Solid Analytic Geometry

Table of contents :
COORDINATES AND LINES
PLANES
SURFACES AND CURVES
SPHERES
QUADRIC SURFACES
THEORY OF MATRICES
ROTATIONS OF AXES AND APPLICATIONS
SPHERICAL COORDINATES
ELEMENTS OF PROJECTIVE GEOMETRY

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8]m > a: 73

OUP

88 15-8-74

1

5.000

OSMANIA UNIVERSITY LIBRARY Call

Accession No.

No. ^""16

Author

"^^ i &

\

A'M S

Title This bonk should be

re^rned

t

n or before the date

last

marked below.

In this

new

position

of

text the author presents an exanalytic geometry of threedimensional space. The material covers the standard topics of space analytic geometry but provides a treatment of the subject which permits immediate generalization to n dimensions. This treatment ties the subject to modern mathematics, and, in particular, to modern algebra. The use of the theory of vector spaces and matrices permits a major simplification in the proofs and in the exposition in general. Thus the aim of the book is to provide a modern and simpler treatment of the subject matter which permits easy generalization and fits the subject into its proper place in modern mathematics.

the

The

early part of the text is simplified by the use of the concepts of inner and scalar product. brief but adequate chapter on the theory of matrices provides a full, clear exposition of the principal axis transformation in the n-dimensional case. An additional feature is provided by the chapter on spherical coordinates where the mathematics of the approximations used in actual physical measurements of direction and distance in rotated coordinate systems is presented. This material is r,?/,) 2 + (z/j/,-) 2 - ^x^XfUi = (xy/ < j ^ HJ and must be nonnegative. The numbers P P, Q Q, and (P Q) 2 are all have shown that

an angle

6

is

expressions

The square x^yS + xfy?

the form

2o: t ?/ t x ; 7//

for

of all expressions

of the

It follows that there exists

+ yS} all

x/y

2 t

)

for

positive,

1

j

=

y'

+

2/0,

z

=

z'

+

ZQ,

sets of coordinates.

z9

FIG.

Formula axes, which

7.

sometimes used

(16) is

simplifies

what we

in

making a translation

shall later call

of

an equation of a

surface.

EXERCISES

What

1.

surfaces?

,. W (6)

-

(^

2

l)

+

-

2

(e) )

(z-

2)

3)*

_ ~

9~~

e

- 3) + (z + I) = 5 - 12 = x 2x + 3y 6?/ + 42 2 = 9 z + Go; 2x + 3y 12z/ + z 2 = 6a; 3a; 3?/ + Oz 9y + 24z 3x - 2/ 2 + s 2 = 6 2(x

2

2)

-

2

2

2

(y 2

2

(d)

+

(y

,

-I

4

2

(c)

translation of axes will simplify the following equations of

2

2

1

COORDINATES AND LINES

Sec. 12]

15

Find the

2.

there

is

#', y', z' coordinates of each of the following points a translation of axes with origin at 0'.

P

if

P=

(1, -1,1), 0'= (2,2, -1) = (b)P (0,0,0), 0'=(1, -1,1) = (-1, -1,1), 0'= (-1, -1,2) ( C)P (d)P = (3,1,1),0' = (-2,2,1) (e)P = (l,2, -3),0' = (1,0, -1) = (2,4,0),0' = (0, -1,0) (/) P

(a)

3.

Let the vectors

responding points. 4. Find the x',

P

in Exercise 2

Give their y', z'

x, y, z

be the

a/,

?/',

2'

coordinates of cor-

coordinates.

equation of the surfaces given by the equamoving the origin to

tions of Exercise 1 after a translation of axes

0'

=

(1,

12.

P*

=

-1,2).

Geometric addition of vectors. Let P\ (xi, y\, Zi) and Then Pi#P 2 determines a plane (#2, 2/2, 22) be two vectors.

FIG.

