Download Translation and rotation (optional)

Summary 6
Translation and rotation
TRANSLATION:
If B=0, the general quadratic equation can be transformed to one of the standard
forms of a conic section. This can be accomplished by using the translation
equations x=x'+h and y=y'+k (x'=x-h and y'=y-k).
Replacing x by x - h in the standard form produces a horizontal shift (phase
shift) of the graph, h units to the right if h>0 or h units to the left if h<0.
Replacing y by y - k produces a vertical shift of the graph, k units up if k>0 or k
units down if k<0.
Method I. Use completing the square, let x - h = x' and y - k = y'.
Method II.
1) In the original equation substitute x = x' + h and y = y' + k.
2) Find the value of h and k that will eliminate the x-term and the y-term in the
equation
3) Find the desired equation by using the values found for h and k.
ROTATION:
If B≠0 we need to use a rotation to transform the general equation equation to
one of the standard forms.  is the angle or rotation.
B
Step 1: Let tan 2 
(if A  C or let   45 if A  C)
AC
Step 2. Using trigonometric relations and the triangle below find cos 2
which has to agree in sign with tan 2
B
2
A-C
Step 3. Use the formulas
1  cos 2
1  cos 2
and cos  
to find sin  and cos 
2
2
sin 
Note : m  tan  
is the slope of x 
cos 
 x  x cos   y  sin 
Step 4. Let 
 y  x sin   y  cos 
sin  
Step 5. Substitution of the results in step 4 into the original equation
eliminates the xy term
B
sin 
Step 6. Using Use either tan 2 
to find the
or tan  
AC
cos 
angle  of rotation
Step 7. Graph in the x'-y' coordinate system.
1
Translation
I.
Translation: a point P has coordinates (x, y) with respect to the x-y system. If a new
system (x, y) is obtained by translated the system to a new origin with coordinates (h, k)
then the coordinates of P with respect to the new system is given by the ordered pair (x-h,
y-k).
y
y
 x  x  h

 y  y  k
 x  x  h
and 
 y  y  k
P
h
x
k
x
Example: Find the values of h and k that will eliminate the x and y terms from the equation
25x 2  9 y 2  100 x  54 y  44  0 . Use the necessary translation.
Let x  x  h and y  y   k
25x 2  9y 2  100x  54y  44  0
 25x   h 2  9y   k 2  100x   h   54y   k   44  0


 25 x 2  2hx   h 2  9( y  2  2ky   k 2 )  100x   100h  54y   54k  44  0
 25x 2  9y  2  (50h  100)x  (18k  54)y   25h 2  9k 2  100h  54k  44  0
50h  100  0  h  2
x  x  2  x  x  2
 To eliminate x  and y  let 



 18k  54  0
k   3  y  y   3  y   y  3
The resulting equation is
25x  2  9y  2  25( 4)  9(9)  100( 2)  54( 3)  44  0
25x  2  9y  2  100  81  200  162  44
25x  2  9y  2  225

x 2 y  2

 1 (an ellipse )
9
25
( x  2) 2 y  32
or

1
9
25
2
Example: Find the values of h and k that will eliminate the x and y terms from the equation xy2x+3y=5
Let x  x   h and y  y   k
xy  2 x  3 y  5   x   h  y   k   2 x   h   3 y   k   5
 x y   hy   kx   hk  2 x   2h  3 y   3k  5
 x y   k  2x   h  3 y   hk  2h  3k  5
k  2  0  k  2
 x  x  3
 x  x  3







h  3  0
h  3  y  y   2  y   y  2
x y   (3)(2)  2(3)  3(2)  5
x y   1
Rotation:
Coordinates- Rotation
II.
Rotation of axis.
Theorem: If a system x-y is rotated counterclockwise through and angle  to obtain the
new system x-y, then the unit vectors in the xand yaxis of the new system are given by
 i  i cos  j sin 
 i  i cos  j sin 
and 

 j  i sin   j cos
 j  i sin   j cos
y
y

x

X
Theorem: The coordinates (x, y) of a point P in the rotated system x - y are given by
 x  x  cos   y  sin 
 x   x cos   y sin 
and 

 y  x  sin   y  cos 
 y    x sin   y cos 
3
Example 2 - Rotate the axis to eliminate the xy term. Sketch, showing both axis.
x2 - 2xy + y2 -1 =0.
Step1. A  C  1    45
Step 2 :
Step 3 : sin  
Step 4 : x 
x
2
1
2

and cos  
y
2

x  y 
2
1
2
and 
x
2

y
2

x  y 
2
2
2
 x  y  
 x   y   x   y    x   y  
Step 5 : x  2 xy  y  1  0  
  2


 1
2 
2 
2  
2 


1
2
1
 x  2  2 x y   y  2  x  2  y  2  x  2  2 x y   y  2  1
2
2
2
1
1
 4 y2  2  y2   y  
 two parallel lines
2
2
2
2

 
 

Y
Y’
X’
x
4
Example 2 - Rotate the axis to eliminate the xy term. Sketch, showing both axis.
5x2 - 4xy + 8y2 -36=0
B
4
4


AC 58 3
3
Step 2 : cos 2 
5
Step1. tan 2 
3
1  
1  cos 2
5  2  1
Step 3 : sin  

2
2
10
5
3
1  
1  cos 2
5  4  2
cos  

2
2
10
5
2 x y  2 x  y 
x 2 y  x  2 y 
Step 4 : x 


and 


5
5
5
5
5
5
2
2
 2 x  y  
 2 x   y   x   2 y    x   2 y  
Step 5 : 5 x  4 xy  8 y  36  0  5
  4

  8
  36
5 
5 
5  
5 


5
4
8
 4 x  2  4 x y   y  2  2 x  2  3 x y   2 y  2  x  2  4 x y   4 y  2  36
5
5
5
2
2

 
 20 x  2  45 y  2  180 
 
x2 y 2

 1  an ellipse
9
4
1
tan  
sin 
5 1


2
cos 
2
5
Y
Y’
x’
x
5

6