
Given \(a = 7\), the vertices are found at \((2 \pm 7, -7)\), which computes to \((9, -7)\) and \((-5, -7)\).
Foci for a horizontal hyperbola stretch beyond the vertices along the transverse axis. Their locations are determined by \((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\).
Plugging in the values \(c = \sqrt{98} = 7\sqrt{2}\), the foci become \((2 \pm 7\sqrt{2}, -7)\).This places them further out than the vertices, illustrating the hyperbola's "opening" along the x-direction.
Asymptotes of a Hyperbola
Asymptotes are lines that the hyperbola approaches but never touches. They help delineate the trajectory of the hyperbola's branches. In the case of a horizontal hyperbola, the asymptotes are described by the equations:
These yield the asymptote lines: \(y = x - 9\) and \(y = -x - 5\).Asymptotes form an "X" pattern around the hyperbola's center, guiding its overall shape and ensuring that, as x or y extend to infinity, the hyperbola moves closer to, but never crosses, these lines.
- \(y = k \pm \frac{b}{a}(x-h)\)
These yield the asymptote lines: \(y = x - 9\) and \(y = -x - 5\).Asymptotes form an "X" pattern around the hyperbola's center, guiding its overall shape and ensuring that, as x or y extend to infinity, the hyperbola moves closer to, but never crosses, these lines.