8.

and, as in Fig. 8, we can determine a point Pa in this plane such that OP 3 is parallel to PiP 2 and in a corresponding direction, |OP 3 = |PiP 2 |. The coordinates of P 2 relative to an x' y', z 1

,

,

|

coordinate system with origin at Pi, will be (x 2 z 2 ~ Zi). But P 3 is located with respect to the exactly where P 2 is located with respect to the x' y', ',

Xi,

y(Z2

be computed by the use of

+

[\(xi

(2/1

-

-

y2)

x2)

2

+

+

(zi

M(?/I

~

-

z>)*

2/2)

+

numbers are given rather than direction cosines, it is necessary to compute direction cosines before formula (29) can If direction

be applied. EXERCISES

Compute the

distances between points and lines in the following cases:

P,- (0,0,0);^ =

^=^

P,= (1,0,0);*"- =

V

(6)

(c)

P =

=

(a)

2

(l,

-1,0);-

~

-

=

7

~

=

= 62 + 12 2)/ = 2 7/- l=2e-4 (1,1, -1);2* (e) ~ g " 1 _ _ z -t( 11 IV ~ +J ~ ( r\P (J) * 2 1, i, i),

(d)

P2 = P =

(0, 1, 2);

3z

=

2

M w

P r 2

re 7 2V w, /, ^;,

x

+

2

3

-

^il

2 l

-

1-

2

CHAPTER PLANES

2

If we draw a ray from the origin O per1. The normal form. pendicular to a given plane, the ray is called the ray normal to the plane and the line of which it is a part the normal line. The will direction cosines have which are deterX, /z, v, ray uniquely

mined unless the plane passes through the case the selection of X,

ju,

v,

rather than

But

origin. X,

v,

ju,

in this

will

not

affect our results.

Let p be the distance from to the given plane so that p ^ Then the normal will intersect the plane in the point P

0.

A

translation of axes, which moves the origin to the coordinates x, y, z of the arbitrary point P by its replaces Po, z' coordinates #', y', (pX, pp., pv).

(x

-

y

pX,

-

*

p/z,

-

pv).

The ray through P and the vector whose transformed coordinates Then P is on the plane (X, /z, v) is normal to the given plane. if and only if the vector just defined is orthogonal to PoP, that is, are

if

and only

if (X, M,

v)

pX, y see that

(x

pp, z

pv)

=

0.

We com-

P is on the given plane if and pute this inner product and 2 2 2 ^ ) = 0. vz if \x Since X 2 + M 2 + M py p(X only + u 2 = 1, we have proved that P (#, y, z) is on the given plane if and only if

+

+

\x

(1)

We shall

call

formula

+

y

+

+

vz

=

p.

equation of a plane in normal form. p > 0. When p == 0, the equation

(1) the

completely unique when is equally valid and the selection of one set vz = M?/ of direction cosines rather than their negatives may be regarded as implying that a positive direction on the line normal to the plane has been selected. This is of little importance. It is

\x

EXERCISE

Give the equation in normal form of the two planes p units from the origin and normal to the line joining Pi to Pi in the following cases: 24

PLANES

Sec. 2]

P = P! = = 1, Px = iP = 7, P!

p = p = = (c) p = (d) p () p =

(a)

5,

(6)

(-1, (-2,

l

2,

2

-1),

1,

(0, 0, 0),

2

P2 =

az

(2)

+

(1, 3,

(2, 2, 1)

-1,2) -3, 8)

(0,

=

(6,

Every

equation.

by

-1) -2)

(1, 2,

P =

(2,l,4),P 2 = (4, -1, 6), P,

1

The general

2.

P -

2, 3),

25

+

cz

linear equation

+

=

d

0,

d are real numbers and a, b, c are not all zero, is an equation of a plane; for we can write this equation in an equivalent form

which

in

a, b, c,

tax

(3)

where

=

e

1

if

d

^

and t

(4)

+ e

if it

(2) is

(6)

=

ecz

1

if

d

>

\d\,

(3)

by

t

The number

0.

