7.2 A classi¯cation algorithm

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There are several possible degenerate cases that can arise from the general

second degree equation. For example x2+y2 = 0 represents the point (0; 0);

x2 + y2 = ¡1 de¯nes the empty set, as does x2 = ¡1 or y2 = ¡1; x2 = 0

de¯nes the line x = 0; (x + y)2 = 0 de¯nes the line x + y = 0; x2 ¡ y2 = 0

de¯nes the lines x ¡ y = 0; x + y = 0; x2 = 1 de¯nes the parallel lines

x = §1; (x + y)2 = 1 likewise de¯nes two parallel lines x + y = §1.

We state without proof a complete classi¯cation 1 of the various cases

1This classi¯cation forms the basis of a computer program which was used to produce

the diagrams in this chapter. I am grateful to Peter Adams for his programming assistance.

142 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

that can possibly arise for the general second degree equation

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0: (7.10)

It turns out to be more convenient to ¯rst perform a suitable translation of

axes, before rotating the axes. Let

¢ =

¯¯¯¯¯¯

a h g

h b f

g f c

¯¯¯¯¯¯

; C = ab ¡ h2; A = bc ¡ f2; B = ca ¡ g2:

If C 6= 0, let

® =

¡

¯¯¯¯

g h

f b

¯¯¯¯

C

; ¯ =

¡

¯¯¯¯

a g

h f

¯¯¯¯

C

: (7.11)

CASE 1. ¢ = 0.

(1.1) C 6= 0. Translate axes to the new origin (®; ¯), where ® and ¯ are

given by equations 7.11:

x = x1 + ®; y = y1 + ¯:

Then equation 7.10 reduces to

ax2

1 + 2hx1y1 + by2

1 = 0:

(a) C > 0: Single point (x; y) = (®; ¯).

(b) C < 0: Two non{parallel lines intersecting in (x; y) = (®; ¯).

The lines are

y ¡ ¯

x ¡ ®

= ¡h § p¡C

b

if b 6= 0;

x = ® and

y ¡ ¯

x ¡ ®

= ¡

a

2h

; if b = 0:

(1.2) C = 0.

(a) h = 0.

(i) a = g = 0.

(A) A > 0: Empty set.

(B) A = 0: Single line y = ¡f=b.

7.2. A CLASSIFICATION ALGORITHM 143

(C) A < 0: Two parallel lines

y = ¡f § p¡A

b

(ii) b = f = 0.

(A) B > 0: Empty set.

(B) B = 0: Single line x = ¡g=a.

(C) B < 0: Two parallel lines

x = ¡g § p¡B

a

(b) h 6= 0.

(i) B > 0: Empty set.

(ii) B = 0: Single line ax + hy = ¡g.

(iii) B < 0: Two parallel lines

ax + hy = ¡g § p¡B:

CASE 2. ¢ 6= 0.

(2.1) C 6= 0. Translate axes to the new origin (®; ¯), where ® and ¯ are

given by equations 7.11:

x = x1 + ®; y = y1 + ¯:

Equation 7.10 becomes

ax2

1 + 2hx1y1 + by2

1 = ¡

¢

C

: (7.12)

CASE 2.1(i) h = 0. Equation 7.12 becomes ax2

1 + by2

1 = ¡¢

C .

(a) C < 0: Hyperbola.

(b) C > 0 and a¢ > 0: Empty set.

(c) C > 0 and a¢ < 0.

(i) a = b: Circle, centre (®; ¯), radius

q

g2+f2¡ac

a .

(ii) a 6= b: Ellipse.

144 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

CASE 2.1(ii) h 6= 0.

Rotate the (x1; y1) axes with the new positive x2{axis in the direction

of

[(b ¡ a + R)=2; ¡h];

where R =

p

(a ¡ b)2 + 4h2.

Then equation 7.12 becomes

¸1x2

2 + ¸2y2

2 = ¡

¢

C

: (7.13)

where

¸1 = (a + b ¡ R)=2; ¸2 = (a + b + R)=2;

Here ¸1¸2 = C.

(a) C < 0: Hyperbola.

Here ¸2 > 0 > ¸1 and equation 7.13 becomes

x2

2

u2 ¡

y2

2

v2 = ¡¢

j¢j

;

where

u =

s

j¢j

C¸1

; v =

s

j¢j

¡C¸2

:

(b) C > 0 and a¢ > 0: Empty set.

