Chapter VII. Classification of Quadric Surfaces

• 65. Intersection of a quadric and a line.General form and its matrix representation.

1

1),,(

)1(0222

222),,( 222

z

y

x

dnml

ncfg

mfbh

lgha

zyxzyxF

dnzmylx

hxygzxfyzczbyaxzyxF

– Intersection of a given line

– With the quadric form, substitute the value from (2), we get the following form:

– Let

TT VzyxP

rVPP

)0,,,(,)1,,,(

(2) ,

00000

00

)3(022 SRrQr

1

,

0

,

10

0

0

00 z

y

x

PVz

y

x

P

– Then the related quantities are in matrix form

)(2

1

222

),,(

000

0000

222

00

00000

z

F

y

F

x

F

AVPAPVR

hgfcba

AVVQ

zyxFAPPS

TT

T

T

– The roots in r of equation (3) are the distances from point on line (2) to the points in which this line intersects the quadric. (parametric form is useful)

– If , Eq. (3) is a quadratic in r. – If Q=0, but R and S are not both zero, (3) was a

quadratic, with one or more infinite roots.– If Q=R=S=0, (2) is satisfied for all values of r.

TzyxP )1,,,( 0000

0Q

– Theorem I. Every line which does not lie on a given quadric surface has two (distinct or coincident) points in common with the surface.

– Theorem II. If a given line has more than two points in common with a given quadric, it lies entirely on the quadric.

• 66. Diametral planes, center. Let , be the intersection points of line (2) with the quadric. The segment is called a chord of the quadric.– Theorem I. The locus of the middle point of a

system of parallel chords of a quadric is a plane.

(* if it’s real, more accurately, on a plane, or the intersection of a plane with the quadric)

2P1P

21PP

– Proof. Let be root of (3), then

the condition that is the middle point of the chord is that

(5) 0

1

)0,,,(

,0

0

0

,

0

0

0

00

21

0201

z

y

x

A

AXV

R

rr

PPPP

T

21, rr

0202

0101

VrPP

VrPP

21PP0P

– Equation (5) is linear in , and in fact a plane equation. It is called the diametral plane.

– Theorem II. All the diametral planes of a quadric have at least one (finite or infinite) point in common.

– Proof. For any unit vector V0, the plane (5) passes through the intersection of the planes

000 ,, zyx

0/// iff 0),,,(of Projection:*

(6) 0

10000

00

wzwywxwzyx

z

y

x

ncfg

mfbh

lgha

AXV T

– Let G be the coefficient matrix, and D=|G|, the determinant.

– If , the plane (6) intersect in a single finite point. If this point does not lie on the surface, it is called the center of the quadric. Otherwise, it is called a vertex of the quadric.

– If D=0, but L,M,N are not all be zero, the plane (6) intersect in a single infinitely distant point.

cfg

fbh

gha

G

0D

– Exercises. P.77, No.1, 2,6,8.

• Equation of a quadric referred to its center.– If a quadric has a center , its equation,

referred to its center as origin, may be obtained in the following way.

),,( 000 zyx

01

'

'

'

1

z

y

x

or

',','

0

0

0

000

z

y

x

z

y

x

zzzyyyxxx

– Since is the center, it satisfies equation (6), and it means that the linear part in quadric disappears. We thus have:

– By eliminating , we have

),,(

(9) 0222

000

222

zyxFS

Shxygzxfyzczbyax

),,( 000 zyx

),,( 000 zyx

0 oft determinan

Sdnml

ncfg

mfbh

lgha

dnml

ncfg

mfbh

lgha

S

cfg

fbh

gha

det.det

Therefore, DS=|A|

(1) Or, if , S=det A/D,

when det A=0, from (9), the quadric is a cone.

(2) If D=0 and , then we have D=0. Since the center was assumed to be a finite point, so that it follows that L=M=N=0, and the surface has a line or plane of centers.

0D

0S

– is called the discriminant of the given quadric. If , the quadric is said to be singular, otherwise, non-singular.

• 68. Principal Planes.– A diametral plane which is perpendicular to the

chords it bisects – Theorem. If the coefficient of a quadric are

real, and if the quadric does not have the plane at infinity as a component, the quadric has at least one real, finite, principal plane.

Adet0

– The condition that the diametral plane (5)

– Is perpendicular to the chords it bisects is that the normal of plane (5) coincides with the direction cosine of the chord.