0,

to obtain

the equation of formula (2) if and But then formula the equation of formula (5). an equation of the plane for which

point

only

P

=

+

Va r+W+~c'- >

=

and we may divide formula

A

eby

(x, y, z) satisfies

satisfies

p

= M,

x-,

=

=

,

the plane p units from the origin and having a normal ray whose direction cosines are the numbers X, /z, v of formula (6).

i.e.,

It is

important to observe that the original coefficients a, 6, c numbers of the normal line. abed ? 0, the general equation may be written in the form

are a set of direction If

This equation

is

called the equation of a plane in intercept

and the numbers /o\ (8)

=

dj,d /--^ ,

g

are called the intercepts of the given plane.

=

form

~ d

They

are indeed the

SOLID ANALYTIC GEOMETRY

26

[Chap. 2

nonzero coordinates of the three intercept points (6,0,0),

(0,0,0)

(0,/,0),

where the given plane cuts the coordinates axes. If a plane defined by formula (2) passes through a point and Pi = (xi, y\, zi), then ax\ + byi + cz\ + d = a(x

(9)

-

xi)

+

-

b(y

y,}

+

-

c(z

zi)

=

0.

This equation involves the coordinates of PI and a set of direction numbers of the normal line. We shall call it the paint, direction number form of an equation of a plane. EXERCISES 1.

Use formula

(9) to

write an equation of a plane through Pi perP 2 to PS in each of the following cases:

pendicular to the line joining (a)

=

P!

(-1,

2, 3),

P = 2

(-3,

1, 2),

P =

(6) (c)

(d) (e)

(/) (g) 2.

(a) (6) (c)

(d) 3.

4.

P = P =

P =

(-5,

3

P =

(-1,

(1,2,

1

(2,

-1,4),P 2 = (-3,1, -2),P 3 =

= = P = P! =

(-1, (-2,

1

(1, 4,

Find

(3, 2,

p, X,

-

ju,

P = P 1, 3), = P -1), = P -2),

2,

-3),

2

2

v for

(4,

-1,

2),

3

+

20

=

0.

4,

P 4 = (9, 10, 12) -8, 0), P 4 = (6, -8,

2

(6, 7, 8),

2

(-9,

7)

each of the following planes:

+z = 6 3x 6y + 7z = -1 3 + 2y + 2z = -6 -3z + 6y + 22 = 14 2x

4z

-1) -4) = (-1, 4, 3), P 3 = (0, -3, -4) = (3, -4, 5) (4, -6, 7), P 3

(1,

-2, -1),

Pa P,

-

3y

l

2

4, 6)

2x

Ans.

2y

Find the intercepts of each Find a set of values of X,

(e)

(/) (g)

(h)

+ 2x - 3z = 28 + 4y + 62 = -49 - 2z = 6 x + -x + y + 2z = -3 6y 12x

T/

of the planes in Exercise 2. v for each of the following planes /x,

through the origin:

(b)

+y= 3z + 4y =

(c)

3z

(a)

x

-

4?/

=

(d) (e)

62

(/)

+ y = 2z -x + y + 2z = 3o: + &y + 2z =

2x

Write an equation of each of two planes parallel to a corresponding plane of Exercise 4 and one unit from the origin. 6. Write a set of equations in symmetric form of a line perpendicular to the given plane and through the point ( 1, 2, 3) in each of the cases 5.

of Exercise 2.

PLANES

Sec. 4]

3,

=

Planes

three

through

and P =

27

=

Let Pi

points.

P

(xi, y\, Zi),

2

be three distinct points. By Sec. 9 of Chap. 1 the three points are collinear if and only if a scalar multiple of (x 2 is (x* yi, 23 xi, 2/3 Zi) #j, (#2, yzj zz),

2/2

22

2/1,

If

2/3, z-6 )

(#3,

3

21).

a plane passes through Pi,

b(y

+

2/0

P we 3,

c(z

=

Zi)

its

equation

If it also

0.

is

a(x

+

#1)

passes through

P

and

2

have the relations

(10)

a(z 3

-

+

a?i)

-

6(2/3

+

2/1)

c(z 3

~

zi)

=

0.

do not have proportional coeffiare not Pi, collinear, and they can be solved of direction numbers (a, 6, c) = (0, 0, 0) of the to yield a set normal to the given plane. Then formula (9) is an equation of These two equations

in a, 6, c

P P

when

cients

2,

3

the required plane.

