/*! 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 51 Show that the asymptotes of the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that the asymptotes of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{a^{2}}=1\) are perpendicular to each other.

Short Answer

Expert verified
Answer: Yes, the asymptotes of the hyperbola are perpendicular to each other.

Step by step solution

01

Find the equations of the asymptotes

The general equation for a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). The asymptotes of this hyperbola have the equations \(y = \pm \frac{b}{a}x\). Since the given hyperbola has its equation as \(\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1\), its asymptotes are \(y = \pm \frac{a}{a}x\), which simplifies to \(y = \pm x\).
02

Calculate the product of the slopes

The slopes of the asymptotes are the coefficients of \(x\) in their equations. The slope of the first asymptote is \(m_1 = 1\), and the slope of the second asymptote is \(m_2 = -1\). To check if they are perpendicular, compute the product of the slopes: \(m_1 \times m_2 = 1 \times -1 = -1\).
03

Conclude if the asymptotes are perpendicular

Since the product of the slopes of the asymptotes is -1, it can be concluded that the asymptotes of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1\) are perpendicular to each other.

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 91Ó°ÊÓ!

Key Concepts

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

Asymptotes
Asymptotes are invisible lines that a hyperbola extends closer to as it stretches toward infinity. They do not meet the hyperbola but give a helpful visual guide to its shape. Generally, for a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we derive the equations of its asymptotes from the expressions \(y = \pm \frac{b}{a}x\). These lines reflect the direction the branches of the hyperbola extend. For instance, when \(a = b\) in the equation \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{a^{2}}=1\), the asymptotes simplify to \(y = \pm x\). This means each asymptote has an origin point and moves diagonally in opposite directions.
Perpendicular Slopes
The concept of perpendicular slopes is essential for recognizing the geometry of lines. Two lines are considered perpendicular if the product of their slopes equals -1. Slope represents how steep a line is, and it's often labeled \(m\) in equations. For hyperbolas like \(\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1\), examining the slopes \(m_1\) and \(m_2\) of the simpler equations \(y = \pm x\) reveals their perpendicular nature. This is because \(m_1 = 1\) and \(m_2 = -1\), and their product \(1 \times -1 = -1\), showing conclusively they angle at 90 degrees to each other. This geometric property assures their perpendicularity, enhancing the understanding of hyperbola behavior.
Equations of Hyperbola
Understanding the equation of a hyperbola is key to grasping its shape and behavior. A hyperbola is one of the conic sections, and its standard equation can appear in two forms: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\). These alter whether the hyperbola opens along the x-axis or y-axis. An equation such as \(\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1\) indicates equal stretching in both directions, framing a hyperbola with perpendicular asymptotes \(y = \pm x\).
Full comprehension of these equations involves recognizing constants \(a\) and \(b\), which guide the dimensions and orientation of the hyperbola. When \(a = b\), as in the mentioned hyperbola, symmetries simplify, making it easier to solve related geometric problems.

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

Consider the ellipse whose equation is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .\) Show that if \(a=b,\) then the graph is actually a circle.

A comet travels in a parabolic orbit with the sun as focus. When the comet is 60 million miles from the sun, the line segment from the sun to the comet makes an angle of \(\pi / 3\) radians with the axis of the parabolic orbit. Using the sun as the pole and assuming the axis of the orbit lies along the polar axis, find a polar equation for the orbit.

Use the information given in Special Topics 10.3. A and summarized in the endpapers at the beginning of this book to find a parameterization of the conic section whose rectangular equation is given. Confirm your answer by graphing. $$\frac{(x+5)^{2}}{25}-\frac{(y-3)^{2}}{50}=1$$

Find a rectangular equation that is equivalent to the given polar equation. $$r=3$$

Let \(P\) be a point at distance \(k\) from the center of a circle of radius \(r .\) As the circle rolls along the \(x\) -axis, \(P\) traces out a curve called a trochoid. [When \(k \leq r\), it might help to think of the circle as a bicycle wheel and \(P\) as a point on one of the spokes. \(]\) (a) Assume that \(P\) is on the \(y\) -axis as close as possible to the \(x\) -axis when \(t=0,\) and show that the parametric equations of the trochoid are $$x=r t-k \sin t \quad \text { and } \quad y=r-k \cos t$$ Note that when \(k=r,\) these are the equations of a cycloid. (b) Sketch the graph of the trochoid with \(r=3\) and \(k=2\) (c) Sketch the graph of the trochoid with \(r=3\) and \(k=4\)

See all solutions

Recommended explanations on Math Textbooks

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.