/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Exercises \(25-28\) give the ecc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 25

Exercises \(25-28\) give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation in Cartesian coordinates. $$ \begin{array}{l}{\text { Eccentricity: } 3} \\ {\text { Vertices: }(0, \pm 1)}\end{array} $$

Short Answer

Expert verified
The equation of the hyperbola is \(y^2 - \frac{x^2}{8} = 1\).

Step by step solution

01

Understand the Structure of a Hyperbola

A hyperbola centered at the origin can be defined as \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), depending on its orientation. Since the vertices are at \((0, \pm 1)\), the hyperbola is vertical, using the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
02

Use the Given Vertices

The distance from the center to the vertices is \(a\), which means \(a = 1\) since the vertices are \((0, \pm 1)\). Thus, \(a^2 = 1^2 = 1\).
03

Use the Eccentricity

Eccentricity \(e\) for a hyperbola is defined as \(e = \frac{c}{a}\). Given \(e = 3\) and \(a = 1\), solve for \(c\) by substituting: \(3 = \frac{c}{1}\), hence \(c = 3\).
04

Find \(b^2\) Using the Relationship Between "a", "b", and "c"

For a hyperbola, \(c^2 = a^2 + b^2\). Substitute \(c = 3\) and \(a = 1\): \(9 = 1 + b^2\). Solve for \(b^2\): \(b^2 = 8\).
05

Write the Hyperbola's Equation

Now that we have \(a^2 = 1\) and \(b^2 = 8\), substitute into \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) to get: \[\frac{y^2}{1} - \frac{x^2}{8} = 1.\] Thus, the equation of the hyperbola is \[y^2 - \frac{x^2}{8} = 1.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Eccentricity
Eccentricity is a measure that describes how "stretched out" the conic section is. For a hyperbola, the eccentricity (denoted as \(e\)) is always greater than 1. This value tells us the shape of the hyperbola, specifically how it diverges from being a perfect circle.

Think of the formula for eccentricity like this:
  • \(e = \frac{c}{a}\)
Here, \(c\) is the distance from the center to each focus of the hyperbola, and \(a\) is the distance from the center to each vertex.

In the context of the given problem:
  • The eccentricity is given as 3, which immediately conveys that this hyperbola is more elongated compared to a unit circle.
  • Using the equation \(3 = \frac{c}{1}\), you quickly determine that \(c = 3\).
This calculation aids in identifying other parameters of the hyperbola.
Vertices
Vertices are key points on the hyperbola that define its main axis. For a hyperbola, the vertices are the points where the hyperbola intersects its transverse axis, which is the axis through which it opens.

In this exercise:
  • The vertices are \((0, \pm 1)\), indicating that the hyperbola opens vertically.
  • This implies that along the y-axis, the most distant points from the center (0) where the hyperbola intersects are at y = 1 and y = -1.
The distance from the center to the vertices is \(a\). For this exercise, \(a = 1\). This is derived from the distance calculation \(a = 1\), which confirms the vertical alignment of the hyperbola.
Cartesian Coordinates
Cartesian coordinates provide a way to express the position of a point using two values, generally referred to as \((x, y)\). They offer a method to describe every point in a plane using these two numbers.

For hyperbolas, Cartesian coordinates describe key characteristics like vertices and foci. These points are crucial for crafting the equation of a hyperbola:
  • The origin or center of the hyperbola is \((0, 0)\).
  • The vertices given as \((0, \pm 1)\) relate to the y-coordinates, indicating a vertical direction.
The Cartesian coordinate framework is essential in developing a coherent understanding of the hyperbola's structure, as it allows algebraic representation of geometric shapes.
Standard Form
The standard form of a hyperbola's equation is a formula that provides a neat and structured way to express the hyperbola mathematically. There are two basic standard forms based on the hyperbola's orientation: horizontal or vertical.
  • For a vertical hyperbola, the standard form is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
  • For a horizontal hyperbola, the form switches to \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
In our exercise case, because the vertices were found at \((0, \pm 1)\), the hyperbola is vertical, and thus we use the first form. With values \(a^2 = 1\) and calculated \(b^2 = 8\), the equation becomes:
  • \(y^2 - \frac{x^2}{8} = 1\)
This standard form equation is an invaluable tool for plotting a hyperbola and understanding its geometry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Length is independent of parametrization To illustrate the fact that the numbers we get for length do not depend on the way we parametrize our curves (except for the mild restrictions preventing doubling back mentioned earlier), calculate the length of the semicircle \(y=\sqrt{1-x^{2}}\) with these two different parametrizations: $$\begin{array}{ll}{\text { a. } \quad x=\cos 2 t,} & {y=\sin 2 t, \quad 0 \leq t \leq \pi / 2} \\ {\text { b. } x=\sin \pi t,} & {y=\cos \pi t, \quad-1 / 2 \leq t \leq 1 / 2}\end{array}$$

The area of the region that lies inside the cardioid curve \(r=\cos \theta+1\) and outside the circle \(r=\cos \theta\) is not $$\frac{1}{2} \int_{0}^{2 \pi}\left[(\cos \theta+1)^{2}-\cos ^{2} \theta\right] d \theta=\pi$$ Why not? What is the area? Give reasons for your answers.

Find the areas of the regions in Exercises \(9-18\) Inside the circle \(r=4 \sin \theta\) and below the horizontal line \(r=3 \csc \theta\)

In Exercises \(51-54,\) use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(11.16 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ x=\frac{1}{3} t^{3}, \quad y=\frac{1}{2} t^{2}, \quad 0 \leq t \leq 1 $$

Graph the lines and conic sections in Exercises \(65-74\) $$ r=1 /(1-\sin \theta) $$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.