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10.5: Rotation of Axes

  • Page ID
    114100
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    Learning Objectives

    In this section, you will:

    • 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. See Figure 1.

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    Figure 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 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.

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

    where A,B,A,B, and CC 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 xyxy term that we have not seen in any of the standard form equations. As we will discuss later, the xyxy term rotates the conic whenever BB is not equal to zero.

    Conic Sections 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 CONIC SECTIONS

    A conic section has the general form

    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 sections where B=0,B=0, and  A   A  and  C   C  are nonzero real numbers. This indicates that the conic has not been rotated.

    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 are positive
    parabola Ax2+Dx+Ey+F=0or Cy2+Dx+Ey+F=0Ax2+Dx+Ey+F=0or Cy2+Dx+Ey+F=0

    Table 2

    HOW TO

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

    1. Rewrite the equation in the general form, Ax2+Bxy+Cy2+Dx+Ey+F=0.Ax2+Bxy+Cy2+Dx+Ey+F=0.
    2. Identify the values of AA and CC from the general form.
      1. If AA and CC are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse.
      2. If AA and CC are equal and nonzero and have the same sign, then the graph may be a circle.
      3. If AA and CC are nonzero and have opposite signs, then the graph may be a hyperbola.
      4. If either AA or CC 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,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,Ax2+By2=0, when A and B have opposite signs.
      On the other hand, the equation, Ax2+By2+1=0,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 1

    Identifying a Conic from Its General Form

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

    1. ⓐ4x2−9y2+36x+36y−125=04x2−9y2+36x+36y−125=0
    2. ⓑ 9y2+16x+36y−10=09y2+16x+36y−10=0
    3. ⓒ 3x2+3y2−2x−6y−4=03x2+3y2−2x−6y−4=0
    4. ⓓ −25x2−4y2+100x+16y+20=0−25x2−4y2+100x+16y+20=0
    Answer
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    • f540864039c318ae25734b2c4aa019e046908fb9
    • e6ef8ccb98f8c2cc049f09318d885e86e16c3876
    • 3d7556137d9d649571fd3d0f1b05390d0e64573a
    TRY IT #1

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

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

    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 xyxy term, which aligns the graphs with the x- and y-axes. When we add an xyxy term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, say θ,θ, then every point on the plane may be thought of as having two representations: (x,y)(x,y) on the Cartesian plane with the original x-axis and y-axis, and (x′,y′)(x′,y′) on the new plane defined by the new, rotated axes, called the x'-axis and y'-axis. See Figure 3.

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

    We will find the relationships between xx and yy on the Cartesian plane with x′x′ and y′y′ on the new rotated plane. See Figure 4.

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    Figure 4 The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle θ.θ.

    The original coordinate x- and y-axes have unit vectors ii and j .j  . The rotated coordinate axes have unit vectors i′i′ and j′.j′. The angle θθ is known as the angle of rotation. See Figure 5. We may write the new unit vectors in terms of the original ones.

    i′=cosθi+sinθjj′=−sinθi+cosθji′=cosθi+sinθjj′=−sinθi+cosθj

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    Figure 5 Relationship between the old and new coordinate planes.

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

    u=x′i′+y′j′u=x′(icosθ+jsinθ)+y′(−isinθ+jcosθ)u=ix'cosθ+jx'sinθ−iy'sinθ+jy'cosθu=ix'cosθ−iy'sinθ+jx'sinθ+jy'cosθu=(x'cosθ−y'sinθ)i+(x'sinθ+y'cosθ)jSubstitute.Distribute.Apply commutative property.Factor by grouping.u=x′i′+y′j′u=x′(icosθ+jsinθ)+y′(−isinθ+jcosθ)Substitute.u=ix'cosθ+jx'sinθ−iy'sinθ+jy'cosθDistribute.u=ix'cosθ−iy'sinθ+jx'sinθ+jy'cosθApply commutative property.u=(x'cosθ−y'sinθ)i+(x'sinθ+y'cosθ)jFactor by grouping.

