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Magazine Chapter 10: Problem 4
Find the vertices, the focl, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$\frac{x^{2}}{49}-\frac{y^{2}}{16}=1$$
Short Answer
Expert verified
Vertices: (7,0), (-7,0); Foci: (卤\(\sqrt{65}\),0); Asymptotes: \(y=\pm\frac{4}{7}x\).
Step by step solution
01
Identify the Form of the Hyperbola
The given equation is \( \frac{x^2}{49} - \frac{y^2}{16} = 1 \), which is in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). This is the equation of a hyperbola centered at the origin with a horizontal transverse axis.
02
Find the Values of a, b, and c
From the equation, \( a^2 = 49 \) and \( b^2 = 16 \). Therefore, \( a = 7 \) and \( b = 4 \). The value of \( c \) can be found using \( c^2 = a^2 + b^2 \), which gives \( c^2 = 49 + 16 = 65 \), so \( c = \sqrt{65} \).
03
Determine the Vertices
For a hyperbola with the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are located at \( (\pm a, 0) \). Thus, the vertices are \( (7,0) \) and \( (-7,0) \).
04
Calculate the Foci
The foci for a hyperbola with a horizontal transverse axis are at \( (\pm c, 0) \). Since \( c = \sqrt{65} \), the foci are \( (\pm \sqrt{65}, 0) \).
05
Write the Equations of the Asymptotes
For the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations of the asymptotes are \( y = \pm \frac{b}{a}x \). Substituting the given values, the asymptotes are \( y = \pm \frac{4}{7}x \).
06
Sketch the Hyperbola
Draw the hyperbola, showing the vertices at \((7, 0)\) and \((-7, 0)\), and the foci at \((\sqrt{65}, 0)\) and \((-\sqrt{65}, 0)\). The asymptotes \( y = \frac{4}{7}x \) and \( y = -\frac{4}{7}x \) should be shown as dashed lines intersecting at the center \((0,0)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertices of a Hyperbola
In the study of hyperbolas, vertices are crucial as they represent the points where the hyperbola crosses the transverse axis. For the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), which describes a hyperbola centered at the origin with a horizontal transverse axis, the vertices are determined by the value of \( a \).
- The formula for locating the vertices is \( (\pm a, 0) \).
- From the equation \( \frac{x^2}{49} - \frac{y^2}{16} = 1 \), we found that \( a^2 = 49 \), leading to \( a = 7 \).
- Therefore, the vertices are \( (7, 0) \) and \( (-7, 0) \).
Foci of a Hyperbola
The foci are vital to understanding the shape and orientation of a hyperbola. They lie along the transverse axis outside of the vertices.
- To find the foci, you use the formula \( c^2 = a^2 + b^2 \), where \( c \) represents the distance from the center to each focus.
- Given \( a^2 = 49 \) and \( b^2 = 16 \), we calculated \( c^2 = 65 \), and hence \( c = \sqrt{65} \).
- The foci of the hyperbola are then at \( (\pm c, 0) \), which translates to \( (\sqrt{65}, 0) \) and \( (-\sqrt{65}, 0) \).
Equations of Asymptotes
Asymptotes give direction and boundary guidance to the hyperbola as it stretches to infinity. They are lines that the branches of a hyperbola approach but never actually touch.
- The equations for the asymptotes of a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are \( y = \pm \frac{b}{a}x \).
- For our specific equation, \( a = 7 \) and \( b = 4 \), so the equations become \( y = \pm \frac{4}{7}x \).
- These equations describe two intersecting lines through the origin (0,0) where the hyperbola appears to stretch out infinitely closer to as \( x \) gets larger in absolute value.
Hyperbola Graphing
Graphing a hyperbola involves bringing together our understanding of its vertices, foci, and asymptotes. This ensures the shape is accurate and reflects the equation.
- Start by marking the vertices at \( (7, 0) \) and \( (-7, 0) \) on the coordinate plane, since this hyperbola has a horizontal transverse axis.
- Next, locate the foci at \( (\sqrt{65}, 0) \) and \( (-\sqrt{65}, 0) \), slightly further from the center than the vertices.
- Draw the asymptotes as dashed lines corresponding to \( y = \frac{4}{7}x \) and \( y = -\frac{4}{7}x \). These guide the hyperbola's path as it curves away from the center.
