/*! 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 11 Graph each hyperbola. $$\frac{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 11

Graph each hyperbola. $$\frac{4 y^{2}}{9}-81 x^{2}=1$$

Short Answer

Expert verified
Plot the hyperbola with center at (0,0), asymptotes \( y = \pm \frac{27}{2} x \), and branches opening vertically.

Step by step solution

01

Identify the Standard Form

First, we need to recognize the form of the hyperbola equation. The given equation is \( \frac{4y^2}{9} - 81x^2 = 1 \). This equation is in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), which represents a vertical hyperbola centered at the origin.
02

Determine the Values of a and b

To find \( a \) and \( b \), we compare the given equation with the standard form. Here, \( \frac{y^2}{\frac{9}{4}} - \frac{x^2}{\frac{1}{81}} = 1 \) implies that \( a^2 = \frac{9}{4} \) and \( b^2 = \frac{1}{81} \). Thus, \( a = \frac{3}{2} \) and \( b = \frac{1}{9} \).
03

Find the Asymptotes

The equations of the asymptotes for a vertical hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) are given by \( y = \pm \frac{a}{b} x \). With \( a = \frac{3}{2} \) and \( b = \frac{1}{9} \), the asymptotes are \( y = \pm \frac{27}{2} x \).
04

Plot the Center and Asymptotes

The center of the hyperbola is at the origin \((0,0)\). Draw the asymptote lines \( y = \frac{27}{2} x \) and \( y = -\frac{27}{2} x \) through the center.
05

Sketch the Hyperbola

Using the center, \( a \), and the asymptotes as guides, sketch the branches of the hyperbola. The branches open vertically because the \( y^2 \) term is positive in the equation.

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 Hyperbola
The standard form of a hyperbola equation is essential for determining its characteristics. In this exercise, the equation given is \( \frac{4y^2}{9} - 81x^2 = 1 \). To recognize the type of hyperbola, we first rewrite it to align more visibly with the standard form:
  • \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) for vertical opening
  • \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for horizontal opening
This shows that our hyperbola opens vertically since the \( y^2 \) term comes first. By comparing this with \( \frac{y^2}{(3/2)^2} - \frac{x^2}{(1/9)^2} = 1 \), we identify that \( a^2 = \frac{9}{4} \) and \( b^2 = \frac{1}{81} \). This form helps us understand that the major axis is vertical.
Asymptotes of Hyperbola
Asymptotes are lines that the hyperbola approaches but never touches. They are crucial in sketching the shape of the hyperbola by showing the directions in which the branches open. For a hyperbola in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the asymptotes are given by the equations:
  • For a vertical hyperbola: \( y = \pm \frac{a}{b} x \)
  • For a horizontal hyperbola: \( y = \pm \frac{b}{a} x \)
In our case, where \( a = \frac{3}{2} \) and \( b = \frac{1}{9} \), the asymptotes are \( y = \pm \frac{27}{2} x \). These lines assist in accurately sketching the hyperbola by forming a guide within which the branches lie.
Vertices and Center of Hyperbola
The center and vertices of a hyperbola serve as vital points in its geometry. The center of the hyperbola acts as a pivot point for its opening direction and is situated where the asymptotes intersect. With the equation \( \frac{4y^2}{9} - 81x^2 = 1 \), the hyperbola is centered at the origin
  • Coordinate of the center: \((0, 0)\)
The vertices are points that lie along the axis of symmetry of the hyperbola. Since the equation's \( y^2 \) term is first, it indicates a vertical alignment around the y-axis, with the vertices located a distance \( a \) from the center along the y-axis:
  • Vertices at \((0, \frac{3}{2})\) and \((0, -\frac{3}{2})\)
Understanding these points aids in the accurate placement and drawing of the hyperbola.

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

In calculus, when finding the area between two polar curves, we need to find the points of intersection of the two curves. Find the values of \(\theta\) where the two conic sections intersect on \([0,2 \pi]\) $$r=\frac{2}{2+\sin \theta}, r=\frac{2}{2+\cos \theta}$$

You decided to race Jeremy Wariner for 800 meters. At that distance, Jeremy runs approximately twice as fast as you. He gave you a 1 -minute head start and crossed the finish line 20 seconds before you. What were each of your average speeds?

To use a TI-83 or TI-83 Plus (function-driven software or graphing utility) to graph a general second-degree equation, you need to solve for \(y\). Let us consider a general second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) Group \(y^{2}\) terms together, \(y\) terms together, and the remaining terms together.$$\begin{array}{c} A x^{2}+B x y+C y^{2}+D x+E y+F=0 \\ C y^{2}+(B x y+E y)+\left(A x^{2}+D x+F\right)=0 \end{array}$$ Factor out the common \(y\) in the first set of parentheses. $$C y^{2}+y(B x+E)+\left(A x^{2}+D x+F\right)=0$$ Now this is a quadratic equation in \(y: a y^{2}+b y+c=0\) Use the quadratic formula to solve for \(y\) $$\begin{array}{c}C y^{2}+y(B x+E)+\left(A x^{2}+D x+F\right)=0 \\\a=C, b=B x+E, c=A x^{2}+Dx+F\end{array}$$$$\begin{aligned} &y=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \quad y=\frac{-(B x+E) \pm \sqrt{(B x+E)^{2}-4(C)\left(A x^{2}+D x+F\right)}}{2(C)}\\\&y=\frac{-(B x+E) \pm \sqrt{B^{2} x^{2}+2 B E x+E^{2}-4 A C x^{2}-4 C D x-4 C F}}{2 C}\\\&y=\frac{-(B x+E) \pm \sqrt{\left(B^{2}-4 A C\right) x^{2}+(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}\end{aligned}$$ Case I: \(B^{2}-4 A C=0 \rightarrow\) The second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) is a parabola. \(y=\frac{-(B x+E) \pm \sqrt{(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}\) Case II: \(B^{2}-4 A C<0 \rightarrow\) The second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) is an ellipse.$$y=\frac{-(B x+E) \pm \sqrt{\left(B^{2}-4 A C\right) x^{2}+(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}$$Case III: \(B^{2}-4 A C>0 \rightarrow\) The second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) is a hyperbola. \(y=\frac{-(B x+E) \pm \sqrt{\left(B^{2}-4 A C\right) x^{2}+(2 B E-4 C D) x+\left(E^{2}-4 C F\right)}}{2 C}\) Use a graphing utility to explore the second-degree equation \(A x^{2}+B x y+C y^{2}+3 x+5 y-2=0\) for the following values of \(A, B,\) and \(C:\) a. \(A=1, B=-4, C=4\) b. \(A=1, B=4, D=-4\) Show the angle of rotation to the nearest degree. Explain the differences.

Use a graphing utility to solve the systems of equations. $$\begin{aligned}4 x^{4}+2 x y+3 y^{2} &=60 \\\x^{4} y &=8-3 x^{4}\end{aligned}$$

A circle and a line have at most two points of intersection. A circle and a parabola have at most four points of intersection. What is the greatest number of points of intersection that a circle and an \(n\)th-degree polynomial can have?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

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.