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Chapter 7: Problem 68

Find an equation for each hyperbola. Vertices \((-10,0)\) and \((10,0)\); asymptotes \(y=\pm 5 x\)

Short Answer

Expert verified
The equation is \(\frac{x^2}{100} - \frac{y^2}{2500} = 1\).

Step by step solution

01

Identify Center and Axes

The vertices of the hyperbola are \((-10, 0)\) and \(10, 0)\). This indicates the hyperbola is centered at the origin \(0, 0)\) and opens horizontally, as the x-coordinates vary.
02

Determine the Distance to Vertices

The distance from the center to a vertex is the value \(a\). Since the vertices are at \((-10, 0)\) and \(10, 0)\), this distance is \(|10 - 0| = 10\). Thus, \(a = 10\).
03

Find Asymptote Slope

The given asymptotes are \(y = \pm 5x\). The slope \((\pm m)\) of the asymptotes provides information to find \(b\).
04

Calculate \(b\) Using Asymptote Slope

For a hyperbola opening horizontally, the relationship between \(a\), \(b\), and the slope \(\pm m\) of the asymptotes is \(\frac{b}{a} = m\). Thus, \( \frac{b}{10} = 5 \,\) giving \(b = 50\).
05

Write the Hyperbola Equation

A hyperbola centered at \(0, 0\) that opens horizontally has the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.\) Using \(a = 10\) and \(b = 50\), the equation \becomes \(\frac{x^2}{100} - \frac{y^2}{2500} = 1\).

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Key Concepts

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

Vertices of Hyperbola
Vertices are key features of a hyperbola, representing the points where the hyperbola intersects its transverse axis. In our exercise, the vertices are located at
  • (-10, 0)
  • (10, 0)
This indicates a horizontal hyperbola centered at the origin (0, 0).

The distance between the center and a vertex along the transverse axis is denoted by \(a\). For this example:
  • The distance from the origin to each vertex is \(|10 - 0| = 10\), so we have \(a = 10\).
Asymptotes
Asymptotes of a hyperbola are straight lines that the hyperbola approaches but never touches. They provide information about the slope and orientation of the hyperbola. For the given problem, the asymptotes are:
  • \(y = \pm 5x\)
This tells us about the direction the branches of the hyperbola open, which is horizontal.

The slope \(m\) of these asymptotes (\( \pm 5 \)) helps us determine the value of \(b\) using the relationship \(\frac{b}{a} = m\). Inserting \(a = 10\), we find \(b = 50\).

Thus, the asymptotes guide us in understanding the "spread" of the hyperbola.
Equation of Hyperbola
The equation of a hyperbola is crucial in defining its structure mathematically. For a horizontal hyperbola centered at the origin, the standard form is:
  • \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
Given \(a = 10\) and \(b = 50\), we substitute these into the equation:

\[ \frac{x^2}{100} - \frac{y^2}{2500} = 1 \]

This equation allows us to describe the precise shape and location of the hyperbola on a coordinate plane.
Axes of Hyperbola
The axes of a hyperbola are lines that help define its shape. The transverse axis is the axis along which the vertices lie, and the conjugate axis is perpendicular to it. In this case:
  • The transverse axis is horizontal, along the x-axis.
  • The conjugate axis is vertical, along the y-axis.
The lengths of these axes are determined by the values of \(a\) and \(b\):
  • Transverse Axis Length = \(2a = 20\)
  • Conjugate Axis Length = \(2b = 100\)
These lengths help us visualize how wide and tall the hyperbola is on the graph, providing a fuller picture of its geometry.

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Most popular questions from this chapter

Particle When an alpha particle (a subatomic particle) is moving in a horizontal path along the positive \(x\) -axis and passes between charged plates, it is deflected in a parabolic path. If the plate is charged with 2000 volts and is 0.4 meter long, then an alpha particle's path can be described by the equation \(y=-\frac{k}{2 v_{0}} x^{2}\) where \(k=5 \times 10^{-9}\) is constant and \(v_{0}\) is the initial velocity of the particle. If \(v_{0}=10^{7}\) meters per second, what is the deflection of the alpha particle's path in the \(y\) -direction when \(x=0.4\) meter? (Source: Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.)

Find an equation for each hyperbola. Vertices \((-3,0)\) and \((3,0) ;\) passing through \((6,1)\)

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$y=-2(x+3)^{2}+2$$

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$y=(x-5)^{2}-4$$

Graph each hyberbola by hand. Give the domain and range. Do not use a calculator. $$x^{2}-y^{2}=1$$
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