, FP = O'P sin PO'F = r sin (^ + cp) ;
hence, substituting, we have,
X = a -\- r cos (6 -\- (f))\
y = b -\- r sin (^ + /o\
y ^ r siu )
will be the required equations of transformation.
35. To find the equations of transformation from a polar
system to a rectangular system.
1°. When the pole and origin are coincident, and when the
initial line coincides with the X-axis.
From equations (3), Art. 34, we have, by squaring and
adding r"^ = x^ -\- y^ ; and,
by division tan r=y- .
X
for the required equations. We have, also, from the same
equations,
cos ^ = - = ^ : sin ^ = ^ = y
2°. When the pole and origin are non-coinciderit, and when
the initial line is parallel to the X-axis.
From equations (2) of the same article, we have, by a simi-
lar process,
r^ = (a; — a)" -\- {y — by
tan. 6 = ^L^^ j also
cos d =
X — a
X — a
r V(x - af + {y- by ;
sin 6
__y —b y — b
^- V(a; — ay + (y — by
for the required equations.
TRANSFORMATION OF CO-ORDINATES. 57
EXAMPLES.
1. The rectangular equation of the circle is x^- -{- 7/ = a^ ■
what is its polar equation when the origin and pole are coin-
cident and the initial line coincides with the X-axis ?
Ans. r = a.
2. The equation of a curve is (x^ + 2/^) " = *" (^^ ~ V"^) '■> es-
quired its polar equation, the pole and initial being taken as
in the previous example.
Ans. r^ = a^ cos 2 0.
Deduce the rectangular equation of the following curves,
assuming the origin at the pole and the initial line coincident
with the X-axis.
3. r = a tan ^ sec 6 5. r^ ^ a^ sin 2
3. 1
Ans. x'^ ^ a- y. Ans. (x^ -\- y'^Y = 2 a^xy.
4. r^ = a^ tan 6 sec^ 6 6. r = a (cos — sin 0)
Ans. x^ = a^y. Ans. x^ -{- y^ ^ a (x — y)
GENERAL EXAMPLES.
Construct each of the following straight lines, transfer
the origin to the point indicated, the new axes being parallel
to the old, and reconstruct :
1. ?/ = 3 cc -f 1 to (1, 2). 5. y = sx -{- b to (c, d).
2. 2 y - a; - 2 = to (- 1, 2). 6. y -\- 2 x = to (2, — 2).
3. i ?/ + X — 4 = to (- 2, — 1). 7. y = nix to (Z, n).
4. y -f ^ -}- 1 = to (0, 2). 8. ?/ — 4x -f c = Oto(cZ,o).
What do the equations of the following curves become when
referred to a parallel (rectangular) system of co-ordinates
passing through the indicated points ?
9. 3 ^2 + 2 ?/2 = 6, (V2, 0).
11. 9 2/^ -4:^2 =-36 (3,0). ^^-y-^^H 2'^
68 PLANE ANALYTIC GEOMETRY.
13. What does the equation x^ -{- y'^ = 4: become when the
X-axis is turned to the left tlirough an angle of 30° and the
Y-axis is turned to the right through the same angle ?
14. What does the equation x'^ — ?/2 _ ^2 become when the
axes are turned through an angle of — 45° ?
15. What is the polar equation of the curve 3/^ = 2 px, the
pole and origin being coincident, and the initial line coincid-
ing with the X-axis ?
16. The polar equation of a curve is r = a (1 -|- 2 cos 6) ;
required its rectangular equation, the origin and pole being
coincident and the X-axis coinciding with the initial line.
Ans. (x^ -\- y^ — 2 ax)^ = a^ (x^ -f y-).
Required the rectangular equation of the following curves,
the pole, origin, initial line, and X-axis being related as in
Example 16 ..
17. r2 = ^ . 20. r = a sec^ -^
Ans. x^ — ?/^ = al
18. r = a sin 0. 21. r = a sin 2 e.
19. r = ae. 22. r^ - 2 r (cos ^ + V3 sin 0) = 5.
Find the polar equations of the loci whose rectangular
equations are :
23. x^ = y^(2a — x). 25. aY = «-^* - ^^•
24. 4.a^x = y- (2 a- x). 26. x^ ^ y^ = a^.
THE CIRCLE.
59
CHAPTER V.
THE CIRCLE.
36. The circle is a curve generated by a point moving in
tlie same plane so as to remain at the same distance from a
fixed point. It will be observed that the circle as here de-
fined is the same as the circumference as defined in plane
geometry.
37. Given the centre of a circle and its radius to deduce its
equation.
Y
Fig. 21.
Let C (x', y') be the centre of the circle, and let P be any
point on the curve. Draw CA and PM || to OY and CIST || to
OX; then
(OA, AC) = (x', y') are the co-ordinates of the centre C.
(OM, MP) = (x, y) are the co-ordinates of the point P.
60 PLANE ANALYTIC GEOMETRY.
Let CF = a. From the figure, we have,
CN2 + NP2 =. CP2 ; ... (1)
But QW = (OM - 0A)2 ={x- x'Y,
NP2 = (MP - ACy = (y- y'f, and
CP2 = a"
Substituting tliese values in (1), we have,
(^ _ :^'y + (y _ y'y = «2 . . . (2)
for the required equation. For equation (2) expresses the
relation existing between the co-ordinates of any point (P) on
the circle ; hence it expresses the relation between the co-
ordinates of every point. It is, therefore, the equation of the
circle.
If in (2) we make a?' = and y' = 0, we have,
x^ -^y'' = a' . . . (3)
or, SAanmetrically, \- ^ = 1 . . . (4)
a^ a^
for the equation of the circle when referred to rectangular
axes passing through the centre.
Let the student discuss and construct equation (3). See
Art. 11.
Cor. 1. If we transpose cc^ in (3) to the second member and
factor, we have,
y"^ = (a -\- x) (a — x) ;
i.e., in the circle the ordinate is a mean "projiortional between
the segments into lohich it divides the diameter.
CoR. 2. If we take £, Fig, 21, as the origin of co-ordinates,
and the diameter Z,H as the X-axis, we have,
£0 = x' = a and ?/' = 0.
These values of x' and y' in (2) give
{x — ay -|- ?/2 = a^,
or, after reduction, x^ -{- y"^ — 2 ace = . . . (5)
for the equation of the circle when referred to rectangular
THE CIRCLE. 61
axes taken at the left hand extremity of the horizontal
diameter.
38. Every equation of the second degree between two varia-
bles, in which the coefficients of the second jiowers of the
variables are equal and the term in xy is missing, is the equa-
tion of a circle.
The most general equation of the second degree in which
these conditions obtain is
ax^ -\- aif -\- ex -\- dy -\- f = ^ . . . . (1)
Dividing through by a and re-arranging, we have,
x^ -\- - X -{- y- ^ - y = —-'- .
a a a
If to both members we now add
c2 d^
4 a^ 4 (^2 '
the equation may be put under the form
, cW f . d Y c^ + d^-4.af
2 a J \ 2 a J 4za^
Comparing this with (2) of the preceding article, we see
that it is the equation of a circle in which
G d
2^' "~2^
are the co-ordinates of the centre and
^ c^ -\- d^ - 4. af ig the radius.
2a
Cor. 1. If ax ^ + ay^ -\- cx -\- dy -\- m = be the equation
of another circle, it must be concentric with the circle repre-
sented by (1) ; for the co-ordinates of the centre are the same.
Hence, when the equations of circles have the variables in
62 PLANE ANALYTIC GEOMETRY.
their terms affected with equal coefficients, each to each, the
circles are concentric. Thus
2 X- -^ 2 if + 3 X + 4.tj -{- 25 =
are the equations of concentric circles.
EXAMPLES.
What is the equation of the circle when the origin is
taken.
1. At D, Fig. 21 ? Ans. x" ^ y" — 2 ay = 0.
2. At K, Fig. 21 ? Ans. x"^ ■^y'^ + 2 ay = 0.
3. At H, Fig. 21 ? Ans. x- + y- + 2 ax = {).
What are the co-ordinates of the centres, and the values of
the radii of the following circles ?
4. 4a;2 + 4 2/2-8x-8?/ + 2 = 0.
Ans. (1, 1), a = yi
5. ^2 ^ 2/^ -I- 4 cc _ 6 ?/ — 3 = 0.
Ans. (- 2, 3), a = 4.
6. 2 x2 + 2 2/2 _ 8 a; = 0.
Ans. (2, 0), a = 2.
7. ^2 _^ 2/2 _ 6 ^ ^ 0.
Ans. (3, 0), a = 3.
8. a;2 + ?/2 — 4 X + 8 ?/ — 5 = 0.
^?zs. (2, — 4) a = 5.
9. ic^ -|- 2/^ — '''^^ -\- n y -\- c = 0.
10. a;'^ + 2/^ = ■'^•
11. x^ — ^x = — y- — my.
12. ^2 + y2 ^ c2 + )
•IS the polar equation of the circle.
This equation might have been obtained directly from the
triangle OO'P.
CoR. 1. If 6' = 0, the initial line OX passes through the cen-
tre and the equation becomes
^2 _j_ j/2 _ 2 rr' cos = a^.
Cor. 2. If & = 0, and / = a, the pole lies on the circum-
ference and the equation becomes
r = 2 a cos 9.
CoE. 3. If 9' = 0, and r' = 0, the pole is at the centre and
the equation becomes
40. To show that the supplemental chords of the circle are
perpendicular to each other.
The supplemental chords of a circle are those chords which
pass thi'ough the extremities of any diameter and intersect each
other on the circicmference.
65
BIG. 23.
Let PB, PA be a pair of supplemental chords. We wish to
prove that they are at right angles to each other.
The equation of a line through B (— «, o) is
y = s (x + a).
For a line through A (a, o), we have
y ^ s' (x — a).
Multiplying these, member by member, Ave have
y^ = ss' (x- — rt.^) . . . (ci)
for an equation which expresses the relation between the
co-ordinates of the point of intersection of the lines.
Since the lines must not only intersect, but intersect 07i the
circle whose equation is
y^ = ^2 — ^2^
this equation must subsist at the same time with equation (a)
above ; hence, dividing, we have
1 = — ss',
or, 1 + ss' = . . . (1)
Hence the supplemental chords of a circle are perpendicular
to each other.
Let the student discuss the proposition for a pair of chords
passing through the extremities of the vertical diameter.
66
PLANE ANALYTIC GEOMETRY.
41. To deduce the equation of the tangent to the circle.
Fig. 24.
Let CS be any line cutting the circle in the points P' (x', y')^
P" {x", y"). Its equation is
y-y' - 'i' I 'C' (^ - ^')- (^^*- 26, (4) ).
Since the points {x', y'), (x", y") are on the circle, we have
the equations of condition
x"" + ?/'" = ft" ... (1)
x"^ + /'2 = a^ . . . (2)
These three equations must subsist at the same time ; hence,
subtracting (2) from (1) and factoring, we have,
(^' + ^") (^' - ^") + {y' + y") {y' - y") = o ;
y' — y" ^ x' + x"
" x' — x" ~ / + y" '
Substituting in the equation of the secant line it becomes
x' + x"
y -y
(x — x')
(3)
y' + y'
If we now revolve the secant line upward about P" the
point P' will approach P" and will finally coincide with it
when the secant CS becomes tangent to the curve. But when
THE CIRCLE. . 67
F' coincides with V", x' = x" and ?/ = y"; hence, substituting
in (3) we have,
y-.f^-'^ix-x"), ... (4)
or, after reduction,
xx" + yij" = a?; . .. (5)
or, symmetrically,
^^ I yy = 1 ... (6)
or a^
for the equation of the tangent.
ScHOL. The Sub-tangent for a given point of a curve is the
distance from the foot of the ordinate of the point of tangency
to the point in which the tangent intersects the X-axis ; thus,
in Fig. 24, AT is the sub-tangent for the point F". To find
its value make y = in the equation of the tangent (5) and
we have,
OT == cc = — .
x"
But AT = OT - OA ^ ^ - x"
x"
.-. sub-tangent = ; = ^ .
x'' x '
42. To deduce the equation of the normal to the circle.
The normal to a curve at a given point is a line perpen-
dicular to the tangent drawn at that point.
The equation of any line through the point F" (x", y") Fig.
24, is y - 2j" = s (x - x") ... (1)
In order that this line shall be perpendicular to the tangent
F"T, we must have
1 -j_ S5' = 0.
x" v"
But Art. 41, (4) s' = — ; hence, we must have s = -i— .
y" x"
68 PLANE ANALYTIC GEOMETRY.
Therefore, substituting in (1), we have,
y -y" = ^'^ri^ - ^") • • • (2) ;
X
or, after reduction,
yx" - xy" = ... (3)
for the equation of the normal.
We see from the form of this equation that the normal to
tlie circle passes through the centre.
ScHOL. The SuB-NOKMAL for a given point on a curve is
the distance from the foot of the ordinate of the point to the
point in which the normal intersects the X-axis. In the circle,
we see from Fig. 24 that the
Sub-normal = x" .
43. By methods precisely analogous to those developed in
the last two articles, we may prove the equation of the tangent
to
ix — x'y + (^ — y'Y = a}
to be
{X - x') ix" - x') + {y- y') {y" - y') = a? . . . (1)
and that of the normal to be
{y -y") {x" - x') - (^ - ^") {y" -y') = o . . . (2)
Let the student deduce these equations.
EXAMPLES.
1. What is the polar equation of the circle ax^ + ay^ -\- ex -\-
dy +/= 0, the origin being taken as the pole and the X-axis
as the initial line ?
Ans. ?'^ + ( - cos ^ -| — sin ^ )?•-)- =^ = 0.
^a a a
2. What is the equation of the tangent to the circle
x^ -\- y^ = 25 at the point (3, 4) ? The value of the sub-
tangent ? A71S. 3x -\- 4,y = 25; J/.
3. What is the equation of the normal to the circle
x^ -\- y^ = 37 at the point (1, 6) ? What is the value of the
sub-normal ? Ans. y = 6x; 1.
THE CIRCLE. 69
4. What are the equations of the tangent and normal to the
circle a?^ + t/^ = 20 at the point whose abscissa is 2 and ordi-
nate negative ? Give also the values of the sub-tangent and
sub-normal for this point.
Ans. 2x -4:ij = 20; 2?/ + 4x = 0;
Sub-tangent = 8 ; sub-normal = 2.
Give the equations of the tangents and normals, and the
values of the sub-tangents and sub-normals, to the following
circles :
5. x^-{-i/ = 12, at (2, + V8).
6. a;2 4- 2/2 = 25, at (3, - 4).
7. X- -{-if = 20, at (2, ordinate +).
8. x^ -\- f = 32, at (abscissa -f, — 4).
9. x^ -{- f = a^, Sit (b, c).
10. x'^ -\- f = m, at (1, ordinate -|-).
11. ic- -)-?/- = k, at (2, ordinate — ).
12. :k- -f- y2 = 18, at (m, ordinate -(-),
13. Given the circle x^ -\- y^ = 45 and the line 2-1/ -\- x = 2;
required the equations of the tangents to the circle which are
parallel to the line.
J (3x + 62/ = 45.
^*- \3x-\-6y = -45.
14. What are the equations of the tangents to the circle
x^ -{- f ^= 45 which are perpendicular to the line 2 y -\- x = 2?
, \oy — Qx = 45.
\ Q X — S y = 45.
16. The point (3, 6) lies outside of the circle x^ -\- y^ ^ 9;
required the equations of the tangents to the circle which
pass through this point.
, ( X = 3.
^^^- Uy-3x = W.