    Because u=x′i′+y′j′,u=x′i′+y′j′, we have representations of xx and yy in terms of the new coordinate system.

    x=x′cosθ−y′sinθandy=x′sinθ+y′cosθx=x′cosθ−y′sinθandy=x′sinθ+y′cosθ

    EQUATIONS OF ROTATION

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

    HOW TO

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

    1. 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.
    2. Substitute the expression for xx and yy into in the given equation, then simplify.
    3. Write the equations with x′x′ and y′y′ in standard form.

    EXAMPLE 2

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

    Find a new representation of the equation 2x2−xy+2y2−30=02x2−xy+2y2−30=0 after rotating through an angle of θ=45°.θ=45°.

    Answer
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    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=0Ax2+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′x′ and y′y′ coordinate system without the x′y′x′y′ term, by rotating the axes by a measure of θθ that satisfies

    cot(2θ)=A−CBcot(2θ)=A−CB

    We have learned already that any conic may be represented by the second 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 term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θθ where cot(2θ)=A−CB.cot(2θ)=A−CB.

    • If cot(2θ)>0,cot(2θ)>0, then 2θ2θ is in the first quadrant, and θθ is between (0°,45°).(0°,45°).
    • If cot(2θ)<0,cot(2θ)<0, then 2θ2θ is in the second quadrant, and θθ is between (45°,90°).(45°,90°).
    • If A=C,A=C, then θ=45°.θ=45°.
    HOW TO

    Given an equation for a conic in the x′y′x′y′ system, rewrite the equation without the x′y′x′y′ term in terms of x′x′ and y′,y′, where the x′x′ and y′y′ axes are rotations of the standard axes by θθ degrees.

    1. Find cot(2θ).cot(2θ).
    2. Find sinθsinθ and cosθ.cosθ.
    3. 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.
    4. Substitute the expression for xx and yy into in the given equation, and then simplify.
    5. Write the equations with x′x′ and y′y′ in the standard form with respect to the rotated axes.

    EXAMPLE 3

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

    Rewrite the equation 8x2−12xy+17y2=208x2−12xy+17y2=20 in the x′y′x′y′ system without an x′y′x′y′ term.

    Answer
    A right triangle in the first quadrant of the x y plane. The horizontal side is length 3 and is on the x-axis. The vertical side is length 4. The hypotenuse is length h and originates at the Origin. The acute angle at the origin is 2 theta.
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    TRY IT #2

    Rewrite the 13x2−63–√xy+7y2=1613x2−63xy+7y2=16 in the x′y′x′y′ system without the x′y′x′y′ term.

    EXAMPLE 4

    Graphing an Equation That Has No x′y′ Terms

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

    x2+12xy−4y2=30x2+12xy−4y2=30

    Answer
    A line with positive slope passing through the origin of the x y pane is shown. The x value of 5 is shown on the x-axis. The y value of 12 is shown on the y-axis. The angle the line makes with the x-axis is 2theta. The line is labeled cotangent (2 theta) = 5/12.
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    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=0Ax2+Bxy+Cy2+Dx+Ey+F=0

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

    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 shown that B2−4AC=B′2−4A′C′.B2−4AC=B′2−4A′C′. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2−4AC,B2−4AC, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.

    USING THE DISCRIMINANT TO IDENTIFY A CONIC

    If the equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax2+Bxy+Cy2+Dx+Ey+F=0 is transformed by rotating axes into the equation 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′.

    The equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax2+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, 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 without Rotating Axes

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

    1. ⓐ 5x2+23–√xy+2y2−5=05x2+23xy+2y2−5=0
    2. ⓑ 5x2+23–√xy+12y2−5=05x2+23xy+12y2−5=0
    Answer
    TRY IT #3

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

    1. ⓐ x2−9xy+3y2−12=0x2−9xy+3y2−12=0
    2. ⓑ 10x2−9xy+4y2−4=010x2−9xy+4y2−4=0
    MEDIA

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

    10.4 Section Exercises

    Verbal

    1.

