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REDUCTION OF A GENERAL SECOND DEGREE EQUATION TO ONE OF THE 9 CANONICAL FORMS
Given a second degree equation
1) F(x,y) = ax2 + 2bxy + cy2 + 2dx + 2ey + f = 0
we wish to reduce it to canonical form. It may represent any one of the following 9 conics:
Our equation represents some conic as referred to an x -y coordinate system. See Figure 1. Here
a conic (an ellipse in this case) is shown located at some point in the plane. Figure 1 shows the original x-y coordinate system along with two other coordinate systems – an intermediate x'- y' coordinate system and the xc - yc canonical coordinate system. We wish to know the location and orientation of the canonical coordinate system and the exact equation of the conic as referred to that system. Figure 1 shows the x'- y' coordinate system as a system obtained by rotating the x-y system by θ degrees about its origin. θ represents the rotation required to eliminate the xy term in the original equation as computed from the formula
A rotation of this amount will put its axes parallel to the axes of the canonical system. Upon performing the rotation the equation of the conic in the x'-y' system becomes
2) G(x',y') = a'x'2 + c'y'2 + 2d'x' + 2e'y' + f = 0
where
a' = a cos2 θ + 2 b sin θ cos θ + c sin2 θ
c' = a sin2 θ - 2b sin θ cos θ + c cos2 θ
d' = d cos θ + e sin θ
e' = e cos θ - d sin θ
Let us now assume that the origin of the xc - yc canonical coordinate system is located at coordinates (h, k) of the x'-y' system and let us do a translation to the xc - yc system. The equation of the conic in the xc - yc system would then be given by
3) a'(xc + h)2 + c'(yc + k)2 + 2d'(xc + h) + 2e'(yc + k) + f = 0
or, on removing parenthesis and collecting similar terms,
4) a'xc2 + c'yc2 + 2(a'h + d')xc + 2(c'k + e')yc + f ' = 0
where
f ' = G(h, k) = a'h2 + c'k2 + 2d'h + 2e'k + f
We consider the following five cases:
Case I. Neither a' nor c' of equation 4) are zero. This case corresponds to canonical forms 1 - 5. Taking
h = - d'/a'
k = - e'/c'
we eliminate the terms in xc and yc and obtain the equation
5) a'xc2 + c'yc2 + f ' = 0
If f ' 0 , we can rewrite 5) as
6)
which is one of the canonical forms 1, 2 or 4.
If f ' = 0 , equation 5) can be written as
7)
which corresponds to canonical form 3 or 5.
The origin of the xc - yc canonical coordinate system is at location (h, k) of the x'-y' system where
h = - d'/a'
k = - e'/c' .
The value of f ' is given by
f ' = a'h2 + c'k2 + 2d'h + 2e'k + f
Case II. a' 0, c' = 0, e' 0 in equation 4). This case corresponds to a parabola of the form
x2 - 4py = 0 .
We will translate the x'-y' system to its final position in a two-step process, first translating in the x' direction to eliminate the xc term of equation 4), and then translating in the y' direction. For the first translation let
h = - d'/a'
k = 0
and equation 4) becomes
8) a'x''2 + 2e'y'' + f ' = 0
or, equivalently,
9) a'x''2 + 2e'(y'' + f ' /2e') = 0
where
f ' = a'h2 + 2d'h + f
h = - d'/a' .
Equation 9) is the equation of the conic as referred to an x''- y'' coordinate system with origin at point (- d'/a', 0) of the x'- y' system and axes parallel to the corresponding x'-y' axes.
Now we substitute
xc = x''
yc = y'' + f '/2e'
into equation 9), an act which is equivalent to a translation in the y'' direction by an amount
k' = - f '/2e' , a translation that puts us into the xc - yc canonical coordinate system .
This gives
10) a'xc2 + 2e'yc = 0
or, equivalently,
11) xc2 + 2e'yc/a' = 0
which is the equation of our parabola in canonical form.
Letting
12) p = - e'/2a'
equation 11) can be written as
13) xc2 - 4pyc = 0 ,
the equation of a parabola in canonical form.
The origin of the xc - yc canonical coordinate system is at location (h, k') of the x'-y' system where
h = - d'/a'
k' = - f '/2e'
and
f ' = a'h2 + 2d'h + f .
Case III. a' = 0, c' 0, d' 0 in equation 4). This case corresponds to a parabola of the form
y2 - 4px = 0 .
As in the previous case we will translate the x'- y' system to its final position in a two-step process. We will first translate in the y' direction to eliminate the yc term of equation 4), and then translate in the x' direction. For the first translation let
h = 0
k = - e'/c'
and equation 4) becomes
14) c'y''2 + 2d'x'' + f ' = 0
or, equivalently,
15) c'y''2 + 2d'(x'' + f ' /2d') = 0
where
f ' = c'k2 + 2e'k + f
k = - e'/c' .
Equation 15) is the equation of the conic as referred to an x''- y'' coordinate system with origin at point (0, - e'/c') of the x'- y' system and axes parallel to the corresponding x'-y' axes.
Now we substitute
xc = x'' + f '/2d'
yc = y''
into equation 15), an act which is equivalent to a translation in the x'' direction by an amount
h' = - f '/2d' , a translation that puts us into the xc - yc canonical coordinate system .
This gives
16) c'yc2 + 2d'xc = 0
or, equivalently,
17) yc2 + 2d'yc/c' = 0
which is the equation of our parabola in canonical form.
Letting
18) p = - d'/2c'
equation 11) can be written as
19) yc2 - 4pxc = 0 ,
the equation of a parabola in canonical form.
The origin of the xc - yc canonical coordinate system is at location (h', k) of the x'-y' system where
h' = - f '/2d'
k = - e'/c'
and
f ' = c'k2 + 2e'k + f
Case IV. a' 0, c' = 0, e' = 0 in equation 4). This case corresponds to one of the canonical forms 7, 8 or 9. Taking
h = - d'/a'
k = 0
we obtain the equation
20) a'xc2 + f ' = 0
or equivalently,
21) xc2 + f '/a' = 0
which corresponds to one of the equations 7, 8 or 9.
The origin of the xc - yc canonical coordinate system is at location (h, k) of the x'-y' system where
h = - d'/a'
k = 0
and
f ' = c'k2 + 2e'k + f .
Case V. a' = 0, c' 0, d' = 0 in equation 4). This case corresponds to one of the canonical forms 7, 8 or 9. Taking
h = 0
k = - e'/c'
we obtain the equation
22) c'yc2 + f ' = 0
or equivalently,
23) yc2 + f '/c' = 0
which corresponds to one of the canonical forms 7, 8 or 9.
The origin of the xc - yc canonical coordinate system is at location (h, k) of the x'-y' system where
h = 0
k = - e'/c'
and
f ' = a'h2 + 2d'h + f .
Reference.
Mathematics, Its Content, Methods and Meaning. Vol. I, p. 210 - 213
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