70 PLANE ANALYTIC GEOMETRY.
17. What is the equation of the tangent to the circle
(x - 2)2 + (y - 3)2 = 5 at the point (4, 4) ?
Ans. 2 X -\- y = 12.
18. Tlie equation of one of two supplementary chords of
the circle a;^ + y^ = 9 is t/ = § a; + 2, what is the equation
of the other ?
Ans. 2 ?/ + 3 .X = 9.
19. Find the equations of the lines which touch the circle
(x — a)'^ ^ (fj — by ^= r- and which are parallel to y = sa: + c.
20. The equation of a circle is a:;^ + 2/^ — 4 x + 4 ?/ = 9 ;
required the equation of the normal at the point whose
abscissa = 3, and whose ordinate is positive.
Ans. 4:X — y = 10.
44. To find the length of that portion of the tangent lying
bettveen any point on it and the point of tangency.
Let (xi, yi) be the point on the tangent. The distance of
this point from the centre of the circle whose equation is
(x — x'Y + (y — y'Y = o?- is evidently
V(a;, - xy + (2/1 - y'f. See Art. 27, (1).
But this distance is the hypothenuse of a right angled tri-
angle whose sides are the radius a and the required distance
d along the tangent ; hence
d' = ix, - xy- + (yi - yj - a^ ... (1)
Cor. 1. If x' = and y' = 0, then (1) becomes
d' = x^ + y,2 _ a^ . . . (2)
as it ought.
45. To deduce the equation of the radical a.xis of tivo given
circles.
The Radical axis of two circles is the locus of a point
from which tangents drawn to the two circles are equal.
THE CIRCLE. 71
Fig. 25.
Let {x — x'Y + (i/ — y'Y = a%
(x — x"y -\- (jj — y") = b^ be the given circles.
Let P (xi, iji) be any point on the radical axis ; then from
the preceding article, we have,
d^ = (xi -X'Y + (7/1 - 2/')
l\2
a^
d'- = (Xi — x"y + (yi - y"y — P
.-. by definition (x^ — x')- + (yi — ?/')- — a" = (x^ — x"Y
+ (2/1 ~ y'Y' ~ ^^j hence, reducing, we have,
2 (a;" — x') a;i + 2 (?/" — y') y^ = x'"'^ — x'- + y"^ — y'^
-\- a- — b~.
Calling, for brevity, the second inember m, we see that
(iCi, ?/i) will satisfy the equation.
2 {x" — x')x +2 (y" — y')y = m . . . (1)
But (ccj, ?/i) is awy point on the radical axis ; hence every
point on that axis will satisfy (1). It is, therefore, the re-
quired equation.
CoK. 1. If c = and c' = be the equation of two circles,,
then, c — c' =
is the equation of their radical axis.
72 PLANE ANALYTIC GEOMETRY.
Cor. 2. From the method of deducing (1) it is easily seen
that if the two circles intersect, the co-ordinates of their points
of intersection must satisfy (1) ; hence the radical axis of two
intersecting circles is the line joining their points of intersection,
PA, Fig. 25.
Let the student prove that the radical axis of any two
circles is perpendicular to the line joining their centres.
46. To shoio that the radical axes of three given circles in-
tersect in a common ■point.
Let c = 0, c' = 0, and c" =
be the equations of the three circles.
Taking the circles two and two we have for the equations of
their radical axes
c _ c' = . . . (1)
c-c" = . . . (2)
c' - c" = . . . (3)
It is evident that the values of x and i/ which simultaneously
satisfy (1) and (2) will also satisfy (3) ; hence the proposition.
The intersection of the radical axes of three given circles is
called The Kadical Cejs^tre of the circles.
EXAMPLES.
Find the lengths of the tangents drawn to the following
circles :
1. (x - 2)2 + {y- 3)- = 16 from (7, 2).
2. a;' + (y + 2)^ = 10 from (3, 0).
3. (x — af ^y- = V1 from (h, c).
4. a;' + ?/- - 2 ic + 4 2/ = 2 from (3, 1).
5. ^2 + 2/' = 25 from (6, 3).
Ans. d = VlO.
Ans. d = V3 .
Ans. d = V20.
THE CIRCLE. 73
6. a;2 + 3/2 _ 2 x = 10 from (5, 2).
Ans. d = 3.
7. (x. - ay -^(ij -hy = c from {d, /).
8. x2 + 2/2 _ 4 7/ = 10 from (0, 0).
Give the equations of the radical axis of each of the follow-
ing pairs of circles :
9. Ux- 2)2 + (y - 3)2 - 10 = 0.
t(a:+3)2 + (2/ + 2)^-6 = 0.
Ans. 5 X -\- o y -{■ 2 = 0.
10. J- a;2 + 2/2 — 4 y = 0.
\(x-3y + t/-^ = Q. Ans. 3x = 2ij.
11. r(a;+3)2 + 2/2_2y_8=0.
\x- -\- y- — 2 y = 0. Aris. x = — J.
12. f (cc + ay + 2/2 _ c2 = 0.
|:r2 + (2/-3)2-16 = 0.
13. ( a;2 + 2/2 = 16.
I (.r — 1)' + r = «'•
14. j a;2 + (2/ - ay = g\
X(x-2y + y'- = d\
Find the co-ordinates of the radical centres of each of the
following systems of circles :
15. {{x- 3)2 + 2/2 = 16.
x' -\- y' = 9.
x2 4- (2/ - 2)2 = 25. Ans. (1,-3).
16. r cc2 _j_ 2/2 — 4 cc -)- 6 2/ — 3 = 0,
■.x--\-y'^ — 4iX^= 12.
(a;2 + 2/2 + 62/ = 7. ^ws. (1, -i).
17.
18.
«^ + r = «^-
(^ - 1)2 + f=9.
x" -\-y- — 2x -\-^y =
= 10.
a;2 -|- 2/^ "" ^-"^ = <^-
a-2 -j- 2/2 = m.
^^ + //" "~ <^'Z/ = ^•
74
PLANE ANALYTIC GEOMETRY.
47. To find the condition that a straight line y = sx -{- h
must fulfil in order that it may touch the circle x' -\- y- = a\
In order that the line may touch the circle the perpendicu-
lar let fall from the centre on the line must be equal to the
radius of the circle.
From Art. 21, Fig. 13, we have
sec )' VI + tan.-' y
P
Vi + «'
hence, a^ (1 4- s') = b'' . . . (1)
is the required condition.
CoR. 1. If we substitute the value of b drawn from (1) in
the equation y = sx -\- b, we have
y ^ sx -^a Vl + s'^ . . . (2)
for the equation of the tangent in terms of its slope.
48. Two tangents are drawn from a- point ivithout the circle ;
requii'ed the equation of the chord joining the points of tangency.
Fig. 26.
Let P' {x', y') be the given point, and let P'P", PT, be the
tangents through it to the circle.
THE CIRCLE. 75
It is required to deduce the equation of PP''.
The equation of a tangent through P" (x", y") is
xx" . yy" _ -I
Since P' (x', y') is on this line, its co-ordinates must satisfy
tlie equation ; hence
^'^'' I y'v" ^ 1
The point {x", y"), therefore, satisfies the equation
^ + ^ = 1; •■•(!)
.-. it is a point on the locus represented by (1). A similar
course of reasoning will show that P is also a point of this
locus. But (1) is the equation of a straight line ; hence, since
it is satisfied for the co-ordinates of both P" and P, it is the
equation of the straight line joining them. It is, therefore,
the required equation.
49. A chord of a given circle is revolved about one of its
points ; required the equation of the locus generated by the
point of intersection of a pair of tangents drawn to the circle at
the points in which the chord cuts the circle.
Let P' (x', i/), Fig. 27, be the point about which the chord
P'AB revolves. It is required to find the equation of the
locus generated by Pi {xi, yi), the intersection of the tangents
APi, BPi, as the line P'AB revolves about P'.
Prom the preceding article the equation of the chord AB is
^1^ I yiV __ 1
a
Since P' (x', y') is on this line, we have
^\^ I yai
+
a' a'
hence ^ + "^ = 1 • • • (1)
a^ a^
76
PLANE ANALYTIC GEOMETRY.
is satisfied for the co-ordinates of Pi (x^, y^ ; hence Pj lies on
the locus represented by (1). But Pi is the intersection of
any pair of tangents drawn to the circle at the points in
Fig. 27.
which the chord, in any position, cuts the circle ; hence (1)
will be satisfied for the co-ordinates of the points of intersec-
tion of every pair of tangents so drawn.
Equation (1) is, therefore, the equation of the required
locus. We observe that equation (1) is identical with (1) of
the preceding article ; hence the chord PP'' is the locus whose
equation we sought.
The point P' (x', y') is called the pole of the line PP"
(^\yly = \\ and the line PP" /^ + ^ = 1 Vs called
\a^ a^ J ya^ a^ J
THE POLAR of the poiut P' {x', y') with regard to the circle
a^ a^
= 1.
THE CIRCLE. 77
As the principles here developed are perfectly general, the
pole may be tvithoiit^ on, or within the circle.
Let the student prove that the line joining the pole and the
centre is perpendicular to the polar.
]SroTE. — The terms ^ole and i:)olaT used in this article have
no connection with the same terms used in treating of polar
co-ordinates, Chapter I.
50. If the polar of the point P' (x',y'), Fig. 27, passes through
Pi (^i> Vi), then the polar of F^ (x^, y^) ivill pass through F'
(x', y').
The equation of the polar to P' (x', y') is
x'x . y'y -|
a^ a-
In order that Pi {x^, t/i) may be on this line, we must have,
^'^1 I y!]h ^ 1,
tt"' a
But this is also the equation of condition that the point
P' (x', y') may lie on the line whose equation is
^1^ ^^ii/ = 1
9 I 9
a'^ a^
But this is the equation of the polar of P^ (xi, y^) ; hence
the proposition.
51. To ascertain the relationship hetiveen the conjugate diam-
eters of the circle.
A pair of diameters are said to he conjugate when they are
so related that when the curve is referred to them as axes its
equation will contain only the second powers of the variables.
Let cc- -f ?/2 = a2 ... (1)
be the equation of the circle, referred to its centre and axes.
To ascertain what this equation becomes when referred to
OY', OX', axes making any angle with each other, we must
substitute in the rectangular equation the values of the old
78
PLANE ANALYTIC GEOMETRY.
co-ordinates in terms of the new. Prom Art. 33, Cor. 1, we
have
X = x' cos 6 -{- y' cos go
y = a;'' sin 6 -\- y' sin go
for tlie equations of transformation. Substituting these
values in (1) and reducing, we have,
y'^ + 2x'y' cos {(f — 9) -\- x'^ = a^ . . . (2)
Now, in order that OY', OX' may be conjugate diameters
they must be so related that the term containing x'y' in (2)
must disappear ; hence the equation of condition,
cos (go — ^) = ;
.-. cp _ ^ = 90°, or cp - e = 270°.
The conjugate diameters of the circle are therefore perpen-
dicular to each other. As there are an infinite number of
pairs of lines in the circle which satisfy the condition of being
at right angles to each other, it follows that in the circle there
are an infinite number of conjugate diameters.
THE CIRCLE. 79
EXAMPLES.
1. Prove that the line y = V3 x -\-10 touches the circle
.-r' -\- y'^ = 25, and find the co-ordinates of the point of tangency.
/ 5—5
Ans. Point of tangency _ -^-y/^, —
2. What must be the value of h in order that the line
y = 2 X -{- b mB,y touch the circle x^ + ?/' = 16 ?
Ans. b = ^ V80.
3. What must be the value of s in order that the line
y = sx — 4: may touch the circle x- -\- y"^ = 22 _
Ans. s = J- V7.
4. The slope of a pair of parallel tangents to the circle
X- -\- y- = 1Q> is 2 ; required their equations.
Ans. W = 2x + Vi0.
]^y = 2x — V80.
Two tangents are drawn from a point to a circle ; required
the equation of the chord joining the points of tangency in
each of the following cases :
Ans. 4 X + 2 ?/ = Q.
Ans. 3 a; -f 4 2/ = 8.
Ans. X -\-o y =^ 16.
5. From (4, 2) to a;- + / = 9.
6. From (3, 4) to x"" -\- if = 8.
7. From (1, 5) to x^ + f~ = 16.
8. From (a, b) to x^ + ^f = c".
Ans. ax -\- by ^= c^.
What are the equations of the polars of the following points :
9. Of (2, 5) with regard to the circle x^ -{- y- = 16?
Ans. ^+^ = 1.
16 16
10. Of (3, 4) with regard to the circle x^ -\- y- = 9?
Ans. S X -(- 4 3/ = 9.
80 PLANE ANALYTIC GEOMETRY.
11. Of (a, h) with regard to the circle x- -\- y^ = m?
Ans. ax -\- by = m..
What are the poles of the following lines :
12. Of 2 a; + 3 2/ = 5 with regard to the circle x"^ -\- y- = 2?> 2
Ans. (10, 15).
13. Of — [- y = 4 with regard to the circle
o
16+16 = ^' ^- (2'*)-
14. Oi y = sx -\- h with regard to the circle
X
~2
a^ a^
Ans. (_^ ^
h ' b
15. Pind the equation of a straight line passing through
(0, 0) and touching the circle x'^ -\- y^ — 3 x -{- 4 y = 0.
Ans. y = — X.
^ 4
GENERAL EXAMPLES.
1. Find the equation of that diameter of a circle which
bisects all chords drawn parallel to y := sx -\- b.
Ans. sy -\- X ^ 0.
2. Eequired the co-ordinates of the points in which the
line 2y — x-{-l = intersects the circle
^^1/1 = 1
4 "^ 4
3. Find the co-ordinates of the points in which two lines
drawn through (3, 4) touch the circle
^ + i! = 1
9 9
[The points are common to the chord of contact and the
circle.]
THE CIRCLE. 81
4. The centre of a circle which touches the Y-axis is at
(4, 0) ; required its equation.
Ans. {x — 4)- -\- y- = 16.
5. Find the equation of the circle whose centre is at the
origin and to which the line y = x -\- S is tangent.
Ans. 1x' -^Itf = 9.
6. Given x~ ^ if = 16 and (x — 5)^ + ?/^ = 4 ; required the
equation of the circle which has their common chord for a
diameter.
7. Required the equation of the circle which has the dis-
tance of the point (3, 4) from the origin as its diameter.
Ans. x^ -\- y"' — o X — 4 ?/ = 0.
8. Find the equation of tlie circle which touches the lines
represented by cc = 3, ?/ ^ 0, and y ■= x.
9. Find the equation of the circle which passes through the
points (1, 2), (- 2, 3), (- 1, - 1).
10. Required the equation of the circle which circumscribes
the triangle whose sides are represented by 3/ = 0, 3 ?/ = 4 a^,
and 3^/= — 4a:;-j-6.
Ans. xJ -\- y^ — & X — 11 ?/ = 0.
11. Required the equation of the circle whose intercepts
are a and b, and which passes through the origin.
Ans. x^ + y- — ax — by = 0.
12. The points (1, 5) and (4, 6) lie on a circle whose centre
is in the line y = x — 4 ; required its equation.
Ans. 2 x^ + 2 y'- - 17 X - y = SO.
13. The point (3, 2) is the middie point of a chord of the
circle x'^ -]-?/- = 16 ; required the equation of the chord.
14. Given x^ -{- y^ ^ 16 and the chord ?/ — 4 a; = 8. Show
that a perpendicular from the centre of the circle bisects the
chord.
15. Find the locus of the centres of all the circles which
pass through (2, 4), (3, - 2).
