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Chapter 11: Problem 29

Give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. \(y^{2}-x^{2}=8\)

Short Answer

Expert verified
The equations are in the form \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \), asymptotes are \( y = \pm x \), and the foci are (0,4) and (0,-4).

Step by step solution

01

Identify the Standard Form

The given equation is \( y^2 - x^2 = 8 \). A hyperbola's standard form can be either \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) or \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, the equation can be converted to the standard form of \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) by dividing every term by 8.
02

Convert to Standard Form

Divide the entire equation \( y^2 - x^2 = 8 \) by 8 to achieve the form \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \). This matches the standard form for a vertical hyperbola, where \( a^2 = 8 \) and \( b^2 = 8 \).
03

Determine the Asymptotes

For a vertical hyperbola in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the equations of the asymptotes are \( y = \pm \frac{a}{b}x \). Here, \( a = \sqrt{8} = 2\sqrt{2} \) and \( b = \sqrt{8} = 2\sqrt{2} \), so the asymptotes are \( y = \pm x \).
04

Identify the Foci

The distance of the foci from the center (0,0) is \( c = \sqrt{a^2 + b^2} \). Since \( a^2 = 8 \) and \( b^2 = 8 \), \( c = \sqrt{16} = 4 \). Thus, the foci are located at \( (0,4) \) and \( (0,-4) \).
05

Sketch the Hyperbola

Plot the center at the origin (0,0). Draw the asymptotes \( y = x \) and \( y = -x \). Since \( a = 2\sqrt{2} \), mark points at (0, \( \pm 2\sqrt{2} \)). The hyperbola will open vertically along the y-axis, approaching but never touching the asymptotes. Mark the foci at (0,4) and (0,-4) as reference points.

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Key Concepts

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

Standard Form of Hyperbola
The standard form of a hyperbola can seem tricky at first, but it's key to understanding its properties. For any hyperbola, the equation takes one of two forms:
  • Vertical hyperbola: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
  • Horizontal hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
Whether vertical or horizontal, it helps to remember that the terms are set up as a difference between two squared terms equaling 1. This equation reflects the fundamental nature of the hyperbola's shape and orientation. In this exercise, transforming the original equation \( y^2 - x^2 = 8 \) into its standard form required dividing by 8, resulting in \( \frac{y^2}{8} - \frac{x^2}{8} = 1 \), indicating a vertical hyperbola.
Hyperbola Asymptotes
Asymptotes of a hyperbola provide a guide to its steepness and direction as it stretches towards infinity. For a vertical hyperbola in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the asymptotes are represented by the equations \( y = \pm \frac{a}{b} x \). This provides the lines towards which the hyperbola approaches but never intersects.In our derived standard equation, both \( a \) and \( b \) equate to \( 2\sqrt{2} \). Thus, the asymptote equations simplify to \( y = \pm x \), perfectly bisecting the quadrants and guiding the hyperbola's opening.
Hyperbola Foci
The foci of a hyperbola reveal much about its focal properties. For this vertical hyperbola, the foci are positioned a distance \( c \) from the center, calculated by the formula \( c = \sqrt{a^2 + b^2} \). This distance reflects how wide or narrow the hyperbola appears relative to its center.Here, with \( a^2 = 8 \) and \( b^2 = 8 \), the value of \( c \) becomes \( \sqrt{16} = 4 \). Therefore, the foci find themselves at coordinates \( (0, 4) \) and \( (0, -4) \), symmetrically placed along the y-axis.
Vertical Hyperbola
A vertical hyperbola is identifiable by its focus and primary stretch, aligned along the y-axis. In this setup, the vertices, foci, and center all fall along a vertical line, depicting how the hyperbola opens up and down. The standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) confirms the vertical orientation.The exercise's equation, \( y^2 - x^2 = 8 \), reinforces this vertical nature. As the equation reforms into the standard form, the vertical nature is spotlighted with vertices and cylindrical cross-sectional curves aligned above and below the center, marked at \( (0, \pm 2\sqrt{2}) \).
Sketching Hyperbolas
Sketching a hyperbola combines several pieces of information: the center, vertices, asymptotes, and foci. The center, in our example, is the origin (0,0). Using the vertices, which in this case are at \( (0, \pm 2\sqrt{2}) \), one can start to shape the hyperbola.Asymptotes, given by the lines \( y = \pm x \), act as guidelines, and should be lightly drawn as dashed lines. These lines form a cross through the origin, helping visualize where the hyperbola will approach. Finally, mark the foci at \( (0,4) \) and \( (0,-4) \) to complete the picture, sketching the hyperbola so it hugs these components while inevitably curving outward, following the shape of the asymptotes.

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Most popular questions from this chapter

The parabola \(x^{2}=-4 y\) is shifted left 1 unit and up 3 units to generate the parabola \((x+1)^{2}=-4(y-3).\) a. Find the new parabola's vertex, focus, and directrix. b. Plot the new vertex, focus, and directrix, and sketch in the parabola.

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