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

    2.

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

    3.

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

    4.

    Given the equation ax2+4x+3y2−12=0,ax2+4x+3y2−12=0, what can we conclude if a>0?a>0?

    5.

    For the equation Ax2+Bxy+Cy2+Dx+Ey+F=0,Ax2+Bxy+Cy2+Dx+Ey+F=0, the value of θθ that satisfies cot(2θ)=A−CBcot(2θ)=A−CB gives us what information?

    Algebraic

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

    6.

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

    7.

    x2−10x+4y−10=0x2−10x+4y−10=0

    8.

    2x2−2y2+4x−6y−2=02x2−2y2+4x−6y−2=0

    9.

    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 the given 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 the angle θθ that will eliminate the xyxy term and write the corresponding equation without the 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 angle based on the given equation. Give the new equation and graph the original 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∘

    37.

    x=(y−1)2,θ=30∘x=(y−1)2,θ=30∘

    38.

    x29+y24=1,θ=30∘x29+y24=1,θ=30∘

    For the following exercises, graph the equation relative to the x′y′x′y′ system in which the equation has no x′y′x′y′ term.

    39.

    xy=9xy=9

    40.

    x2+10xy+y2−6=0x2+10xy+y2−6=0

    41.

    x2−10xy+y2−24=0x2−10xy+y2−24=0

    42.

    4x2−33–√xy+y2−22=04x2−33xy+y2−22=0

    43.

    6x2+23–√xy+4y2−21=06x2+23xy+4y2−21=0

    44.

    11x2+103–√xy+y2−64=011x2+103xy+y2−64=0

    45.

    21x2+23–√xy+19y2−18=021x2+23xy+19y2−18=0

    46.

    16x2+24xy+9y2−130x+90y=016x2+24xy+9y2−130x+90y=0

    47.

    16x2+24xy+9y2−60x+80y=016x2+24xy+9y2−60x+80y=0

    48.

    13x2−63–√xy+7y2−16=013x2−63xy+7y2−16=0

    49.

    4x2−4xy+y2−85–√x−165–√y=04x2−4xy+y2−85x−165y=0

    For the following exercises, determine the angle of rotation in order to eliminate the xyxy term. Then graph the new set of axes.

    50.

    6x2−53–√xy+y2+10x−12y=06x2−53xy+y2+10x−12y=0

    51.

    6x2−5xy+6y2+20x−y=06x2−5xy+6y2+20x−y=0

    52.

    6x2−83–√xy+14y2+10x−3y=06x2−83xy+14y2+10x−3y=0

    53.

    4x2+63–√xy+10y2+20x−40y=04x2+63xy+10y2+20x−40y=0

    54.

    8x2+3xy+4y2+2x−4=08x2+3xy+4y2+2x−4=0

    55.

    16x2+24xy+9y2+20x−44y=016x2+24xy+9y2+20x−44y=0

    For the following exercises, determine the value of  k   k  based on the given equation.

    56.

    Given 4x2+kxy+16y2+8x+24y−48=0,4x2+kxy+16y2+8x+24y−48=0, find kk for the graph to be a parabola.

    57.

    Given 2x2+kxy+12y2+10x−16y+28=0,2x2+kxy+12y2+10x−16y+28=0, find kk for the graph to be an ellipse.

    58.

    Given 3x2+kxy+4y2−6x+20y+128=0,3x2+kxy+4y2−6x+20y+128=0, find kk for the graph to be a hyperbola.

    59.

    Given kx2+8xy+8y2−12x+16y+18=0,kx2+8xy+8y2−12x+16y+18=0, find kk for the graph to be a parabola.

    60.

    Given 6x2+12xy+ky2+16x+10y+4=0,6x2+12xy+ky2+16x+10y+4=0, find kk for the graph to be an ellipse.


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