/*! 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 34 Find an equation for the hyperbo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 34

Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. \(y\) -intercepts \(\pm 2\) asymptotes \(y=\pm \frac{1}{4} x\)

Short Answer

Expert verified
The equation of the hyperbola is \(\frac{y^2}{4} - \frac{x^2}{64} = 1\).

Step by step solution

01

Understand the Standard Form of a Hyperbola

A hyperbola has the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) when the transverse axis is vertical. This form is applicable because the \(y\)-intercepts are given. Here, \(2a\) represents the distance between the \(y\)-intercepts.
02

Determine 'a' using the y-intercepts

The \(y\)-intercepts of the hyperbola are \(\pm 2\). This implies that \(a = 2\), resulting in \(a^2 = 4\). Thus, the equation form is \(\frac{y^2}{4} - \frac{x^2}{b^2} = 1\).
03

Use Asymptote Equation to Find 'b^2'

The equations of the asymptotes are \(y = \pm \frac{1}{4}x\). For a hyperbola with vertical transverse axis, the slopes of asymptotes are \(\pm \frac{a}{b}\). Hence, \(\frac{a}{b} = \frac{1}{4}\). Solve for \(b\): \(b = 4a\). Since \(a = 2\), \(b = 8\), and \(b^2 = 64\).
04

Write the Equation of the Hyperbola

Substitute \(a^2\) and \(b^2\) back into the hyperbola standard form equation. The equation is \(\frac{y^2}{4} - \frac{x^2}{64} = 1\).

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.

Standard Form of a Hyperbola
Hyperbolas are fascinating curves, and their equations can be neatly written in a standard form. When a hyperbola is aligned such that the transverse axis is vertical, its standard form appears as \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). This formula is tailored to suit hyperbolas where the main axis runs up and down, fitting when the \( y \)-intercepts are noted.
Each term in this equation serves a purpose:
  • \( y^2 \) and \( a^2 \): This part focuses on the vertical spread, crucial because the transverse axis is vertical. Here, \( a \) is half the distance between the \( y \)-intercepts.
  • \( x^2 \) and \( b^2 \): This part addresses the stretch along the horizontal.
The equation follows a clear pattern. The positive term on the vertical axis dominates as the transverse axis directs upwards or downwards.
Asymptotes of a Hyperbola
Asymptotes in a hyperbola act like invisible guide lines, showing the hyperbola's direction as they stretch to infinity. Hyperbolas without the asymptotic support would stray wildly!
When a hyperbola's transverse axis is vertical, its asymptotes are equations of lines. They follow the form \( y = \pm \frac{a}{b}x \), where the plus or minus indicates the two lines moving in opposing directions.
  • In our example, asymptotes are \( y = \pm \frac{1}{4} x \).
  • This inform us that \( \frac{a}{b} = \frac{1}{4} \), directing us to a helpful ratio between \( a \) and \( b \).
Understanding asymptotes not only helps in sketching the hyperbola but also aids in determining the relationship between \( a \) and \( b \), essential for forming the hyperbola equation accurately.
Center of a Hyperbola
The center of a hyperbola is the point in the middle of its two distinct branches, a crucial pivot that affects the entire shape. For simplification, hyperbolas are often expressed with the center at the origin, \( (0,0) \). In such cases, the hyperbola equation aligns in a form without additional terms that push the graph around the coordinate plane.
The central location serves as a base for symmetry:
  • All other points on the hyperbola are measured against this central hub.
  • Both axes of the hyperbola cross at the center.
In our exercise, placing the center at the origin supports the symmetry and allows clear calculations without added complexity. This foundational point ensures each step precisely measures the hyperbola's dimensions and direction from a known benchmark.
Transverse Axis of a Hyperbola
The transverse axis of a hyperbola is no ordinary line; it is the main axis around which the hyperbola centers its curves. The transverse axis tells you the direction where the hyperbola swells the most.
In hyperbolas with a vertical transverse axis:
  • The equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) affirms the focus is on the \( y \)-axis.
  • \( a \): represents half the distance between the vertices along this vertical path.
For the exercise's hyperbola, the transverse axis is vertical since it has \( y \)-intercepts. Noting the values of \( a \) allows us to draw nearer to specifying how the hyperbola curves and extends. This insight helps us grasp not only the curve鈥檚 form but also its geometric focus, contributing to a deep understanding of its graph and properties.

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

Sketch the graph of the polar equation. \(r \theta=1, \theta>0\) (spiral)

Curves \(C_{1}, C_{2}, C_{3},\) and \(C_{4}\) are given parametrically, for \(t\) in \(\mathbb{R}\). Sketch their graphs, and indicate orientations. $$\begin{aligned}&C_{1}: x=t, \quad y=1-t\\\&C_{2}: x=1-t^{2}, \quad y=t^{2}\\\&C_{3}: x=\cos ^{2} t, \quad y=\sin ^{2} t\\\&C_{4}: x=\ln t-t, \quad y=1+t-\ln t, t>0\end{aligned}$$

Sketch the graph of the polar equation. $$r=-2 \sin \theta$$

Sketch the graph of the polar equation. $$r=2 \sin 4 \theta$$

Sketch the graph of the polar equation. $$r=-3 \sec \theta$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Calculus

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.