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Chapter 10: Problem 32

Find an equation for the hyperbola that satisfies the given conditions. Vertices \((0, \pm 6),\) asymptotes \(y=\pm \frac{1}{3} x\)

Short Answer

Expert verified
The equation of the hyperbola is \(\frac{y^2}{36} - \frac{x^2}{324} = 1\).

Step by step solution

01

Identify the center and orientation

The vertices of the hyperbola are given as \((0, \pm 6)\). This means the center of the hyperbola is at the origin \((0,0)\), and it is oriented vertically since the y-values of the vertices change.
02

Determine 'a' and 'b'

The distance from the center to the vertices is 6, hence \(a = 6\). The slopes of the asymptotes \(y = \pm \frac{1}{3}x\) imply that, for a vertical hyperbola, \(\frac{a}{b} = \frac{1}{3}\). From \(a = 6\), solve \(\frac{6}{b} = \frac{1}{3}\) for \(b\), getting \(b = 18\).
03

Determine 'c'

Use the relation \(c^2 = a^2 + b^2\) for hyperbolas. Substitute \(a = 6\) and \(b = 18\) to get \(c^2 = 36 + 324 = 360\). Therefore, \(c = \sqrt{360} = 6\sqrt{10}\).
04

Form the equation of the hyperbola

The standard form of a vertical hyperbola is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Substitute \(a = 6\) and \(b = 18\) to get the equation \(\frac{y^2}{36} - \frac{x^2}{324} = 1\).

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Key Concepts

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

Vertices of a Hyperbola
Vertices of a hyperbola are crucial points that define its shape and orientation. They are points where the hyperbola is closest to its center. In a vertical hyperbola, which opens up and down, the vertices lie along the y-axis if the center is at the origin. For the given hyperbola with vertices at \((0, \pm 6)\), it tells us that:
  • The center of the hyperbola is at the origin \((0,0)\).
  • The hyperbola is vertically oriented because the vertices move along the y-axis from the center.
The value of 'a' is the distance from the center to each vertex. In this case, that distance is 6. Therefore, \(a = 6\). Understanding the position and distance of the vertices helps in forming the general equation for the hyperbola.
Asymptotes of a Hyperbola
Asymptotes are the lines that a hyperbola approaches but never touches. They give a hyperbola its characteristic open shape. For a hyperbola centered at the origin, the equations of the asymptotes provide essential insights into its orientation and dimensions.Given the asymptotes of this hyperbola are \(y=\pm \frac{1}{3} x\), it lets us know:
  • The slope of the asymptotes of a hyperbola is \(\frac{a}{b}\), where the hyperbola is vertical.
  • Here, \(\frac{6}{b} = \frac{1}{3}\), solving gives \(b = 18\).
Knowing the asymptotes' slopes helps to determine the relationship and size of the distances \(a\) and \(b\), thus aiding in writing the hyperbola's equation properly.
Standard Form of a Hyperbola Equation
The standard form of a hyperbola's equation is instrumental in describing its geometric properties. For a vertical hyperbola centered at the origin, the equation is given by:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]Substitute the vertical hyperbola's specific values, where \(a = 6\) and \(b = 18\), resulting in:\[\frac{y^2}{36} - \frac{x^2}{324} = 1\]This equation succinctly captures the hyperbola's essential features, including the distances to its vertices and its asymptotes. Understanding each part of the equation helps visualize the hyperbola's curvature and span, making solving complex problems more manageable.

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Most popular questions from this chapter

This exercise deals with confocal parabolas, that is, families of parabolas that have the same focus. (a) Draw graphs of the family of parabolas $$x^{2}=4 p(y+p)$$ for \(p=-2,-\frac{3}{2},-1,-\frac{1}{2}, \frac{1}{2}, 1, \frac{3}{2}, 2\). (b) Show that each parabola in this family has its focus at the origin. (c) Describe the effect on the graph of moving the vertex closer to the origin.

Find parametric equations for the line with the given properties. Find parametric equations for the circle \(x^{2}+y^{2}=a^{2}\).

Sketch the curve given by the parametric equations. $$x=\sin t, \quad y=\sin 2 t$$

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$x^{2}+16=4\left(y^{2}+2 x\right)$$

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$16 x^{2}-9 y^{2}-96 x+288=0$$
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