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8.5: Rotation of Axes

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Learning Objectives
  • Identify nondegenerate conic sections given their general form equations.
  • Use rotation of axes formulas.
  • Write equations of rotated conics in standard form.
  • Identify conics without rotating axes.

As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone (Figure 8.5.1).

imageedit_2_4675119068.jpg

Figure 8.5.1: The nondegenerate conic sections

Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 8.5.2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines.

imageedit_6_5854995623.jpg

Figure 8.5.2: Degenerate conic sections

Identifying Nondegenerate Conics in General Form

In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below.

Ax2+Bxy+Cy2+Dx+Ey+F=0

where A, B, and C are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation.

You may notice that the general form equation has an xy term that we have not seen in any of the standard form equations. As we will discuss later, the xy term rotates the conic whenever B is not equal to zero.

Table 8.5.1
Conic Sections Example
ellipse 4x2+9y2=1
circle 4x2+4y2=1
hyperbola 4x29y2=1
parabola 4x2=9y or 4y2=9x
one line 4x+9y=1
intersecting lines (x4)(y+4)=0
parallel lines (x4)(x9)=0
a point 4x2+4y2=0
no graph 4x2+4y2=1
GENERAL FORM OF CONIC SECTIONS

A conic section has the general form

Ax2+Bxy+Cy2+Dx+Ey+F=0

where A, B, and C are not all zero. Table 8.5.2 summarizes the different conic sections where B=0, and A and C are nonzero real numbers. This indicates that the conic has not been rotated.

Table 8.5.2
ellipse Ax2+Cy2+Dx+Ey+F=0, AC and AC>0
circle Ax2+Cy2+Dx+Ey+F=0, A=C
hyperbola Ax2Cy2+Dx+Ey+F=0 or Ax2+Cy2+Dx+Ey+F=0, where A and C are positive
parabola Ax2+Dx+Ey+F=0 or Cy2+Dx+Ey+F=0
How to: Given the equation of a conic, identify the type of conic
  1. Rewrite the equation in the general form (Equation ???), Ax2+Bxy+Cy2+Dx+Ey+F=0
  2. Identify the values of A and C from the general form.
    • If A and C are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse.
    • If A and C are equal and nonzero and have the same sign, then the graph may be a circle.
    • If A and C are nonzero and have opposite signs, then the graph may be a hyperbola.
    • If either A or C is zero, then the graph may be a parabola.

If B=0, the conic section will have a vertical and/or horizontal axes. If B does not equal 0, as shown below, the conic section is rotated. Notice the phrase “may be” in the definitions. That is because the equation may not represent a conic section at all, depending on the values of A, B, C, D, E, and F. For example, the degenerate case of a circle or an ellipse is a point:

Ax2+By2=0,

when A and B have the same sign.

The degenerate case of a hyperbola is two intersecting straight lines: Ax2+By2=0, when A and B have opposite signs.

On the other hand, the equation, Ax2+By2+1=0, when A and B are positive does not represent a graph at all, since there are no real ordered pairs which satisfy it.

Example 8.5.1: Identifying a Conic from Its General Form

Identify the graph of each of the following nondegenerate conic sections.

  1. 4x29y2+36x+36y125=0
  2. 9y2+16x+36y10=0
  3. 3x2+3y22x6y4=0
  4. 25x24y2+100x+16y+20=0
Solution
  1. Rewriting the general form (Equation ???), we have Ax2+Bxy+Cy2+Dx+Ey+F=04x2+0xy+(9)y2+36x+36y+(125)=0
    with A=4 and C=9, so we observe that A and C have opposite signs. The graph of this equation is a hyperbola.
  2. Rewriting the general form (Equation ???), we have Ax2+Bxy+Cy2+Dx+Ey+F=00x2+0xy+9y2+16x+36y+(10)=0
    with A=0 and C=9. We can determine that the equation is a parabola, since A is zero.
  3. Rewriting the general form (Equation ???), we have Ax2+Bxy+Cy2+Dx+Ey+F=03x2+0xy+3y2+(2)x+(6)y+(4)=0
    with A=3 and C=3. Because A=C, the graph of this equation is a circle.
  4. Rewriting the general form (Equation ???), we have Ax2+Bxy+Cy2+Dx+Ey+F=0(25)x2+0xy+(4)y2+100x+16y+20=0
    with A=25 and C=4. Because AC>0 and AC, the graph of this equation is an ellipse.
Exercise 8.5.1

Identify the graph of each of the following nondegenerate conic sections.

