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Chapter 11: Problem 41

In transforming an equation of the form $$A x^{2}+B x y+C y^{2}+D x+E y+F=0$$ to an \(x^{\prime}-y^{\prime}\) equation without an \(x^{\prime} y^{\prime}\) term using rotation of axes, explain why there is always a value of \(\theta\) in the interval \(0^{\circ}<\theta<90^{\circ}\) for which \(\cot 2 \theta=\frac{A-C}{B}\).

Short Answer

Expert verified
A suitable \(\theta\) exists because \(\cot\) covers all real numbers within its periodic range in \(0^\circ < \theta < 90^\circ\).

Step by step solution

01

Understanding the Problem

We need to transform the quadratic equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) to remove the \(x'y'\) term by rotating the axes. The goal is to find an angle \(\theta\) such that \(\cot 2\theta = \frac{A-C}{B}\).
02

Axis Rotation Formula

When rotating the axes by an angle \(\theta\), the new coordinates \((x', y')\) are related to the old ones \((x, y)\) by: \[ x = x'\cos \theta - y'\sin \theta \] \[ y = x'\sin \theta + y'\cos \theta \] This rotation alters the coefficients of the equation.
03

Transformation of Coefficients

The strategy is to choose \(\theta\) such that the \(x'y'\) term vanishes. The coefficient of \(x'y'\) in the transformed equation is expressed as: \[ \frac{B}{2}\sin 2\theta + (A-C)\sin^2 \theta \] To eliminate the \(x'y'\) term, we set this equation to zero.
04

Setting the Condition for Vanishing Term

Set \( \frac{B}{2}\sin 2\theta + (A-C)\sin^2 \theta = 0 \) which simplifies to: \[ B\cos 2\theta = A - C \] This implies \(\cot 2\theta = \frac{A-C}{B}\), which is the desired condition to set \(\theta\).
05

Finding Possible Theta Values

The equation \(\cot 2\theta = \frac{A-C}{B}\) describes a relationship that always has solutions for \(\theta\) as long as the right side is defined. Since \(\cot\) is periodic and defined for all non-zero values in the interval, there is always at least one such \(\theta\) in \(0^\circ < \theta < 90^\circ\).
06

Conclusion

Thus, there is always a \(\theta\) for which the rotation removes the \(x'y'\) term if \(\cot 2\theta = \frac{A-C}{B}\). This transformation is possible due to the defined range and periodic nature of \(\cot\).

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Key Concepts

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

Conic Sections
Conic sections are figures formed by the intersection of a plane with a double napped cone. These figures include ellipses, parabolas, and hyperbolas. Understanding these shapes is essential for various fields in mathematics and science. Each conic section can be represented by a simple quadratic equation format of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\).

  • **Ellipse:** All the points where the sum of the distances from two fixed points (foci) is constant.
  • **Parabola:** All points equidistant from a fixed point (focus) and a fixed straight line (directrix).
  • **Hyperbola:** All the points where the difference of the distances from two fixed points (foci) is constant.
These conics can be rotated on a graph, and the way they transform under rotation is a fundamental aspect of their study. Removing the cross product term \(xy\) with rotation often makes it easier to identify the conic type and analyze its properties.
Quadratic Equation Transformation
Transformation of a quadratic equation involves changing its axis of orientation in such a way that it simplifies or alters the equation into a new form that is more standardized. The given equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) can be transformed by rotating the axes to remove the cross product term \(xy\).

This is achieved through a rotation formula. The formula involves changing from old coordinates \((x, y)\) to new rotated coordinates \((x', y')\):

\[ x = x'\cos \theta - y'\sin \theta \]
\[ y = x'\sin \theta + y'\cos \theta \]

By carefully choosing an angle \(\theta\), we can ensure that the coefficient of \(x'y'\) is zero, simplifying the equation to its canonical form. This helps in directly identifying the type of conic section being dealt with and is one of the key steps in transforming the quadratic equation.
Angle of Rotation
The angle of rotation, \(\theta\), is crucial in transforming a quadratic equation such that it eliminates the term \(x'y'\) in the equation's expanded form. Finding the correct \(\theta\) involves understanding trigonometric identities, particularly those involving \(\cot 2\theta\).

Setting \(\cot 2\theta = \frac{A-C}{B}\) enables the elimination of the \(x'y'\) term. This condition arises because the trigonometric expression simplifies to obtain an angle \(\theta\) that when applied, balances the equation effectively simplifying it.

The periodic nature of the cotangent function ensures there is always an angle \(\theta\) in the interval \(0^\circ<\theta<90^\circ\), making this method reliably applicable for any given quadratic equation with real coefficients. This rotation, simplification, and transformation process is crucial in deriving and understanding the properties of conic sections and their graphical representations.

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Most popular questions from this chapter

Determine the equation of the hyperbola satisfying the given conditions. Write each answer in the form \(A x^{2}-B y^{2}=\) Cor in the form \(A y^{2}-B x^{2}=C\). Vertices (0,卤7)\(;\) graph passes through the point (1,9)

(a) Let \(e_{1}\) denote the eccentricity of the hyperbola \(x^{2} / 4^{2}-y^{2} / 3^{2}=1,\) and let \(e_{2}\) denote the eccentricity of the hyperbola \(x^{2} / 3^{2}-y^{2} / 4^{2}=1 .\) Verify that \(e_{1}^{2} e_{2}^{2}=e_{1}^{2}+e_{2}^{2}\) (b) Let \(e_{1}\) and \(e_{2}\) denote the eccentricities of the two hyperbolas \(x^{2} / A^{2}-y^{2} / B^{2}=1\) and \(y^{2} / B^{2}-x^{2} / A^{2}=1\) respectively. Verify that \(e_{1}^{2} e_{2}^{2}=e_{1}^{2}+e_{2}^{2}\)

In this exercise we graph a hyperbola in which the axes of the curve are not parallel to the coordinate axes. The equation is \(x^{2}+4 x y-2 y^{2}=6\) (a) Use the quadratic formula to solve the equation for \(y\) in terms of \(x .\) Show that the result can be written \(y=x \pm \frac{1}{2} \sqrt{6 x^{2}-12}\) (b) Graph the two equations obtained in part (a). Use the standard viewing rectangle. (c) It can be shown that the equations of the asymptotes are \(y=(1 \pm 0.5 \sqrt{6}) x .\) Add the graphs of these asymptotes to the picture that you obtained in part (b). (d) Change the viewing rectangle so that both \(x\) and \(y\) extend from -50 to \(50 .\) What do you observe?

In designing an arch, architects and engineers sometimes use a parabolic arch rather than a semicircular arch. (One reason for this is that, in general, the parabolic arch can support more weight at the top than can the semicircular arch.) In the following figure, the blue arch is a semicircle of radius \(1,\) centered at the origin. The red arch is a portion of a parabola. As is indicated in the figure, the two arches have the same base and the same height. Assume that the unit of distance for each axis is the meter. GRAPH CANT COPY (a) Find the equation of the parabola in the figure. (b) Using calculus, it can be shown that the area under this parabolic arch is \(\frac{4}{3} \mathrm{m}^{2} .\) Assuming this fact, show that the area beneath the parabolic arch is approximately \(85 \%\) of the area beneath the semicircular arch. (c) Using calculus, it can be shown that the length of this parabolic arch is \(\sqrt{5}+\frac{1}{2} \ln (2+\sqrt{5})\) meters. Assuming this fact, show that the length of the parabolic arch is approximately \(94 \%\) of the length of the semicircular arch.

Graph the equations. $$(x+y)^{2}+4 \sqrt{2}(x-y)=0$$
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