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25/05/2018, 15:48

Rotation of 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 ...

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 [link].

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 [link]. 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.

Degenerate conic sections

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.

A x 2 +Bxy+C y 2 +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.

Conic Sections Example
ellipse 4 x 2 +9 y 2 =1
circle 4 x 2 +4 y 2 =1
hyperbola 4 x 2 −9 y 2 =1
parabola 4 x 2 =9y or 4 y 2 =9x
one line 4x+9y=1
intersecting lines ( x−4 )( y+4 )=0
parallel lines ( x−4 )( x−9 )=0
a point 4 x 2 +4 y 2 =0
no graph 4 x 2 +4 y 2 = − 1
A General Note
General Form of Conic Sections

A nondegenerate conic section has the general form

A x 2 +Bxy+C y 2 +Dx+Ey+F=0

where  A,B, and  C  are not all zero.

[link] 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.

ellipse A x 2 +C y 2 +Dx+Ey+F=0, A≠C and AC>0
circle A x 2 +C y 2 +Dx+Ey+F=0, A=C
hyperbola A x 2 −C y 2 +Dx+Ey+F=0 or −A x 2 +C y 2 +Dx+Ey+F=0, where  A  and  C  are positive
parabola A x 2 +Dx+Ey+F=0 or C y 2 +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, A x 2 +Bxy+C y 2 +Dx+Ey+F=0.
  2. Identify the values of  A  and  C  from the general form.
    1. If  A  and  C  are nonzero, have the same sign, and are not equal to each other, then the graph is an ellipse.
    2. If  A  and  C  are equal and nonzero and have the same sign, then the graph is a circle.
    3. If  A  and  C  are nonzero and have opposite signs, then the graph is a hyperbola.
    4. If either  A  or  C  is zero, then the graph is a parabola.
Identifying a Conic from Its General Form

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

  1. 4 x 2 −9 y 2 +36x+36y−125=0
  2. 9 y 2 +16x+36y−10=0
  3. 3 x 2 +3 y 2 −2x−6y−4=0
  4. −25 x 2 −4 y 2 +100x+16y+20=0
  1. Rewriting the general form, we have

    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, we have

    A=0  and  C=9.  We can determine that the equation is a parabola, since  A  is zero.

  3. Rewriting the general form, we have

    A=3  and  C=3.  Because  A=C, the graph of this equation is a circle.

  4. Rewriting the general form, we have

    A=−25  and  C=−4.  Because  AC>0  and  A≠C, the graph of this equation is an ellipse.

Try It

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

  1. 16 y 2 − x 2 +x−4y−9=0
  2. 16 x 2 +4 y 2 +16x+49y−81=0
  1. hyperbola
  2. 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. See [link].

The graph of the rotated ellipse   x 2 + y 2 –xy–15=0

We will find the relationships between  x  and  y  on the Cartesian plane with   x ′   and   y ′   on the new rotated plane. See [link].

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  i  and  j  . The rotated coordinate axes have unit vectors   i ′   and   j ′ . The angle  θ  is known as the angle of rotation. See [link]. We may write the new unit vectors in terms of the original ones.

i ′ =cos θi+sin θj j ′ =−sin θi+cos θj
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= x ′ i ′ + y ′ j ′ u= x ′ (i cos θ+j sin θ)+ y ′ (−i sin θ+j cos θ) Substitute. u=ix' cos θ+jx' sin θ−iy' sin θ+jy' cos θ Distribute. u=ix' cos θ−iy' sin θ+jx' sin θ+jy' cos θ Apply commutative property. u=(x' cos θ−y' sin θ)i+(x' sin θ+y' cos θ)j Factor by grouping.

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

x= x ′ cos θ− y ′ sin θ and y= x ′ sin θ+ y ′ cos θ
A General Note
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= x ′ cos θ− y ′ sin θ

and

y= x ′ sin θ+ y ′ cos θ
How To

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

  1. Find  x  and  y  where  x= x ′ cos θ− y ′ sin θ  and  y= x ′ sin θ+ y ′ cos θ.
  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.
Finding a New Representation of an Equation after Rotating through a Given Angle

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

Find  x  and  y, where  x= x ′ cos θ− y ′ sin θ  and  y= x ′ sin θ+ y ′ cos θ.