(c) C > 0 and a¢ < 0: Ellipse.

Here ¸1; ¸2; a; b have the same sign and ¸1 6= ¸2 and equa-

tion 7.13 becomes

x2

2

u2 +

y2

2

v2 = 1;

where

u =

r

¢

¡C¸1

; v =

r

¢

¡C¸2

:

(2.1) C = 0.

(a) h = 0.

(i) a = 0: Then b 6= 0 and g 6= 0. Parabola with vertex

µ

¡A

2gb

; ¡

f

b

:

7.2. A CLASSIFICATION ALGORITHM 145

Translate axes to (x1; y1) axes:

y2

1 = ¡

2g

b

x1:

(ii) b = 0: Then a 6= 0 and f 6= 0. Parabola with vertex

µ

¡

g

a

; ¡B

2fa

:

Translate axes to (x1; y1) axes:

x2

1 = ¡

2f

a

y1:

(b) h 6= 0: Parabola. Let

k =

ga + bf

a + b

:

The vertex of the parabola is

µ

(2akf ¡ hk2 ¡ hac)

d

;

a(k2 + ac ¡ 2kg)

d

:

Now translate to the vertex as the new origin, then rotate to

(x2; y2) axes with the positive x2{axis along [sa; ¡sh], where

s = sign (a).

(The positive x2{axis points into the ¯rst or fourth quadrant.)

Then the parabola has equation

x2

2 = ¡2st

pa2 + h2

y2;

where t = (af ¡ gh)=(a + b).

REMARK 7.2.1 If ¢ = 0, it is not necessary to rotate the axes. Instead

it is always possible to translate the axes suitably so that the coe±cients of

the terms of the ¯rst degree vanish.

EXAMPLE 7.2.1 Identify the curve

2x2 + xy ¡ y2 + 6y ¡ 8 = 0: (7.14)

146 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

Solution. Here

¢ =

¯¯¯¯¯¯

2 1

2 0

1

2 ¡1 3

0 3 ¡8

¯¯¯¯¯¯

= 0:

Let x = x1 + ®; y = y1 + ¯ and substitute in equation 7.14 to get

2(x1 + ®)2 + (x1 + ®)(y1 + ¯) ¡ (y1 + ¯)2 + 4(y1 + ¯) ¡ 8 = 0: (7.15)

Then equating the coe±cients of x1 and y1 to 0 gives

4® + ¯ = 0

® + 2¯ + 4 = 0;

which has the unique solution ® = ¡2

3 ; ¯ = 8

3 . Then equation 7.15 simpli¯es

to

2x2

1 + x1y1 ¡ y2

1 = 0 = (2x1 ¡ y1)(x1 + y1);

so relative to the x1; y1 coordinates, equation 7.14 describes two lines: 2x1¡

y1 = 0 or x1 + y1 = 0. In terms of the original x; y coordinates, these lines

become 2(x+ 2

3 )¡(y ¡ 8

3 ) = 0 and (x+ 2

3 )+(y ¡ 8

3 ) = 0, i.e. 2x¡y +4 = 0

and x + y ¡ 2 = 0, which intersect in the point

(x; y) = (®; ¯) = (¡

2

3

;

8

3

):

EXAMPLE 7.2.2 Identify the curve

x2 + 2xy + y2 + +2x + 2y + 1 = 0: (7.16)

Solution. Here

¢ =

¯¯¯¯¯¯

1 1 1

1 1 1

1 1 1

¯¯¯¯¯¯

= 0:

Let x = x1 + ®; y = y1 + ¯ and substitute in equation 7.16 to get

(x1+®)2+2(x1+®)(y1+¯)+(y1+¯)2+2(x1+®)+2(y1+¯)+1 = 0: (7.17)

Then equating the coe±cients of x1 and y1 to 0 gives the same equation

2® + 2¯ + 2 = 0:

Take ® = 0; ¯ = ¡1. Then equation 7.17 simpli¯es to

x2

1 + 2x1y1 + y2

1 = 0 = (x1 + y1)2;

and in terms of x; y coordinates, equation 7.16 becomes

(x + y + 1)2 = 0; or x + y + 1 = 0:

7.3. PROBLEMS 147

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