82 PLANE ANALYTIC GEOMETRY.
16. Show that if the polars of two points meet in a third
point, then that point is the pole of the line joining the first
two points.
17. Required the equation of the circle whose sub-tangent
= 8, and whose sub-normal = 2.
Ans. x^ -\- y^ = 20.
18. Eequired the equation of ihe circle whose sub-normal
= 2, the distance of the point in which the tangent intersects
the X-axis from the origin beinr = 8.
A71S. X- -\- 1/^ ^ 16.
19. Eequired the conditions in order that the circles
ax^ -f- ay^ -\- ex -\- dy -\- e = Q and ax- -f- ajf -\- kyc. -\- ly -{- m ^
may be concentric.
Ans. c ^^ k, d = I.
20. Required the polar co-ordinates of the centre and the
radius of the circle
■ r"^ — 2r (cos ^ + VS sin 6) = 5.
Ans. (2, 60°) ; r = 3.
21. A line of fixed length so moves that its extremities
remain in the co-ordinate axes ; required the equation of the
circle generated by its middle point.
22. Find the locus of the vertex of a triangle having given
the base = 2a and the sum of the squares of its sides = 2 b^.
Ans. x"^ -\- y"^ = h^ — a^.
23. Find the locus of the vertex of a triangle having given
the base =2 a and the ratio of its sides
= — . Ans. A circle.
n
24. Find the locus of the middle points of chords drawn
from the extremity of any diameter of the circle
^ _|- l!_ = 1.
THE PARABOLA.
83
CHAPTER VI.
THE PARABOLA.
52. The parabola is the locus generated by a point moving
in the same plane so as to remain always equidistant from a
fixed point and a fixed line.
The fixed point is called the Focus ; the fixed line is called
the Directrix ; the line drawn through the focus perpendic-
ular to the directrix is called the Axis ; the point on the axis
midway between the focus and directrix is called the Vertex
of the parabola.
53. To find the equation of the j^arabola, given the focus and
directrix.
R
Y
^^„,,--^'^
B
1
D^.^-""''^
^^
D
//
\ "^ '
^
C
Fig. 29.
Let EC be the directrix and let F be the focus. Let OX,
the axis of the curve, and the tangent OY drawn at the vertex
84 PLANE ANALYTIC GEOMETRY.
0, be the co-ordinate axes. Take any point P on the curve
and draw PA ||.to OY, PB || to OX and join P and F. Then
(OA, AP) = {x, y) are the co-ordinates of P.
From the right angled triangle FAP, we liave
tf = AP2 = FP2 - FA^; ... (1)
But from the mode of generating the curve, we have
FP2 = BP- = (AO -f OD)- = (x-\- 0D)2,
and from the figure, we have
FA- = (AO - 0F)2 = (x- OF)l
Substituting these values in (1), we have
■if = (x + ODf - (x- 0F)2. . . (2)
Let DF = 2), then OD = OF = -g ; hence
y^ = ix +^-
lA" U ^P
X —
or, after reduction, ?/^= 2^j).x ... (3)
As equation (3) is true for cmy point of the parabola it is
true for every point ; hence it is the equation of the curve.
CoR. 1. If (x', y') and (x'^, y") are the co-ordinates of any
two points on the parabola, we have,
tf = 2px' and y"^ = 2 px" ;
hence y'^ : y"^ :: x' : x" ;
i.e., the squares of the ordinates of any two points on the para-
bola are to each other as their abscissas.
ScHOL. By interchanging x and y, or changing the sign of
the second member, or both in (3), we have
y"^ = ~ 2px for the equation of a parabola symmetrical
with respect to X and extending to the left of Y;
cc^ = 2 py for the equation of a parabola symmetrical with
respect to Y and extending above X.
£c^ = — 2py for the equation of a parabola symmetrical
with respect to Y and extending below X.
Let the student discuss each of these equations. See
Art. 13.
THE PARABOLA. 85
54. To construct the parabola, given the focus and directrix.
Fig. 30.
First Method. — Let DR be the directrix and let F be the
focus.
From F let fall the perpendicular FD on the directrix ; it
will be the axis of the curve. Take a triangular ruler ADC
and make its base and altitude coincide with the axis and
directrix, respectively. Attach one end of a string, whose
length is AD, to A ; the other end to a pin fixed at F. Place
the point of a pencil in the loop formed by the string and
stretch it, keeping the point of the pencil pressed against the
base of the triangle. Now, sliding the triangle up a straight
edge placed along the directrix, the point of the pencil will
describe the arc OP of the parabola ; for in every position of
the pencil point the condition of its being equally distant
from the focus and directrix is satisfied. It is easily seen, for
instance, that when the triangle is in the position A'D'C that
FP = PD'.
Second Method. — Take any point C on the axis and erect
86 PLANE ANALYTIC GEOMETRY.
the periDendicular P'CP. Measure the distance DC. With F
as a centre and DC (= FP) as a radius describe the arc of a
circle, cutting P'CP in P and P'. P and P' will be points of
the parabola. By taking other points along the axis we may,
by this method, locate as many points of the curve as may be
desired.
55. To find the Latiis-rectum, or parameter of the jjarabola.
The Latus-Rectum, or Parameter of the parahola, is the
double ordinate 2)cissin(/ through the focus.
The abscissa of the points in which the latus-rectum pierces
the parabola is x =-^- .
Making this substitution in the equation
if = 2px
we have
■ 2/^ = 2p|
= F
Hence
2y=2p.
CoR. 1. Forming a proportion from the equation
2/2 = 2];jx, .
we have x : y :: y : 1 p ;
i.e., the latxis-rectum of the jparahola is a third p>'^'02^ortional to
any abscissa and its corres^yonding ordinate.
EXAMPLES.
Find the latus-rectum and write the equation of the parab-
ola which contains the point :
1. (2,4). 3. (a,b).
Ans. 8, y^ = S x. Ans. —, y'^ = — x.
a a
2. (-2,4). 4. {-a, 2).
Ans. — 8, ?/2 = — 8 X. Ajis. ? 2/" = ^•
a a
5. What is the latus-rectnm of the parabola x'^^2py?
How is it defined in this case ?
THE PARABOLA. 87
6. What is the equation of the line which passes through
the vertex and the positive extremity of the latus-rectum of
any parabola whose equation is of the form ?/^ = 2 px ?
Ans. y =^ 2 X.
7. The focus of a parabola is at 2 units' distance from the
vertex of the curve ; what is its equation
(a) when symmetrical with respect to the X-axis ?
\b) " " " " " " Y-axis?
Ans. (a) y^ = 8 cc, (b) cc^ = 8 y.
Construct each of the following parabolas by three differ-
ent methods.
8. ?/2 = 8 X. 10. x^ = 6i/.
9. y" = — 4: X. 11. ic- = — 10 y.
12. What are the co-ordinates of the points on the parabola
y^ = Q X where the ordinate and abscissa are equal ?
Ans. (0, 0), and (6, 6).
13. Required the co-ordinates of the point on the parabola
x^ — 4:y whose ordinate and abscissa bear to each other the
ration 3 : 2. Ans. (6, 9).
14. What is the equation of the parabola when referred to
the directrix and X-axis as axes ? Ans. y'^ = 2px — iJ^.
Find the points of intersection of the following :
15. y" ^= 4x and 2 y — cc = 0.
Ans. (0, 0), (16, 8).
16. 03^ = 6 y and ?/ — a; — 1 = 0.
17. ?/2 = — 8 a; and x + 3 = 0.
18. ?/- = 2 ic and x" -\- y"- := 8.
Ans. (2, 2), (2, - 2).
19. a;2 = _ 4 ?/ and 3 x^ + 2 y'-^ = 6.
20. x- = -iy and ?/- = 4 x.
88 PLANE ANALYTIC GEOMETRY.
56. To deduce the polar equation of the ijarabola, the focus
being taken as the pole.
The equation of the parabola referred to OY, OX, Fig. 29, is
y- = 22:)x ... (1)
To refer the curve to the initial line FX and the pole F
-2 , ) we have for the equations of transformation, Art. 34,
Cor. 1,
ic = -2 + r cos
y = r sm 6.
Substituting these values in (1), we have
r^ sin' =^ p' -\- 2 pr cos 6.
But sin- = 1 — cos- 6 ;
.-. 7- := j;;2 _j_ 2 237' COS 6 + '"^ cos^ ^ — (^ -j- /• cos Oy,
,-. r ^ p -\- r cos 0,
or, solving,
r = -^ ... (2)
1 - cos ^ ^ ^
is the required equation.
We might have deduced this value directly as follows :
Let P (?', &) Fig. 29 be any point on the curve ; then
FP = DA = DF + FA =p + r cos ^ ;
i. e., ?* = j9 -j- r cos Q.
Hence r = ^ .
1 — cos
Cor. 1. If 6 = 0, r = co.
li e = 90°, r=p.
If 6 = 180°, r = I .
If ^ = 270°, r = p.
li d = 360°, r = oo.
An inspection of the figure will verify these results.
THE PARABOLA. 89
57. To deduce the equation of the tangent to the •parabola.
If {x\ y'), {x", y") be the points in which a secant line cuts
the parabola, then
y-y'= y^-^^, (x-x') ... (1)
will be its equation. Since (x', y'), (x", y") are points of the
parabola, we have
y'^ = 2px' ... (2)
y"^ = 2px" ... (3)
These three equations must subsist at the same time ;
hence, subtracting (3) from (2) and factoring, we have
?/ - y" _ 2^
i.e., x' — x" y' + y"
Substituting this value in (1), the equation of the secant
becomes
y-y' = -^^ (x-x') . . . (4)
y +y
When the secant, revolved about (x", y"), becomes tangent
to the parabola {x', y') coincides with {x", y") ; hence x' = x",
y' = y". Making this substitution in (4), we have,
y - 2/" = 4 (^ - ^") (5)
y
or, simplifying, recollecting that y""^ = 2 ^^x", we have
^Jy" =p(x + x") ... (6)
for the equation of the tangent to the parabola.
58. To deduce the value of the sub-tangent.
Making ?/ = in (6), Art. 57, we have
x = -x" = OT, (Fig. 31)
for the abscissa of the point in which the tangent intersects
the X-axis. But the sub-tangent CT is the distance of this
point from the foot of the ordinate of the point of tangency ;
i.e., twice the distance just found; hence
Sub-tangent = 2 x" \,
90
PLANE ANALYTIC GEOMETRY.
i.e., the sub-tangent is equal to double the abscissa of the point
of tangency.
59. The preceding principle affords us a simple method of
constructing a tangent to a parabola at a given point.
Let P" (x", y") be any point of the curve. Draw the ordi-
nate P"C, and measure OC. Lay off OT = OC.
Fig. 31.
A line joining T and P" will be tangent to the parabola
at P".
60. To deduce the equation of the normal to the parabola.
The equation of any line through P" (x", y") Fig. 31, is
y-y" = s(x- x") ... (1)
We have found Art. 57, (5) for the slope of the tangent P"T
hence, for the slope of the normal P^IST, v/e have
p '
THE PARABOLA. 91
Substituting this value of s in (1), we have
y -y" = - ^ (X - X") ... (2)
for the equation of the normal to the parabola.
61. To deduce the value of the sub-normal.
Making y = in (2) Art. 60, we have, after reduction,
x=p-\-x" = 01^; Eig. 31,
.-. Sid)-normal = ISTC =p -]- a?" — x'' = p.
Hence the sub-normal in the parabola is constant and equal
to the semirparameter FB.
62. To show that the tangents drawn at the extremities of
the latiis rectum, are perpendicular to each other.
The co-ordinates of the extremities of the latus-rectum are
~ , p\ for the upper point, and [-^ ,— p j for the lower point.
Substituting these values successively in the general equa-
tion of the tangent line. Art. 57 (6), we have
yp =p{x +1
or, cancelling,
y = x-\-P...(l)
y=-x-^ . . . (2)
for the equations of the tangents. As the coefficient of x
in (2) is minus the reciprocal of the coefficient of x in (1), the
lines are perpendicular to each other.
Cor. 1. Making ?/ = in (1) and (2), we find in each case
that X ^ — ^ ; hence, the tangents at the extrem,ities of the
-yp=p{x+l
92
PLANE ANALYTIC GEOMETRY.
latus-rectum and the directrix meet the axis of the parabola
in the same point.
The values of the coefficients of x in (1) and (2) show that
these tangent lines make angles of 45° with the X-axis.
63. To deduce the equation of the parabola when referred to
the tangents at the extremities of the latus-rectum as axes.
Fig. 32.
The equation of the parabola when referred to OY, OX, is
2/2 = 2px . . . (1).
We wish to ascertain what this equation becomes when the
curve is referred to DY', DX', as axes.
Let P' (x', y') be any point of the curve ; then, Fig. 32
(OC, CP') = (x, y), and (DC, C'P') = {x', y').
Prom the figure, we have,
OC -- DC - DO = DK + CM - DO ;
THE PARABOLA. 93
but
DK = x'gos 45° = -^, CM = /cos 45° =^, DO =|
V2 V2 2
mce
X
X
V2
V2
V
2
We
have,
also,
11
CP' =
_ y'
= MP'-
x'
C'K;
I.e.,
V2 V2
Substituting the values of x and y in (1), we have,
\{:U' -x'Y ^^{x' ^y') -p'^ . . . (2)
In order to simplify this expression let DP = a ; then from
the triangle DPF, we have,
DF = j9 = a cos 45° = -^ .
V2i
Substituting this value of j) in (2) and multiplying through
by 2, we have, {y' — x'Y = 2a (x' -\- y') — a-,
or, tj'^ + x'-^ - 2 x'y' - 2 ax' -2aij' + a^ = 0.
Adding 4 x'y' to both members, the equation takes the form
(x' +y' -ay = 4.x'y',
or x' -\- y' — a ^ ^2 a;'^ t/''^ ;
.-. transposing, x' -}^2 x'^ y'^ -\- y' ^ a;
.: x'^ ^y'}^= ^a^-, . . . (3)
or, symmetrically, dropping accents,
^;±2^=±l...(4)
a^ a^
is the required equation.
94 PLANE ANALYTIC GEOMETRY.
EXAMPLES.
1. What is the polar equation of the parabola, the pole
being taken at the vertex of the curve ?
A71S. r ^= 2i 2) cot 6 cosec 6.
Find the equation of the tangent to each of the following
parabolas, and give the value of the subtangent in each case :
2. 7/2 = 4 £c at (1, 2). Ans. y = x + l; 2.
3. cc^ = 4 ?/ at (— 2, 1). Ans. x + y -i-l=0; 2.
4. y^ = — 6xat (— 6, ord +). Ans. 2y -\-x = 6', 12.
5. x^ = — 8 y a.t (abs -{-, — 2). Ans. x -\- y = 2; 4.
6. y^ = 4 ax at (a, — 2 a).
7. 7/2 = 7JIX at (m, ?7i).
8. iC' = —2^1/ ^t (^bs +, — p).
2py at ^a^-s — , I j
Write the equation of the normal to each of the following
parabolas :
10. To 7/2 = 16 ic at (1, 4).
11. To x' = — lOy at (abs +, - 2).
12. To 7/2 = — mx at (— m, vi).
13.
To ic2 = 2 7n,y at ( a^s — , — ]
14. The equation of a parabola is x^ J- y^ = J- a^ ; what
are the co-ordinates of the vertex of the curve ?
Ans. na^-a
\4: '4
15. Given the parabola y^ = 4x and the line y — x = ;
required the equation of the tangent which is,
(a) parallel to the line,
(b) perpendicular to the line.
Ans. (a) y = x -\- 1, (b) y -\- x -\- 1 = 0.