0)(

)()(

nmlzcfg

yfbhxgha

kcfg

kfbh

kgha

– The condition that these linear equation has none zero solution about direction cosine is its determinant of coefficient matrix is zero.

– Thus k is the characteristic eigenvalue of matrix D

(14) 0

kcfg

fkbh

ghka

• 69. Reality of the roots of the discriminant. – Theorem I. The roots of the discriminanting

cubic are real.– Notice: the coefficient matrix is real and

symmetric, from linear algebra, it has “three real roots, countering on the multiple roots.”

– Theorem II. Not all the roots of the discriminating cubic are equal to zero.

– * if they are all zero, D is then equivalent to a zero matrix, implies that a=b=c=f=g=h=0.

• 70. Simplification of the equation of a quadric.– Let the axes be transformed in such a way that a

real, finite principal plane of quadric F(x,y,z)=0 is taken as x=0.

– Then the surface is symmetric with respect to x=0, the coefficients of terms of first degree in x must all be zero. Hence the equation has the form:

0

11

,

0

0

0

000

0222222

z

y

x

Az

y

x

dnm

ncf

mfb

a

A

dnzmyfyzczbyaxT

– Moreover, , since otherwise x=0 would not be a principal plane.

– Now let the planes y=0,z=0 be rotated about X-axis through the angle defined by

– This rotation reduces the coefficient of yz to zero., now the equation has the form:

cb

f

22tan

0a

(16) 0''2'2''' 222 dznymzcybxa

• 71. Classification of quadric surfaces.– Since the equation of a quadric can be always

be reduced to the form of (16), a completed classification can be made

– By considering the possible values of the coefficients. Here 0a

– I. Let both b’ and c’ be different from zero. By a translation of the axes in such a way that

– Is the new origin, the equation reduces to:

)'

',

'

',0(

c

n

b

m

"''' 222 dzcybxa

elliposoidImaginary ,1

sheet twoof dHyperboloi,1

sheet one of dHyperboloi ,1

Elliposoid ,1

0" if (1)

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

c

z

b

y

a

xc

z

b

y

a

xc

z

b

y

a

xc

z

b

y

a

x

d

cone real ,0

coneImaginary ,0

0" if (2)

2

2

2

2

2

2

2

2

2

2

2

2

c

z

b

y

a

xc

z

b

y

a

x

d

– II. Let – (1) if , by a translation of axes, the

equation may be reduced to

– This equation takes the form

0',0' cb0'n

0'2'' 22 znybxa

paraboloid Hyperbolic,2

paraboloid Elliptic,2

2

2

2

2

2

2

2

2

nzb

y

a

x

nzb

y

a

x

– (2) if n’=0, this may be reduced to

0"'' 22 dybxa

imaginary)(may

planes ngintersecti ofPair ,0

0,d" if (2.b),

cylinder. Quadric .01

,0" if),.2(

2

2

2

2

2

2

2

2

b

y

a

x

b

y

a

x

da

– III. Let b’=c’=0. The equation (16) is in this case

– (1) if m’ and n’ are not both zero, since the plane 2m’y+2n’z+d’=0 is perpendicular to x=0, we may rotate and translate the axes so that the plane is now y=0. The equation becomes

0''2'2' 2 dznymxa

cylinder Parabolic ,22 myx

– (2) if m’ and n’ are both zero, we have

0' if,0: twomultiples of plane One

0' if,0:planes parallel Two2

22

dx

dkx

• 72. Invariants under motion.– Invariant under motion

• A function of the coefficients of the equation of a surface,

• the value of which is unchanged when the axes are rotated and translated.

– E.g. • I=a+b+c, 222 hgfabcabcJ

dnml

ncfg

mfbh

lgha

A

cfg

fbh

gha

D det,

• 73. Proof that I,J, and D are invariant.– *hint. Using matrix representation, the result

can be obtained much easily.– I, J, D is exactly the coefficients of the equation

of eigenvalues. It is unchanged under motion.

0

:

0

23

DJkIkk

kcfg

fkbh

ghka

• 74. Proof that is invariant– Using matrix representation.– Let U be a transformation, X=UX’, then

– the coefficient matrix becomes

AUU T

'' AUXUXAXX TTT

AUAUAUU TT detdetdetdet)det(

• Exercises: P.89, No. 2,4,6,8,18.