<meta http-equiv="refresh" content="1; url=/nojavascript/"> General Algebraic Forms | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Go to the latest version.

6.5: General Algebraic Forms

Created by: CK-12
 0  0  0 Share To Groups

Learning Objectives

  • Understand how the cross sections of a cone relate to degree two polynomial equations.
  • Understand what happens to the algebraic forms of conic sections when they are neither horizontally or vertically oriented.
  • Recognize the algebraic form of different types of conic sections.

The Cross Sections of the Cone Are Degree Two Polynomial Equations

Let’s examine all the equations of the conic sections we’ve studied in this chapter.

Ellipses: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, where a and b are any positive numbers (the circle is the specific case when a = b).

Parabolas: y = ax^2 or x = ay^2 where a is any non-zero number.

Hyperbolas: \frac{x^2}{a^2} - \frac{y^2}{b^2} =1 or \frac{y^2}{a^2} - \frac{x^2}{b^2} =1, where a and b are any positive numbers.

All these equations have in common that they are degree-2 polynomials, meaning the highest exponent of any variable—or sum of exponents of products of variables—is two. So for example, here are some degree two polynomial equations in a more general form:

2x^2 + 5y^2 - 3y + 4 & = 0 \\x^2 - 3y^2 + x - y + 3 & = 0 \\10x^2 - y - 5 & = 0 \\xy & = 2

Some of these probably already look like conic sections to you. For example, in the first equation, we can complete the square to remove the -3y term and we will see that we have an ellipse. In the second equation we can complete the square twice to remove both the x and -y terms and we will have a hyperbola. This is a hyperbola, not an ellipse, because the coefficient of the x^2 and y^2 terms have opposite signs.

The third equation is a parabola since there is an x^2 term and y term but not a y^2 term. Do you see how you can solve for y, putting the equation in the standard form for a vertically oriented parabola?

But what about the fourth equation? Like the others, it is a degree-2 polynomial, since the exponents of the x and y term sum to 2. But the fourth equation looks nothing like any of the forms for conic sections that we’ve examined so far. Nonetheless, as we saw in the last section, xy =2 appears to be a hyperbola with foci (2,2) and (-2,-2). The reason it doesn’t fit either of the standard forms for hyperbolas is because it is diagonally oriented, rather than horizontally or vertically oriented (do you see how its two foci lie on a diagonal line, rather than a horizontal or vertical line?)

In order to see how such differently-oriented conic sections fit into our standard forms, we need to rotate them so that they are either horizontally or vertically oriented.

Rotation of Conics

Remember in the section on parabolas we discussed rotating objects in the plane. In that section we showed that if we take a point P = (x,y) and rotate it \theta degrees, it changes the x-coordinate to x' \cos (\theta) - y' \sin(\theta) and the y-coordinate changes to x' \sin (\theta) + y' \cos (\theta). (Note: in that section, because we were looking at a plane embedded in space, we happened to be looking at the xz-plane. We also we’re using the “prime” symbols x' and y'. Also, in the particular case of that section we were rotating the cone clockwise. It is more standard to rotate counter-clockwise, so the signs on the sine functions have changed. So you can simply substitute x' for x and y' for z into the formulas in that section, switch the signs on the sine functions, and come up with these rotation rules for the xy-plane.)

Now suppose we have an equation of degree two polynomials, such as the xy = 2 example discussed above. In order to put it into the more recognizable form of a ellipse, parabola, or a hyperbola, we need to rotate it in such a way so that rotated version has no xy term. So we need to find an appropriate angle \theta such that changing the x-coordinate to x' \cos(\theta)-y' \sin (\theta) and the y-coordinate to x' \sin (\theta) + y' \cos (\theta) results in an equation with no xy term. We need to investigate what happens to a degree-two polynomial equation. Such an equation can be written in the form:

Ax^2 + By^2 + Cxy + Dx + Ey + F = 0

If C=0, such as in the first three examples at the beginning of this section, we are done, as there is no xy term and we already know how to classify these into conics. If C \neq 0, we need to rotate the curve so that we have an equation with no xy term. When x is replaced by x' \cos (\theta)-y' \sin(\theta) and y is replaced by x' \sin(\theta)+y' \cos (\theta), only the first three terms of this equation are in danger of producing an xy term. To see if we can determine if an appropriate angle \theta can always be found, let’s substitute our new variables in for x and y into the first three terms of the equation:

A(x' \cos(\theta)-y' \sin(\theta))^2 + B(x' \sin (\theta) + y' \cos (\theta))^2 + C (x' \cos (\theta) - y' \sin (\theta))(x' \sin (\theta) + y' \cos (\theta))

Then, let’s multiply this expression out, but only examine the terms that are a multiple of x'y', since that is what we’re trying to eliminate.