THE PARABOLA. 95
16. The point (—1, 2) lies outside the parabola y~ = (Sx;
what are the equations of the tangents through the point to
the parabola ?
17. The point (2, 45°) is on a parabola which is symmetri-
cal with respect to the X-axis ; required the equation of the
parabola, the pole being at the focus.
A71S. / = (4 - 2 V2) X.
18. The subtangent of a parabola = 10 for the point (5, 4) ;
required the equation of the curve and the value of the sub-
normal.
Ans. tf = —x\ p .
64. The tangent to the parabola makes equal angles with the
focal line drawn to the j^oint of tangency and the axis of the
curve.
From Fig. 31 we have,
FT = FO -f OT = £ -f x'
2^
We have, also,
P
FP" = DC = DO + OC = ^ + x".
.-. FT = FP".
The triangle FP"T is therefore isosceles and
jrp"T = FTP".
65. To find the condition that the line y = sx -{- c must fulfil
in order to touch the parabola y^ = 2 px.
Eliminating y from the two equations, and solving the
resulting equation with respect to x, we have,
p — sc ^^-\/ {cs — pY — cV
• • (1)
for the abscissae of the points of intersection of the parabola
and line, considered as a secant. When the secant becomes
96 PLANE ANALYTIC GEOMETRY.
a tangent, these abscissas become equal ; but the condition for
equality of abscissas is that the radical in the numerator of
(1) shall be zero ; hence
(cs — p)- — c^s^ = 0,
or, solving c = -^
"" 2 s
is the condition that the line must fulfil in order to touch the
parabola.
Cor. 1. Substituting the value of c in the equation
y = sx -\- c,
we have, y = sx -\- — — ... (2)
^ s
for the equation of the tangent in terms of its slope.
66. To find the locus generated by the intersection of a tan-
gent, and a perpendicular to it from the focus as the point of
tangency moves around the curve.
The equation of a straight line through the focus | -^, ] is
y=s'(x-P\...{l)
In order that this line shall be perpendicular to the tangent
y = sx -\- ^ . . . (2)
^ 2s ^ ^
we must have, s' = — - ;
s
hence y = — - a; + -^ ... (3)
s 2 s
is the equation of a line through the focus perpendicular to
the tangent. Subtracting (3) from (2), we have
s -\ — ) cc ^ 0,
or, ic = 0,
for the equation of the required locus. But cc = is the
equation of the Y-axis ; hence, the perjyendiculars from the
THE PARABOLA. 97
focus to the tangents of a parabola intersect the tangents on the
Y-axis.
67. To find the locus genei'ated by the inter'section of ttvo tan-
gents which are. 'perpendicular to each other as thepoints of tan-
gency moves around the curve.
The equation of a tangent to the parabola is, Art. 65 (2),
y = 5^ + -^ • • • (1)
The equation of a perpendicular tangent is
y = _l^_^. . . (2)
Subtracting (2) from (1), we have,
S ) \ S I \L
x = -l.. . (3)
is the equation of the required locus. But (3) is the equa-
tion of the directrix; hence, the intersection of all pjerpendicu-
lar tangents drawn to the parabola are points of the directrix.
68. Two tangents are drawn to the parabola from a point
without ; required the equation of the line joining the points of
tangency.
Let (x', y') be the given point without the parabola, and let
(x", y"), (x2, 2/2) be the points of tangency. Since (x', ?/) is
on both tangents, its co-ordinates must satisfy their equations ;
hence, the equations of condition,
y'y"=p(x' + x''),
y% =v ix' ^xo^.
The two points of tangency {x" , if), (xo, y^ must therefore
satisfy
y'y =p (x' -\-x'),
or yif =p {x +x') . . . (1)
Since (1) is the equation of a straight line, and is satisfied
for the co-ordinates of both points of tangency, it is the
equation of the line joining those points.
98
PLANE ANALYTIC GEOMETRY.
69. To find the equation of the polar of the pole (x,' y') with
regard to the parabola ^/^ = 2 px.
The pjolar of a pole with regard to a given curve is the line
generated by the point of intersection of a pjair of tangents
drawn to the curve at the p)oints in which a secant line through
the pjole intersects the curve as the secant line revolves about the
pole.
By a course of reasoning similar to that of Art. 49, we may
prove tlie required equation to be
yy' =p {x ^x') . . . (1)
As the reasoning by means of whicli (1) is deduced is per-
fectly general, the pole may be without, on, or within the
parabola.
Cor. 1, If we make, in (1), {x' , ?/') = | ^ , j, we have
— _-2 •
^ — 2 '
hence, the directrix is the p>olar of the focus.
70. To ascertain the position and direction of the axes,
other than the axis of the parabola and the tangent at the
vertex, to which if the piarabola be referred its equation will
remain unchanged in form.
Fig. 33.
THE PARABOLA. 99
Since the equation is to retain the form
•?/2 _ 2pcc . . . (1)
let y'-' = 2p'x' ... (2)
be the equation of the parabola when referred to the axes,
whose position and direction we are now seeking. It is
obvious at the outset that whatever may be the position of
the axes relatively to each other, the new Y'-axis must be
tangent to the curve, and the new origin must be 07i the
curve ; for, if in (2) we make x' = 0, Ave have ?/' = -|- 0, a
result whicli we can only account for by assuming the Y'-axis
and the new origin in the positions indicated. This conclu-
sion, we shall see, is fully verified by the analysis which
follows.
Let us refer the curve to a pair of oblique axes, making
any angle with each other, the origin being anywhere in the
plane of the curve. The equations of transformation are.
Art. 33 (1),
X = a -\-x^ cos 9 -{- ■}/ cos go
y = b -\- x^ sin -{- y' sin go.
Substituting these values in (1), we have,
y'^ sin"^ and AN = p, Art.
61. Hence in the triangle FO'A
AO' = FO' sin 2cp = FO' 2 sin (jp cos q>.
In the triangle NO' A,
, ^, . T,x , cos (p
AO = AN cot cp =p -^ — — ;
sm cp
cos QP
hence FO' 2 sin go cos qp = ^^ —. ;
sm qp
THE PARABOLA.
101
But
,. YO' = — ^ —
2/ = ^J^--
Sin.-' qp
.-. 2/ = 4 FO'.
72. To ^wcZ ^7ie equatiori of any diameter in terms of the
slope of the tangent and the semi-parameter.
The equation of any diameter as O'X', Fig. 33, is
y = AO'= b.
But from the triangle AO'N, we have,
b = A^GOtq) = -^—=^',
tan (jp s
hence y = - ■ ■ ■ Q-)
s
is the required equation.
73. To show that the tangents draivn at the extremities of
any chord meet in the diameter which bisects that chord.
p:
R
Y
J^
^
M^
^
o'X^
/
\
1/
X
f//
/
y
\
\
X'
Fig. 34.
Let P' {x', y'), P" (x", y") be the extremities of the chord
then
y-y' = l^i---^ ■■■(>->
102 PLANE ANALYTIC GEOMETRY.
is its equation. The equation of the tangents at P' (x', y'),
P" (x", y") are
yy' =- x> (x -\- x') . . . (2)
yy" =p (x + x") ... (3)
Eliminating x from (2) and (3) by subtraction, we have,
for tlie ordinate of tlie point of intersection of the tangents.
x' — x"
But is the reciprocal of the slope of chord P'P",
y' - y"
(see (1) ). Hence, since the chord PT" and the tangent Y'T
are parallel, we have,
x' -x" ^ 1
y'-y"~~s'
Substituting in (4) it becomes
V
y = -•
s
Comparing this value of y with (1) of the preceding
article, we see that the point of intersection is on the diameter.
EXAMPLES.
1. What must be the value of c in order that the line
y ^ 4zX -{- c may touch the parabola ?/^ = 8 cc ?
Ans. \.
2. What is the parameter of the parabola which the line
y = 3x -\-2 touches ?
Ans. 24.
3. The slope of a tangent to the parabola y"^ = Q x \& = 3.
What is the equation of the tangent ?
Ans. y = 3 0? + i-
4. The point (1, 3) lies on a tangent to a parabola ; required
the equation of the tangent and the equation of the parabola,
the slope of the tangent = 4.
Ans. y = 4:X — 1; y^ = — 16 x.
THE PARABOLA. 103
5. In the parabola y- = 8 a; what is the parameter of the
diameter whose equation is ?/ — 16 = ?
Ans. 136.
6. Show that if two tangents are drawn to the parabola
from any point of the directrix they will meet at right angles.
7. From the point (—2, 5) tangents are drawn to ?/^ = 8 a; ;
required the equation of the chord joining the points of
tangency. Ans. 5y — 4a; + 8 = 0.
8. What are the equations of the tangents to y- = Qx
which pass through the point (— 2, 4) ?
Find the equation of the polar of the pole in each of the
following cases :
9. Of (- 1, 3) with regard to y"- = ^x.
Ans. 3?/ — 2a;+2 = 0.
10. Of (2, 2) with regard to i/- = - 4 a;.
Ans. 2?/ + 2x + 4 = 0.
11. Of («, h) with regard to ?/^ = 4 x.
Ans. by — 2 x — 2 a ^ 0.
12. Given the parabola y"^ ^ x and the point (— 4, 10) ; to
find the intercej)ts of the polar of the point.
Ans. a = 4, b = .
5
13. The latus-rectum of a parabola = 4 ; required the pole
of the line y — 8a:; — 4 = 0.
Ans. (1 1).
14. Given y^ = 10 x and the tangent 2y — x = 10; required
the equation of the diameter passing through the point of
tangency.
Ans. y = 10.
GENERAL EXAMPLES.
1. Assuming the equation of the parabola, prove that every
point on the curve is equally distant from the focus and
directrix.
104 PLANE ANALYTIC GEOMETRY.
2. Find the equation of the parabola which contains the
points (0, 0), (2, 3), (- 2, 3).
Ans. 3x^ = 4:y.
3. What are the parameters of the parabolas which pass
through the point (3, 4) ?
A71S. J/, and |.
4. Find the equation of that tangent to y^ = 9 x which is
parallel to the line y — 2x — 4z = 0.
Ans. 8y — 16x — 9 = 0.
5. The parameter of a parabola is 4 ; required the equation
of the tangent line which is perpendicular to the line
y = 2 X -\- 2. Give also the equation of the normal which is
parallel to the given line.
6. A tangent to y^ = 4:X makes an angle of 45° with the
X-axis ; required the point of tangency.
A71S. (1, 2).
Show that tangents drawn at the extremities of a focal
chord
7. Intersect on the directrix.
8. Meet at right angles.
9. That a line joining their point of intersection with the
focus is perpendicular to the focal chord.
10. Find the equation of the normal in terms of its slope.
11. Show that from any point within the parabola three
normals may be drawn to the curve.
12. Given the parabola r = to construct the tan-
^ 1 + cos ^
gent at the point whose vectorial angle = 60°, and to find the
angle which the tangent makes with the initial line.
A71S. 6 = 60°.
13. Find the co-ordinates of the pole, the normal at one
extremity of the latus-rectum being its polar.
THE PARABOLA. 105
14. Ill the parabola y^ = 4 ck what is the equation of the
chord which the point (2, 1) bisects ?
Ans. y = 2 X — 3.
15. The polar of any point in a diameter is parallel to the
ordinates of that diameter.
16. The equation of a chord of y"^ = 10 x \s y = 2 x — 1;
required the equation of the corresponding diameter.
17. Show that a circle described on a focal chord of the
parabola touches the directrix.
18. The base of a triangle = 2 a and the sum of the tan-
gents of the base angles = b. Show that the locus of the
vertex is a parabola.
19. Required the equation of the chord of the parabola
y^ =:2px whose middle point is (m, n).
. n X — m
Ans. — = .
p y — n
20. A focal chord of the parabola y"^ = 2px makes an
angle = / cos q)
y = x' sin -\- y' sin cp
for the equations of transformation. Substituting in (1), we
have
(a^ sin^ 6 -\-b'^ cos^ 6) x"^ + (^^ sin^ -\- b' cos 9 cos qp = . . . (3)
is the condition that a pair of axes must fulfil in order to be
conjugate diameters of the ellipse.
Making the co-efficient of x'y' equal to zero in (2), we have
after dropping accents
(o-^ sin^ 6 -\-b'^ cos^ 6) x'^ -\- (cr sin^ ;
area CTC'T' = area BB'H'H.
I.e.,
99, To shoiv that the ordinate of any point on the ellipse is
to the ordinate of the corresponding point on the circumscribing
circle as the semi-conjugate axis of the ellipse is to the semi-
transverse axis.
R
/-^
B
A
?"
«M
/
Xv
X
D
X
[
D'
J
Jj
B'
Fig. 45.
Let DP', DP'' be the ordinates of the corresponding points
P' (x', y') and P" (x", y").
Since P' (x', y') is on the ellipse, we have
y- = — (a^ — X-).
a-
Since Y" (x", y") is on the circle whose radius is a, we have
y"^- = a2 _ a;"l
Dividing these equations, member by member, we have
fO 7 9
^ = — , (since X =x ) ;
y -^ a-
.: y' : y" :: b : a.
134
PLANE ANALYTIC GEOMETRY.
Similarly we may prove that
where Xi is the abscissa of any point on the ellipse, and x^ is
the corresponding abscissa of a point on the inscribed circle.
100. The principles of the preceding article give us a
method of describing the ellipse by points when the axes are
given.
From 0,. Fig. 45, as a centre with radii equal to the semi-
axes OA, OB describe the circles A'RA, BOB'. Draw any
radius OR of the larger circle, cutting the smaller circle in M ;
draw MIST || to OA', cutting the ordinate let fall from E in jST ;
N is a point of the ellipse. Since MN is parallel to the base
of the triangle ED'O, we have
D'N : D'R :: OM : OR ;
i. e., y' -.y" ■.-.h-.a;
hence, the construction.
101. To show that the area of the ellipse is to the area of
the circumscribing circle as the semi-minor axis of the ellijjse is
to its semi-major axis.
Pa
Pi
Fig. 46.
THE ELLIPSE. 135
Inscribe in the ellipse any polygon ARRiE,2E'3l^4A', and
from its vertices draw the ordinates RD,RiDi, etc., producing
them upward to meet the circle in P, P^, Pg, etc. Joining
these points we form the inscribed polygon APP1P2P3P4A' in
the circle.
Let (x, y^), (x', y{), {x", y<^ etc., be the co-ordinates of
P, Pi, Pg, etc., and let (a;, ?/), (x' , y'), (x", y"), etc., be the co-ordi-
nates of the corresponding points R, E-i, Eg? etc., of the ellipse.
Then Area RDDiRj = (x - x') ^ ^^ ^
Area PDDiPi = {x - x') l^+ll .
hence Area RDDiRi _ y +/
Area PBD^Pi 2/0 + 2/i '
But, Art. 99, ^ = - and i^ == - ;
yo a yi a
2/0+2/1 » '
Hence Area RDDiRi _ ^
Area PDDjPi ~ a '
We may prove in like manner that every corresponding pair
of trapezoids bear to each other this constant ratio ; hence,
by the Theory of Proportion, the sum of all the trapezoids in
the ellipse will bear to the sum of all the trapezoids in the
circle the same ratio. Representing these sums by 'M and
ST, respectively, we have
^t b
ST a
As this relationship holds true for any number of trape-
zoids, it holds true for the limits to which the sum of the
trapezoids of the ellipse and the sum of the trapezoids of
the circle approach as the number of trapezoids increase.
