Reduction of the general equation of a conic section and classification

• Purpose
• Reduction in the case of delta not zero.
• The translation
• The rotation
• Example
• Reduction with delta = zero.
• Rotation
• Translation
• Example:
• Classification of all affine conic sections

Purpose

In this chapter, we'll work with orthonormal axes. We'll show how we can simplify the general equation of a conic section by means of a translation and a rotation. Further, we assume that the conic section is not degenerated.

Reduction in the case of delta not zero.

We start from the general equation of a conic section.

 
 F(x,y,1) =
 
 a x2 + 2 b" x y + a' y2 + 2 b' x + 2 b y + a" = 0

The translation

We start with translation of the axes to the point (x_o,y_o). The transformation formulas are

 
 x = x' + xo
 y = y' + yo

In the new system of axes the equation is

 
 
 a (x' + xo)2 + 2 b" (x' + xo) (y' + yo) + a' (y' + yo)2
 
  + 2 b' (x' + xo) + 2 b (y' + yo) + a" = 0
 
<=>
 a x'2 + 2 b" x' y' + a' y'2 + 2(a xo + b" yo + b')x'
 
 + 2( b" xo + a' yo + b)y'
 
 + a xo2 + 2 b" xo yo + a' yo2 + 2 b' xo + 2 b yo + a" = 0

To simplify this, we choose x_o and y_o such that

 
 / a xo + b" yo + b' = 0
 \ b" xo + a' yo + b = 0
 
<=>
 xo and yo is a solution of the system
 
 / Fx' (x,y,z) = 0
 \ Fy' (x,y,z) = 0

Since delta is not zero, this system has always one unique solution.

With these special values of x_o and y_o, the equation of the conic section becomes

 
 a x'2 + 2 b" x' y' + a' y'2 +
 + a xo + 2 b" xo yo + a' yo + 2 b' xo + 2 b yo + a" = 0
 
<=>
 a x'2 + 2 b" x' y' + a' y'2 + F(xo,yo,1) = 0
 

The rotation

Suppose the translation is done and then the equation of the conic section has the form

 
 
 a x2 + 2 b" x y + a' y2 + a" = 0
 

To reduce this further we use a rather tricky method.
(From a total different approach of the problem, this method is not tricky at all. See later)

Consider the linear transformation of R x R with matrix

 
 [a b"]
 [b" a']

We calculate the eigenvalues and characteristic vectors.
The eigenvalues are the roots of

 
 |a - r b"|
 |b" a' - r| = 0
<=>
 (a - r)(a' - r) - b"b" = 0
<=>
 r2 -(a+a')r + aa' -b"2 = 0
<=>
 2
 r -(a+a')r + delta = 0
 
 

The discriminant of this quadratic equation is never < 0. Say r₁ and r₂ are these real roots. These roots are not 0 because their product is not 0. Now we can calculate the characteristic vectors. We know that each real multiple of such characteristic vector is also a characteristic vector. So, it is possible to calculate a suitable characteristic vector such that this vector is a unit vector.
Denote (cos(t),sin(t)) as coordinates of the characteristic unit vector corresponding with r₁. Then we have

 
 [ a b"][cos(t)] [cos(t)]
 [ b"  a'][sin(t)] = r1 [sin(t)]
 
<=>
 acos(t) + b"sin(t) = r1 cos(t)
 b"cos(t) + a'sin(t) = r1 sin(t)

Now we rotate the coordinate system by an angle = t. The transformation matrix is

 
 [cos(t) -sin(t) 0]
 M = [sin(t) cos(t) 0]
 [ 0 0 1]

The new cubic matrix C1 of the conic section is given by the formula

 
 T
 C1 = M C M
<=>
 [cos(t) sin(t) 0][ a b" 0 ][cos(t) -sin(t) 0]
 [-sin(t) cos(t)  0][ b" a' 0 ][sin(t) cos(t) 0]
 [ 0 0 1][ 0 0 a"][ 0 0 1]
<=>
 
 [cos(t) sin(t) 0][acos(t) + b"sin(t) -asin(t)+b"cos(t) 0 ]
 [-sin(t) cos(t) 0][b"cos(t) + a'sin(t) -b"sin(t)+a'cos(t) 0 ]
 [ 0 0 1][ 0 0 a"]
<=>
 [cos(t) sin(t) 0][r1 cos(t) -asin(t)+b"cos(t) 0 ]
 [-sin(t) cos(t) 0][r1 sin(t) -b"sin(t)+a'cos(t) 0 ]
 [ 0  0 1][ 0 0 a"]
<=>
 [r1 0 (*) 0 ]
 [0 r2(*) 0 ]
 [0 0 a"]

