Translation and Rotation of Axes

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                                         Translation and Rotation of Axes

Consider a seconddegree equation
(1)             Ax 2  Bxy  Cy 2  Dx  Ey  F  0
in x and y. If B  0, we may use translation of axes if necessaryto put the equation in a standard form for
 a parabola or an ellipse or a hyperbola.
Example. Consider the seconddegree equation
(2)             9 x 2  16 y 2  54 x  64 y  127  0
in x and y. We begin by completing the squares in x and y.
        9 x 2  54x   16 y 2  64 y   127.
                               
        9 x 2  6 x  16 y 2  4 y  127.   
        9x   2
                                                
                   6 x  9  16 y 2  4 y  4  127  81  64.
        9( x  3) 2  16( y  2) 2  144.
        ( x  3) 2 ( y  2) 2
                              1.
            16         9
We then use translation of axes to introduce new (u,v)-coordinate axes defined by
(3)                   u = x + 3, v = y - 2
or equivalently by
(4)                   x = u - 3, y = v + 2.
In the new (u,v)-coordinate system the curve here has the equation
                      u2 v2
                            1,
                      16 9
which is an equation of a hyperbola with center at the origin, vertices at (±4,0), foci at
                 
  16  9 ,0   5,0  with eccentricity e  , with the directrices the straight lines
                                                 5
                                                 4
                                    16
defined by the equations u   , and with asymptotes the straight lines defined by the
                                     5
                       3
the equations v   u . In the original (x,y)- coordinate system, the center is at the point
                       4
(-3,2), the vertices are at the points (1,2) and (-7,2), the foci are at the points (2,2) and
                                                                                    16
(-8,2), the directrices are the straight lines defined by the equations x  3  , and the
                                                                                     5
asymptotes are the straight lines defined by the equations (y – 2) =   x  3 .
                                                                             3
                                                                             4
Example. Consider here the second degree equation
(5)             4 x 2  16 x  16 y  32  0
                                                                                               2


in x and y. We begin by completing the square in x.
                             
              4 x 2  4 x  16 y  2  0.
               4x   2
                                  
                          4 x  4  16 y  2  16.
               4  x  2  16   y  1.
                             2



                              x  2 .
                           1
               ( y  1) 
                                       2

                           4
We then use translation of axes to introduce new (u,v) –coordinate axes defined by
(5)            u = x + 2, v = y – 1
or equivalently by
               x = u – 2, y = v + 1.
In the new (u,v) – coordinate system the curve has the equation
                     1
                v  u2,
                     4
which is an equation of a parabola with the vertex at the origin. We now discuss finding
the focus and the directrix of this parabola. Put the above equation in the form
                u 2  4v  4 pv
where p=1. In the (u,v) – coordinate system, the focus of this parabola is the point (0,p)
= (0,1), and the directrix is the straight line defined by the equation v= -p = -1. In the
original (x,y) – coordinate system, the vertex of this parabola is at the point (-2,1), the
focus is at the point (-2,2), and the directrix is the straight line defined by the equation
y=0. We now discuss this in terms of our previous study of a parabola defined by an
equation
                 y  f ( x)  ax 2  bx  c
where a ≠ 0. Here we have
                16 y  4 x 2  16 x  32
or
                      1
                 y  x 2  x  2.
                       4
Let
                      1
                 a  , b  1, c  2.
                      4
The vertex is the point
                  b 4ac  b 2 
                  ,
                  2a                .
                             4a    
Note that
                    b          1
                                 2
                    2a           1
                             2
                                 4
and
                                                                                           3


                                1
                             4   2  12
                 4ac  b2
                            4             1,
                    4a              1
                                 4
                                    4
so the vertex is the point (-2,1). The focus is the point

                 b 4ac  b 2 1 1 
                
                 2a , 4a  4  a        
                                         
                                  
                                  
                    2,1  1  1 
                            4 1
                                  
                                4
                 ( 2,2).
The directrix is the straight line defined by the equation
                      4ac  b 2 1 1         1 1
                 y                 1    0.
                         4a        4 a      4 1
                                                4
_____________________________________________________________________
Consider again the second degree equation
(1)              Ax 2  Bxy  Cy 2  Dx  Ey  F  0
in x and y. Suppose now that B≠0. We discuss here use of rotation of axes to get rid of
the xy-term. Consider a new (u,v) – coordinate system. Let  be an angle from the
positive x-axis to positive u-axis. Let O be the point which is the origin for both
coordinate systems. Then, let P be some given point other than O. Let OP denote the
vector with the initial point O and with terminal point P. Let r denote the length of this
vector OP , and let  be an angle from the positive u-axis to OP . Let (x,y) be the
coordinates of P in the original coordinate system, and let (u,v) be the coordinates of P in
the new coordinate system.

                            y-axis
             v-axis
                                            P
                                     r               u-axis

                                     
                                      
                       O                                                 x-axis
                                                                                                  4


We have that
         x  r  cos   
          r  cos  cos   sin  sin  
          r cos   cos   r sin    sin  
          u  cos   v  sin  
and
         y  r  sin    
          r  sin   cos   cos sin  
          r cos   sin    r sin    cos 
          u  sin    v  cos .
Thus we have the equations
          x  u cos   v sin  
(6)      
          y  u sin    v cos 
for rotation of axes. We now make the substitutions (6) in the equation (1). We get:

         A  u cos   v sin    B  u cos   v sin    u sin    v cos 
                                    2


          C u sin    v cos   D  u cos   v sin  
                                     2


          E  u sin    v cos   F  0.
            
