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Chapter 10: Problem 10

Find the vertices, the focl, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$y^{2}-16 x^{2}=1$$

Short Answer

Expert verified
Vertices: (0, 1), (0, -1). Foci: \((0, \frac{\sqrt{17}}{4}), (0, -\frac{\sqrt{17}}{4})\). Asymptotes: \(y = 4x\), \(y = -4x\).

Step by step solution

01

Determine the Standard Form

The given equation of the hyperbola is \( y^{2} - 16x^{2} = 1 \). This is already in the form \( \frac{y^2}{1} - \frac{x^2}{\frac{1}{16}} = 1 \). Here, \( a^2 = 1 \) and \( b^2 = \frac{1}{16} \). Therefore, \( a = 1 \) and \( b = \frac{1}{4} \).
02

Identify the Vertices

For a hyperbola in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices are located at \((0, a)\) and \((0, -a)\). Since \( a = 1 \), the vertices are \((0, 1)\) and \((0, -1)\).
03

Find the Foci

The foci of the hyperbola are calculated using the formula \( c = \sqrt{a^2 + b^2} \). Here, \( c = \sqrt{1 + \frac{1}{16}} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4} \). The foci are \((0, \frac{\sqrt{17}}{4})\) and \((0, -\frac{\sqrt{17}}{4})\).
04

Determine the Asymptotes

The equations of the asymptotes for the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) are \( y = \pm \frac{a}{b}x \). Substituting \( a = 1 \) and \( b = \frac{1}{4} \), the asymptotes are \( y = \pm 4x \).
05

Sketch the Hyperbola

Draw the transverse axis vertically through the vertices \((0, 1)\) and \((0, -1)\). Draw the asymptotes, \( y = 4x \) and \( y = -4x \), intersecting at the center \((0, 0)\). The hyperbola opens along the \( y \)-axis, approaching the asymptotes at a distance and passing through the vertices. Mark the foci at \((0, \frac{\sqrt{17}}{4})\) and \((0, -\frac{\sqrt{17}}{4})\).

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Key Concepts

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

Vertices of Hyperbola
One of the key features of a hyperbola are its vertices. To understand the vertices of a hyperbola, let's start with the standard form of a hyperbola equation. In the standard form, \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] the vertices are determined by the values of \( a \) and \( b \). For this type of vertical hyperbola, which opens upwards and downwards, the vertices are located on the \( y \)-axis, which is the transverse axis.
  • The coordinates of the vertices are \((0, a)\) and \((0, -a)\).
In our exercise, \( a = 1 \), so the vertices are \((0, 1)\) and \((0, -1)\). These points are significant because they are the closest points of the hyperbola to its center, serving as key reference points in the graph.
Foci of Hyperbola
The foci (or focal points) of a hyperbola play a crucial role in defining its unique shape. Located along the transverse axis, the foci of a hyperbola are determined using the relationship: \[ c = \sqrt{a^2 + b^2} \] where \( c \) represents the focal distance from the center. For instance, if the transverse axis is vertical as in our exercise equation, the foci are at \((0, c)\) and \((0, -c)\).
  • In our example: \( c = \sqrt{1 + \frac{1}{16}} = \frac{\sqrt{17}}{4} \).
  • Thus, the foci are at \((0, \frac{\sqrt{17}}{4})\) and \((0, -\frac{\sqrt{17}}{4})\).
Foci make hyperbolas unique because they are used to plot the loci of the hyperbola, enhancing the understanding of its drastically open ends.
Equations of Asymptotes
Asymptotes are imaginary lines that hyperbolas approach but never actually reach. They are essential for sketching a hyperbola as they guide its shape and direction. For a hyperbola given in the form: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] The equations of the asymptotes are: \[ y = \pm \frac{a}{b}x \] For our particular hyperbola, after substituting \( a = 1 \) and \( b = \frac{1}{4} \), the asymptotes become:
  • \( y = 4x \)
  • \( y = -4x \)
These lines intersect at the center of the hyperbola, directing its path and forming a cross that helps define the boundary of the hyperbola's arms.
Standard Form of Hyperbola
The standard form of a hyperbola is crucial in identifying and graphing these curves. This form helps us easily determine various components such as the vertices, foci, and asymptotes. The general equation of a hyperbola is:
  • \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
  • Or, \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
The distinction between these two forms lies in the orientation of the hyperbola.
  • If the \( y^2 \) term is positive, the transverse axis is vertical.
  • If the \( x^2 \) term is positive, the transverse axis is horizontal.
In our example of \( y^2 - 16x^2 = 1 \), it follows the first form, meaning it opens upwards and downwards. This ensures we see a hyperbola extending along the \( y \)-axis.
Graphing Hyperbolas
Graphing hyperbolas involves integrating all the components: vertices, foci, and asymptotes. Steps to graph a hyperbola effectively include:
  • Identify the standard form of the hyperbola equation to determine its type and direction.
  • Plot the center of the hyperbola, which is at the origin for this example.
  • Locate and plot the vertices, which are \((0, 1)\) and \((0, -1)\), marking the ends of the transverse axis.
  • Mark the foci points, \((0, \frac{\sqrt{17}}{4})\) and \((0, -\frac{\sqrt{17}}{4})\), as additional reference points.
  • Draw the asymptotes \( y = 4x \) and \( y = -4x \), extending through the origin and providing a guide for the hyperbola's arms.
Finally, sketch the curve of the hyperbola, ensuring it approaches but never crosses the asymptotes and passes through the vertices. This comprehensive approach makes graphing hyperbolas systematic and intuitive.

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