EXAMPLE

ILLUSTRATIVE an

Find

the

of

equation

plane through

1,2,3),

(

3,1,2),

(

(-5,4,6). Solution

Equations (10) become

-2a 4a

Then -46 = a(z

+

1)

+

oc,

6(y

8a

-

=

2)

c,

+

-

=

and

x

-

10t/

+

82 4-

1

26

so that

c(z

+

20

b

-

-

c

3c

-

-46 = 40,

3)

a(x

+

1)

-

24

=

x

equation of the plane.

= -10a.

b

10a(y

-

IQy Ans.

-

+

Then

2)

+

Sa(z

Sz

-

3

=

IQy

+

x

-

3)

= an

is

82

=

3.

EXERCISE Find an equation of the plane through Pi,

P

2,

PS

for each set of points

listed in Exercise 1 of Sec. 2.

Parallel planes have the same normal line. 4. Parallel planes. Then every plane parallel to ax + by + cz + d = is obtained if we leave a, &, c fixed and vary d. If we put an equation of a plane in normal form \x + ny + vz = p, we obtain all planes parallel to the given plane and on the same side of the origin by varying p ^ 0. The distance

between the given plane and a second plane \x

+

Mj/

+

vz

=

p\

SOLID ANALYTIC GEOMETRY

28

then clearly

is

\p

p\\.

[Chap.

however, we consider

If,

all

2

planes

parallel to the given plane and on the other side of the origin, we see that their equations are (\x ny vz) p lw Then the

distance between Xx is

p

+

+

ny

+

vz

=

+

+

= p and

(\x

+

+

ny

=

vz)

p\

pi.

ILLUSTRATIVE EXAMPLES I.

+

4z

Find an equation of a plane which = 7 and which contains the point

is

parallel to the plane 3#

P =

(

2y

2, 3, 3).

Solution

The equation may be taken

to have the form 3x

P

2y

+

4z

=

k

y

on the plane. Then 6 6 + 12 = = k and the required equation is 3x 2y + 4z = 0. II. Find the distance between the plane 3x 2y + 4z = 7 and the = 4z 12. 3;r + 2y plane

where k

is

to be determined so that

is

Solution

The normal forms of these two planes are obtained by multiplying equations by (9 + 4 + 16)~* = (29)""*. The distance is then

the

(29)-*12

-

=

5(29)"*. (29)"*7 III. Find the distance between the plane 3x 15. 82 = the plane 60; 4?/

2y

+

4z

=

7

and

+

Solution

distance is 7(29-*) Y(29)"* = ^(29)-* = \ \/29. IV. Find the equations in normal form of the two planes that are

+

The

five units

from the plane

x

+

2y

-{

2z

+

3

=

and

parallel to

it.

Solution

The normal form

of an equation of the given plane is obtained by 4 the (1 J and thus is 4)~* = given equation by multiplying

+ +

x-2y -

2z

3

4 units from the origin where we planes are then 6 units and as equivalent to the replacement of X, ju, v interpret a negative distance 2z = 18 for the v. The answers are then x 2y n, by X,

The two

value 6 and x

2y

2z

=

12 for the value

4.

EXERCISES 1. Find an equation 6f a plane parallel to the given plane and through the given point in the following cases:

PLANES

Sec. 5]

(a) (6) (c)

(d) (e)

(/) (g)

3x 2x

-

2y

+z=

1,

89

(-1, -2,1)

+ 2y-z = 0, (-1,1,1) z + + 3z = 4, (-2,1,1) 2x-y-2z = 7, (3, -1,3) 3* + 4y - 62 = 183, (4, 5, 2) I8x - Uy + I7z = 987, (1, 1, 0) Ite - 7y + 6z = 111,968, (2, 3, 1) 2/

Find an equation of a plane parallel to the given plane and having

2.

x-intercept 3 in each of the cases of Exercise 1. 3. Find an equation of a plane parallel to the given plane and two units further from the origin in each case of Exercise 1 except (e), (/),

and

(0).

