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Chapter 12: Problem 26

Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. $$ \text { foci } F(0,\pm 10), \quad \text { asymptotes } y=\pm \frac{1}{3} x $$

Short Answer

Expert verified
The equation of the hyperbola is \( \frac{y^2}{90} - \frac{x^2}{10} = 1 \).

Step by step solution

01

Understand the Structure of a Hyperbola

A hyperbola with a center at the origin is generally expressed in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). Since the foci are along the y-axis, the equation will be in the form \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).
02

Identify the Distance to the Foci

The foci of this hyperbola are given as \( (0, \pm 10) \), which means the distance from the center to a focus is \( c = 10 \). This will be used to find \( b \).
03

Use the Asymptote Slope to Find \(a\) and \(b\)

The asymptotes of a hyperbola in the form \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \) have slopes \( \pm \frac{a}{b} \). Given \( y = \pm \frac{1}{3}x \), we have \( \frac{a}{b} = \frac{1}{3} \).
04

Use the Relationship between \(a\), \(b\), and \(c\)

For hyperbolas of the form \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the relationship \( c^2 = a^2 + b^2 \) holds. We know \( c = 10 \), and \( \frac{a}{b} = \frac{1}{3} \), thus \( a = \frac{b}{3} \).
05

Setup Equations to Solve for \(b\)

From \( \left( \frac{b}{3} \right)^2 + b^2 = 10^2 \), simplify to \( \frac{b^2}{9} + b^2 = 100 \). Combine like terms: \( \frac{10b^2}{9} = 100 \).
06

Solve for \(b\)

Solve \( \frac{10b^2}{9} = 100 \) by multiplying both sides by 9: \( 10b^2 = 900 \). Divide by 10 to get \( b^2 = 90 \). Therefore, \( b = \sqrt{90} \).
07

Calculate \(a\) from \(b\)

Using \( a = \frac{b}{3} \) and knowing \( b = \sqrt{90} \), calculate \( a^2 = \left( \frac{\sqrt{90}}{3} \right)^2 = \frac{90}{9} = 10 \), so \( a = \sqrt{10} \).
08

Write the Equation of the Hyperbola

The equation of the hyperbola is \( \frac{y^2}{90} - \frac{x^2}{10} = 1 \), using the values found for \( a^2 = 10 \) and \( b^2 = 90 \).

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Key Concepts

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

Foci of Hyperbola
The foci of a hyperbola are crucial points that help define its shape and orientation. For a hyperbola centered at the origin with its foci on the y-axis, the coordinates are given as
  • \( F(0, \, \pm c) \)
For the hyperbola in our exercise, the foci are located at
  • \( (0, \, \pm 10) \)
This means the distance from the center of the hyperbola to each focus is 10 units. This distance \( c \) plays a vital role in determining the shape of the hyperbola and is used in the formula
  • \(c^2 = a^2 + b^2\)
in order to relate it to the parameters \( a \) and \( b \). Understanding the foci helps in visualizing where the hyperbola stretches towards. The foci lie outside the hyperbola on its transverse axis.
Asymptotes of Hyperbola
Asymptotes are imaginary lines that the hyperbola approaches but never actually touches or intersects. They give a skeleton or guide for the hyperbola鈥檚 open ends. For hyperbolas in the form
  • \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
the slopes of the asymptotes are given by
  • \( \pm \frac{a}{b} \)
In this particular hyperbola, the asymptotes are
  • \( y = \pm \frac{1}{3}x \)
This indicates that the ratio \( \frac{a}{b} = \frac{1}{3} \). Asymptotes are essential because they anchor the hyperbola鈥檚 direction in the coordinate plane, guiding how it opens up along its axes.
Center of Hyperbola
The center of a hyperbola is the central point from which the various parts of the hyperbola are equidistant. For hyperbolas centered at the origin, the equation takes a simplified form. In this case, the center coordinates are
  • \((0, 0)\)
This central point allows for easy measurement of distances to the foci and vertices, as well as determining the hyperbola鈥檚 orientation. From the center at the origin, this hyperbola spreads its branches outward along the axes dictated by the foci. The center is vital in equation formulation, as it serves as the starting point for finding distances to other key characteristics.
Hyperbola Properties
Hyperbolas are fascinating geometric structures characterized by two symmetric 鈥榖ranches鈥�. They have several defining properties:
  • Shape and Equation: Hyperbolas have two parts that open outward, depicted in equations like
    • \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
  • Axes: A hyperbola鈥檚 transverse axis extends between the two branches.
  • Vertices and Foci: The hyperbola鈥檚 branches contain points called vertices, and its foci are located along the transverse axis.
  • Equation Elements: It derives its properties from components such as \( a, b, \) and \( c \), which allow calculation of distances between various pivotal elements like foci and vertices.
Hyperbolas are distinguished by their ability to illustrate important mathematical and real-world phenomena through these properties.

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Most popular questions from this chapter

One section of a suspension bridge has its weight uniformly distributed between twin towers that are 400 feet apart and that rise 90 feet above the horizontal roadway. A cable strung between the tops of the towers has the shape of a parabola, with center point 10 feet above the roadway. Suppose coordinate axes are introduced. as shown in the figure. (a) Find an equation for the parabola. (b) Set up an integral whose value is the length of the cable. (c) If nine equispaced vertical cables are used to support the parabolic cable, find the total length of these supports.

Exer. \(17-26:\) Find an equation for the conic that satisfies the given conditions. hyperbola, with foci \(F(\pm 10,0)\) and vertices \(V(\pm 5,0)\)

Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. \(x\) -intercepts \(\pm 2, \quad y\) -intercepts \(\pm \frac{1}{3}\)

Exer \(1-16\) : Find the vertices and foci of the conic, and sketch its graph. $$ y^{2}-2 x^{2}+6 y+8 x-3=0 $$

Graph, on the same coordinate axes, the given hyperbolas. (a) Estimate their first-quadrant point of intersection. (b) Set up an integral that can be used to approximate the area of the region in the first quadrant bounded by the hyperbolas and a coordinate axis. $$ \begin{array}{l} \frac{(y-0.1)^{2}}{1.6}-\frac{(x+0.2)^{2}}{0.5}=1 ; \\ \frac{(y-0.5)^{2}}{2.7}-\frac{(x-0.1)^{2}}{5.3}=1 \end{array} $$
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