Section 11

General theory of surfaces of the second order

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Problem 11.1

List the surfaces of the second order for which:

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(1)

R=4R = 4

(2)

R=3R = 3

(3)

R=2R = 2

(4)

R=1R = 1

(5)

r=3r = 3

(6)

r=2r = 2

(7)

r=1r = 1.

Problem 11.2

Characterize by means of invariants the "main" group of real surfaces of the second order: ellipsoids, hyperboloids, paraboloids.

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Problem 11.3

Characterize by means of invariants the following groups of surfaces of the second order:

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(1)

paraboloids and parabolic cylinders

(2)

surfaces consisting of planes

(3)

"imaginary" surfaces: "imaginary ellipsoids", "imaginary cones", "imaginary elliptic cylinders", "pairs of imaginary intersecting planes", "pairs of imaginary parallel planes".

Problem 11.4
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(1)

Which of the "imaginary" surfaces of the second order (see problem 11.3,3)11. 3, 3)) have no real points? Characterize by means of invariants this group of surfaces.

(2)

Which of the "imaginary" surfaces of the second order have real points, and what do these surfaces look like? Characterize by means of invariants this group of surfaces.

Problem 11.5

Characterize by means of invariants the surfaces of the second order that do not degenerate into the empty set, a point, a line, a plane, or a pair of planes.

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Problem 11.6

Characterize by means of invariants the real surfaces that have:

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(1)

two families of rectilinear generators

(2)

one family of rectilinear generators.

Problem 11.7

List the surfaces of the second order whose canonical equations contain a nonzero constant term. Characterize these surfaces by means of invariants.

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Problem 11.8

Let the equation of the surface be written in matrix form (2), and its transform in form (4) (see the introduction to § 11). Express the coefficients A′,a′,k′A^{\prime }, \mathbf{a}^{\prime }, k^{\prime } in terms of A,a,kA, \mathbf{a}, k, if:

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(1)

ξ=ξ′+b\xi = \xi^{\prime } + \mathbf{b}

(2)

ξ=Sξ′+b\xi = S\xi^{\prime } + \mathbf{b}.

Problem 11.9
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(1)

Verify that there exist second-degree equations in which, by passing to a new Cartesian coordinate system, it is impossible to eliminate all terms with first powers of the variables. List all types of such surfaces and characterize them by means of invariants.

(2)

Verify that there exist second-degree equations in which, by passing to a new Cartesian coordinate system, it is possible to eliminate all terms of degree below the second. List all types of such surfaces and characterize them by means of invariants.

Problem 11.10
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(1)

Which real surfaces of the second order have a center of symmetry?

(2)

How many centers of symmetry can a surface of the second order have?

(3)

Prove that for a surface of the second order only the following possibilities can occur. The surface:

(1) has no center of symmetry;

(2) has a unique center of symmetry;

(3) has a line consisting of centers of symmetry;

(4) has a plane consisting of centers of symmetry.

Problem 11.11

List and characterize by means of invariants the types of surfaces of the second order:

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(1)

having no center

(2)

having a unique center

(3)

having infinitely many centers.

Problem 11.12

Prove the statements:

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(1)

if a surface of the second order has a center, and it is located at the origin, then the equation of the surface contains no linear terms

(2)

if the equation of a surface of the second order contains no linear terms, then the surface has a center located at the origin.

Problem 11.13
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(1)

Using the results of problems 11. 8 and 11. 12, obtain the system of equations for the center of a surface of the second order (i.e., equation (10) from the introduction to § 11).

(2)

How will the constant term of equation (2) change if the origin is placed at the center of the surface of the second order?

Problem 11.14

Prove that the condition det⁡A≠0\operatorname {det} A \neq 0 is necessary and sufficient for the existence of a unique center of a surface of the second order given by equation (2).

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Problem 11.15

Justify the second method for simplifying the equation of a surface of the second order, presented in the introduction to § 11, i.e., prove each of the statements in items а) and б).

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Problem 11.16
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(1)

The equation of a surface of the second order is written in expanded form (1) and in matrix form (2). All coefficients of the expanded equation are multiplied by a number μ≠0\mu \neq 0. What happens to the matrix AA? How do the roots of the characteristic equation ∣A−λE∣=0\left|A - \lambda E\right| = 0 change as a result?

