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Chapter 5: Problem 73

Use a rotation of axes to put the conic in standard position. Identify the graph, give its equation in the rotated coordinate system, and sketch the curve. $$x^{2}+x y+y^{2}=6$$

Short Answer

Expert verified
The graph is a circle with equation \((x')^2 + (y')^2 = 12\).

Step by step solution

01

Identify the coefficients

The given quadratic equation is \( x^2 + xy + y^2 = 6 \). Here, we identify the coefficients for the standard conic form \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \). Thus, \( A = 1 \), \( B = 1 \), \( C = 1 \), \( D = 0 \), \( E = 0 \), and \( F = -6 \).
02

Calculate rotation angle

To eliminate the \( xy \) term, we use the formula \( \tan(2\theta) = \frac{B}{A-C}\). Substitute \( A = 1 \), \( B = 1 \), \( C = 1 \): \( \tan(2\theta) = \frac{1}{1-1} = \frac{1}{0} \). Because this results in undefined division, \( 2\theta = 90^\circ \), so \( \theta = 45^\circ \).
03

Find new variables

With \( \theta = 45^\circ \), use the rotation formulas: \( x = x'\cos(45^\circ) - y'\sin(45^\circ) \) and \( y = x'\sin(45^\circ) + y'\cos(45^\circ) \). Simplify since \( \cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \), resulting in \( x = \frac{\sqrt{2}}{2}(x' - y') \) and \( y = \frac{\sqrt{2}}{2}(x' + y') \).
04

Substitute into the original equation

Substitute \( x = \frac{\sqrt{2}}{2}(x' - y') \) and \( y = \frac{\sqrt{2}}{2}(x' + y') \) into the original equation: \( x^2 + xy + y^2 = 6 \). Simplify using the identities \( (x')^2 + (y')^2 = r'^2 \): \( (x')^2 + (y')^2 = 12 \).
05

Identify and sketch the conic

The equation \( (x')^2 + (y')^2 = 12 \) is in the standard form of a circle with radius \( \sqrt{12} \) and centered at the origin. Sketch a circle with radius \( 2\sqrt{3} \).
06

Summary of the transformed system

The conic is a circle, and its equation in the rotated \((x', y')\) system is \( (x')^2 + (y')^2 = 12 \).

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Key Concepts

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

Rotation of Axes
The rotation of axes is a technique used to simplify conic sections by eliminating the cross-product term, or the \( xy \) term, from the quadratic equation. It's like spinning the graph around so that it aligns more neatly with the x or y axis. This makes it easier to recognize and work with the conic shape.
  • Start by identifying the coefficients in the equation you have.
  • You'll use the formula \( \tan(2\theta) = \frac{B}{A-C} \) to find the angle \( \theta \) to rotate.
  • If you end up with something undefined like \( \frac{1}{0} \), it means you've got a special angle, like 90掳, for rotation.
These steps help you reorient your equation into a friendlier form, often revealing circles, ellipses, parabolas, or hyperbolas in disguise. By knowing the rotation angle, you adjust your coordinates to make the conic's features easier to see.
Quadratic Equations
Quadratic equations in conic sections take the form \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \). These equations describe curves like circles and ellipses. Here's how you deal with them:
  • Identify the coefficients \( A, B, \) and \( C \) to understand the curve's nature.
  • The presence of an \( xy \) term usually means you need to perform a rotation of axes to simplify.
  • Simplify the equation by eliminating the \( xy \) term to see the true form.
Quadratic equations are powerful because they allow us to explore a range of curves beyond basic lines and circles. By manipulating them, you unveil the pure geometry of these shapes.
Circle Equation
The equation of a circle in a coordinate plane is typically written as \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the center, and \( r \) is the radius. After rotating axes, you can transform more complex equations into this familiar form.
  • When using the rotation of axes on an equation like \( x^2 + xy + y^2 = 6 \), you find something closer to \((x')^2 + (y')^2 = 12 \).
  • This tells you there's a circle centered at the origin in the new rotated plane.
  • The radius can be found as \( r = \sqrt{12} \), which simplifies to \( 2\sqrt{3} \).
Understanding this conversion is crucial because circles are one of the simplest and most symmetrical forms, making them easier to sketch and analyze.
Trigonometry in Conics
Trigonometry often comes into play when dealing with conic sections, especially during coordinate rotation. Knowing some basic trigonometric identities helps in rewriting the coordinates:
  • Using \( \sin(\theta) \) and \( \cos(\theta) \), you modify the \( x \) and \( y \) variables.
  • At \( 45^\circ \), \( \sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2} \).
  • This simplifies substitution into transformed equations, so you see the circle's format more easily.
Trigonometry is the bridge that allows you to neatly rotate and analyze these conics without losing the intricacies of the shape. This makes the abstract concept of rotation tangible and applicable.

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Most popular questions from this chapter

Identify the graph of the given equation. $$x^{2}-y-1=0$$

Let \(\mathrm{q}\) be a unit vector in \(\mathbb{R}^{n}\) and let \(W\) be the subspace spanned by q. Show that the orthogonal projection of a vector v onto \(W\) (as defined in Sections 1.2 and 5.2 ) is given by \\[\operatorname{proj}_{w}(\mathbf{v})=\left(\mathbf{q} \mathbf{q}^{T}\right) \mathbf{v}\\] and that the matrix of this projection is thus \(\mathbf{q q}^{T}\) (Hint: Remember that, for \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^{n}, \mathbf{x} \cdot \mathbf{y}=\mathbf{x}^{T} \mathbf{y} .)\)

In Exercises \(19-22\), find the orthogonal decomposition of v with respect to \(W\) $$\mathbf{v}=\left[\begin{array}{l} 2 \\ 1 \\ 5 \\ 3 \end{array}\right], W=\operatorname{span}\left(\left[\begin{array}{r} 1 \\ -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 1 \\ 1 \end{array}\right]\right)$$

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite, negative semidefinite, or indefinite \(x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}\)

Sometimes the graph of a quadratic equation is a straight line, a pair of straight lines, or a single point. We refer to such a graph as a degenerate conic. It is also possible that the equation is not satisfied for any values of the variables, in which case there is no graph at all and we refer to the conic as an imaginary conic. Identify the conic with the given equation as either degenerate or imaginary and, where possible, sketch the graph. $$3 x^{2}+y^{2}=0$$
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