/*! 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 894 Discuss the graph of \(\left(\ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 43: Problem 894

Discuss the graph of \(\left(\mathrm{x}^{2} / 9\right)-\left(\mathrm{y}^{2} / 9\right)=1\).

Short Answer

Expert verified
The given equation represents a hyperbola with a horizontal major axis centered at the origin (0,0): \(\frac{x^2}{9} - \frac{y^2}{9} = 1\). The vertices are at (卤3, 0), and the asymptotes are given by y = 卤x. To sketch the graph, plot the center, vertices, and asymptotes, and draw the branches of the hyperbola opening horizontally, approaching the asymptotes.

Step by step solution

01

Identify the hyperbola's standard equation

The given equation is \(\frac{x^2}{9} - \frac{y^2}{9} = 1\). This equation represents a hyperbola because of the subtraction between the two squared variables. We can write it in the standard form for hyperbolas as \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where (h,k) is the center of the hyperbola, and a and b are related to its shape and orientation. In this case, the equation already is in standard form.
02

Identify the center

The center of a hyperbola is given by the point (h,k). From the standard form, we can see that h = 0 and k = 0. Therefore, the center of the given hyperbola is at the origin, (0,0).
03

Identify the vertices

The vertices of a hyperbola are the points along its major axis, a units away from the center. In this case, a = 3, and the major axis is horizontal because the positive term corresponds to the x^2 term. Therefore, the vertices are at (卤3, 0).
04

Identify the asymptotes

The asymptotes of a hyperbola are the lines which the hyperbola approaches as it extends to infinity. For a hyperbola in the given standard form, the equations of the asymptotes are given by \(y = \pm \frac{b}{a}(x - h) + k\). In this case, a = 3, b = 3, h = 0, and k = 0, so the asymptotes are given by y = 卤x.
05

Sketch the graph

Begin by plotting the center at the origin, (0,0). Then, plot the vertices at (卤3, 0). Next, draw the asymptotes with the equations y = x and y = -x. Finally, sketch the two branches of the hyperbola, which open horizontally, with the vertices as their closest points to the center, and approach the asymptotes as the branches extend away from the center. Your graph should have a hyperbola opening to the left and right, centered at the origin, with its branches approaching the lines y = 卤x.

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
In mathematics, a hyperbola is represented by an equation of the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Here,
  • \(h\) and \(k\) are the coordinates of the center of the hyperbola.
  • \(a\) and \(b\) are constants that define its dimensions and orientation.
Consider the equation \(\frac{x^2}{9} - \frac{y^2}{9} = 1\). This equation is already in standard form. The center of this hyperbola is at (0,0) because there is no horizontal or vertical shift, i.e., \(h = 0\) and \(k = 0\). The values of \(a\) and \(b\) are both equal to 3 since \(a^2 = 9\) and \(b^2 = 9\). Thus, the hyperbola is symmetric with respect to the center.
Asymptotes of Hyperbola
Asymptotes are lines that a curve approaches as it heads towards infinity. They are critical in shaping the branches of a hyperbola.For hyperbolas of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the equations of the asymptotes are given by:\[ y = \pm \frac{b}{a}(x - h) + k \]In our specific example, we have \(h = 0\), \(k = 0\), \(a = 3\), and \(b = 3\). Therefore, the asymptote equations simplify to:
  • \(y = \pm x\)
These lines reflect the directions toward which the hyperbola extends. They cross through the center of the hyperbola and give a crucial guide for sketching its shape.
Vertices of Hyperbola
The vertices of a hyperbola are the points where the hyperbola intersects its major axis. For our horizontal hyperbola with the form of \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are located \(a\) units from the center along the x-axis. In this case:
  • The center is at (0,0), and \(a = 3\).
  • Thus, the vertices are located at the points (卤3, 0).
These vertices mark the closest points of the hyperbola to the center and are crucial in defining the shape's extend along the major axis. The hyperbola opens outwards starting from these vertices.
Graphing Hyperbolas
Graphing a hyperbola involves sketching its center, vertices, and asymptotes, helping to visualize its shape. Start by marking the center of the hyperbola at the origin, (0,0).
  • Plot the vertices at (卤3, 0), which are located on the horizontal axis.
  • Draw the asymptotes: the lines moving through the center with slopes forming the lines \(y = x\) and \(y = -x\).
  • Sketch the branches of the hyperbola that open left and right from these vertices, approaching the asymptotes as they move outward.
The graph of this hyperbola illustrates a curve that tends toward the two asymptotes without ever touching them. This depiction helps in understanding how the hyperbola behaves mathematically with its symmetrical, open-ended branches.

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

Show that if the coordinates \((\mathrm{x}, \mathrm{y})\) of a point \(\mathrm{P}\) satisfy $$ \left(\mathrm{x}^{2} / 9\right)-\left(\mathrm{y}^{2} / 16\right)=1 $$ then \(\mathrm{F}_{1} \mathrm{P}-\mathrm{F}_{2} \mathrm{P} \mid=6\), where \(\mathrm{F}_{1}(-5,0)\) and \(\mathrm{F}_{2}(5,0)\) are the foci.

Find the equation for the set of points the difference of whose distances from \((5,0)\) and \((-5,0)\) is 6 units.

Prove that the ratio of the areas of any polygon circumscribed about a given circle is equal to the ratio of their perimeters.

Sketch the graph of the equation \(\mathrm{y}=2 / \mathrm{x}\).

Find the equation of the hyperbola with vertices \(\mathrm{V}_{1}(8,0)\), \(\mathrm{V}_{2}(2,0)\) and eccentricity \(\mathrm{e}=2\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

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.