8.5: Rotation of Axes (2024)

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Learning Objectives

In this section, you will:

Identify nondegenerate conic sections given their general formequations.
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 planeintersects two right circular cones aligned tip to tip andextending infinitely far in opposite directions, which we also calla cone. The way in which weslice the cone will determine the type of conic section formed atthe intersection. A circle is formed by slicing a cone with a planeperpendicular to the axis of symmetry of the cone. An ellipse isformed by slicing a single cone with a slanted plane notperpendicular to the axis of symmetry. A parabola is formed byslicing the plane through the top or bottom of the double-cone,whereas a hyperbola is formed when the plane slices both the topand bottom of the cone. See Figure1.

Figure 1 Thenondegenerate conic sections

Ellipses, circles, hyperbolas, and parabolas are sometimescalled the nondegenerate conic sections, incontrast to the degenerate conic sections,which are shown in Figure2. A degenerate conic results when a plane intersects thedouble cone and passes through the apex. Depending on the angle ofthe plane, three types of degenerate conic sections are possible: apoint, a line, or two intersecting lines.

Figure 2 Degenerateconic sections

Identifying Nondegenerate Conics inGeneral Form

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

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

where A,B,A,B, and CC are not all zero. Wecan use the values of the coefficients to identify which type conicis represented by a given equation.

You may notice that the general form equation hasan xyxy term that we have not seen in any of the standardform equations. As we will discuss later, the xyxy termrotates the conic whenever BB is not equal to zero.

ConicSections	Example
ellipse	4x2+9y2=14x2+9y2=1
circle	4x2+4y2=14x2+4y2=1
hyperbola	4x2−9y2=14x2−9y2=1
parabola	4x2=9yor 4y2=9x4x2=9yor 4y2=9x
one line	4x+9y=14x+9y=1
intersecting lines	(x−4)(y+4)=0(x−4)(y+4)=0
parallel lines	(x−4)(x−9)=0(x−4)(x−9)=0
a point	4x2+4y2=04x2+4y2=0
no graph	4x2+4y2=−14x2+4y2=−1

Table 1

GENERAL FORM OF CONICSECTIONS

A conic section has the generalform

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

where A,B,A,B, and CC are not all zero.

Table 2 summarizes the different conic sectionswhere B=0,B=0, and A A and C C arenonzero real numbers. This indicates that the conic has not beenrotated.

ellipse	Ax2+Cy2+Dx+Ey+F=0,A≠Cand AC>0Ax2+Cy2+Dx+Ey+F=0,A≠Cand AC>0
circle	Ax2+Cy2+Dx+Ey+F=0,A=CAx2+Cy2+Dx+Ey+F=0,A=C
hyperbola	Ax2−Cy2+Dx+Ey+F=0or −Ax2+Cy2+Dx+Ey+F=0,Ax2−Cy2+Dx+Ey+F=0or −Ax2+Cy2+Dx+Ey+F=0, where AA and CC arepositive
parabola	Ax2+Dx+Ey+F=0or Cy2+Dx+Ey+F=0Ax2+Dx+Ey+F=0or Cy2+Dx+Ey+F=0

Table 2

HOWTO

Given the equation of a conic, identify the type ofconic.

Rewrite the equation in the generalform, Ax2+Bxy+Cy2+Dx+Ey+F=0.Ax2+Bxy+Cy2+Dx+Ey+F=0.
Identify the values of AA and CC from thegeneral form.
1. If AA and CC are nonzero, have the samesign, and are not equal to each other, then the graph may be anellipse.
2. If AA and CC are equal and nonzero and havethe same sign, then the graph may be a circle.
3. If AA and CC are nonzero and have oppositesigns, then the graph may be a hyperbola.
4. If either AA or CC is zero, then the graphmay be a parabola.
If B = 0, the conicsection will have a vertical and/or horizontal axes.If B does not equal 0, asshown below, the conic section is rotated. Notice the phrase “maybe” in the definitions. That is because the equation may notrepresent a conic section at all, depending on the valuesof A, B, C, D, E,and F. For example, thedegenerate case of a circle or an ellipse is a point:
Ax2+By2=0,Ax2+By2=0, when A and B have the same sign.
The degenerate case of a hyperbola is two intersecting straightlines: Ax2+By2=0,Ax2+By2=0, when A and B have oppositesigns.
On the other hand, theequation, Ax2+By2+1=0,Ax2+By2+1=0, when A and B arepositive does not represent a graph at all, since there are no realordered pairs which satisfy it.