Find the distances between the following pairs of planes: 2x + 2y - z = 3, 2x + 2y - z = 18 - y - 2z = 6, -2x + y + 2z = 12 (6) 2x

4.

(a)

(c)

(d) (e)

3x 6x

+

2y

+

6z

=

14,

3z

+

2y

+

6z

=

21

- 2y + 32 = 28, 12x - 4y + 62 = 35 18x - 6y - 9z = 35, -Go; + 2y + 82 = 21

Find equations of pairs of planes four units from the

5.

first

of the

planes in Exercise 4.

Distance from a plane to a point. If an equation ax + by d = of a plane is given and a point P = (zi, yi, z\) is cz given, we require a formula for the distance d from the plane to P. We first observe that we may divide the equation of the plane by c 2 to convert it to the normal form 62 \/a 2 5.

+

+

+

+

\x

Then the equation

+

ny

is

on this

8

(11)

>

If 5

0,

if 8

=

Xzi

+

is

\x

M2/i

+

+

P and the origin < 0, the point P and

the point

and

=

vz

plane, side of the plane.

p.

of the plane parallel to tue given plane

from the origin plane, we have

units further

P

+

py

*i

+

~

vz

= p

+

5.

and

d

Since

P-

are on opposite sides of the the origin are on the same

the given plane passes through the origin, the meaning of the sign of 5 is not clear, and we shall simply use the formula If

ILLUSTRATIVE EXAMPLES Find the distance from the plane 2x -1,5).

I.

(2,

+

2y

z

=

3 to the point

SOLID ANALYTIC GEOMETRY

30

[Chap. 2

SoliUion

The

distance

is

+

[2(2)

Find the distance from 2x

II.

-

2(-l)

+

(5)

2y

z

3]

= -2.

=

to

1, 5).

(2,

Solution

The

distance

is

|i(4

-

-

2

=

5)|

1.

EXERCISES

Find the distances from the following planes to the corresponding

1.

points:

(d)

+ 2y - z = 3, (5, 3, -2) + y + 2z = -2, (2, -2) 3z + 2y + 6z = 14, (1, -2, 1) 12x-4y + 6z = 35, (-1, -1,1)

(e)

18*

(/)

-6*

(a) (6) (c)

2x

-2*

1,

-

9y 2

-

+

Find the the plane 3x 2.

+

(-1,

2,

-3)

+

Solve Exercise 2

3.

(1, 1, 3)

coordinate of a point P = Qz = 7 to the point 2y

z

+

= 14, 3z = 0,

9z

if

the distance

is

3, 2, 2) if

(

P

the distance from

is 2.

4.

Angle between two planes. The angle 6 between two planes normal to the defined to be the angle between the rays from 6.

is

two

planes.

It is not

through the origin. Xix

+ my +

v\z

=

uniquely defined

if

either plane passes

+ p,y + vv\.

the equations are \x = XXi MMi pi, then cos If

+

equations are given in general form, they into normal form.

must

-f~

first

= p and When the

vz

be converted

EXERCISES 1.

Compute the

planes:

+ 2y - z = 3,3x - 2y + 6z = + y + z = 2x - y - 2z = 6 - y + z = 2, 2x + y = 4 + = 3, 3z + 4y = 2

(a)

2x

(6)

x

(c)

x

(d) 2.

(a) (6) (c)

7.

a?

cosine of the angle between the following pairs of

Show

T/

that the following pairs of planes are perpendicular.

+ 2y - z = 3, x - 2y - 2z = + y + z = 1, x - z = 6 3x - 27/ + 4^ = 1, 2x + y - z =

2x *

The

14

1,

line.

If

1

two planes. Two nonparallel the planes have as equations

line of intersection of

planes intersect in a

1

PLANES

Sec. 7]

+

+ +

biy b zy

31

+ c z +

ciz

di

2

d*

= =

0,

then their line of intersection is orthogonal to both of their normals and the direction cosines of the line satisfy the equations

a 2X

+ +

+ &2M + bin

= =

civ c 2 j>

0.