136 PLANE ANALYTIC GEOMETRY.
But these limits are the area of the ellipse and the area of
the circle ; hence
area of ellipse h
area of circle a
Cor. Since the area of the circle is tt a^, we have
area of ellipse h
TT a^ a
.*. area of ellipse = tt ab.
Since -k a^ : -k ah : : ir ah : tt V^ ,
we see that the area of the ellipse is a mean proportional be-
tween the areas of the circumscribed and inscribed circles.
EXAMPLES.
1. What must be the value of c in order that the line
y = 2 X -\- c may touch the ellipse
Ans. G = 5.
2. The semi-transverse of an ellipse is 10 ; what must be
the value of the semi-conjugate axis in order that the ellipse
may touch the line 2?/ + a; — 14 = 0?
Ans. b = V24.
3. What are the equations of the tangents to the ellipse
^'' _i_ f" — 1
whose inclination to X-axis = 45° ?
4. The locus of the intersection of the tangents to the
2 2
ellipse -— + ^ = 1
a- b'^
drawn at the extremities of conjugate diameters is an ellipse ;
required its equation.
^^^- -^ + li = 2.
a^ b^
THE ELLIPSE. 137
5. Tangents are drawn from the point (0, 8) to tlie ellipse
^ + y^ = 1 ;
required the equation of the line joining the points of
tangency. Ans. 8 ?/ — 1 = 0.
Eequired the polar of the point (5, 6) with respect to the
following ellipses :
6. x^ + 3,f = (Fig. 45) is called the eccentric
angle of the point P' {x\ y') on the ellipse. Show that (x', y')
= (ct cos q>, h sin gp) and from these values of the co-ordinates
deduce the equation of the ellipse.
23. Express the equation of the tangent at (x", y") in terms
of the eccentric angle of the point.
Ans. - cos (p -\- — sin (p = 1.
a b
24. If (x', y'), (x", y") are the ends of a pair of conjugate
diameters whose eccentric angles are (p and tp', show that
q>' -q> = 90°.
THE HYPERBOLA.
141
CHAPTER VIII.
THE HYPERBOLA.
102. The hyperbola is the locus of a point so moving in a
plane that the difference of its distances from two fixed points
is always constant and equal to a given line. The fixed
points are called the Foci of the hyperbola. If the points
are on the given line produced and equidistant from its
extremities, then the given line is called the Transverse Axis
of the hyperbola.
103. To deduce the equation of the hyperbola, given the foci
and the transverse axis.
Fig. 47.
Let F, F' be the foci, and AA' the transverse axis. DraAv
OYlto A A' at its middle point, and take OY, OX as the
142 PLANE ANALYTIC GEOMETRY.
co-ordinate axes. Let P be any point of the curve. Draw
PF, PF' ; draw also PD || to OY.
Then (OD, DP) = {x, y) are the co-ordinates of P.
Let AA' =2 a, FF' = 2 OF = 2 0F'= 2 c, FP = r and FT
= /.
From the right angled triangles FPD and F'PD, we have
r = Vy^ + (^ — <^y and / = ■yjy'^ -\- (x -\- cy . . . (a)
From the mode of generating the curve, we have
r' -r = 2a.
Hence, substituting,
^f + (^ + cf -^y^J^(x-cy = 2a', . . . (1)
or, clearing of radicals and reducing, we have
(c" — oF) x^ — a^y'^ = a? {c~ — or) . . . (2)
for the required equation. This equation, like that of the
ellipse (see Art. 75), may be put in a simpler form.
Let c" -a"" =h^ . . . (3)
This value in (2) gives, after changing signs,
ahf- - Vx' = - o?h-, ... (4)
or, symmetrically,
^-f^ = l. • • (5)
a- b-
for the equation of the hyperbola when referred to its centre
and axes.
Let the student discuss this equation. (See Art. 14)
Cor. 1. li h = a in (5), we have
x^ -,f = a? , . . (6)
The curve represented by this equation is called the Equi-
lateral Hyperbola. Comparing equation (6) with the equation
of the circle
a?^ + 2/^ = «^
%ve see that the equilateral hyperbola bears the same relation to
the comvion hyperbola that the circle bears to the ellipse.
THE HYPERBOLA. 143
Cor. 2. If (x', y') and (x", y") are the co-ordinates of two
points on the curve, we have from (4)
y'^ = il (x'2 - a") and y"^ = -^ (cc"^ _ a^) •
Oj cc
hence y'^ : y"" :: (x' — a) (x -{- a) : {x" — a) (x" + «) ;
i.e., the squares of the ordinates of any two points on the
hyperbola are to each other as the rectangles of the segments in
tvhich they divide the ti-ansverse axis.
Cob. 3. By making x = x' — a and y =■]/ va. (4) we have
after reducing and dropping accents,
aY -V'x^-\-2 aWx = ... (7)
for the equation of the hyperbola, A' being taken as origin.
104. From equation (3) Art. 103, we have
^» = J_ Vc2 - a'.
Laying this distance off above and below the origin on the
Y-axis, we have the points B, B', Fig. 47, Art. 103. The line
BB' is called the Co^mugate Axis of the hyperbola. The
jDoints A and A' are called the Vertices of the curve. The
point bisects all lines drawn through it and terminating in
the curve ; for this reason it is called the Centre of the
hyperbola.
The ratio -^Ti^jITfl r
^'' +^ =~ = e. See (3) Art. 103 .. . (1)
a a
is called the Eccentricity of the hype rbola. This ratio is
evidently > 1. The value of c = -|- Va^ + b^ measures the
distance of the foci F, F' from the centre.
If b = a in (1), we have e = V2 for the eccentricity of the
equilateral hyperbola.
105. To find the values of the focal radii, r, r' of a point
on the hyperbola in terms of the abscissa of the point.
From equations (a) Art. 103, we have
?' = V 2/^ + (^ — (^Y '
144
PLANE ANALYTIC GEOMETRY.
From the equation of the hjqoerbola, (4) Art. 103, we have
y^ = — - (x^ — a^) = — x^ — b^.
Hence, substituting
=v/:
p
x"^ — b^ -\- x^ — 2 ex -\- c%
V a-
x^ — 2 ex -}- c^ — b^,
= d^x^ -2cx + a'-, Art. 104 (1),
= —X — a\
a
hence r = ex — a . . . (V)
Similarly, we find
r' = ex -{- a . . . (2)
106. To construct the hyperbola having given the transverse
axis and the foci of the curve.
Fig. 48.
First Method. — Let A A' be the transverse axis and F, F', the
foci. Take a straisfht-edge ruler whose length is L and attach
THE HYPERBOLA. 145
one of its ends at F' so that the ruler can freely revolve about
that point. Cut a piece of cord so that its length shall be
= L — 2 a, and attach one end to the free end of the ruler,
and the other end to the focus F. Place the ruler in the
position indicated by the full lines, Fig. 48, and place the
point of a pencil in the loop formed by the cord. Stretch
the cord, keeping the point of the pencil against the edge of
the ruler. If we now revolve the ruler upward about F', the
point of the pencil, kept firmly pressed against the ruler,
will describe the arc AP' of the hyperbola. By fixing the
end of the ruler at F, we may describe an arc of the other
branch. It is evident in this process that the difference of
the distances of the point of the pencil from the foci F',F,
is always equal to 2 a.
Second Method. — Take any point D on the transverse axis.
Measure the distances A'D, AD. With F' as a centre and A'D
as a radius describe the arc of a circle ; with F as a centre and
AD as a radius describe another arc. The intersection of
these arcs will determine two points, Pi, Po, of the curve. By
interchanging centres and radii we may locate the points Pi,
Rs) on the other branch. In this manner we may determine as
many points as the accuracy of the construction may require.
107. To find the latus-rectuvi or parameter of the hyperbola.
The Latus-Pectum, or Parameter of the hyperbola, is the
double ordinate passing through either focus.
Making x^= ^^ Vft^ -|- b'^ in the equation of the hyperbola
IP-
2/" = ^r (^^ — «^)»
a-
ip- 2 b"^
we have y = — .-. 2 y
a a
Forming a proportion from this equation, we have
2y:2b::b:a;
.:2y:2h::2b:2a;
i.e, the latus-rectum of the hyperbola is a third pjroportional to
the axes.
146
PLANE ANALYTIC GEOMETRY.
108. The equation of the ellipse when referred to its centre
and axes is
a'^y'^ + ^^^^ = f'^^"-
The equation of the hyperbola when referred to its centre
and axes is
ay — b-x"^ = — a~h'^.
Comparing these equations, we see that the only difference
is in the sign of b"^. If, therefore, in the various analytical
expressions we have deduced for the ellipse, we substitute
— b'^ for b^, or, what is the same thing, + & V— 1 for b, we
will obtain the corresponding analytical expressions for the
hyperbola.
109. To deduce the equation of the conjugate hyperbola.
T\vo hyperbolas are Conjugate when the transverse and con-
jugate axes of one are respectively the conjugate and trans-
verse axes of the other.
Thus in Fig. 49, if AA' be the transverse axis of the hyper-
bola which has BB' for its conjugate axis, then the hyperbola
which has BB' for its transverse axis and AA' for its conjugate
THE HYPERBOLA. 147
axis is its conjugate; and, conversely, the hyperbola whose
transverse axis is BB' and conjugate axis is AA' has for its
conjugate the hyperbola whose transverse axis is AA' and
whose conjugate axis is BB'.
We have deduced. Art. 103, (5),
£! - l! = 1 . . . (1)
for the equation of the hyperbola whose transverse axis lies
along the X-axis. We wish to find the equation of its conju-
gate. It is obvious from the figure that the hyperbola which
has BB' for its transverse axis and AA' for its conjugate axis
bears the same relation to the Y-axis as the hyperbola whose
transverse axis is AA' and conjugate axis is BB' bears to the
X-axis ; hence, changing a to h and b to a, x to y and y to x
in (1), we have
li _ ^ = 1
or 4 - ^ = - 1 • • • (2)
for the equation of the conjiigate hyperbola to the hyperbola
whose equation is (1).
Comparing (1) and (2) we see that the equation of any
hyperbola and that of its conjugate differ only in the sign of
the constant term.
CoR. Since V^^ + ^-^ = V«^ -f- b^, the focal distances of
any hyperbola and those of its conjugate are equal.
The eccentricities of conjugate hyperbolas, however, are
not equal. For the hyperbola whose semi-transverse axis is
a and semi-conjugate axis is h, we have
Art. 104, (1) e=^^' + ^'.
a
For its conjugate hyperbola, we have
,_ Vfr + b-"
148 PLANE ANALYTIC GEOMETRY.
EXAMPLES.
Find the semi-axes, the eccentricity and the latus-rectum
of each of the following hyperbolas :
1. 9 ^2 _ 4 ^2 _ _ og_ b. 3^/ -2x^ = 12.
6. ay- — hx^ = — ab.
4 ^
8. y' — mx' = n.
Write the equation of the h3^perbola having given :
9. The transverse axis = 12 ; the distance between the
foci = 16.
Ans. ^ = 1
36 28
2.
X- 2/- _ 1
4 9
3.
y'^ — 16 X' = — 16.
4.
4 cc^ — 16 y- = — 64,
10. The transverse axis = 10; parameter = 8.
/2
Ans. i^ ^ 1.
25 20
11. Semi-conjugate axis = 6 ; the focal distance = 10.
Ans. -^ = 1.
64 36
12. The equation of the conjugate hyperbolatoa;^— 3 y- = 6.
Ans. cc2 — 3 2/' + 6 = 0.
13. The conjugate axis is 10, and the transverse axis is
double the conjugate.
. x^ %r -,
Ans. -^ =1 1.
100 25
14. The transverse axis is 8, and the conjugate axis = \
distance between foci.
Ans. i^-V = i.
16 16
THE HYPERBOLA. 149
15. Given the hyperbola
^ _ J^ = 1-
10 4 '
required the co-ordinates of the point whose abscissa is double
its ordinate.
-• (Vf ' \/f )
16. Write the equation of the conjugate hyperbola to each
of the hyperbolas given in the first eight examples above.
17. Given the hyperbola 9 ?/^ — 4 a;^ = — 36 ; required the
focal radii of the point whose ordinate is = 1 and abscissa
positive.
18. Determine the points of intersection of
— — -^^— = 1, and -— + -^ = 1.
4 9 ' 16 16
110. To deduce the iwlar equation of the hyperbola, either
focus being taken as the pole.
Let us take F as the pole, Fig. 47.
Let (FP, PFD) = (?•, 6) be the co-ordinates of any point P
on the curve. From Art. 105, (1), we have
1^^ = r = ex — a . . . (1)
From Fig. 47, OD = OF + FD ;
i.e., X = ae -\- r cos 6.
Substituting this value in (1) and reducing, we have
r = _ '' (^ - ^') ... (2)
1 _ e cos ^ ^ ^
for the polar equation of the hyperbola, the right hand focus
being taken as the pole.
Similarly from Art. 105, (2), we have
a(l- e^)
1 — e cos 9
for the polar equation, the left hand focus being the pole.
r- = ^^ ^ ■ • • (3)
1 — e cos 9
150
PLANE ANALYTIC GEOMETRY.
Cor.
If e = 0, r = - a - ae = - FA',
r' = a -\- ae = F'A.
If ^ =
90°,
, 2 a"e^ — a^ b^ . , ^
r = — a -\- ae = = — = semi-Latus rectum.
a a
r' ^ a — ae^ = = = semi-latus rectum.
a a
lie = 180°, r = - a + ae = FA,
r' = a — ae= — F'A'.
If 6 = 270°, ?• = — a -\- ae^ = — = semi-latus rectum.
a
b^
r' ^ a — ae^ = = semi-latus rectum.
a
111. To deduce the equation of condition for the stipple-
mentary chords of the hyperbola.
By a method similar to that of Art. 81, or by placing — h'^
for y^ in (3) of that article, we have
ss' = ^;... (1)
hence, the product of the slopes of any pair of supplementary
chords of an hyperbola is the same for every pair.
CoR. If a. = 6, we have
ss = 1, or, s = — ,
s
.'. tan « = cot «' ;
hence, the sum of the ttvo acute angles ivhich any p)air of sup-
pjlementary chords of an equilateral hyperbola make with the
X-axis is equal to 90°.
112. T'o dedaice the equation of the tangent to the hyperbola.
By a method entirely analogous to that adopted in the
circle, or ellipse, or parabola, Arts. 41, 82, 57 ; or substituting
— ^2 for 7.2 in (5) of Art. 82, we find
'- yl= 1 ... (1)
to be the equation of the tangent to the hyperbola.
THE HYPERBOLA. 151
113. 2^0 deduce the value of the sub-tangent.
By operating on (1) of the preceding article (see Art. 83),
we find
Sub-tangent = x" — = .
114. Tlie slope of a line passing through the centre of an
hyperbola (0, 0) and the point of tangency {x", y") is
t = y:L.
x"
The slope of the tangent is, Art. 112, (1)
t' = ^ ^
~ a^ ' y"'
Multiplying these equations, member by member, we have
ttf^^. . . (1)
a^
Comparing (1) of this article with (1) of Art. Ill, we find
ss' = tt' . . . (2)
Hence, the line from the centre of the hyperbola to the
point of tangency and the tangent enjoy the property of being
the supplemental chords of an hyperbola whose semi-axes
bear to each other the ratio - •
a
CoR. If s ^ t, then / = t' ; i.e., if one supplementary
chord of an hyperbola is parallel to a line drawn through the
centre, then the other supplementary chord is parallel to the
tangent drawn to the curve at the point in which the line
through the centre cuts the curve.
115. The preceding principle affords us a simple method
of drawing a tangent to the hyperbola at any given point of
the curve.