The elements marked by (*) are rather difficult to calculate. Here you see how it is done:

 
 -a cos(t) sin(t) + b" cos2 (t) - b" sin2 (t) + a' sin(t) cos(t)
 
 = - sin(t) (a cos(t) + b" sin(t) ) + cos(t) (b" cos(t) + a' sin(t) )
 
 = - sin(t) r1 cos(t) + cos(t) r1 sin(t)
 
 = 0
and
 a sin2 (t) - b" sin(t) cos(t) - b" sin(t) cos(t) + a' cos2 (t)
 
 = a(1 - cos2 (t)) - b"sin(t)cos(t) - b"sin(t) cos(t) + a'(1 - sin2 (t) )
 
 = a + a' - (a cos2 (t) + b" sin(t)cos(t) + b"sin(t)cos(t) + a' sin2 (t))
 
 = a + a' - (cos(t)(a cos(t) + b"sin(t))+ sin(t)(b"cos(t) + a'sin(t)))
 
 = a + a' - r1
 
 = r1+ r2 - r1
 
 = r2

The equation of the conic section in the new coordinate system is

 
 r1 x'2 + r2 y'2 + a" = 0

This is the reduced equation of the conic section.

Example

Take the conic section with equation

 
  3 x2 + 2xy + 3 y2 -32 y + 92 = 0

We start with translation of the axes to the point (x_o,y_o).

 
 xo and yo is a solution of the system
 
 / Fx' (x,y,z) = 0
 \ Fy' (x,y,z) = 0
<=>
 / 6 x + 2 y = 0
 \ 2 x + 6 y = 32
<=>
 x = -2 and y = 6

After the translation the equation becomes

 
 a x'2 + 2 b" x' y' + a' y'2 + F(xo,yo,1) = 0
 
<=>
 3 x'2 + 2x'y' + 3 y'2 - 4 = 0
 

This is the equation of the conic section after the translation to (-2,4). Now we start with the rotation from the equation

 
 3 x2 + 2xy + 3 y2 - 4 = 0

The eigenvalues are the roots of

 
 |a - r b"|
 |b" a' - r| = 0
<=>
 |3-r 1|
 | 1 3-r| = 0
<=>
 r2 - 6 r + 8 = 0
<=>
 r1 = 4 and r2 = 2
 

The equation of the conic section after the rotation is

 
 
 r1 x'2 + r2 y'2 + a" = 0
 
<=>
 4 x'2 + 2 y'2 - 4 = 0
 
<=>
 x'2 y'2
 ----- + ----- = 1
 1  2
and this is an ellipse

If you want to calculate the rotation angle t :

 
 [ a b"][cos(t)] [cos(t)]
 [ b" a'][sin(t)] = r1 [sin(t)]
 
<=>
 acos(t) + b"sin(t) = r1cos(t)
 b"cos(t) + a'sin(t) = r1sin(t)
<=>
 3 cos(t) + sin(t) = 4 cos(t)
 cos(t) + 3 sin(t) = 4 sin(t)
<=>
 cos(t) - sin(t) = 0
<=>
 tan(t) = 1
 
 choose t = pi/4

Reduction with delta = zero.

We start from the general equation of a conic section.

 
 F(x,y,1) =
 
  a x2 + 2 b" x y + a' y2 + 2 b' x + 2 b y + a" = 0

Rotation

Consider again the linear transformation of R x R with matrix

 
 [a b"]
 [b" a']

We calculate the eigenvalues and characteristic vectors. The eigenvalues are the roots of

 
 |a - r b"|
 |b" a' - r| = 0
<=>
 (a - r)(a' - r) - b"b" = 0
<=>
 r2 -(a+a')r + aa' -b"2 = 0
<=>
 r2 -(a+a')r + delta = 0
<=>
 r2 -(a+a')r = 0
<=>
 r1 = 0 and r2 = a+a'

Denote (cos(t),sin(t)) as coordinates of the characteristic unit vector corresponding with r₁=0. Then we have

 
 [ a b"][cos(t)] [cos(t)]
 [ b" a'][sin(t)] = r1 [sin(t)]
 
<=>
 acos(t) + b"sin(t) = 0
 b"cos(t) + a'sin(t) = 0

Now we rotate the coordinate system by an angle = t. The transformation matrix is

