Copyright © 2011 Pearson, Inc.

8.4Translation and

Rotation of Axes

Slide 8.4 - 2 Copyright © 2011 Pearson, Inc.

What you’ll learn about

Second-Degree Equations in Two Variables Translating Axes versus Translating Graphs Rotation of Axes Discriminant Test

… and whyYou will see ellipses, hyperbolas, and parabolas as members of the family of conic sections rather than as separate types of curves.

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Example Graphing a Second-Degree Equation

Solve for y, and use a funcion grapher to graph the conic

4x2 −4xy+ 4y2 +6x−12y−3=0

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Example Graphing a Second-Degree Equation

Solve for y, and use a funcion grapher to graph the conic

4x2 −4xy+ 4y2 +6x−12y−3=0

Compare the given equation to the general second-

degree equation

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

A=4,B=−4,C =4;B2 −4AC =−48 < 0The graph is an ellipse.

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Example Graphing a Second-Degree Equation

4x2 −4xy+ 4y2 +6x−12y−3=0

Rewrite the equation as a quadratic in y.

4y2 + −4xy−12y( ) + 4x2 +6x−3( ) =0

Then solve for y using the quadratic formula.

y=4x+12± −4xy−12y( )

2−16 4x2 +6x−3( )

8or

y1 =x+3+ 12−3x2

2 and y2 =

x+3− 12−3x2

2

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Example Graphing a Second-Degree Equation

The graph of this ellipse is shown.

Notice the axis is not parallel to either axis

(because of the –4xy term).

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Translation-of-Axes Formulas

The coordinates (x, y) and ( ′x , ′y ) based on parallel

sets of axes are related by either of the following

translations formulas:′x =x+ h and ′y =y+ k or x= ′x −h and y= ′y −k.

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Example Translation Formula

Prove that 9x2 + 4y2 +18x−16y−11=0 is the equation

of an ellipse. Translate the coordinate axes so that the

origin is at the center of this ellipse.

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Example Translation Formula

Complete the square for both the x and y.

9x2 +18x+ 4y2 −16y=11

9(x2 + 2x+1) + 4(y2 −4y+ 4) =11+9 +16

9(x+1)2 + 4(y−2)2 =36

(x+1)2

4+(y−2)2

9=1

Prove that 9x2 + 4y2 +18x−16y−11=0 is the equation

of an ellipse. Translate the coordinate axes so that the

origin is at the center of this ellipse.

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Example Translation Formula

This is a standard equation of an ellipse.

If we let ′x =x+1 and ′y =y−2,then the equation of the ellipse becomes

′x( )2

4+( ′y )2

9=1.

Prove that 9x2 + 4y2 +18x−16y−11=0 is the equation

of an ellipse. Translate the coordinate axes so that the

origin is at the center of this ellipse.

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Rotation-of-Axes Formulas

The coordinates (x, y) and ( ′x , ′y ) based on rotated

sets of axes are related by either of the following

rotation formulas:′x =xcosα + ysinα and ′y =−xsinα + ycosα, or

x= ′x cosα − ′y sinα and y= ′x sinα + ′y cosα,where α, 0 <α <π / 2, is the angle of rotation.

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Rotation of Cartesian Coordinate Axes

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Example Rotation of Axes

Prove that 2xy −25=0 is the equation of a hyperbola

by rotating the coordinate axes through an angle α =π / 4.

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Example Rotation of Axes

The rotation equations are

x = ′x cosπ / 4− ′y sinπ / 4 and

y= ′x sinπ / 4 + ′y cosπ / 4

x=′x − ′y

2 and y=

′x + ′y

2.

Prove that 2xy −25=0 is the equation of a hyperbola

by rotating the coordinate axes through an angle α =π / 4.

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Example Rotation of Axes

The equation 2xy −25=0 becomes

2′x − ′y

2

⎛

⎝⎜⎞

⎠⎟′x + ′y

2

⎛

⎝⎜⎞

⎠⎟−25=0

′x( )2− ′y( )

2=25

′x( )2

25−

′y( )2

25=1

Prove that 2xy −25=0 is the equation of a hyperbola

by rotating the coordinate axes through an angle α =π / 4.

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Coefficients for a Conic in a Rotated System

If we apply the rotation formulas to the general second-

degree equation in x and y, we obtain a second-degree

equation in ′x and ′y of the form

′A ′x 2 + ′B ′x ′y + ′C ′y 2 + ′D ′x + ′E ′y + ′F =0,where the coefficients are

′A =Acos2α + Bcosα sinα +Csin2α′B =Bcos2α + (C−A)sin2α′C =Ccos2α −Bcosα sinα + Asin2α′D =Dcosα + Esinα′E =Ecosα −Dsinα ′F =F

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Angle of Rotation to Eliminate the Cross-Product Term

If B ≠0, an angle of rotation α such that

cot2α =A−C

B and 0 <α <π / 2 will

eliminate the term ′B x'y' from the

second degree equation in the rotated

′x ′y coordinate system.

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Discriminant Test

The second-degree equation

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

g a hyperbola if B2 −4AC > 0,

g a parabola if B2 −4AC =0,

g an ellipse if B2 −4AC < 0,except for degenerate cases.

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Conics and the EquationAx2 + Bxy + Cy2 + Dx + Ey + F = 0

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Quick Review

Assume 0 ≤α<π /2.1. Given cot2α =3/ 4, find cos2α.

2. Given cot2α =1/ 3, find cos2α.3. Given cot2α =1, find α.

4. Given cot2α =1/ 3, find α.5. Given cot2α =3/ 4, find cosα.

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Quick Review Solutions

Assume 0 ≤α<π /2.1. Given cot2α =3/ 4, find cos2α. 3/5

2. Given cot2α =1/ 3, find cos2α. 1/23. Given cot2α =1, find α. π /8

4. Given cot2α =1/ 3, find α. π /6

5. Given cot2α =3/ 4, find cosα. 2/ 5