  1. 16y2x2+x4y9=0
  2. 16x2+4y2+16x+49y81=0
Answer a

hyperbola

Answer b

ellipse

Finding a New Representation of the Given Equation after Rotating through a Given Angle

Until now, we have looked at equations of conic sections without an xy term, which aligns the graphs with the x- and y-axes. When we add an xy term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, say θ,then every point on the plane may be thought of as having two representations: (x,y) on the Cartesian plane with the original x-axis and y-axis, and (x,y) on the new plane defined by the new, rotated axes, called the x'-axis and y'-axis (Figure 8.5.3).

CNX_Precalc_Figure_10_04_003.jpg

Figure 8.5.3: The graph of the rotated ellipse x2+y2xy15=0

We will find the relationships between x and y on the Cartesian plane with x and y on the new rotated plane (Figure 8.5.4).

CNX_Precalc_Figure_10_04_004.jpg

Figure 8.5.4: The Cartesian plane with x- and y-axes and the resulting x− and y−axes formed by a rotation by an angle θ.

The original coordinate x- and y-axes have unit vectors ˆi and ˆj. The rotated coordinate axes have unit vectors ˆi and ˆj.The angle θ is known as the angle of rotation (Figure 8.5.5). We may write the new unit vectors in terms of the original ones.

ˆi=cosθˆi+sinθˆj

ˆj=sinθˆi+cosθˆj

CNX_Precalc_Figure_10_04_005.jpg

Figure 8.5.5: Relationship between the old and new coordinate planes.

Consider a vector u in the new coordinate plane. It may be represented in terms of its coordinate axes.

u=xi+yj=x(icosθ+jsinθ)+y(isinθ+jcosθ)Substitute.=ixcosθ+jxsinθiysinθ+jycosθDistribute.=ixcosθiysinθ+jxsinθ+jycosθApply commutative property.=(xcosθysinθ)i+(xsinθ+ycosθ)jFactor by grouping.

Because u=xi+yj, we have representations of x and y in terms of the new coordinate system.

x=xcosθysinθ

and

y=xsinθ+ycosθ

EQUATIONS OF ROTATION

If a point (x,y) on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle θ from the positive x-axis, then the coordinates of the point with respect to the new axes are (x,y). We can use the following equations of rotation to define the relationship between (x,y) and (x,y):

x=xcosθysinθ

and

y=xsinθ+ycosθ

How to: Given the equation of a conic, find a new representation after rotating through an angle
  1. Find x and y where x=xcosθysinθ and y=xsinθ+ycosθ.
  2. Substitute the expression for x and y into in the given equation, then simplify.
  3. Write the equations with x and y in standard form.
Example 8.5.2: Finding a New Representation of an Equation after Rotating through a Given Angle

Find a new representation of the equation 2x2xy+2y230=0 after rotating through an angle of θ=45°.

Solution

Find x and y, where x=xcosθysinθ and y=xsinθ+ycosθ.

Because θ=45°,

x=xcos(45°)ysin(45°)x=x(12)y(12)x=xy2

and

y=xsin(45°)+ycos(45°)y=x(12)+y(12)y=x+y2

Substitute x=xcosθysinθ and y=xsinθ+ycosθ into 2x2xy+2y230=0.

2(xy2)2(xy2)(x+y2)+2(x+y2)230=0

Simplify.

2(xy)(xy)2(xy)(x+y)2+2(x+y)(x+y)230=0FOIL methodx22xy+y2(x2y2)2+x2+2xy+y230=0Combine like terms.2x2+2y2(x2y2)2=30Combine like terms.2(2x2+2y2(x2y2)2)=2(30)Multiply both sides by 2.4x2+4y2(x2y2)=60Simplify. 4x2+4y2x2+y2=60Distribute.3x260+5y260=6060Set equal to 1.