(2)

The equation of a surface of the second order is written in a rectangular coordinate system, and a passage to another rectangular system is made in it. Prove that the characteristic equation ∣A−λE∣=0\left|A - \lambda E\right| = 0 does not change as a result, and hence its roots do not change. Does det⁡A\operatorname {det} A change?

Problem 11.17
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(1)

A second-order equation is given in matrix form (2). Express the matrix BB of the large quadratic form of the surface given by this equation in terms of A,a,kA, \mathbf{a}, k.

(2)

In the equation of the surface a change of coordinates ξ=Sξ′+b\xi = S\xi^{\prime } + \mathbf{b} is made. Write out the matrix TT by which the large quadratic form of the surface is transformed, and prove that, under a passage from one rectangular coordinate system to another, det⁡B\operatorname {det} B does not change.

(3)

Prove that, under an orthogonal change of coordinates that leaves the origin fixed, the characteristic equation ∣B−λE∣=0\left|B - \lambda E\right| = 0 does not change, and hence its coefficients and roots do not change.

Problem 11.18

Suppose that in some general Cartesian coordinate system the equation of a surface of the second order coincides in form with one of the canonical equations. Prove that there exists a rectangular Cartesian coordinate system in which the equation of this surface is canonical of the same type.

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Problem 11.19

Using the result of problem 11. 18, determine the type of the surface given, in a general Cartesian coordinate system, by the equation:

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(1)

(x+y+z)(x−y+125z)=1(x + y + z)(x - y + 125z) = 1

(2)

(x+y)(x+y+1)=1(x + y)(x + y + 1) = 1

(3)

(x+y)(x+y+1)=x−y(x + y)(x + y + 1) = x - y

(4)

(x+y+z+1)(x−y+z)=x+z+1(x + y + z + 1)(x - y + z) = x + z + 1

(5)

(x+y+z+1)(x−y+z)=x+2z+1(x + y + z + 1)(x - y + z) = x + 2z + 1

(6)

(x+y)(x−y)=z(x + y)(x - y) = z

(7)

(x+y+z)(x−y+z)=0(x + y + z)(x - y + z) = 0

(8)

(x+2y)(x+2y+1)=z(x + 2y)(x + 2y + 1) = z

(9)

(x+y)(x−75y)=z2(x + y)(x - 75y) = z^{2}

(10)

(x+y)2=3x+z(x + y)^{2} = 3x + z

(11)

(x+2y+3z)(2x+3y+4z)+(3x+4y+75z)2=1(x + 2y + 3z)(2x + 3y + 4z) + (3x + 4y + 75z)^{2} = 1

(12)

(x+2y+3z)(2x+3y+4z)−(3x+4y+75z)2=1(x + 2y + 3z)(2x + 3y + 4z) - (3x + 4y + 75z)^{2} = 1

(13)

(x+y)z−x2−x=0(x + y) z - x^{2} - x = 0

(14)

(x+y+z)2+(x+2y+3z)2+(2x−y+z)2=0(x + y + z)^{2} + (x + 2y + 3z)^{2} + (2x - y + z)^{2} = 0

(15)

(x+y+z)2+(x+2y+3z)2+(2x+3y+4z)2=0(x + y + z)^{2} + (x + 2y + 3z)^{2} + (2x + 3y + 4z)^{2} = 0

(16)

(x+y+z)2+(x−2y+z)2=0(x + y + z)^{2} + (x - 2y + z)^{2} = 0

(17)

(x+y+z)2+(x+y)2+(y+z)2=1987(x + y + z)^{2} + (x + y)^{2} + (y + z)^{2} = 1987

(18)

(x+y+z)2+(x+y)2=1987(x + y + z)^{2} + (x + y)^{2} = 1987

(19)

(x+y+z)2+(x+y)2+(2x+2y+z)2=1(x + y + z)^{2} + (x + y)^{2} + (2x + 2y + z)^{2} = 1

(20)

(x+y)2+z2+1=0(x + y)^{2} + z^{2} + 1 = 0.

Problem 11.20

In a general Cartesian coordinate system a hyperboloid is given by the equation (x+y+z)(x−y+z)−(2x−y+2z)2=1(x + y + z)(x - y + z) - (2x - y + 2z)^{2} = 1. Find the equation of its asymptotic cone.

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Problem 11.21

A surface is given in a general Cartesian coordinate system by an equation containing a parameter kk. Determine the type of the surface for all values of kk.