152
PLANE ANALYTIC GEOMETRY.
Fig. 50.
Let P' be auy point at which we wish to draw a tangent-
Join P' and 0, and from A' draw A'C || to P'O ; join C and A.
The line P'T, drawn from P' || to CA will be the required
tangent.
116. To deduce the equation of the 7iormal to the hyperbola^
We can do this by operating on the equation of the tangent,
as in previous cases, or by changing h- into — b'-^ in the equa-
tion of the normal to the ellipse. Art. 86, (3). By either
method, we obtain
y-y =-
cry
1^
{x - x")
(1)
for the required equation.
117. To deduce the value of the sub-normal.
By a course of reasoning similar to that of Art. 87, we have-
sub-normal = — 5- ^ '
a'
CoK. lib = a,
sub-no7'mal = x" ;
i.e., in the equilateral hyperbola the sub-normal is equal tO'
the abscissa of the point of tangency.
THE HYPERBOLA. 153
EXAMPLES.
1. Deduce the polar equation of the hyperbola, the pole
being at the centre.
.2 ^ o^
a^ sin^ 6 — &^ cos ^ 6
Write the equation of the tangents to each of the follow-
ing hyperbolas, and give the value of the sub-tangent in
each case.
2. 9 y' - 4 a;2 = — 36, at (4, ord. +).
3. yi _ i^ = _ 1 at (5, ord. +).
9 16 ' V . -ry
4. ^ _ id = 1 at (4, ord. +).
9 16 ' ^ ' ^
5. 2/^ -- 4 a;2 = — 36, at (abs. +, 6).
6. ay' — bx^ = — ab, at (-Vab, ord. -|-)-
7. ^ - ll = 1, at (Vm, 0).
8. Write the equation of the normal to each of the above
liyperbolas, and give the value of the sub-normal in each
case.
9. The equation of a chord of an hyperbola is y — x — 6
= ; what is the equation of the supplemental chord, the
axes of the hyperbola being 12 and. 8 ?
Ans. y = -X — - .
-^9 3
10. Given the equations
^^ = — 1 and y — OS = :
9 4 ' ^ '
required the equations of the tangents to the hyperbola at the
points in which the line pierces the curve.
154 PLANE ANALYTIC GEOMETRY.
11. One of the supplementary chords of the hyperbola
9 2/^ — 16 rK^ = — 144 is parallel to the line y = x; what are
the equations of the chords ?
Ans. \ 16 16.
(-^9 3
12. Given the hyperbola 2 a;^ — 3 ^/^ = 6 ; required the
equations of the tangent and normal at the positive end of
the right hand focal ordinate.
13. What is the equation of a tangent to
^_ jT _-,
4 6 ~ '
which is parallel to the line 2 y — a; + l = 0?
118. The angle formed by the focal lines drawn to any point
of the hyperbola is bisected by the tangent at that point.
Making ?/ = o in the equation of the tangent line, Art.
112, (1), we have
x = ^ = OT. Fig. 50.
x'
From Art. 104, (1) OF = OF' = ae ;
hence OF - OT = FT = ae - ^ = ^ {ex" - a).
OF' + OT = F'T = ae + -^ = -^ (ex" + a);
XX
.: FT : F'T :: ex" - a : ex" + a.
But from Art. 105 we have
FP' = ex" - a
FT' = ex" + a ;
. : FF : FT' :: ex" - a : ex" + a.
Hence FT : F'T :: FP' : F'P' ;
i.e., the tangent P'T divides the base of the triangle FP'F'
into two segments, which are proportional to the adjacent
sides ; it must therefore bisect the angle at the vertex.
THE HYPERBOLA. 155
Cor. Since the normal P'N, Fig. 50, is perpendicular to
the tangent, it bisects the external angle formed by the focal
radii.
ScHOL. The principle of this article gives us another
method of drawing a tangent to the hyperbola at a given
point. Let P' be the point. Fig. 50. Draw the focal radii
FP', F'P'. The line P'T drawn so as to bisect the angle
between the focal radii will be tangent to the curve at P'.
119. To find the condition that the line y = sx -\- c must
fulfil 171 order that it may touch the hyperbola
a"" b-
By a method similar to that employed in Art. 89, we find
s2 a'' -h'' = (-' . . . (1)
for the required condition.
Cob. 1. Substituting the value of c drawn from (1) in the
equation of the line, we have
?/ = sx -1- ^ s^a^ — b'^ ... (2)
for the equation of the tangent to the hyperbola in terms of its
slope.
120. To find the locus generated by the intersection of a
tangent to the hyperbola and a perpendicular to it from a focus
as the point of tangency inoves around the curve.
x^+ tf^a'' . . . (1)
is the equation of the required locus. (See Art. 90.)
121. To find the locus generated by the intersection of tivo
tangents ivhich are perpendicular to each other as the points of
tangency move around the curve.
x^ -]- y^ = a^ - b'' . . . (1)
is the equation of the required locus. (See Art. 91.)
156 PLANE ANALYTIC GEOMETRY.
122. Tivo tangents are draivn to the hyperbola from, a point
without ; required the equation of the line joining the points of
tangency.
^ ^ yy_ ^-^^ (i\
O 7 9 ' ' ' \ /
a^ 0-
is the required equation. (See Art. 92.)
123. To find the equation of the polar of the pole (x', y'),
with regard to the hypjerhola
a^ b-' '
^ ^ _ yy __ -^ /-^
a^ b^
is the required equation. (See Arts. 49 and 93.)
124. To deduce the equation of the hyperbola tvhen referred
to a pair of conjugate diameters.
A pair of diameters are said to be conjugate when they are so
related that the equation of the hyperbola, when the cu7've is
referred to them as axes, contains only the second p)owers of the
variables.
— - i^ = 1 ... (1)
a"' b'^' ^ ^
is the required equation, and
ct^ sin 6 sin cp — b"^ cos 6 cos go = 0,
or tan 6 tan qo = — ... (2)
a'^
is the condition for conjugate diameters. (See Art. 94.)
Cor. Erom the form of (1) we see that all chords drawn
parallel to one of two conjugate diameters are bisected by the other.
ScHOL. From Art. Ill, (1) we have
b^
a^
hence ss' = tan d tan (jd.
If, therefore, s = tan 6, we have s' = tan cp ; i.e., if one of
tivo conjugate diameters is parallel to a chord, the other conjxi-
gate diameter is parallel to the supplement of that chord.
THE HYPERBOLA.
157
From Art. 114 we have
a-
hence tt' = tan 6 tan qp.
If, therefore, t = tan 6, we have t' = tan q> ; i.e., */ one of
two conjugate diametei's is parallel to a tangent of the liyper-
bola, the other conjugate diameter coincides with the line joining
the point of tangency and the centre.
125. Prom the condition for conjugate diameters,
tan 6 tan cp =
b""
we see that the products of the slopes of any pair of conju-
gate diameters is positive ; hence, the slopes are both positive
or both negative. It appears, therefore, that any two conju-
gate diameters must lie in the same quadrant.
126. To find the equation of a conjugate diameter.
Fig. 51.
158 PLANE ANALYTIC GEOMETRY.
Let V'R" be any diameter ; then P'E', drawn through the
centre parallel to the tangent at P" (P^N') will be its eon-
jugate diameter. Art. 124, Schol.
The equation of the tangent at P" (x", y") is
• (1)
XX yy _^
a? V^
hence, the
equation of P'R' is
^^" yy" __
or
V' x"
y = ^r- —TT^ • ■
«*• y
But
^ = cot P"OX =
1
y
s
hence
y = -^x . . . (£
')
(2)
as
is the equation of a diameter in terms of the slope of its conju-
gate diameter.
127. To find the co-ordinates of either extremity of a
diameter, the co-ordinates of one extremity of its co7ijugate
diameter being given.
Let the co-ordinates of P" (x", y"), Fig. 61, be given.
By a course of reasoning similar to that of Art. 96, we find
«=-[-- y", rf = 4--X .
a
The upper signs correspond to the point P' {x', y') ; the
loAver signs to the point R' (— x', — y').
128. To shoiv that the difference of the squares of any pair
of semi-conjugate diameters is equal to the difference of the
squares of the semi-axes.
By a course of reasoning similar to that of Art. 97, or, by
substituting — b^ for b^, — b'^ for b'^ in (4) of that article, we
find
a'2 _ S'2 _ ^2 _ 52 _ _ _ ^ij
THE HYPERBOLA. 159
Cor. If a = b, then a' =b' ; i.e., the equilateral hyperbola
has equal conjugate diameters.
129. To show that the 'parallelogram constructed on any two
conjugate diameters is equivalent to the rectangle constructed
on the axes.
By a method similar to that of Art. 98, we can show that
4 a'b' sin (go - ^) = 4 ab ;
i.e., Area MNM'N' = Area CDC'D'. Fig. 51.
EXAMPLES.
1. The line y = 2x -\-c touches the hyperbola
x^ y^ 1
what is the value of c .^
Ans. c = i V32.
2. A tangent to the hyperbola
x^ _ ?/^ _ -1
10 12 ~
has its Y-intercept = 2 ; required its slope and equation.
Ans. VlTe ; y = ViTG a; + 2.
3. A tangent to the hyperbola 4?/^ — 2x^ = 6 makes an
angle of 45° with the X-axis ; required its equation.
4. Two tangents are drawn to the hyperbola A y^ — 2 x^ ^
— 36 from the point (1, 2) ; required the equation of the chord
of contact.
Ans. 9 cc — 8 y ^ 36.
5. What is the equation of the polar of the right-hand
focus ? Of the left-hand focus ?
6. What is the polar of (1, ^) with regard to the hyperbola
4y^ — x'^ = — 4? Ans. x — 2 y = 4.
7. Find the diameter conjugate to y = x in the hyperbola
-^ _ i^ = 1
9 16
Ans. y = ^ X.
160
PLANE ANALYTIC GEOMETRY.
8. Given the chord y — 2x -\- Q oi the hyperbola
required the equations of the supplementary chord.
Ans. y = ^ X — §.
9. In the last example find the equation of the pair of
conjugate diameters which are parallel to the chords.
Ans. y = 2x, 9 y = 2x.
10. The point (5, ^^) lies on the hyperbola 9y- — IQx^ =
— 144; required the equation of the diameter passing through
it; also the co-ordinates of the extremities of its conjugate
diameter.
130. To deduce the equations of the rectilinear asymptotes
of the hyperbola.
An Asymptote of a curve is a line passing within a finite
distance of the origiii which the curve continually approaches^
and to which it becomes tangent at an infinite distance.
Fig. 52.
THE HYPERBOLA. 161
The equation of the hyperbola whose transverse axis lies
along the X-axis may be put under the form
The equations of the diagonals, DD', CC, of the rectangle
constructed on the axes AA', BB' are
y' = ^-x,
a
or, squaring, y''^ = ^x^ . . . (2)
where y' represents the ordinates of points on the diagonals.
Let P' {x, y) be any point on the X-hyperbola ; and let D"
(x, y') be the corresponding point on the diagonal DD'. Sub-
tracting (1) from (2) and factoring, we have
hence / - 2/ = ^"^' = /' • • • (3)
As the points D", P' recede from the centre, 0, their ordi-
nates D"]Sr, P'N increase and become infinite in value when
D" and P' are at an infinite distance. But as the ordinates
increase the value of the fraction (3), which represents their
difference, decreases and becomes zero when y' and y are
infinite ; hence, the points D'' and P' are continually approach-
ing each other as they recede from the centre until at infinity
they coincide. But the locus of D" during this motion is the
infinite diagonal DD' ; hence, the diagonals of the rectangle
constructed on the axes of the hyperbola are tha asymptotes of
the curve.
Therefore y z= -\ — x and tj = x
a a
are the reqiiired equations.
CoR. 1. If a- = h, then
y = -\- X and y =. — x\
i.e., the asymptotes of the equilateral hyperbola make angles of
45° with the IL-axis.
162
PLANE ANALYTIC GEOMETRY.
Cor. 2. The equation of the hyperbola conjugate to (1)
may be put under the form
^.2
- (x'^ + a^) . . . (4)
Subtracting (1) from (4), we have
y
y = PiP' =
2 b''
y" + y
hence, an hyperbola and its conjugate are curvilinear asymp-
totes of each other.
CoK. 3. Subtracting (2) from (4), we have
b-
y
y' = P,D" =
y"^y"
hence, the rectilinear asymptotes of an hyperbola and of its con-
jugate are the same.
131. To deduce the equation of the hyperbola when referred
to its rectilinear asym,ptotes as axes.
Y
Fig. 53.
The equation of the hyperbola when referred to OY, OX,
IS
1 .
(1)
THE HYPERBOLA. 163
We wish to ascertain what this equation becomes when
OY', OX' the rectilinear asymptotes are taken as axes.
Let P' be any point of the curve ; let Y'OX = XOX' = 6.
Then (OC, OF) = {x, y) ; (OD, DP') = {x', ij').
Prom the figure, OC = OK + DP ; CP' = RP' - DK ;
i.e., x = {x' + y') cos 6 \ y = {y — x') sin Q.
But from the triangle OAB, we have
AB h
sin B
cos it =
OB ^ci? + ^2
OA a
OB Va- + y"
hence, x = ix' -{- y') — ==:^^- ; y = iv' — ^') — : .
Substituting these values in (1), we have
(^' + yj - in' - xy = «' + ^^
or, reducing and dropping accents,
xy = — J— ... (2)
for the equation of the hyperbola referred to its asymptotes.
In a similar manner we may show that
xy = ^ ■ • • (3)
is the equation of the hyperbola conjugate to (1), when re-
ferred to its asymptotes as axes.
Cor. Multiplying (2) by sin 2 6 we may place the result in
the form
vx sin 2 ^ = Va^_+1' . ^/^!_±i! siu 2 ^ •
2 2 '
that is DP'. P'H' = ON. AH ;
therefore area ODP'P = area OMAN ;
hence, the area of the parallelogravi constructed upon the co-
ordinates of any point of the hyperbola, the asymptotes being
axes, is constant and equal to the area of the rhombus con-
structed upon the co-ordinates of the vertex.
164
PLANE ANAL YTIC GEOME TR Y.
132. To deduce the equation of the tangent to the hyperbola
when the curve is referred to its rectilinear asymptotes as axes.
Fig. 54.
By a course of reasoning similar to that employed in ArtSo
41, 57, 82, we find the required equation to be
y — y
or, symmetrically.
x" ^ y"
y^^x- X")
■ ■ (2)
(1)
CoR. If we make ?/ = in (2), we have
x = 2x" = OT. Fig. 54.
But OM = x", .: OM = MT .-. T'D = TD ;
hence, the point of tangency in the hyperbola bisects that por-
tion of the tangent included between the asymptotes.
THE HYPERBOLA. 165
133. Since D (x", y") is a point of the hyperbola, we have
(see Fig. 54)
4 x"y" =a^ + b%
or 2x" .2 y" = a- + h"- ;
i.e., OT . OT' = a' ^y . . . (1)
hence, the rectangle of the intercepts of a tangent on the asymp-
totes is constant and equal to the sum of the squares on the
semi-axes.
134. From (1) of the last article we have, after multiply-
,1 , , sin 2 6
mg through by — - — ,
^'^ • ^^' sin 2 ^ = ^J—±Jl. sin 2 ^ = {a^ + h'-) sin 9 cos 6.
But, Art. 131,
sm ii = — , cos 9 =
hence sm 2 9 ^ ab ;
i.e., area OTT' = area OAD'B.
.-. the triangle formed by a tangent to the hyperbola and its
asymptotes is equivalent to the rectangle on the serai-axes.