-2 Ax'y' \cos (\theta) \sin (\theta) + 2 Bx'y' \cos (\theta) \sin (\theta) + Cx'y' (\cos^2 (\theta) - \sin^2 (\theta))

This reduces to:

(-2 A \cos(\theta) \sin(\theta)+2B \cos(\theta)\sin(\theta)) + C(\cos^2(\theta) - \sin^2(\theta)))x'y'

So we are interested in whether or not there is an angle \theta such that the above coefficient of x'y' is zero:

(-2 A \cos(\theta) \sin(\theta) + 2 B \cos(\theta) \sin(\theta)) + C (\cos^2 (\theta) - \sin^2(\theta)) & = 0 \\C(\cos^2(\theta) - \sin^2(\theta)) & = (2 A \cos(\theta) \sin(\theta) - 2 B \cos (\theta) \sin (\theta)) \\C (\cos^2(\theta) - \sin^2(\theta)) & = 2 (A-B)(\cos(\theta)\sin(\theta))

Separating the \theta terms from the A, B, and C terms, we have:

\frac{2 \cos(\theta) \sin(\theta)}{\cos^2(\theta)-\sin^2(\theta)} = \frac{C}{A-B}

But remember the double angle formulas \sin(2 \theta) = 2 \cos(\theta)\sin(\theta) and \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) (see http://authors.ck12.org/wiki/index.php/Trigonometric_Identities#Double-Angle_Identities for more on the double angles). This means that we need to find an angle such that:

\frac{\sin(2 \theta)}{\cos(2 \theta)} = \frac{C}{A-B}

or:

\tan(2 \theta) = \frac{C}{A-B}

Can we find such an angle? Well remember the graph of tangent (see http://authors.ck12.org/wiki/index.php/Circular_Functions#y_.3D_tan.28x.29.2C_The_Tangent_Graph for more on the tangent graph). It spans all values, so no matter what C, A, and B equal, we can find the appropriate tangent value. The only time this expression gives us trouble is if the denominator is zero, or A=B. But in this case there’s an easy solution that you will find in the exercise below.

For the other cases, you can use the inverse tan function on your calculator or computer program to find out the value of 2 \theta, and then divide by 2 to find the value of \theta.

Example: Rotate the following conic section so that it is oriented either horizontally or vertically, and then analyze the result:

79x^2 + 37y^2 + 42 \sqrt{3}xy + \left (-200 \sqrt{3} - 8 \right ) x + \left (8 \sqrt{3} - 200 \right ) y + 4 = 0

Solution: To find how much to rotate this conic we need to solve for \theta in the equation \tan(2 \theta) = \frac{C}{A-B}. We have C = 42 \sqrt{3},A=79, and B=37, so we have:

\tan(2 \theta) & = \frac{42 \sqrt{3}}{79-37} \\\tan (2 \theta) & = \sqrt{3}

The angle that solves this equation is \theta = \frac{\pi}{6} (or 30^\circ).

So we replace x with x' \cos(30) - y' \sin (30) = \frac{\sqrt{3}}{2}x' - \frac{1}{2}y', and y with x' \sin(30)+y'\cos(30) = \frac{1}{2}x' + \frac{\sqrt{3}}{2}y', giving us:

& 79 \left (\frac{\sqrt{3}}{2}x' - \frac{1}{2}y' \right )^2 + 37 \left (\frac{1}{2}x' + \frac{\sqrt{3}}{2}y' \right )^2 + 42 \sqrt{3} \left (\frac{\sqrt{3}}{2}x' - \frac{1}{2}y' \right ) \left (\frac{1}{2}x' + \frac{\sqrt{3}}{2}y' \right ) + \\& \left (-200 \sqrt{3} -8 \right ) \left (\frac{\sqrt{3}}{2}x' - \frac{1}{2}y' \right ) + \left (8 \sqrt{3} - 200 \right ) \left (\frac{1}{2}x' + \frac{\sqrt{3}}{2}y' \right ) + 4 = 0

Multiplying through by 4 we have:

& 79 \left (\sqrt{3}x' - y' \right )^2 + 37 \left (x' + \sqrt{3}y' \right )^2 + 42 \sqrt{3} \left (\sqrt{3}x' - y' \right ) \left (x' + \sqrt{3}y' \right ) + \\& \left (-200 \sqrt{3} - 8 \right ) \left (2 \sqrt{3}x' - 2y' \right ) + \left (8 \sqrt{3} - 200 \right ) \left (2x' + 2\sqrt{3}y' \right ) + 16 = 0