Because  θ=45°,

x= x ′ cos( 45° )− y ′ sin( 45° ) x= x ′ ( 1 2 )− y ′ ( 1 2 ) x= x ′ − y ′ 2

and

y= x ′ sin(45°)+ y ′ cos(45°) y= x ′ ( 1 2 )+ y ′ ( 1 2 ) y= x ′ + y ′ 2

Substitute  x= x ′ cosθ− y ′ sinθ  and  y= x ′ sin θ+ y ′ cos θ  into  2 x 2 −xy+2 y 2 −30=0.

2 ( x ′ − y ′ 2 ) 2 −( x ′ − y ′ 2 )( x ′ + y ′ 2 )+2 ( x ′ + y ′ 2 ) 2 −30=0

Simplify.

2 ( x ′ − y ′ )( x ′ − y ′ ) 2 − ( x ′ − y ′ )( x ′ + y ′ ) 2 + 2 ( x ′ + y ′ )( x ′ + y ′ ) 2 −30=0 FOIL method             x ′ 2 −2 x ′ y ′ + y ′ 2 − ( x ′ 2 − y ′ 2 ) 2 + x ′ 2 +2 x ′ y ′ + y ′ 2 −30=0 Combine like terms.                                                              2 x ′ 2 +2 y ′ 2 − ( x ′ 2 − y ′ 2 ) 2 =30 Combine like terms.                                                        2( 2 x ′ 2 +2 y ′ 2 − ( x ′ 2 − y ′ 2 ) 2 )=2(30) Multiply both sides by 2.                                                               4 x ′ 2 +4 y ′ 2 −( x ′ 2 − y ′ 2 )=60 Simplify.                                                                  4 x ′ 2 +4 y ′ 2 − x ′ 2 + y ′ 2 =60 Distribute.                                                                                      3 x ′ 2 60 + 5 y ′ 2 60 = 60 60 Set equal to 1.

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

x ′ 2 20 + y ′ 2 12 =1

This equation is an ellipse. [link] shows the graph.

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  A x 2 +Bxy+C y 2 +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   x ′ y ′   term, by rotating the axes by a measure of  θ  that satisfies

cot( 2θ )= A−C B

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

A x 2 +Bxy+C y 2 +Dx+Ey+F=0

where  A,B, and  C  are not all zero. However, if  B≠0, 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 )= A−C B .

  • 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   x ′ y ′   system, rewrite the equation without the   x ′ y ′   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  sin θ  and  cos θ.
  3. Substitute  sin θ  and  cos θ  into  x= x ′ cos θ− y ′ sin θ  and  y= x ′ sin θ+ y ′ cos θ.
  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.
Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term

Rewrite the equation  8 x 2 −12xy+17 y 2 =20  in the   x ′ y ′   system without an   x ′ y ′   term.

First, we find  cot(2θ).  See [link].

8 x 2 −12xy+17 y 2 =20⇒A=8, B=−12 and C=17                     cot(2θ)= A−C B = 8−17 −12                     cot(2θ)= −9 −12 = 3 4
cot( 2θ )= 3 4 = adjacent opposite

So the hypotenuse is

3 2 + 4 2 = h 2 9+16= h 2 25= h 2 h=5   

Next, we find  sin θ and cos θ.

sin θ= 1−cos(2θ) 2 = 1− 3 5 2 = 5 5 − 3 5 2 = 5−3 5 ⋅ 1 2 = 2 10 = 1 5 sin θ= 1 5 cos θ= 1+cos(2θ) 2 = 1+ 3 5 2 = 5 5 + 3 5 2 = 5+3 5 ⋅ 1 2 = 8 10 = 4 5 cos θ= 2 5

Substitute the values of  sin θ  and  cos θ  into  x= x ′ cos θ− y ′ sin θ  and  y= x ′ sin θ+ y ′ cos θ.