Write the equations with x and y in the standard form.

x220+y212=1

This equation is an ellipse. Figure 8.5.6 shows the graph.

imageedit_10_2436746571.png

Figure 8.5.6

Writing Equations of Rotated Conics in Standard Form

Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form Ax2+Bxy+Cy2+Dx+Ey+F=0 into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the x and y coordinate system without the xy term, by rotating the axes by a measure of θ that satisfies

cot(2θ)=ACB

We have learned already that any conic may be represented by the second degree equation

Ax2+Bxy+Cy2+Dx+Ey+F=0

where A, B,and C are not all zero. However, if B0, then we have an xy term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ where cot(2θ)=ACB.

  • If cot(2θ)>0, then 2θ is in the first quadrant, and θ is between (0°,45°).
  • If cot(2θ)<0, then 2θ is in the second quadrant, and θ is between (45°,90°).
  • If A=C, then θ=45°.
How to: Given an equation for a conic in the xy system, rewrite the equation without the xy term in terms of x and y,where the x and y axes are rotations of the standard axes by θ degrees
  1. Find cot(2θ).
  2. Find sinθ and cosθ.
  3. Substitute sinθ and cosθ into x=xcosθysinθ and y=xsinθ+ycosθ.
  4. Substitute the expression for x and y into in the given equation, and then simplify.
  5. Write the equations with x and y in the standard form with respect to the rotated axes.
Example 8.5.3: Rewriting an Equation with respect to the x and y axes without the xy Term

Rewrite the equation 8x212xy+17y2=20 in the xy system without an xy term.

Solution

First, we find cot(2θ).

8x212xy+17y2=20A=8, B=12 and C=17

imageedit_14_9742345318.png

Figure 8.5.7

From Figure 8.5.7:

cot(2θ)=ACB=81712=912=34

cot(2θ)=34=adjacentopposite

So the hypotenuse is

32+42=h29+16=h225=h2h=5

Next, we find sinθ and cosθ.

sinθ=1cos(2θ)2=1352=55352=53512=210=15sinθ=15cosθ=1+cos(2θ)2=1+352=55+352=5+3512=810=45cosθ=25

Substitute the values of sinθ and cosθ into x=xcosθysinθ and y=xsinθ+ycosθ.

x=xcosθysinθ=x(25)y(15)=2xy5

and

y=xsinθ+ycosθ=x(15)+y(25)=x+2y5

Substitute the expressions for x and y into in the given equation, and then simplify.

8(2xy5)212(2xy5)(x+2y5)+17(x+2y5)2=208((2xy)(2xy)5)12((2xy)(x+2y)5)+17((x+2y)(x+2y)5)=208(4x24xy+y2)12(2x2+3xy2y2)+17(x2+4xy+4y2)=10032x232xy+8y224x236xy+24y2+17x2+68xy+68y2=10025x2+100y2=10025100x2+100100y2=100100

Write the equations with x and y in the standard form with respect to the new coordinate system.

x24+y21=1

Figure 8.5.8 shows the graph of the ellipse.

imageedit_18_8476133234.png

Figure 8.5.8

Exercise 8.5.2

Rewrite the 13x263xy+7y2=16 in the xy system without the xy term.

Answer

x24+y21=1

Example 8.5.4 :Graphing an Equation That Has No xy Terms

Graph the following equation relative to the xy system:

x2+12xy4y2=30

Solution

First, we find cot(2θ).

x2+12xy4y2=20A=1, B=12,and C=4

cot(2θ)=ACBcot(2θ)=1(4)12cot(2θ)=512

Because cot(2θ)=512, we can draw a reference triangle as in Figure 8.5.9.

imageedit_22_9251267068.png

Figure 8.5.9

cot(2θ)=512=adjacentopposite

Thus, the hypotenuse is

52+122=h225+144=h2169=h2h=13

Next, we find sinθ and cosθ. We will use half-angle identities.

sinθ=1cos(2θ)2=15132=13135132=81312=213

cosθ=1+cos(2θ)2=1+5132=1313+5132=181312=313

Now we find x and y.

x=xcosθysinθ

x=x(313)y(213)

x=3x2y13

and

y=xsinθ+ycosθ

y=x(213)+y(313)

y=2x+3y13

Now we substitute x=3x2y13 and y=2x+3y13 into x2+12xy4y2=30.