Multiplying out we have:

& 237(x')^2 - 158 \sqrt{3}x'y' + 79(y')^2 + 37(x')^2 + 74 \sqrt{3}x'y' + 111(y')^2 + 126(x')^2 + 84 \sqrt{3}x'y' - 126(y')^2 \\& - 1200x' - 16\sqrt{3} x' + 400 \sqrt{3}y' + 16y' + 16\sqrt{3}x' - 400 x' + 48y' -400 \sqrt{3}y' + 16 = 0

Which simplifies to:

400(x')^2 + 64(y')^2 - 1600x' + 64y' + 16 = 0

Which, divided by 16 is:

25(x')^2 + 4(y')^2 - 100x' + 4y' + 1 = 0

Grouping the x' and y' terms and completing the squares, we have:

25((x')^2-4x' + 4) - 100 + 4\left ((y')^2 + y' + \frac{1}{4} \right )-1 + 1 & = 0 \\25((x')^2-2)^2 + 4 \left ((y')^2 + \frac{1}{2} \right )^2 & = 100 \\\frac{((x')^2-2)^2}{4} + \frac{\left ((y')^2 + \frac{1}{2} \right )^2}{25} & = 1

We recognize this as an ellipse, centered at the point \left (2, - \frac{1}{2} \right ).

Review Questions

  1. What’s the easy solution to the above problem when A=B?
  2. Use the above answer or a calculator or computer program to determine how much you would have to rotate the following conic to eliminate the xy term and orient it either horizontally or vertically: x^2 + y^2 + 6x -3y + 10xy + 100= 0.
  3. Use the above answer or a calculator or computer program to determine how much you would have to rotate the following conic to eliminate the xy term and orient it either horizontally or vertically: 5x^2-3x+4xy+21=0.
  4. Rotate the following conic section so that it is oriented either horizontally or vertically, and then analyze the result: 3x^2+3y^2+ \left (4 \sqrt{2} + 2 \right)xy+4\sqrt{2}x+4 \sqrt{2}y - 4 =0

Review Answers

  1. \theta = \frac{\pi}{2}
  2. \theta = \frac{\pi}{2}
  3. \theta \approx 31.72^\circ
  4. To figure out the angle of rotation, since A=B we have \theta = \frac{\pi}{4} (or 45^\circ). After shifting the equation by this amount, we have a relation, 4x^2 + 8x - y^2 + 8y - 4 = 0. Completing the square, this results in a shifted hyperbola: \frac{(x+1)^2}{1} - \frac{(y-2)^2}{4} =1.

General Algebraic Forms

How can we look at a degree-2 polynomial equation and determine which conic section it depicts?

Ax^2 + By^2 + C xy + Dx + Ey + F=0

When C=0 we have already discussed how to determine which conic section the equation refers to. In summary, if A and B are both positive, the conic section is an ellipse. This is also true of A and B are both negative, as the entire equation can be multiplied by -1 without changing the solution set. If A and B differ in sign, the equation is a hyperbola, and if A or B equals zero the equation is a parabola.

There are a few new, more general, rules I will show you that give more information about the case when C \neq 0 and hence the conic section needs to be rotated to achieve horizontal or vertical orientation.

If C^2 < 4AB, the equation is an ellipse (note when C=0 this holds whenever A and B are the same sign, which is consistent with our simpler rule stated above.)

If C^2 > 4AB, the equation is an hyperbola (note when C=0 this holds whenever A and B are the opposite sign, which is consistent with our simpler rule stated above.)

If C^2 = 4AB, the equation is a parabola (note when C=0, either A or B equals zero, which is consistent with our simpler rule stated above.)

Review Questions

  1. State what type of conic section is represented by the following equation: 5x^2+6y^2+2x-5y+xy=0.
  2. State what type of conic section is represented by the following equation: x^2+3y-20xy+20=0.
  3. The rules above do not account for “degenerate” conic sections, that is the conic section that looks like an X made by the intersection of a plane containing the line at the center of the cone. Explain the conditions on the coefficients that lead to the degenerate conic sections.

Review Answers

  1. Ellipse
  2. Hyperbola
  3. C=0, and either A=0 or B=0 (or both).

Image Attributions

Description

Categories:

Concept Nodes:

Grades:

Date Created:

Feb 23, 2012

Last Modified:

Jul 09, 2014
Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Math-Analysis.6.5
ShareThis Copy and Paste

Original text