Rotation and Systems of Quadratic Equations

Conic sections may also be classified by their eccentricities:

(a)  Circle or ellipse:  0 < e < 1
(b)  Parabola:  e = 1
(c)  hyperbola:  e> 1
 

Conics that are not parallel to the x- or y-axes:

General form of the equation:  Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

The xy term in the equation causes the conic to be non-parallel to the axes.
To graph the conic, rotate the axes so the conic is parallel to one of the axes.
The new equation will have the form A' (x')2 + B' x' y' + C' (y')2 + D' x' + E' y' + F' = 0
To find the new coordinates, use
    tan 2q =      B          and       x = x' cos q - y' sin q
                    A - C                    y = x' sin q + y' cos q

Example:
Rotate the axes and graph 7x2 - 6\/3 xy + 13y2 - 16 = 0

tan 2q   -6\/3   =  -6\/3   =    \/3
              7 - 13         -6

      2q = p/3
        q = p/6

x = x' cos p/6 - y' sin p/6                    y = x' sin p/6 + y' cos p/6
   = x' (\/3 /2) - y' (1/2)                        y = x' (1/2) + y' (\/3 /2)

Plug these into the equation 7x2 - 6\/3 xy + 13y2 - 16 = 0,
7(x' (\/3 /2) - y' (1/2))2 - 6\/3 (x' (\/3 /2) - y' (1/2))(x' (1/2) + y' (\/3 /2)) + 13(x' (1/2) + y' (\/3 /2) )2 - 16 = 0
Multiplying,
7((3/4)(x' )2 - (2\/3 /4)x' y' + (1/4)(y' )2 - 6\/3((\/3 /4)(x' )2 + (3/4)x' y' - (1/4)x' y' - (\/3 /4)(y' )2 + 13((1/4)(x' )2
             + (2\/3 /4)x' y' + (3/4)(y')2 - 16 = 0
Combining like terms and factoring out the variables,
(21/4  -  18/4  +  13/4)(x' )2  +  (-14\/3 /4  -  12\/3 /4  +  26\/3 /4)x' y'  +  (7/4  +  18/4  +  39/4)(y' )2  -  16  =  0
Simplifying,
4(x' )2 + 16(y' )2 = 16
Dividing by  16,
(x' )2 + (y' )2 = 1
 4          1

This is an ellipse.
Rotate the axes p/6 radians and draw the ellipse on the new axes.


 

Rotation Invariants:

(Things that don't change when the axes are rotated)
(a)  F = F'
(b)  A + C = A' + C'
(c)  B2 - 4AC = (B')2 - 4 A' C'

This means (c) can be used to define the conic before rotation is done.
(a)  Ellipse or Circle:  B2 - 4AC < 0
(b)  Parabola:  B2 - 4AC = 0
(c)  Hyperbola:  B2 - 4AC  > 0

These can be derived from the properties of AC on the general equations of conics that are parallel to the x or y axes.

Example:
Classify the graph of each of the following equations:
1.  2xy - 4 = 0
2.  3x2 - 4xy + 3y2 - 5x = 0
3.  x2 - 6xy + 9y2 - 3y + 5 = 0
4.  3x2 + 7xy + 4y2 - 8 = 0

   Answers:
   1.  B2 - 4AC = 4 - 4(0)(0) = 4 > 0                     Hyperbola
   2.  B2 - 4AC = 16 - 4(3)(3) = 16 - 36 = -20        Ellipse
   3.  B2 - 4AC = 36 - 4(1)(9) = 36 - 36 = 0           Parabola
   4.  B2 - 4AC = 49 - 4(3)(4) = 49 - 48 = 1 > 0     Hyperbola
 

Solving quadratic systems:

(1)  -x2 y2 8x + 20y - 7 = 0
        x2 + 9y2 + 8x + 4y + 7 = 0

       Using elimination, add the two equations,
       8y2 + 24y = 0
       y(8y + 24) = 0
      y = 0  or  8y + 24 = 0
                              8y = -24
                                y = -3
      Substituting back into one of the original equations,
      When y = 0, -x2 02 8x + 200 - 7 = 0
                          0 = x2 + 8x + 7
                          0 = (x + 7)(x + 1)
                          x = -7, -1
      When y = -3, -x2 (-3)2 8x + 20(-3) - 7 = 0
                            -x2 - 9 - 8x - 60 - 7 = 0
                            -x2 - 8x - 76 = 0
                            0 = x2 + 8x + 76
                           The solutions are imaginary because the discriminant is negative.

     So the solutions are (-7,0) and (-1,0).

(2)  x2 + 4y2 2x 8y + 1 = 0
      2x 4y 1 = 0

      Using substitution, solve the second equation for y:    4y = 2x + 1
                                                                                          y = (1/2)x - 1/4
       Substituting this value for y into the first equation,
         x2 + 4((1/2)x - 1/4)2 2x 8((1/2)x - 1/4) + 1 = 0
        Squaring the first parentheses,
        x2 + 4((1/4)x2 - (1/4)x + 1/16) 2x 8((1/2)x - 1/4) + 1 = 0
       Simplifying,
       x2 + x2 - x + 1/4 2x 4x - 1/4 + 1 = 0
       2x2 - 7x  + 1 = 0
       Using the quadratic formula,
       x = 7 ± \/(49 - 4(2)(1))  =  7 ± \/41
                       4                            4

      Substitute these back into  y = (1/2)x - 1/4
        y = (1/2)( 7 + \/41)  - 1/4  =  (7 + \/41)  - 2  =  5 + \/41
                             4                            8           8           8
        y = (1/2)( 7 - \/41)  - 1/4  =  (7 - \/41)  - 2  =  5 - \/41
                             4                            8           8           8

      So the solutions are  (7 + \/41 , 5 + \/41 ) ,   (7 - \/41 , 5 - \/41 )
                                           4            8                   4            8


Problems (there are no problems here yet)


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