However, the simplest method of determining the line two points on it. The procedure is illustrated below. ILLUSTRATIVE

is

to find

EXAMPLE

Give a set of equations in symmetric form of the

line

L

of intersection

of the planes

2x

+y-

=

22

5x

1,

and determine direction cosines

+

4y

-

6z

82

=

1

=

= ~2

of L.

Solution

We solve for x and y by writing 8x + =

ty

4 therefore 3x ;

6 and

x

Put Put

2 2

= =

0, 3,

=

2s

+

6 >

t/

^

=

2z

2^

+

22

- 9 -g

and obtain P t = (2, -3, 0) on L. Then L has as and obtain P 2 = (4, 1, 3) on L.

-

x

2

_ "

y

+

3

its

equation

_ ~

2

2

2z

3

223

and the corresponding direction cosines are

Vl7'

VT7' EXERCISE

Give a set of equations in symmetric form of the line intersection of the following two planes, and determine its direction cosines, - 2y z = 5, x y z = 1 (a) 3z (ft)

(c)

(d) (e)

(/) (g)

+ + + + +

z = s - 2y 2z Zx 4y - 3s 6s 2y 4x - j/ + 2 = * - V - 32 = - 2z 4x 3y

+

+ +

6,

= =

re

1,

-

z

x

-

=

2

y

+

z

=

6

-y+2 = 6 -5, x - 3^ = 4 2, 2x + y - 2 = -3 2, 2x + 2y = x + 2y - 3z = 4 1

,

SOLID ANALYTIC GEOMETRY

3* 8.

+

cz

Angle between a line and a plane. + d = and the line

x-

xi

(14)

_

-

y

ai

=

yi

[Chap. 2

Let the plane ax

s

bi

-

+

by

z\

ci

between the line and the be given. We require the angle This angle is defined to be the complement of the angle plane.

between the directed normal to the plane and the line and is not unique unless the line is also directed. When both are directed, we will have 6

where we have now made

=

cos

and

X, M> v

Xi, pi, v\

unique.

Then

and

sin

sin

(16)

=

XXi

+

MMI

+

vv\.

EXERCISE Find the sine of the angle between the following lines and planes: - 2y + 4z = 1, 3x + 4 = 2y - 1 = 6z + 7 (a) 3z

+

(6)

x

(c)

3z

2y

+

2y

Pencils

9.

+

biy define

+

CiZ

two

-

=

2z

-

62

of

+

=

di

=

4,

planes.

Let

and g(x,

distinct planes.

(17) is

=

=

=

2,

sf(x, y, 2)

= the

y, z)

2-3 equations f(x, yS$) = a\x = a 2x b*y c& d* =

+

+

+

Then the equation

+

tg(x, y, 2)

=

called the equation of the pencil of planes determined

by the

two given planes. Indeed, formula (17) is an equation of a plane for all real numbers s and t which are such that the vector When the two given planes are s(a,i, 61, Ci) + t(a 2 6 2 02) 5^ 0. ,

,

not parallel, they intersect in a line L; formula (17) represents a plane Q(s,

41

Thus we do not include planes

I.

in the set of

surfaces called cones.

A polynomial f(x,

to be homogeneous

y, z) is said

terms of

if all

have the same degree n. A homogeneous polynomial of is called an nic form. We shall study quadratic forms, forms of degree two, later. If f(x, y, z) is homogeneous of

f(x, y, z)

degree n i.e.,

degree n, then

n

>

7

f (x

+

x', y', z'

(x

1.

n

f(x, y, z).

Also /(O,

0, 0)

=

+

XQ, y'

then

,

,

?/o,

if

yo, z'

+

=

f (x, y, z)

y', z')

homogeneous of degree n > 1 in an equation of a cone with vertex at

ZQ) is

is

z

zo),

which moves the origin to

of axes,

replaces /(z, y, z) by g(x' y', z'}. is on the surface S defined

Since g(Q,

,

V

the point

Q =

has the property that g(x',

f (x, y, z)

//

). y For a translation

(XQ,

t

0.