135. Draw the chord RE,', Fig. 54, parallel to the tangent
T'T. Draw also the diameter OL through D.
Since TD = T'D, we have P/L = RL.
Since OL is a diameter, we have LK = LH ; hence
R'L — LK = RL - LH ;
i.e., R'K = RH ;
hence, the intercepts of a chord between the hyperbola and its
asymptotes cure equal.
166 PLANE ANALYTIC GEOMETRY.
EXAMPLES.
1. What are the equations of the asymptotes of the hyper-
bola ^ _ i^ = 1?
9 16
Ans. y = _1_ |- cc.
What are the equations of the asymptotes of the follow-
ing hyperbolas :
2.
16 ^
Ans.
, X
4.
10
~f-'
3.
3 2/2 _ 2 a;2 =
-6.
5.
mx^ -
- ny^ =
Ans. y ^ i Vf X.
6. What do the equations given in the four preceding ex-
amples become when the hyperbolas which they represent are
referred to their asymptotes as axes ?
7. The semi-conjugate axis of the hyperbola xy = 25 is
6 ; what is the value of the semi-transverse axis ?
A71S. 8.
What are the equations of the tangents to the following
hyperbolas :
8. To xy = 10, at (1, 10).
Ans. y -\-10x = 20.
9. To xy = + 12, at (2, 6).
Ans. y = — 3x + 12.
10. To xy = m, at (— 1, —m).
11. To xy=-p, at (-2,|^
12. Required the point of the hyperbola xy = 12 for which
the sub-tangent = 4.
Ans. (4, 3).
13. The equations of the asymptotes of an hj^perbola
whose transverse axis = 16 are 3 ?/ = 2 x and 3 ?/ + 2 ic = ;
required the equation of the hyperbola.
Ans. ^_^ = i,
64 256
THE HYPERBOLA. 167
14. Prove that the product of the perpendiculars let fall
from any point of the hyperbola on the asymptotes is con-
stant and
«2 -I- &2 •
GENERAL EXAMPLES.
1. The point (6, 4) is on the hyperbola whose transverse is
10 ; required the equation of the hyperbola.
Ans. ^ = 1.
25 400
2. Assume the equation of the hyperbola, and show that
the difference of the distances of any point on it from the
foci is constant and = 2 a.
3. Required the equation of the hyperbola, transverse
axis = 6, which has 5 y = 2 x and 3 ?/ = 13 a; for the equa-
tions of a pair of conjugate diameters.
X
btf_
Ans. _ - ^1^ = 1.
9 78
4. Show that the ratio of the sum of the focal radii of any
point on the hyperbola to the abscissa of the point is con-
stant and = 2 e.
5. What are the conditions that the line y =: sx -\- c must
fulfil in order to touch
— — -^ = 1 at infinity ?
Ans. s = -J- - , c = 0.
a
6. Show that the conjugate diameters of an hyperbola are
also the conjugate diameters of the conjugate hyperbola.
7. Show that the portions of the chord of an hyperbola
included between the hyperbola and its conjugate are equal.
8. What is the equation of the line which passes through
the focus of an hyperbola and the focus of its conjugate
hyperbola ?
Ans. X -\- y ^ -y/ a^ -(- b^.
168 PLANE ANALYTIC GEOMETRY.
9. Show that
e' a
when e and e' are the eccentricities of two conjugate hyper-
bolas. }
10. Find the angle between any pair of conjugate diame-
ters of the hyperbola.
11. Show that in the hyperbola the curve can be cut by
only one of two conjugate diameters.
12. Find whether the line y = ^x intersects the hyperbola
16 y- — 9 a;- — — 144, or its conjugate.
13. Show that the conjugate diameters of the equilateral
hyperbola make equal angles with the asymptotes.
14. Show that lines drawn from any point of the equilat-
eral hyperbola to the extremities of a diameter make equal
angles with the asymptotes.
15. In the equilateral hyperbola focal chords drawn parallel
to conjugate diameters are equal,
16. A perpendicular is drawn from the focus of an hyper-
bola to the asymptote : show
(a) that the foot of the perpendicular is at the distance a
from the centre, and
{b) that the foot of the perpendicular is at the distance h
from the focus.
17. For what point of an hyperbola is the sub-tangent =
the sub-normal ?
18. Show that in the equilateral hyperbola the length of
the normal is equal to the distance of the point of contact
from the centre.
19. Show that the tangents drawn at the extremities of any
chord of the hyperbola intersect on the diameter which
bisects the chord.
THE HYPERBOLA. 169
20. Find the equation of the chord of the hyperbola
^ — J^ =1
9 12
which is bisected at the point (4, 2).
21. Eequired the equations of the tangents to
16 10
which make angles of 60° with the X-axis.
22. Show that the rectangle of the distances intercepted on
the tangents drawn at the vertices of an hyperbola by a
tangent drawn at any point is constant and equal to the
square of the semi-conjugate axis.
23. Given the base of a triangle and the difference of the
tangents of the base angles ; required the locus of the vertex,
24. Show that the polars of (?»., n) with respect to the
hyperbolas
^ — l!- = 1, ll — ^ = 1 are parallel.
25. If from the foot of the ordinate of a point (ic, y) of the
hyperbola a tangent be drawn to the circle constructed on
the transverse axis, and from the point of tangency a line be
drawn to the centre, the angle which this line forms with the
transverse axis is called the eccentric angle of (a;, y). Show
that (cc, ?/) = {a sec qo, b tan qo), and from these values deduce
the equation of the hyperbola.
26. If («', y'), {x", y") are the extremities of a pair of
conjugate diameters whose eccentric angles are qo' and go, show
that (p' + 9 = 90°.
170 PLANE ANALYTIC GEOMETRY.
CHAPTER IX.
THE GENERAL EQUATION OF THE SECOND DEGREE.
136. The most general equation of the second degree be-
tween two variables is
ay~ + bxy -\- cx^ -{- dy -\- ex + f == . . . (1)
in which a, b, c, d, e, f are any constant quantities whatever.
To investigate the properties of the loci which this equation
represents under all possible values of the constants as to
sign and magnitude is the object of this chapter.
137. The equations of the lines in a plane, with which we
have had to do in preceding chapters, are
Ax + By + C = 0. Straight line.
(Ace + B?/ 4" C)^ = 0. Two coincident straight lines.
2/2 _ aj2 _ Q_ rji^^Q straight lines.
2/2 -|- a;^ = a". Circle.
?/2 -j- a;2 = 0. Two imaginary straight lines.
y"^ = 2, px. Parabola.
aY + h^'x^ = a%\ Ellipse.
o?y^ — b^x^ = — a^b^. Hyperbola.
a^y- — b'^x" — a-b^. Hyperbola.
Comparing these equations with the general equation, we
see that all of them may be deduced from it by making the
constants fulfil certain conditions as to sign and magnitude.
We are, therefore, prepared to expect that the lines which
these equations represent will appear among the loci repre-
sented by the general equation of the second degree between
two variables. In the discussion which is to ensue we shall
find that these lines are the only loci represented by this
equation.
EQUATION OF THE SECOND DEGREE. 171
DISCUSSION.
138. To show that the locus represented by a coynjplete equa-
tion of the secoiid degree hetxoeen two variables is also represented
by an equation of the second degree betiveen two variables, in
which the term containing xy is wanting.
Let us assume the equation
ay"^ + ^^y + ^^^ +^y + ^^ + / = • • • (1)
and. refer the locus it represents to rectangular axes, making
the angle 6 with the old axes, the origin remaining the same.
From Art. 33, Cor. 2, we have
X = x' cos 9 — y' sin
y = x sin 6 -{- y' cos 6
for the equations of transformation. Substituting these values
in (1), we have,
aY' + b'x'y' + c'x'^ + d'y + e'x' +/ = . . . (2)
in which
a' = a cos^ 6 -\- c sin^ 9 — b sin 9 cos 9
b' = 2 {a— c) sin ^ cos ^ + ^» (cos^ 9 — sin^ 9)
c' = a sin2 9 ^ c cos^ ^ + 5 sin ^ cos ^ J. ... (3)
d' ^ d cos 9 — e sin 9
e' =^ d sin 9 -\- e cos 9
Since 9, the angle through which the axes have been turned,
is entirely arbitrary, we are at liberty to give it such a value
as will render the value of b' equal to zero. Supposing it to
have that value, we have
2 (a — c) sin 9 cos 9 + b (cos^ 9 — sin^ $) = 0,
or (a — c) sm 2 ^ + 6 cos 2 ^ = . . . (4)
or tan 29 = -^— ... (5)
c — a
Since any real number between -|- oo and — go is the tan-
gent of some angle, equation (5) will always give real value
for 2 9 ; hence the above transformation is always possible.
Making &' = in (2), we have, dropping accents,
a'y'' + cV ^d'y ■\-e'x-^f=^ . . . (6)
for the equation of the locus represented by (1). To this
equation, then, we shall confine our attention.
172 PLANE ANALYTIC GEOMETRY.
CoE. 1. To find the value of a' and c' iii terms of a, b,
and c. Adding and then subtracting the first and third of
the equations in (3), we have
c' -\-a' = c -\- a . . . {!)
c' — a' = (c —a)coB2 6 -\-bB\n26 . . . (8)
Squaring (4) and adding to the square of (8), we have
(c' - a'y = (c - ay + b^ ;
.-. c' — a' = V(c - af -\-b^ . . . (9)
Subtracting and then adding (7) and (9), we have
a' = i\c + a- V(c - af -{- b^ . . . (10)
c' = 1 ^c + a + V(c - ay + b'l . . . (11)
Cor. 2. To find the signs of a' and c'. Multiplying (10)
and (11), we have
a'c' = \\{c + ay - {{c-ay -^b')\;
.: a'c' = -i{b^ -4.ac) . . . (12)
Hence, the sigiis of a' and c' depend upon the sign of the
quantity P — 4: ac.
The following cases present themselves :
1. b^ 4 ac. The sign of the second member of (12) is
negative, .-. a' must be positive and c' negative, or a' must be
negative and c' positive.
139. To transform the equation a't/"^ -\- c'x^ + d'y -\- e'x -\-
y = into an equation in which the first pmwers of the vari-
ables are missing.
EQUATION OF THE SECOND DEGREE. 173
Let us refer the locus to a parallel system of rectangular
axes, the origin being at the point (m, n). From Art. 32, we
have
X = VI -\- x', y ^= n -{- y'.
Substituting these values in the given equation, we have
a'y'2 + c'x"" + d"y'+ e"x'+f" = ... (2)
in which
d" =2a'n + d' 1
6" = 2 c'm + e' !- . . . (3)
f" = a'n^ + c'lv? -\- d'n + e'ln + / J
Since m and n are entirely arbitrary, we may, in general,
give them such values as to make
2 a'?i + cZ' = and 2 c'm + e' = ;
i.e., in general, we may make
and m = - -^ ... (4)
2 a' 2 c'
We see from these values that when a' and c' are not zero,
this transformation also is possible ; and equation (2) becomes,
after dropping accents,
ay+flV+r = ... (5)
Equation (5), we observe, contains only the second power of
the variables ; hence it is satisfied for the points (x, y) and
(— X, — y). But only the equation of curves with centres
can satisfy this condition ; hence, equation (5) is the equa-
tion of central loci. When either a' or c' is zero, then n or m
is infinite and the transformation becomes impossible. Hence
arise two cases which require special consideration.
140. Case 1. a' = o.
Under this supposition equation (6), Art. 138, becomes
c'x- + d'y + e'x +/= ... (1)
Referring the locus of this equation to parallel axes, the
origin being changed, we have for the equations of trans-
formation
X = m -\- x', y = n -\- y'.
174 PLANE ANALYTIC GEOMETRY.
Substituting in (1), we have
c'x'^ + d'y' + (2 dm + e') x' + c'm'' + d'n -\- e'm -\-f =0 . . . (2)
Kow, in general, we may give vi and ?i such values as to
make
2 c'??i -[- e' = 0, and c'm^ + c?'?i + e'm -|-y = ;
i.e., we may make
m = , and
e
c'm^ + e^«?, + / _ e'' - 4/c^
;. . . . (a)
If c^' is not zero (since a' = 0, c' is not zero), this transfor-
mation is possible and (2) becomes, after dropping accents,
c'x^ + d^y = 0,
°^ d'
x^=-^y . . . (3)
CoR. If d' = 0, (1) becomes
cV-f e'a;-f / = ... (4)
or, solving with respect to x,
2 c' ^ ^
141. Case 2. c' = o.
Under this supposition equation (6), Art. 138, becomes
ay + d'y + e'x+f=0 ... (1)
Transforming this equation so as to eliminate y and the
constant term, by a method exactly similar to that of the
preceding article, we find
d'
n = ,
2 a' '
d'^ - 4 a'f
4 a'e' '
and, if e' is not zero, we have (a' is not zero since c' = 0)
f=-^,x . . . (2)
a
EQUATION OF THE SECOND DEGREE.
175
Cob. If e' = 0, equation (1) becomes
y
d' ^ Vd'^ - 4. fa'
(3)
142. Summarizing the results of the preceding articles, we
find that the discussion of the general equation
ay^ -j- hxi/ -^ cx'^ ->r di/ -{- ex + f =
has been reduced to the discussion of the three simple forms :
1. a' if + c'x- -\-f" = 0. Art. 139, (5)
f=-I.tj. Art. 140, (3)
2/2 = _ 1- ic. Art. 141, (2)
_ e' jz ^e'-' - ^fo'
^~ 2 c'
-d' :^ ^d'^ - 4. fa'
2 a'
Art. 140, (5)
Art. 141, (3)
The discussion now involves merely a consideration of the
sign and magnitude of the constants which enter into these
equations.
143. b^ <'i ac.
Under this supposition, since a' and c' are both positive or
both negative, Art. 138, Cor. 2, neither a' nor c' can be zero ;
hence, forms 2 and 3 of the preceding article are excluded
from consideration.
The first form becomes either
ay + c'x^+/"=0, 1 .^^
or _ay-cV+.r=0|
in which a' and c' may have any real value and /' may have
any sign and any value. Hence arise four cases :
176
PLANE ANALYTIC GEOMETRY.
Case 1. If f" has a sign different from that of a' and c',
equations (1) are equations of ellipses whose semi-axes are
v/^-dj=Y//;
Case 2. If /" has the same sign as that of a' and c', equa-
tions (1) represent imaginary curves.
Case 3. If a' = c' and/" has a different sign from that of
a' and c', equations (1) are equations of circles. If f" has
the same sign as a' and c', then the equations represent imagi-
nary curves.
Case 4. If/" = 0, equations (1) are equations of t^vo imagi-
nary straight lines jmssing through the origin.
Hence, when 5^ < 4 ac, every equation of the second degree
bettveen two variables represents an ellipse, an imaginary curve,
a circle, or two imaginary straight lines intersecting at the
origin.
144. ^2 = 4 ac.
Under this supposition, Art. 138, Cor. 2, either a' = 0, or
c' = ; hence, form (1) of Art. 142 is excluded.
Resuming the forms
y = -
d'
e'
(2)
_ e' _J_ Ve'- - 4 fc'
" = 2^^ -~
y
^ -d: ^^ ^d'^- - ^fa'
~ 2 a'
(3)
J
we have four cases depending upon the sign and magnitude
of the constants.
Case 1. If d' and c' in the first form of (2) are not zero,
and if e' and a' in the second form of (2) are not zero, then
equations (2) are equations of parabolas.
EQUATION OF THE SECOND DEGREE. 177
Case 2. Since the first form of (3) is independent of y, it
represents tiuo lines parallel to each other and to the Y-axis.