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(1)

x12+x22+4x32+4x1x3+2x2x3+2x1+2x2+5x3+k=0x_{1}^{2} + x_{2}^{2} + 4x_{3}^{2} + 4x_{1} x_{3} + 2x_{2} x_{3} + 2x_{1} + 2x_{2} + 5x_{3} + k = 0

(2)

2x12+kx22+8x1x2+4x1x3−4x1−8x2−4x3=02x_{1}^{2} + kx_{2}^{2} + 8x_{1} x_{2} + 4x_{1} x_{3} - 4x_{1} - 8x_{2} - 4x_{3} = 0

(3)

3x12+2x22+2x32+4x1x2−2x1x3+4x1+4x2+4x3+k=03x_{1}^{2} + 2x_{2}^{2} + 2x_{3}^{2} + 4x_{1} x_{2} - 2x_{1} x_{3} + 4x_{1} + 4x_{2} + 4x_{3} + k = 0

(4)

kx12+8x22+x32+16x1x2+4x1x3+4x2x3−4x1−4x2+2x3=0kx_{1}^{2} + 8x_{2}^{2} + x_{3}^{2} + 16x_{1} x_{2} + 4x_{1} x_{3} + 4x_{2} x_{3} - 4x_{1} - 4x_{2} + 2x_{3} = 0

(5)

3x22+x32+6x1x2+2x1x3+4x2x3+kx2+x3+1=03x_{2}^{2} + x_{3}^{2} + 6x_{1} x_{2} + 2x_{1} x_{3} + 4x_{2} x_{3} + kx_{2} + x_{3} + 1 = 0

(6)

(р) x12−2x22−3x32−4x1x2−6x1x3−2x1+4x2+6x3+k=0x_{1}^{2} - 2x_{2}^{2} - 3x_{3}^{2} - 4x_{1} x_{2} - 6x_{1} x_{3} - 2x_{1} + 4x_{2} + 6x_{3} + k = 0.

Problem 11.22

A surface is given by an equation in a rectangular Cartesian coordinate system. Find the canonical coordinate system and the canonical equation of this surface. Determine the type of the surface.

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(1)

2x2+9y2+2z2−4xy+4yz−1=02x^{2} + 9y^{2} + 2z^{2} - 4xy + 4yz - 1 = 0

(2)

4y2−3z2+4xy−4xz+8yz=04y^{2} - 3z^{2} + 4xy - 4xz + 8yz = 0

(3)

x2+y2+4z2−2xy+4xz−4yz−2x+2y+2z=0x^{2} + y^{2} + 4z^{2} - 2xy + 4xz - 4yz - 2x + 2y + 2z = 0

(4)

x2+y2+z2−xy+xz+yz+3x+3y−3z=0x^{2} + y^{2} + z^{2} - xy + xz + yz + 3x + 3y - 3z = 0

(5)

x2−3z2−4yz−4y+2z+5=0x^{2} - 3z^{2} - 4yz - 4y + 2z + 5 = 0

(6)

x2+y2+z2−2xy−4x+4y+3=0x^{2} + y^{2} + z^{2} - 2xy - 4x + 4y + 3 = 0

(7)

y2+2xz+2x+2z+1=0y^{2} + 2xz + 2x + 2z + 1 = 0

(8)

x2+2y2+5z2+4yz+20y+20z−10=0x^{2} + 2y^{2} + 5z^{2} + 4yz + 20y + 20z - 10 = 0

(9)

−x2+5y2+5z2+8yz+2x+12y+24z+36=0- x^{2} + 5y^{2} + 5z^{2} + 8yz + 2x + 12y + 24z + 36 = 0

(10)

2x2+5y2+5z2+6yz+4x+16y+16z+10=02x^{2} + 5y^{2} + 5z^{2} + 6yz + 4x + 16y + 16z + 10 = 0

(11)

4x2+4y2−4xy−12x−12y−5z+1=04x^{2} + 4y^{2} - 4xy - 12x - 12y - 5z + 1 = 0

(12)

x2+y2+z2+2xy−12x+4y+6z−3=0x^{2} + y^{2} + z^{2} + 2xy - 12x + 4y + 6z - 3 = 0

(13)

4x2+9y2−12xy+2x+10y+1=04x^{2} + 9y^{2} - 12xy + 2x + 10y + 1 = 0

(14)

6xy−8y2−z2+60y+2z+89=06xy - 8y^{2} - z^{2} + 60y + 2z + 89 = 0

(15)

