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Chapter 10: Problem 4

Rotate the coordinate axes to remove the \(x y\) -term. Then identify the type of conic and sketch its graph. $$ x^{2}-x y+y^{2}-2=0 $$

Short Answer

Expert verified
The conic is a circle; the graph is a circle with radius 2.

Step by step solution

01

Identify the Conic Equation

The given equation is \( x^2 - xy + y^2 - 2 = 0 \), a second-degree equation of the form \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \) where \( A = 1 \), \( B = -1 \), and \( C = 1 \).
02

Determine the Rotation Angle

The angle \( \theta \) needed to eliminate the \( xy \)-term is found using \( \tan(2\theta) = \frac{B}{A-C} \). Here, \( A = 1 \), \( B = -1 \), and \( C = 1 \), so \( \tan(2\theta) = \frac{-1}{1-1} \). This results in an undefined expression, signaling a \( 45^\circ \) rotation since it corresponds to the bisector of the angle between the axes.
03

Substitute New Variables

Substitute the new variables: \( x = x' \cos \theta - y' \sin \theta \) and \( y = x' \sin \theta + y' \cos \theta \). For a \( 45^\circ \) rotation, \( \cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \). Hence, the new variables become \( x = \frac{\sqrt{2}}{2}(x' - y') \) and \( y = \frac{\sqrt{2}}{2}(x' + y') \).
04

Transform the Equation

Substitute the expressions for \( x \) and \( y \) into the original equation. After substituting \( (\frac{\sqrt{2}}{2}(x' - y')) \) and \( (\frac{\sqrt{2}}{2}(x' + y')) \) into \( x^2 - xy + y^2 - 2 = 0 \), the \( xy \)-term is eliminated, resulting in an equation only in terms of \( x'^2 \) and \( y'^2 \). The precise transformed equation simplifies to \( x'^2 + y'^2 = 4 \).
05

Identify the Conic

The transformed equation \( x'^2 + y'^2 = 4 \) is a circle centered at the origin with a radius of 2.
06

Sketch the Graph

Draw a circle with a center at the origin (\( 0,0 \)) and a radius of 2 in the rotated \( x'y' \)-system. If reverted to the original \( xy \)-system without considering numerical evaluation steps, it confirms the graph will still follow a circular pattern.

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Key Concepts

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

Coordinate Rotation and its Importance
When working with conic sections, especially those involving an \( xy \)-term, coordinate rotation is a valuable technique. It helps simplify the equation of the conic, making it easier to recognize and analyze the curve. This rotation suggests a transformation of the axes to eliminate the \( xy \)-term effectively.
  • To find the angle of rotation, use the formula: \[ \tan(2\theta) = \frac{B}{A-C} \] The values \( A = 1 \), \( B = -1 \), and \( C = 1 \) make the formula \[ \tan(2\theta) = \frac{-1}{1-1} \] result in an undefined expression, which implies a \(45^\circ\) angle. This angle is significant as it corresponds to the bisector of the angle in typical arrangement.
  • Once you determine the angle of rotation, the original \( x \) and \( y \) coordinates are replaced by rotated coordinates \( x' \) and \( y' \) using the relationships: \[ x = x' \cos \theta - y' \sin \theta \] \[ y = x' \sin \theta + y' \cos \theta \]
This step paves the way for a simpler form of the equation, free from cumbersome mixed terms.
Conic Identification Through Equation Analysis
The primary goal after performing a coordinate rotation is to identify the conic section type. Conic sections such as circles, ellipses, parabolas, and hyperbolas are represented by second-degree equations. The given equation is a classic form: \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \).
  • When transformed, the given \( x^2 - xy + y^2 - 2 = 0 \) results in\( x'^2 + y'^2 = 4 \), which we can immediately recognize as a standard form of a circle.
  • A circle is identified by the equation\( x'^2 + y'^2 = r^2 \), where \( r \) is the radius. Here \( r = 2 \), indicating a circle centered at the origin with this radius.
Understanding the equation's form post-transform allows for easy identification and categorization of the conic type. This recognition is crucial for graphing and further mathematical inquiries.
Equation Transformation: Steps for Simplification
Transforming equations via substitution plays a key role in simplifying second-degree conic equations. Let's go step-by-step:
  • After determining the rotational relationship, substitute the expressions \( x = \frac{\sqrt{2}}{2}(x' - y') \) and \( y = \frac{\sqrt{2}}{2}(x' + y') \) into the original equation \( x^2 - xy + y^2 - 2 = 0 \).
  • What follows is the elimination of the complex \( xy \)-term, effectively simplifying the expression into \( x'^2 + y'^2 = 4 \). This is achieved through algebraic manipulation of the substituted terms.
  • The resulting equation reveals its structure clearly, showing how coordinate transformations make graphing and analyzing equations more straightforward.
This process not only demystifies the equation in terms of notation but also illuminates the elegant simplicity hidden beneath seemingly complex structures.

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Most popular questions from this chapter

Use the following values, where needed: radius of the Earth \(=4000 \mathrm{mi}=6440 \mathrm{~km}\) year (Earth year) \(=365\) days (Earth days) \(1 \mathrm{AU}=92.9 \times 10^{6} \mathrm{mi}=150 \times 10^{6} \mathrm{~km}\) (a) Let \(a\) be the semimajor axis of a planet's orbit around the Sun, and let \(T\) be its period. Show that if \(T\) is measured in days and \(a\) is measured in kilometers, then \(T=\left(365 \times 10^{-9}\right)(a / 150)^{3 / 2}\) (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is \(a=57.95 \times 10^{6} \mathrm{~km}\) (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is \(e=0.206\). (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c).

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Vertical and horizontal asymptotes of polar curves can sometimes be detected by investigating the behavior of \(x=r \cos \theta\) and \(y=r \sin \theta\) as \(\theta\) varies. This idea is used in these exercises. Show that the spiral \(r=1 / \theta^{2}\) does not have any horizontal asymptotes.
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