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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.
Slide 8.4 - 3 Copyright © 2011 Pearson, Inc.
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
Slide 8.4 - 4 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 5 Copyright © 2011 Pearson, Inc.
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
Slide 8.4 - 6 Copyright © 2011 Pearson, Inc.
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).
Slide 8.4 - 7 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 8 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 9 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 10 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 11 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 12 Copyright © 2011 Pearson, Inc.
Rotation of Cartesian Coordinate Axes
Slide 8.4 - 13 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 14 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 15 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 16 Copyright © 2011 Pearson, Inc.
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
Slide 8.4 - 17 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 18 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 19 Copyright © 2011 Pearson, Inc.
Conics and the EquationAx2 + Bxy + Cy2 + Dx + Ey + F = 0
Slide 8.4 - 20 Copyright © 2011 Pearson, Inc.
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.
Slide 8.4 - 21 Copyright © 2011 Pearson, Inc.
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