5x2+8y2+4xy+2x+44y−36z+65=05x^{2} + 8y^{2} + 4xy + 2x + 44y - 36z + 65 = 0

(16)

(р) −x2+y2+z2−2yz+2x+3y−5z+1=0- x^{2} + y^{2} + z^{2} - 2yz + 2x + 3y - 5z + 1 = 0

(17)

9y2+16z2+24yz+5x+10y+5z+11=09y^{2} + 16z^{2} + 24yz + 5x + 10y + 5z + 11 = 0

(18)

16x2+9y2−z2−24xy−9x−12y+4z+71=016x^{2} + 9y^{2} - z^{2} - 24xy - 9x - 12y + 4z + 71 = 0

(19)

2x2+2y2+z2−10xy+20x−8y+29=02x^{2} + 2y^{2} + z^{2} - 10xy + 20x - 8y + 29 = 0

(20)

−x2+7y2−24yz+2x+120y=0- x^{2} + 7y^{2} - 24yz + 2x + 120y = 0

(21)

x2−4y2−4z2+10yz+2x+2y+2z+3=0x^{2} - 4y^{2} - 4z^{2} + 10yz + 2x + 2y + 2z + 3 = 0

(22)

3x2+4xy+8x+8y−4z=03x^{2} + 4xy + 8x + 8y - 4z = 0

(23)

−x2−9y2+6xy+50x−50y−15z−100=0- x^{2} - 9y^{2} + 6xy + 50x - 50y - 15z - 100 = 0

(24)

(р) 4x2+y2+9z2+4xy−12xz−6yz+2x+6y−6z−5=04x^{2} + y^{2} + 9z^{2} + 4xy - 12xz - 6yz + 2x + 6y - 6z - 5 = 0.

Problem 11.23

A surface is given in a rectangular Cartesian coordinate system by an equation containing a parameter kk. For the given value of kk, find the canonical coordinate system and the canonical equation of the surface. Determine the type of the surface for all possible values of kk. If the surface is a line, a plane, or a pair of planes, find the linear equations of these sets in the original coordinate system.

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(1)

5x2+5y2+3z2+2xy+22xz+22yz+26x+34y+102z+49=05x^{2} + 5y^{2} + 3z^{2} + 2xy + 2\sqrt{2} xz + 2\sqrt{2} yz + 26x + 34y + 10\sqrt{2} z + 49 = 0;

(2)

2x2+9y2+2z2−4xy+4yz+4x+2y−4z−1=02x^{2} + 9y^{2} + 2z^{2} - 4xy + 4yz + 4x + 2y - 4z - 1 = 0

(3)

4y2−3z2+4xy−4xz+8yz+4x−2z+k=04y^{2} - 3z^{2} + 4xy - 4xz + 8yz + 4x - 2z + k = 0;

(1) k=−1k = - 1;

(2) k=5k = 5;

(3) k=11k = 11.

(4)

2x2+2y2+2z2+2xy+2xz+2yz+4x−4y+4=02x^{2} + 2y^{2} + 2z^{2} + 2xy + 2xz + 2yz + 4x - 4y + 4 = 0

(5)

3x2+3y2+3z2−2xy−2xz−2yz−8x+8y+k=03x^{2} + 3y^{2} + 3z^{2} - 2xy - 2xz - 2yz - 8x + 8y + k = 0

(1) k=4k = 4;

(2) k=8k = 8.

(6)

x2+y2+4z2−2xy+4xz−4yz−12x+12y−24z+6=0x^{2} + y^{2} + 4z^{2} - 2xy + 4xz - 4yz - 12x + 12y - 24z + 6 = 0

(7)

x2+y2+4z2−2xy+4xz−4yz+24y−24z+12=0x^{2} + y^{2} + 4z^{2} - 2xy + 4xz - 4yz + 24y - 24z + 12 = 0

(8)

x2+y2+4z2−2xy+4xz−4yz+5x+y−2z=0x^{2} + y^{2} + 4z^{2} - 2xy + 4xz - 4yz + 5x + y - 2z = 0

(9)

2x2+2y2+2z2−2xy+2xz+2yz−6x−6y−12z+k=02x^{2} + 2y^{2} + 2z^{2} - 2xy + 2xz + 2yz - 6x - 6y - 12z + k = 0

(1) k=15k = 15;

(2) k=18k = 18.