 
 [cos(t) -sin(t) 0]
 M = [sin(t) cos(t) 0]
 [ 0 0 1]

The new cubic matrix C1 of the conic section is given by the formula

 
 T
 C1 = M C M

In the same way as in the case delta not zero, you can show that

 
 [0 0 * ]
 C1 = [0 * * ]
 [* * * ]

The * stand for real numbers.
The equation of the conic section in the new coordinate system is of the form

 
 
 a'y2 + 2 b' x + 2 b y + a" = 0

From this form we start with a translation

Translation

The equation of the conic section is

 
 a'y2 + 2 b' x + 2 b y + a" = 0

If you calculate DELTA, you'll find

 
 
 DELTA = - a' b'2

Since the conic section is not degenerated, a' and b' are not zero.
We translate the axes to the point (x_o,y_o). The transformation formulas are

 
 x = x' + xo
 y = y' + yo

In the new system of axes the equation is

 
 a'(y' + yo)2 + 2 b' (x' + xo) + 2 b (y' + yo) + a" = 0
<=>
 a' y'2 + 2 b'x' + 2(a' yo + b)y' + a' yo 2 + 2 b' xo + 2 b yo + a" = 0

Because a' and b' are not zero, we can always choose x_o and y_o such that

 
 a' yo + b = 0
and
 a' yo 2 + 2 b' xo + 2 b yo + a" = 0

then the reduced equation is

 
 a' y'2 + 2 b'x' = 0

This is an equation of a parabola.

Example:

We start with the equation

 
 9 x2 -24 xy + 16 y2 -2x + 4y + 5 = 0

delta = 0.
One of the eigenvalues is r₁ = 0.
Denote (cos(t),sin(t)) as coordinates of the unit characteristic vector corresponding with r₁=0. Then we have

 
 [ a b"][cos(t)] [cos(t)]
 [ b" a'][sin(t)] = r1 [sin(t)]
 
<=>
  acos(t) + b"sin(t) = 0
 b"cos(t) + a'sin(t) = 0
<=>
 9 cos(t) - 12 sin(t) = 0
 -12 cos(t) + 16 sin(t) = 0
then
 tan(t) = 3/4 ; cos(t)= 4/5 ; sin(t) = 3/5
 
 [ 4/5 -3/5 0]
 M = [ 3/5 4/5  0]
 [ 0 0 1]
we find
 
 T [ 0 0 2/5 ]
 C1 = M C M = [ 0 25 11/5]
 [2/5 11/5 5 ]
 
The equation after the rotation is
 
 2
 25 y + 4/5 x + 22/5 y + 5 = 0
 
Now the translation
 x = x' + xo
 y = y' + yo
 
 4 22
 25 (y'+ yo)2 + - (x'+ xo) + -- (y'+ yo) + 5 = 0
 5 5
<=>
 25y'2 + (4/5)x' + (50yo + 22/5)y' + 25yo2 + (4/5)xo + (22/5)yo + 5=0
 
We choose xo and yo such that
 
 (50yo + 22/5) = 0 and 25yo2 + (4/5)xo + (22/5)yo + 5=0
 
We find
 751 11
 xo = - ---- yo = - ----
 125 125
 
The reduced equation is
 
 25 y2 + (4/5) x = 0
 

Classification of all affine conic sections

Take a arbitrary conic section.

 
 a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2 = 0

The ideal points of that conic section are the solutions of

 
 /
 |a x2 + 2 b" x y + a' y2 + 2 b' x z + 2 b y z + a" z2 = 0
 |
 \ z = 0
 
<=>
 /
 | a x2 + 2 b" x y + a' y2 = 0
 |
 \ z = 0

We see that the ideal points of the given conic section are the same points as these of the lines

 
 a x2 + 2 b" x y + a' y2 = 0

From the theory about degenerated affine conic sections, we know the relationship between the nature and number of the ideal points and delta. So, about the given general conic section, we can say:

• if delta > 0
  The conic section has two conjugate imaginary ideal points.
  Since an ellipse has two conjugate imaginary ideal points, we generalize and say that all conic sections with delta > 0 are ellipses.
• if delta = 0
  The conic section has two coinciding ideal points.
  Since a parabola has two coinciding ideal points, we generalize and say that all conic sections with delta = 0 are parabolas.
• if delta < 0
  The conic section has two different ideal points.
  Since a hyperbola has two differente ideal points, we generalize and say that all conic sections with delta < 0 are hyperbolas .