         A  u 2 cos2    2uv cos sin    v 2 sin 2             
               
          B  u cos sin    uv cos    uv sin    v 2 sin   cos 
                    2                           2                   2
                                                                                        
          C  u  sin    2uv sin   cos   v cos  
                    2   2                                   2           2
                                                                                
          D  u cos   v sin    E  u sin    v cos   F  0.
         A cos    B cos sin    C sin   u
                2                                   2               2


           2 A cos sin    B cos    sin   2C sin   cos uv
                                            2               2


          A sin    B sin   cos   C cos    v
                    2                                   2               2


          D cos   E sin    u   D sin    E cos   v  F  0.

We want to have:
                                                               
       2 A cos sin    B  cos2    sin 2    2C  sin  cos   0.
         C  A  sin2   B  cos2   0.
                        AC
(7)      cot2           .
                         B
                                                                                                  5


                                                                                  rm)
Example. Use a suitable rotation of axes to eliminate the xy - term (the product te in the equation
                34 x 2  24 xy  41y 2  250 y  325  0.
Let
             A  34, B  24, C  41, D  0, E  250, F  325.
Let  be an angle of rotation such that
                             A - C 34  41
                   
              cot(2 )                    .
                              B      24
Hence
                              7
                   
              cot(2 )          .
                              24
Note that
                  (7) 2  (24) 2  625  25.
Suppose we take
                             7
                  
              cos(2 )          .
                             25
Then
                             24
              sin(2 )  -      .
                             25
Thus we may take
           3
                 2  2.
             2
Hence
          3
                  .
           4
Therefore
                                             7
                                         1
                                
                        1  cos(2 )         25  9  3
            sin( )                
                             2             2     25 5
and
                                            7
                                       1
                              
                     1  cos(2 )            25   16   4 .
            
         cos( )                 
                          2               2        25     5
Then our equations for rotation of axes are
                                      4     3
                    
         x  u cos( )  v sin( )   5 u  5 v
        
        
         y  u sin( )  v cos( )  3 u  4 v.
                               
        
                                    5     5

We now make the substitutions in the given equation.
                                                                                              6


                     16     24       9          12        7  12 
               34   u 2  uv  v 2   24    u 2  uv  v 2 
                     25     25      25          25       25  25 
                       9       24     16           3   4 
                41   u 2  uv  v 2   250   u  v   325  0.
                       25      25     25           5   5 
               625 2            1250 2
                     u  0uv        v  150u  200v  325  0.
                25               25
               25u 2  50v 2  150u  200v  325  0.
               u 2  2v 2  6u  8v  13  0.
                u   2
                                               
                          6u  2  v 2  4v  13  0.
                u   2
                                                   
                          6u  9  2  v 2  4v  4  13  9  8.
                u  32  2  v  22  4.
                u  32  v  22  1.
                4           2
Next weuse translation of axes to introduce new (s, t) - coordinate axes defined by
               s  u  3, t  v - 2
or equivalently by
             u  s - 3, v  t  2.
In the new (s, t) - coordinatesystem the curvehas the equation
              s2 t 2
                  1,
              4 2
which is an equation of an ellipse with center at the origin, verticesat the points  2,0, minor axis

                                                                             
with endpoints at the points (0, 2 ), foci at the points  4 - 2 ,0   2 ,0 , eccentrici e 
                                                                                         ty
                                                                                                     2
                                                                                                      2
                                                                                                        ,

                                                                      4
and directrices the straight lines defined by the equations s        . In the (u, v) - coordinatesystem,
                                                                     2
the center of the ellipse has coordinates (-3,2), the vertices have coordinates (-1,2)or (-5,2), the
endpoints of the minor axis have coordinates (-3,2 2 ) or (-3,2 2 ) , the foci have coordinates
                                                                                                            4
(-3  2 ,2) or (-3 - 2 ,2), and the directrices are the straight lines defined by the equation u  3 
                                                                                                             2
                                 4
or the equation u  3     .
                          2
Return now to the original (x, y) - coordinatesystem.The coordinates of the center of the ellipse are given
by
                                                                           7


             4     3       4        3       6
       x   5 u  5 v   5 (3)  5  2  5
      
      
       y  3 u  4 v  3 (3)  4  2   17 .
      
           5     5     5        5          5
                                ex
The coordinates of the first vert are given by
              4          3         2
        x   5  (1)  5  2   5
       
       
        y  3  (1)  4  2   11 ,
       
            5          5          5
                                       ex
and the coordinates of the other vert are given by
              4          3      14
        x   5  (5)  5  2  5
       
       
        y  3  (5)  4  2   23 .
       
            5          5         5
The coordinates of the first endpoint of the minor axis are given by
              4          3
                                
        x    (3)   2  2  
              5          5
                                      6 3 2
                                      5
                                            
                                            5
       
       
            5          5
                                       
        y  3  (3)  4  2  2   17  4 2 ,
                                       5      5
and the coordinates of the other endpoint of the minor axis are given by
             4          3
                            
       x    (3)   2  2  
             5          5
                                        
                                     6 3 2
                                     5    5
      
      
           5          5
                                   
       y  3  (3)  4  2  2   17  4 2 .
                                      5     5
                                                          8




                                 y

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•   •   •    • •   •   •   • • • •    •   •   •   •   •   x
            -5                 •              5
                               •
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                            -7 •           6 17 
                                 Center   , 
                                          5 5 


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Jun Wang Jun Wang Dr
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