(10)

2x2+2y2+2z2−2xy+2xz+2yz+18z+18=02x^{2} + 2y^{2} + 2z^{2} - 2xy + 2xz + 2yz + 18z + 18 = 0;

(11)

x2+2y2+z2−2xy−2yz+6x−6y+k=0x^{2} + 2y^{2} + z^{2} - 2xy - 2yz + 6x - 6y + k = 0

(1) k=8k = 8;

(2) k=9k = 9.

(12)

x2+2y2+z2−2xy−2yz+18x−6y+6z−18=0x^{2} + 2y^{2} + z^{2} - 2xy - 2yz + 18x - 6y + 6z - 18 = 0

(13)

3x2−7y2+3z2+8xy−8yz−8xz−4x+6y+8z+k=03x^{2} - 7y^{2} + 3z^{2} + 8xy - 8yz - 8xz - 4x + 6y + 8z + k = 0;

(1) k=−12k = - 12;

(2) k=−3k = - 3;

(3) k=6k = 6.

(14)

2y2−3z2−23xy−4xz+43yz+50z+k=02y^{2} - 3z^{2} - 2\sqrt{3} xy - 4xz + 4\sqrt{3} yz + 50z + k = 0;

(1) k=−75k = - 75;

(2) k=−70k = - 70;

(3) k=−80k = - 80.

(15)

2x2+2y2+2z2+8xy+8xz−8yz−8x−4y+8z+k=02x^{2} + 2y^{2} + 2z^{2} + 8xy + 8xz - 8yz - 8x - 4y + 8z + k = 0;

(1) k=−4k = - 4;

(2) k=2k = 2;

(3) k=8k = 8.

(16)

4x2+4y2−2z2+4xy−8xz+8yz+12x−12y+24z+k=04x^{2} + 4y^{2} - 2z^{2} + 4xy - 8xz + 8yz + 12x - 12y + 24z + k = 0;

(1) k=−42k = - 42;

(2) k=−36k = - 36;

(3) k=−30k = - 30.

(17)

8y2+4xy+2xz+4yz+4x+8y+k=08y^{2} + 4xy + 2xz + 4yz + 4x + 8y + k = 0

(1) k=0k = 0;

(2) k=−9k = - 9.

(18)

8y2+4xy+2xz+4yz+8x+6y+4z+6=08y^{2} + 4xy + 2xz + 4yz + 8x + 6y + 4z + 6 = 0

(19)

4x2+y2+9z2+4xy−12xz−6yz+11x+3y−z+1=04x^{2} + y^{2} + 9z^{2} + 4xy - 12xz - 6yz + 11x + 3y - z + 1 = 0

(20)

4x2+y2+9z2+4xy−12xz−6yz+6x−2y−6z+11=04x^{2} + y^{2} + 9z^{2} + 4xy - 12xz - 6yz + 6x - 2y - 6z + 11 = 0

(21)

4x2+y2+9z2+4xy−12xz−6yz+4x+2y−6z+k=04x^{2} + y^{2} + 9z^{2} + 4xy - 12xz - 6yz + 4x + 2y - 6z + k = 0;

(1) k=1k = 1;

(2) k=−13k = - 13.

(22)

−x2+10y2−z2−8xy+6xz+8yz+24x−8y−16z+k=0- x^{2} + 10y^{2} - z^{2} - 8xy + 6xz + 8yz + 24x - 8y - 16z + k = 0;

(1) k=−26k = - 26;

(2) k=−14k = - 14;

(3) k=−2k = - 2.

(23)

2x2−y2+2z2+4xy−2xz+4yz+10x−2y−2z+k=02x^{2} - y^{2} + 2z^{2} + 4xy - 2xz + 4yz + 10x - 2y - 2z + k = 0;

(1) k=2k = 2;

(2) k=5k = 5;

(3) k=8k = 8.

(24)

x2+y2−2z2+10xy+4xz−4yz+13x+11y+2z+6=0x^{2} + y^{2} - 2z^{2} + 10xy + 4xz - 4yz + 13x + 11y + 2z + 6 = 0.

(25)

x2+y2−2z2+10xy+4xz−4yz+24x−12z+k=0x^{2} + y^{2} - 2z^{2} + 10xy + 4xz - 4yz + 24x - 12z + k = 0;

(1) k=−12k = - 12;

(2) k=−6k = - 6.