(3x2y13)2+12(3x2y13)(2x+3y13)4(2x+3y13)2=30(113)[(3x2y)2+12(3x2y)(2x+3y)4(2x+3y)2]=30Factor.(113)[9x212xy+4y2+12(6x2+5xy6y2)4(4x2+12xy+9y2)]=30Multiply.(113)[9x212xy+4y2+72x2+60xy72y216x248xy36y2]=30Distribute.(113)[65x2104y2]=30Combine like terms.65x2104y2=390Multiply.x264y215=1Divide by 390.

Figure 8.5.10 shows the graph of the hyperbola x264y215=1

imageedit_26_7758236974.png

Figure 8.5.10

Identifying Conics without Rotating Axes

Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is

Ax2+Bxy+Cy2+Dx+Ey+F=0

If we apply the rotation formulas to this equation we get the form

Ax2+Bxy+Cy2+Dx+Ey+F=0

It may be shown that

B24AC=B24AC

The expression does not vary after rotation, so we call the expression invariant. The discriminant, B24AC, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.

HOWTO: USING THE DISCRIMINANT TO IDENTIFY A CONIC

If the equation

Ax2+Bxy+Cy2+Dx+Ey+F=0

is transformed by rotating axes into the equation

Ax2+Bxy+Cy2+Dx+Ey+F=0

then B24AC=B24AC

The equation Ax2+Bxy+Cy2+Dx+Ey+F=0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these. If the discriminant, B24AC, is

  • <0, the conic section is an ellipse
  • =0, the conic section is a parabola
  • >0, the conic section is a hyperbola
Example 8.5.5: Identifying the Conic without Rotating Axes

Identify the conic for each of the following without rotating axes.

  1. 5x2+23xy+2y25=0
  2. 5x2+23xy+12y25=0
Solution

a. Let’s begin by determining A, B, and C.

5Ax2+23Bxy+2Cy25=0

Now, we find the discriminant.

B24AC=(23)24(5)(2)=4(3)40=1240=28<0

Therefore, 5x2+23xy+2y25=0 represents an ellipse.

b. Again, let’s begin by determining A,B, and C.

5Ax2+23Bxy+12Cy25=0

Now, we find the discriminant.

B24AC=(23)24(5)(12)=4(3)240=12240=228<0

Therefore, 5x2+23xy+12y25=0 represents an ellipse.

Exercise 8.5.3

Identify the conic for each of the following without rotating axes.

  1. x29xy+3y212=0
  2. 10x29xy+4y24=0
Answer a

hyperbola

Answer b

ellipse

Key Equations

General Form equation of a conic section Ax2+Bxy+Cy2+Dx+Ey+F=0
Rotation of a conic section

x=xcosθysinθ

y=xsinθ+ycosθ

Angle of rotation θ, where cot(2θ)=ACB

Key Concepts

  • Four basic shapes can result from the intersection of a plane with a pair of right circular cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola.
  • A nondegenerate conic section has the general form Ax2+Bxy+Cy2+Dx+Ey+F=0 where A, B and C are not all zero. The values of A, B, and C determine the type of conic. See Example 8.5.1.
  • Equations of conic sections with an xy term have been rotated about the origin. See Example 8.5.2.
  • The general form can be transformed into an equation in the x and y coordinate system without the xy term. See Example 8.5.3 and Example 8.5.4.
  • An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section. See Example 8.5.5.

This page titled 8.5: Rotation of Axes is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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8.4: The Parabola
8.6: Conic Sections in Polar Coordinates