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Chapter 11: Problem 16

(a) Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola. (b) Use a rotation of axes to eliminate the xy-term. (c) Sketch the graph. $$x y+4=0$$

Short Answer

Expert verified
The graph is a hyperbola; use rotation to derive \\(x'^2 - y'^2 = -8\\). Sketched as a hyperbola.

Step by step solution

01

Identify the Equation Type

The given equation is \(xy + 4 = 0\). To determine the type of conic section, we'll first write this in the standard second-degree form: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). For \(xy + 4 = 0\), we see \(A = 0, B = 1, C = 0, D = 0, E = 0, F = 4\). The discriminant \( ext{\Delta}\) is given by \( ext{\Delta} = B^2 - 4AC\). Substitute the values to find \( ext{\Delta}\).
02

Calculate the Discriminant

Substitute \(A = 0\), \(B = 1\), \(C = 0\) into the discriminant formula: \(\text{\Delta} = B^2 - 4AC = 1^2 - 4(0)(0) = 1\). Since \(\text{\Delta} > 0\), this indicates a hyperbola.
03

Find Rotation Angle to Eliminate xy-Term

For hyperbolas and other conic sections with xy-terms, we use a rotation to eliminate the xy-term. The angle \(\theta\) is found using \(\cot(2\theta) = \frac{A-C}{B} = \frac{0-0}{1} = 0\), leading to \(2\theta = 90^\circ\) or \(270^\circ\). Therefore, \(\theta = 45^\circ\) or \(135^\circ\).
04

Apply the Rotation of Axes

Using the rotation formulas, substitute \(x = x'\cos\theta - y'\sin\theta\), \(y = x'\sin\theta + y'\cos\theta\), and simplify. For \(\theta = 45^\circ\), the rotation equations are \(x' = \frac{x-y}{\sqrt{2}}\), \(y' = \frac{x+y}{\sqrt{2}}\). Substitute these into the original equation and simplify.
05

Derive Transformed Equation

Substitute \(x = \frac{x'-y'}{\sqrt{2}}\) and \(y = \frac{x'+y'}{\sqrt{2}}\) into \(xy + 4 = 0\). After simplification, we have \(x'^2 - y'^2 = -8\), which is a standard form of a hyperbola.
06

Sketch the Graph

Now that we have the simpler form \(x'^2 - y'^2 = -8\), we identify this as a hyperbola. Sketching involves recognizing the symmetry about the axes because the equation in standard form suggests hyperbolas open along x-axis and y-axis. Since it derived from the rotated coordinates, adjust the standard hyperbola graph to align with the axes rotated by \(45^\circ\).

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Key Concepts

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

Discriminant
In the world of conic sections, the discriminant plays a crucial role in identifying the type of conic section represented by a quadratic equation. Conic sections include parabolas, ellipses, and hyperbolas.
To find the discriminant (\(\Delta\)) of a conic, we use the standard quadratic form: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. \]
  • The discriminant formula is:\[ \Delta = B^2 - 4AC. \]
  • If \(\Delta = 0\), the equation represents a parabola.
  • If \(\Delta > 0\), it signifies a hyperbola.
  • If \(\Delta < 0\), the graph is an ellipse.
In the exercise, with the equation \(xy + 4 = 0\), the coefficients are \(A = 0\), \(B = 1\), \(C = 0\), so we substitute these into the formula to get \(\Delta = 1^2 - 4(0)(0) = 1\). Since \(\Delta > 0\), this indicates that the graph is a hyperbola.
Hyperbola
A hyperbola is one of the fascinating shapes found in the family of conic sections. It consists of two symmetrical, open curves. The equation given in the exercise: \( xy + 4 = 0 \), once simplified, represents a hyperbola.
Hyperbolas have distinct features:
  • They are formed when the plane cuts through both nappe (or cones) of the double cone.
  • The equation of a hyperbola in its standard form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( - \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
  • The hyperbola has two branches, each opening indefinitely.
  • Its centering symmetry is usually around the origin or another specific point after translation.
  • It has asymptotes that guide its shape but the graph does not actually touch these lines.
For this exercise, after eliminating the \(xy\)-term through rotation, the hyperbola's standard form helps us interpret and sketch it accurately.
Rotation of Axes
The rotation of axes is a useful technique when dealing with conic sections that contain an \(xy\)-term. This term can make the conic section difficult to interpret and graph, necessitating a transformation. By rotating the axes, we eliminate the \(xy\)-term and simplify the equation for better understanding.
Here's how rotation works:
  • To find the rotation angle \(\theta\), use \(\cot(2\theta) = \frac{A-C}{B}\).
  • For our exercise where \(A = 0\), \(C = 0\), \(B = 1\), the solution gives \(2\theta = 90^\circ\), making \(\theta = 45^\circ\).
  • Substitute into rotation formulas: \(x = x'\cos\theta - y'\sin\theta\) and \(y = x'\sin\theta + y'\cos\theta\).
  • With \(\theta = 45^\circ\), we use: \(x' = \frac{x-y}{\sqrt{2}}\) and \(y' = \frac{x+y}{\sqrt{2}}\).
This transformation makes the original equation \(xy + 4 = 0\) easier to analyze, converting it to \(x'^2 - y'^2 = -8\), a recognizable form of a hyperbola.

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Most popular questions from this chapter

Determine the equation of the given conic in XY-coordinates when the coordinate axes are rotated through the indicated angle. $$x^{2}-y^{2}=2 y, \quad \phi=\cos ^{-1} \frac{3}{5}$$

An equation of a parabola is given. (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. $$8 x^{2}+12 y=0$$

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is closest to the sun is called perihelion, and the point at which it is farthest is called aphelion. These points are the vertices of the orbit. The earth's distance from the sun is \(147,000,000 \mathrm{km}\) at perihelion and \(153,000,000 \mathrm{km}\) at aphelion. Find an equation for the earth's orbit. (Place the origin at the center of the orbit with the sun on the \(x\) -axis.) (IMAGE CAN'T COPY)

An equation of an ellipse is given. (a) Find the center, vertices, and foci of the ellipse. (b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse. $$\frac{x^{2}}{9}+\frac{(y+5)^{2}}{25}=1$$

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