EXAMPLE 1

Identifying a Conic from Its GeneralForm

Identify the graph of each of the following nondegenerate conicsections.

ⓐ4x2−9y2+36x+36y−125=04x2−9y2+36x+36y−125=0
ⓑ 9y2+16x+36y−10=09y2+16x+36y−10=0
ⓒ 3x2+3y2−2x−6y−4=03x2+3y2−2x−6y−4=0
ⓓ −25x2−4y2+100x+16y+20=0−25x2−4y2+100x+16y+20=0

Answer

TRYIT #1

Identify the graph of each of the following nondegenerate conicsections.

ⓐ 16y2−x2+x−4y−9=016y2−x2+x−4y−9=0
ⓑ 16x2+4y2+16x+49y−81=016x2+4y2+16x+49y−81=0

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

Until now, we have looked at equations of conic sections withoutan xyxy term, which aligns the graphs withthe x-and y-axes. When we addan xyxy term, we are rotating the conic about the origin.If the x-and y-axes are rotated throughan angle, say θ,θ, then every point on the plane may bethought of as having two representations: (x,y)(x,y) onthe Cartesian plane with the original x-axis and y-axis, and (x′,y′)(x′,y′) on the newplane defined by the new, rotated axes, calledthe x'-axisand y'-axis.See Figure3.

Figure 3 The graphof the rotated ellipse x2+y2–xy–15=0x2+y2–xy–15=0

We will find the relationshipsbetween xx and yy on the Cartesian planewith x′x′ and y′y′ on the new rotated plane.See Figure4.

EQUATIONS OFROTATION

If a point (x,y)(x,y) on the Cartesian plane isrepresented on a new coordinate plane where the axes of rotationare formed by rotating an angle θθ from thepositive x-axis, then thecoordinates of the point with respect to the new axesare (x′,y′).(x′,y′). We can use the following equationsof rotation to define the relationshipbetween (x,y)(x,y) and (x′,y′):(x′,y′):

x=x′cosθ−y′sinθx=x′cosθ−y′sinθ

and

y=x′sinθ+y′cosθy=x′sinθ+y′cosθ

HOWTO

Given the equation of a conic, find a new representationafter rotating through an angle.

Find xx and yy where x=x′cosθ−y′sinθx=x′cosθ−y′sinθ and y=x′sinθ+y′cosθ.y=x′sinθ+y′cosθ.
Substitute the expressionfor xx and yy into in the given equation, thensimplify.
Write the equations with x′x′ and y′y′ instandard form.

EXAMPLE 2

Finding a New Representation of anEquation after Rotating through a Given Angle

Find a new representation of theequation 2x2−xy+2y2−30=02x2−xy+2y2−30=0 after rotatingthrough an angle of θ=45°.θ=45°.

Answer

Writing Equations of Rotated Conicsin Standard Form

Now that we can find the standard form of a conic when we aregiven an angle of rotation, we will learn how to transform theequation of a conic given in theform Ax2+Bxy+Cy2+Dx+Ey+F=0Ax2+Bxy+Cy2+Dx+Ey+F=0 intostandard form by rotating the axes. To do so, we will rewrite thegeneral form as an equation inthe x′x′ and y′y′ coordinate system withoutthe x′y′x′y′ term, by rotating the axes by a measureof θθ that satisfies

cot(2θ)=A−CBcot(2θ)=A−CB

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

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

where A,B,A,B, and CC are not all zero.However, if B≠0,B≠0, then we have an xyxy termthat prevents us from rewriting the equation in standard form. Toeliminate it, we can rotate the axes by an acuteangle θθ where cot(2θ)=A−CB.cot(2θ)=A−CB.