x= x ′ cos θ− y ′ sin θ x= x ′ ( 2 5 )− y ′ ( 1 5 ) x= 2 x ′ − y ′ 5

and

y= x ′ sin θ+ y ′ cos θ y= x ′ ( 1 5 )+ y ′ ( 2 5 ) y= x ′ +2 y ′ 5

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

                                  8 ( 2 x ′ − y ′ 5 ) 2 −12( 2 x ′ − y ′ 5 )( x ′ +2 y ′ 5 )+17 ( x ′ +2 y ′ 5 ) 2 =20      8( (2 x ′ − y ′ )(2 x ′ − y ′ ) 5 )−12( (2 x ′ − y ′ )( x ′ +2 y ′ ) 5 )+17( ( x ′ +2 y ′ )( x ′ +2 y ′ ) 5 )=20        8( 4 x ′ 2 −4 x ′ y ′ + y ′ 2 )−12( 2 x ′ 2 +3 x ′ y ′ −2 y ′ 2 )+17( x ′ 2 +4 x ′ y ′ +4 y ′ 2 )=100 32 x ′ 2 −32 x ′ y ′ +8 y ′ 2 −24 x ′ 2 −36 x ′ y ′ +24 y ′ 2 +17 x ′ 2 +68 x ′ y ′ +68 y ′ 2 =100                                                                                                   25 x ′ 2 +100 y ′ 2 =100                                                                                                   25 100 x ′ 2 + 100 100 y ′ 2 = 100 100  

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

x ′ 2 4 + y ′ 2 1 =1

[link] shows the graph of the ellipse.

Try It

Rewrite the  13 x 2 −6 3 xy+7 y 2 =16  in the   x ′ y ′   system without the   x ′ y ′   term.

x ′ 2 4 + y ′ 2 1 =1

Graphing an Equation That Has No x′y′ Terms

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

x 2 +12xy−4 y 2 =30

First, we find  cot( 2θ ).

x 2 +12xy−4 y 2 =20⇒A=1, B=12,and C=−4
cot(2θ)= A−C B cot(2θ)= 1−(−4) 12 cot(2θ)= 5 12

Because  cot( 2θ )= 5 12 , we can draw a reference triangle as in [link].

cot( 2θ )= 5 12 = adjacent opposite

Thus, the hypotenuse is

5 2 + 12 2 = h 2 25+144= h 2 169= h 2 h=13

Next, we find  sin θ  and  cos θ.  We will use half-angle identities.

sin θ= 1−cos(2θ) 2 = 1− 5 13 2 = 13 13 − 5 13 2 = 8 13 ⋅ 1 2 = 2 13 cos θ= 1+cos(2θ) 2 = 1+ 5 13 2 = 13 13 + 5 13 2 = 18 13 ⋅ 1 2 = 3 13

Now we find  x  and  y. 

x= x ′ cos θ− y ′ sin θ x= x ′ ( 3 13 )− y ′ ( 2 13 ) x= 3 x ′ −2 y ′ 13

and

y= x ′ sin θ+ y ′ cos θ y= x ′ ( 2 13 )+ y ′ ( 3 13 ) y= 2 x ′ +3 y ′ 13

Now we substitute  x= 3 x ′ −2 y ′ 13   and  y= 2 x ′ +3 y ′ 13   into   x 2 +12xy−4 y 2 =30.

                                         ( 3 x ′ −2 y ′ 13 ) 2 +12( 3 x ′ −2 y ′ 13 )( 2 x ′ +3 y ′ 13 )−4 ( 2 x ′ +3 y ′ 13 ) 2 =30                                     ( 1 13 )[ (3 x ′ −2 y ′ ) 2 +12(3 x ′ −2 y ′ )(2 x ′ +3 y ′ )−4 (2 x ′ +3 y ′ ) 2 ]=30  Factor. ( 1 13 )[ 9 x ′ 2 −12 x ′ y ′ +4 y ′ 2 +12( 6 x ′ 2 +5 x ′ y ′ −6 y ′ 2 )−4( 4 x ′ 2 +12 x ′ y ′ +9 y ′ 2 ) ]=30 Multiply.   ( 1 13 )[ 9 x ′ 2 −12 x ′ y ′ +4 y ′ 2 +72 x ′ 2 +60 x ′ y ′ −72 y ′ 2 −16 x ′ 2 −48 x ′ y ′ −36 y ′ 2 ]=30 Distribute.                                                                                                   ( 1 13 )[ 65 x ′ 2 −104 y ′ 2 ]=30 Combine like terms.                                                                                                            65 x ′ 2 −104 y ′ 2 =390 Multiply.                                                                                                                                    x ′ 2 6 − 4 y ′ 2 15 =1  Divide by 390.

[link] shows the graph of the hyperbola   x ′ 2 6 −

Lê Thị Khánh Huyền

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