Theorem

=

=

f(tx, ty, tz)

7

(xi, yi

=

=

f(x, y, z)

If

0.

and

a point on S, then g(xi, yi, Zi)

is

zi)

,

by

0, 0)

V = = 0,

for every t. But t(xi', */) = 7 7 = z of of is form a set the vector ) parametric (x y VL, Zi) equations of the line joining V to Q, and we have proved that all points of this line are on S, S is a cone.

g(txi', tyi', fe/)

t

n

7

g(xi',

?/i

,

1

,

',

The

cones, which consist of all points on the lines joining the vertex to the points of a fixed plane curve, are of particular Let us then derive the equations of such interest in geometry.

surfaces in the case where the given curve

plane and (XQ,

?/o,

cone

if

is

ZQ)

equation the vertex.

and only

if

P

a point on the curve. (2)

X

-

XQ

=

-

t(Xi

a curve in the

is

is

(x,

Z Q 7* 0.

ZQ(X

P

assume that

y

XQ),

-

y)

7/o

= %l ~

2/o),

is

ZQ

tZQ.

not a point of the x, y plane and thus t and obtain

Then we may eliminate XQ)

=

-

(Xi

XQ)(ZQ

-

z)

,

Z Q (y

y

)

=

(j/1

~

J/o)

(ZQ

We /ox

(3)

y

0.

Z

We

x,

Suppose that PO = Then a point P = (x, y, z) is on the P = t(P\ PO), where Pi = (x\, y\, 0) is Consequently

its

solve for Xi

xi

=

and

^ H x

,

y\ to obtain

ZQ(X

-

XQ)

ZQ

Z

=

z Q (y

__ 9

7/1

t/o

-

y

-\

ZQ

Z

)

-;

^

Z).

SOLID ANALYTIC GEOMETRY

42

[Chap. 3

and substitute these formulas in (xi, y^ = 0. The resulting equation in x y, z is an equation of the given cone. y

ILLUSTRATIVE EXAMPLES I. Find an equation of the elliptic cone with vertex at 2 base the curve 4# 2 y = 1.

1, 3),

(2,

and

+

Solution

We use

formula

(3)

and write

-

3(x

Then

equation

=

z)zi

(3

3#

-

2)

(3

2z,

0)1/1

=

3y

+

3(y

+

z,

1)

so that the required

is

4(3*

-

22)

2

+

(3ff

+

2

2)

=

(3

-

2

2)

.

The

solution should be left in this form since the equation is not essentially simplified when the indicated operations are carried out.

Find an equation of the hyperbolic cone through the point (3, 0, 0) 2 2 = 1 the and (?/ /4) hyperbola whose equations are z = 0, (x /9) II.

.

Solution

y

We use the method of derivation of formula (3) to write x - 8)2,, = ty,, z = tzi. Then -3 1. one line of S through each point of S. :

For if P is any point of and therefore assume that

+ /n -i(s, y) + of degree k.

x

=

ta,

0(0

=

y

/S,

we may

/(O, 0)

=

translate the origin to

Then

0.

f(x, y)

tft,

z

line L through the origin P = ty, and = */n(, 0) + f^Vn-iCa,

P y)

has equations

/?)++

0)

/(te,

f

n (x,

+ fi(x, y) where each/fc(, y) is homogeneous

-

Every

=

=

1 and that f(x, y) is and the only line through P is irreducible. Hence, a = ft = the z axis. EXERCISE that

ftx

f(x, y)

Show that a surface f(x, y) = determined by a homogeneous polynomial f(x, y) of degree n consists of r ^ n planes through the z axis or is a point surface. 11.

+ -Pi is

ty,

=

Tangent z

=

+

Zo

y = yQ in a point

intersect a surface /(x, y, 2) real number

=

if t\ is

+

x

XQ

=

+ = ^o + ,

=

T/O

ft?,

2o

t%,

+

ti

which