If cot(2θ)>0,cot(2θ)>0, then 2θ2θ isin the first quadrant, and θθ isbetween (0°,45°).(0°,45°).
If cot(2θ)<0,cot(2θ)<0, then 2θ2θ isin the second quadrant, and θθ isbetween (45°,90°).(45°,90°).
If A=C,A=C, then θ=45°.θ=45°.

HOWTO

Given an equation for a conic inthe x′y′x′y′ system, rewrite the equation withoutthe x′y′x′y′ term in termsof x′x′ and y′,y′, wherethe x′x′ and y′y′ axes are rotations of thestandard axes by θθ degrees.

Find cot(2θ).cot(2θ).
Find sinθsinθ and cosθ.cosθ.
Substitute sinθsinθ and cosθcosθ into x=x′cosθ−y′sinθx=x′cosθ−y′sinθ and y=x′sinθ+y′cosθ.y=x′sinθ+y′cosθ.
Substitute the expressionfor xx and yy into in the given equation, andthen simplify.
Write the equations with x′x′ and y′y′ inthe standard form with respect to the rotated axes.

EXAMPLE 3

Rewriting an Equation with respect tothe x′ and y′ axes without the x′y′ Term

Rewrite theequation 8x2−12xy+17y2=208x2−12xy+17y2=20 inthe x′y′x′y′ system withoutan x′y′x′y′ term.

Answer

TRYIT #2

Rewrite the 13x2−63–√xy+7y2=1613x2−63xy+7y2=16 inthe x′y′x′y′ system withoutthe x′y′x′y′ term.

EXAMPLE 4

Graphing an Equation That HasNo x′y′ Terms

Graph the following equation relative tothe x′y′x′y′ system:

x2+12xy−4y2=30x2+12xy−4y2=30

Answer

Identifying Conics without RotatingAxes

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

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

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

A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0

It may be shownthat B2−4AC=B′2−4A′C′.B2−4AC=B′2−4A′C′. The expressiondoes not vary after rotation, so we call the expressioninvariant. Thediscriminant, B2−4AC,B2−4AC, is invariant and remainsunchanged after rotation. Because the discriminant remainsunchanged, observing the discriminant enables us to identify theconic section.

USINGTHE DISCRIMINANT TO IDENTIFY A CONIC

If theequation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax2+Bxy+Cy2+Dx+Ey+F=0 istransformed by rotating axes into theequation A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0,A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0, then B2−4AC=B′2−4A′C′.B2−4AC=B′2−4A′C′.

Theequation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax2+Bxy+Cy2+Dx+Ey+F=0 is anellipse, a parabola, or a hyperbola, or a degenerate case of one ofthese.

If the discriminant, B2−4AC,B2−4AC, is

<0,<0, the conic section is an ellipse
=0,=0, the conic section is a parabola
>0,>0, the conic section is a hyperbola

EXAMPLE 5

Identifying the Conic withoutRotating Axes

Identify the conic for each of the following without rotatingaxes.

ⓐ 5x2+23–√xy+2y2−5=05x2+23xy+2y2−5=0
ⓑ 5x2+23–√xy+12y2−5=05x2+23xy+12y2−5=0

Answer

TRYIT #3

Identify the conic for each of the following without rotatingaxes.

ⓐ x2−9xy+3y2−12=0x2−9xy+3y2−12=0
ⓑ 10x2−9xy+4y2−4=010x2−9xy+4y2−4=0

MEDIA

Access this online resource for additional instruction andpractice with conic sections and rotation of axes.

Introduction to ConicSections

10.4 SectionExercises

Verbal

What effect does the xyxy term have on the graph of aconic section?

If the equation of a conic section is written in theform Ax2+By2+Cx+Dy+E=0Ax2+By2+Cx+Dy+E=0 and AB=0,AB=0, whatcan we conclude?

If the equation of a conic section is written in theform Ax2+Bxy+Cy2+Dx+Ey+F=0,Ax2+Bxy+Cy2+Dx+Ey+F=0, and B2−4AC>0,B2−4AC>0, whatcan we conclude?

Given theequation ax2+4x+3y2−12=0,ax2+4x+3y2−12=0, what can weconclude if a>0?a>0?

For theequation Ax2+Bxy+Cy2+Dx+Ey+F=0,Ax2+Bxy+Cy2+Dx+Ey+F=0, thevalue of θθ thatsatisfies cot(2θ)=A−CBcot(2θ)=A−CB gives us whatinformation?

Algebraic

For the following exercises, determine which conic section isrepresented based on the given equation.

9x2+4y2+72x+36y−500=09x2+4y2+72x+36y−500=0

x2−10x+4y−10=0x2−10x+4y−10=0

2x2−2y2+4x−6y−2=02x2−2y2+4x−6y−2=0

4x2−y2+8x−1=04x2−y2+8x−1=0

10.

4y2−5x+9y+1=04y2−5x+9y+1=0

11.

2x2+3y2−8x−12y+2=02x2+3y2−8x−12y+2=0

12.

4x2+9xy+4y2−36y−125=04x2+9xy+4y2−36y−125=0

13.

3x2+6xy+3y2−36y−125=03x2+6xy+3y2−36y−125=0

14.

−3x2+33–√xy−4y2+9=0−3x2+33xy−4y2+9=0

15.

2x2+43–√xy+6y2−6x−3=02x2+43xy+6y2−6x−3=0

16.

−x2+42–√xy+2y2−2y+1=0−x2+42xy+2y2−2y+1=0

17.

8x2+42–√xy+4y2−10x+1=08x2+42xy+4y2−10x+1=0

For the following exercises, find a new representation of thegiven equation after rotating through the given angle.

18.

3x2+xy+3y2−5=0,θ=45°3x2+xy+3y2−5=0,θ=45°

19.

4x2−xy+4y2−2=0,θ=45°4x2−xy+4y2−2=0,θ=45°

20.

2x2+8xy−1=0,θ=30°2x2+8xy−1=0,θ=30°

21.

−2x2+8xy+1=0,θ=45°−2x2+8xy+1=0,θ=45°

22.

4x2+2–√xy+4y2+y+2=0,θ=45°4x2+2xy+4y2+y+2=0,θ=45°

For the following exercises, determine theangle θθ that will eliminate the xyxy term andwrite the corresponding equation withoutthe xyxy term.

23.

x2+33–√xy+4y2+y−2=0x2+33xy+4y2+y−2=0

24.

4x2+23–√xy+6y2+y−2=04x2+23xy+6y2+y−2=0

25.

9x2−33–√xy+6y2+4y−3=09x2−33xy+6y2+4y−3=0

26.

−3x2−3–√xy−2y2−x=0−3x2−3xy−2y2−x=0

27.

16x2+24xy+9y2+6x−6y+2=016x2+24xy+9y2+6x−6y+2=0

28.

x2+4xy+4y2+3x−2=0x2+4xy+4y2+3x−2=0

29.

x2+4xy+y2−2x+1=0x2+4xy+y2−2x+1=0

30.

4x2−23–√xy+6y2−1=04x2−23xy+6y2−1=0

Graphical

For the following exercises, rotate through the given anglebased on the given equation. Give the new equation and graph theoriginal and rotated equation.

31.

y=−x2,θ=−45∘y=−x2,θ=−45∘

32.

x=y2,θ=45∘x=y2,θ=45∘

33.

x24+y21=1,θ=45∘x24+y21=1,θ=45∘

34.

y216+x29=1,θ=45∘y216+x29=1,θ=45∘

35.

y2−x2=1,θ=45∘y2−x2=1,θ=45∘

36.

y=x22,θ=30∘y=x22,θ=30∘