The second form of (3) represents, similarly, tivo li7ies which
are parallel to the X-axis.
Case 3. If e"^ < 4/c' the first form of (3) represents tivo
imaginary lines.
If d'"^ < 4: fa', the second form of (3) represents tivo imagi-
nary lines.
Case 4. If e'^ = 4/c', the first form of (3) represents one
straight line parallel to the Y-axis.
If c^'2 _ 4.faf, the second form of (3) represents one straight
line parallel to the X-axis.
Hence, when &^ := 4 ac, every equation of the second degree
between two variables represents a parabola, two parallel straight
lines, two hnaginary lines, or one straight line,
145. b''> 4. ac.
Under this supposition, Art. 138, Cor. 2, since a' and
c' must have opposite signs, neither a' nor c' can be zero ;
hence forms (2) and (3) of Art. 142 are excluded from con-
sideration under this head. The first form becomes either
ay-cV+/" = 1
or -aY + c'x'^f" =0] " " " (^)
We have here three cases.
Case 1. If f has a different sign from that of a', equations
(1) are equations of hyperbolas whose semi-axes are
« = y//;andi=y'r.
If /" has a different sign from that of c', equations (1) are
still equations of hyperbolas.
Case 2. If a' = c' , equations (1) are equations of equilat-
eral hyperbolas.
Case 3. If f" = 0, equations (1) are equations of two inter-
secting straight lines.
178 PLANE ANALYTIC GEOMETRY.
Hence, when 6^ > 4 ac, every equation of the second degree
between two variables represents an hyperbola, an equilateral
hyperbola, or two intersecting straight lines.
146. Summary. The preceding discnssion has elicited the
following facts :
1. That the general equation of the second degree between
tivo variables represents, under every conceivable value of the
constants which enter into it, an ellipse, a parabola, an hyper-
bola, or one of their limiting cases.
2. When b'^ < 4ac it represents an ellipse, or a limiting case.
3. When h'^ = 4:ac it represents a parabola, or a limiting
case.
4. When b'^ > 4ac it represents an hyperbola, or a limiting
case.
EXAMPLES.
1. Given the equation 3 2/^ + 2a;?/ + 3a;^ — 83/ — 8x = 0;
to classify the locus, transform and construct the equation.
{a) To classify. Write the general equation and just below
it the given equation, thus :
ay^ + bxy -\- cx^ -\- dy -\- ex -}- f =
Sy"" -{-2xy -{-Sx'' -8y -Sx = . . . (1)
Substituting the co-efficients in the class characteristic
52 _ 4 ^(.^ -^g have 5^ — 4 ac = 4 — 36 = — 32 ;
hence &^ <4 ac.
and the locus belongs to the ellipse class. Art. 146.
(b) To refer the locus to axes such that the term containing
xy shall disap)pear.
From Art. 138, (5), we have
b
tan 2 ^ =
c — a
2
hence tan 2 6 = = + 00,
o — o
/. 2^ = 90° .-. ^ = 45° . . . (2)
EQUATION OF THE SECOND DEGREE. 179
i.e., the new X-axis makes an angle of -\- 45° with the old
X-axis. Taking now (10), (11), (3), Art. 138, and substitut-
ing values, we have
a' = ^\c -\- a — V(c — af + b~\ = 2.
c' = X ^c + a -f V(c-ay-\-b'\ = 4.
d' = dcos6 — e sin 6 = ^ -^2 (d — e) = 0.
e' = dsine -{-ecos6 = ^^2 (d -\- e) = —8 V2.
Substituting these values in (6), Art. 138, we have (/
being zero),
2tf-^4.x~-8 V2T a; = ... (3)
(c) To refer the locus to its centre and axes.
Substituting the values found above in (4), Art. 139, we
have ?^ = = 0.
2 a'
m = - -^ = 8 V2 _ ^2
Hence/" = a'n- + c'm"" + d'n + e'm-\-f= - 8, Art. 139,
(3).
Substituting this value of /" together with the values of a'
and c' found above in (5), Art. 139, we have
2 y2 _|. 4 ^2 _ 8 = 0,
or ^' _i_ l! = 1 . . . (4)
2^4 ^ ^
for the reduced equation. The semi-axes of the ellipse are
a = V2 and & = 2.
(d) To construct.
Draw the axis OX', making an angle of 45° with the old
X-axis. See (b). Draw OY' 1 to OX'. The equation of the
curve when referred to these axes is given in (3). Constructing
180
PLANE ANALYTIC GEOMETRY.
the point 0' (V2, 0) we have the centre of the ellipse. See
(c). Draw O'Y" -L to OX' at 0'. The equation of the curve
when referred to O'Y", O'X' as axes is given in (4).
Fig. 55.
Having the semi-axes, V2 and 2, we can construct the
ellipse by either of the methods given in Art. 78.
DISCUSSION.
If 2/ = in (1), we have for the X-intercepts 0, OD,
3
If cc = in (1), we have for the Y-intercepts 0, OC,
7/ == 0, 7/ = - .
If £c = in (3), we have y = j- 0; i.e., the ellipse is tangent
to the Y'-axis.
If y = in (3), we have for the X'-intercepts 0, OB,
X = 0, cc = 2 V2.
If a; = in (4), we have for the Y"-intercepts O'A, O'A'.
2/- ±2.
EQUATION OF THE SECOND DEGREE. 181
If 2^ = ill (4), we have for the X'-intercepts O'B, O'O,
a; = -t- V2.
2. Given the equation t/ — 2 xy -{- x^ — 2 7j — \ = 0, class-
ify the locus, transform and construct the equation.
(a) To classify.
ay"^ + bxy + cx^ -\- dy -\- ex -\- f =
y'' — 2xy -h x--2y-l = ... (1)
hence b"^ — 4= ac = A — 4: = 0,
,'. b- = 4:ac ;
hence the locus belongs to the parabola class, Art. 146.
(b) To refer the locus to axes such that the term containing
xy shall disappear.
From Art. 138, (5), we have
tan 26 = —^ ;
c — a
hence, substituting
tan 2 ^ = —
1 _ 1 -
.-. ^ = -45° . . . (2)
Substituting the values of the coefficients in (10), (11), (3)
of Art. 138, we have
a' = ^\c + a — V(c — ay + b''\= 0.
c' = :L^c^a + V(c - ay + b'^ ^ = 2.
d' = dGosd — esinO = —2 (^ V2) = — V2.
e' = t^ sin ^ + e cos ^ = — 2 (— i V2) = + V2.
Substituting these values in (1), Art. 140 (since a' — 0),
we have 2x^ — V2 y + V2 a; — 1 = . . . (3)
(c) To refer the p>arahola to a tangent at the vertex and the
axis.
Substituting the values of the constants in (a), Art. 140,
we have e' V2 oc^ i
m = — . = = — .60 nearly.
2 c' 4 ^
e'2 — 4 /c' 5
182
PLANE ANALYTIC GEOMETRY.
Substituting the values of d' and c in (3), Art. 140 (since d)
is not zero), we have
a;2 = iV2.2/ . . . (4)
for the reduced equation.
(d) To construct.
V
Fig. 56.
Draw OX' making an angle of — 45° with the X-axis ;
draw OY' ± to OX'. See {b). The equation of the parabola
when referred to these axes is given in (3).
Constructing the point (— .35, — .90), we have the vertex
of the parabola 0'. See (c). Draw O'X" and O'Y" parallel to
the axes OX', OY' respectively. The equation of the parab-
ola referred to these axes is given in (4). The curve can now
be constructed by either of the methods given in Art. 54.
EQUATION OF THE SECOND DEGREE. 183
DISCUSSION.
If ic = in (1), we have for the Y-intercepts OC, OC,
y = 2.4 2/ = - •4-
If y = in (1), we have for the Y-intercept OD, OD',
a^= -t 1.
If cc = in (3), we have for the Y'-intercept OK,
1
y =
= - .707.
V2
If 2/ = in (3), we have for the X'-intercepts OL, OL',
_ _ V2 + VlO _ - V2 - Vlo
If X = in (4), y = 0; if ?/ = in (4), x = -t 0.
3. Given the equation y^ — 2x^ — 2'i/-\-6x — 3 = 0, classify
the locus, transform and construct the equation.
(a) To classify.
ay'^ -\- bxy + cx'^ -\- dy -\- ex + f= 0.
y^ — 2x^-2 y + 6x — 3 = . . . (1)
b^-4:ac = 8 .: b^>4.ac;
hence, the locus belongs to the hyperbola class, Art. 146.
(b) To ascertain the direction of the rectangular axes (xy
being wanting).
tan 26 = — ^— = -^ = ;
c — a — 3
.-. ^ = 0;
i.e., the new X-axis is parallel to the old X-axis.
(c) To refer the hyperbola to its centre and axes, we have,
Art. 139, (4),
n =
d' _ e'
2^''^~~T7'
hence n = \, m ^= -.
2
Substituting in the value of /", Art. 139, (3), we have
/" = a'n'' -f c'm^ + d'n + e'm -\- f = 1 -^ - 2 +9-3;
hence f" = -.
-^ 2
184
PLANE ANALYTIC GEOMETRY.
This value, together with the values of a' and c' in (5), Art.
139, gives 2^f-4.x^ = -l . . . (3)
for the required equation.
(d) To construct.
Fig. 57.
Construct the point 0' (|, 1), and through it draw O'X" || to
OX, and O'Y" || to OY. The equation of the hyperbola
referred to these axes is given in (3). AVe see from this equa-
tion that the semi-transverse axis is -. Laying off this dis-
tance to the right and then to the left of 0', we locate the
vertices of the curve A, A'.
DISCUSSION.
If a; = in (1), we have for the Y-intercepts OC, OC,
y = 3,y = —1.
If 2/ = in (1), we have for the X-intercepts OD, OD',
3 + V3
V3
EQUATION OF THE SECOND DEGREE. 185
If a; = in (3), we have
If ?/ = in (3), we have for the X-intercepts O'A, O'A',
-^\
From this data the student may readily determine the
eccentricity, the parameter, and the focal distances of the
hyperbola.
4. Given the equation T/^-fa;^ — 42/ + 4a;— 1 = 0, class-
ify the locus, transform and construct the equation.
(a) b"^ <4 ac .-. the locus belongs to the ellipse class.
(b) 6 = 0.: new X-axis is || to old X-axis.
(c) (m, n)=(-2,2) and /" = - 9
hence x^ -\- y^ = 9
is the transformed equation of the locus, which from the form
of the equation is evidently a circle.
(d) Locate the point (— 2, 2). With this point as a centre,
and with 3 as a radius, describe a circle ; it will be the re-
quired locus.
5. y^ - 2 xy -\- x^ - 2 = 0.
(a) b^ ^ 4: ac .-. parabola class.
(b) = — 4:5° .: new X-axis inclined at an angle of — 45°
to the old X-axis. We have also
a' = 0, c' = 2, d' = 0,e' =
i.e., X = 1 and x = — 1 . . . (1)
are the equations of the locus when referred to the new axes.
(c) The construction gives the lines OX', OY' as the new
axes of reference.
Equations (1) are the equations of the two lines CM, CM'
drawn || to the Y'-axis and at a unit's distance from it.
186
PLANE ANALYTIC GEOMETRY.
Fig. 58.
We may construct the locus of the given equation without
going through the various steps required by the general
method. Factoring the given equation, we have
hence
{y -x + V2) {y -X- V2) =
y = X — V2 and y = x -\- V2
are the equations of the locus. Constructing these lines
(OY, OX being the axes of reference), we get the two
parallel lines CM, CM'.
Classify, transform, and construct each of the following
equations :
6. 2/' - 2 cKy + a;' + 2 2/ - 2 a; + 1 = 0.
7. 2/' + 2 x?/ + a;2 - 1 = 0.
8. b y^ + 2 xy + 5 x" — 12 X — 12 y
x = ^^2
2 "^ 3
EQUATION OF THE SECOND DEGREE. 187
9. 2 1/ -\- 2 x^ — 4.y — 4.x -\- 1 = 0.
^' + y' = |.
10. y^ + x^ — 2 X -\- 1 = ().
y = x^ -!,{{),()).
11. ?/2 _|_ ^2 _j_ 2 a; + 2 = 0.
Imaginary ellipse.
12. 2/'' - 2 xy + a;2 - 8 a; + 16 = 0.
Parabola.
13. y^- — 2xy -\-x'^ — y -\-2x — l=0.
Parabola.
14. 4 a;y — 2 ic + 2 = 0.
Hyperbola.
15. y~ — 2x' -\-2y + !=().
Two intersecting lines.
16. ?/2 _ a;2 4_ 2 2/ + 2 cc - 4 = 0.
Equilateral hyperbola.
17. 2/^ — 2 XT/ + a;2 + 2 2/ + 1 = 0.
18. 2/^ + 4 xy + 4 a;' — 4 = 0.
19. y"" — 2xy -\-2x'' — 2y -\-2x = 0.
20. 2/2 — 4 a;?/ + 4 a;2 = 0.
21. 2/^ — 2 cc?/ — cc2 _^ 2 = 0.
22. 2/'-ic' = 0.
188 PLANE ANALYTIC GEOMETRY.
CHAPTER X.
HIGHER PLANE CURVES.
147. Loci lying in a single plain and represented by equa-
tions other than those of the first and second degrees are
called Higher Plane Curves. We shall confine our atten-
tion in this chapter to the consideration of a few of those
curves which have become celebrated by reason of the labor
expended upon them by the ancient mathematicians, or which
have become important by reason of their practical value in
the arts and sciences.
EQUATIONS OF THE THIRD DEGREE.
148. The Semi-cubic Parabola.
This curve is the locus generated by the intersection of the
ordinate TT^ of the common parabola with the perpendicular
OP let fall from its vertex upon the tangent drawn at T' as
the point of tangency moves around the curve.
1. To deduce the rectangular equation.
Let T' {x", y") be the point of tangency, and let P (cc , 3/ )
be a point of the curve.
Let y"^ = 4: px be the equation of the common parabola.
Since the equation of the tangent line T'M to the parabola
is Art. 57, (6),
yy" = 2p(x + x"),
the equation of the perpendicular (OM) let fall from the
vertex is
y = — ^ x . . . (V)
HIGHER PLANE CURVES.
189
Fig. 59.
Since TT' is parallel to OY, we have for its equation
x=x" . . . (2)
Combining (1) and (2), we have
y =
- _ y
'p
But
hence
y" = V4:2)x" ;
V4 px" „
y = — ^ • X .
2 p
Squaring and dropping accents, we have
2/^ = ^ . . . (3)
P
for the equation of the semi-cubic parabola.
This curve is remarkable as being the first curve which was
rectified, that is, the length of a portion of it was shown to
190
PLANE ANALYTIC GEOMETRY.
be equal to a certain number of rectilinear units. It derives
its name from the fact that its equation (3) may be written
x^ — p'^ y.
2. To deduce the polar equation.
Making x = r cos 6 and y ^= r sin Q in (3), we have, after
reduction,
r = p tan ^ 6 sec ^ . . . (4)
for the polar equation of the curve.
ScHOL. Solving (3) with respect to y, we have
y
= Wf
An inspection of this value shows
(a) That the curve is symmetrical with respect to the
X-axis ;
(b) That the curve extends infinitely from the Y-axis in
the direction of the positive abscissas.
149. To dupjlicate the cube hy the aid of the parabola.
Let a be the edge of the given cube. We wish to con-
struct the edge of a cube such that the cube constructed on it
shall be double the volume of the given cube ; i.e., that the
condition x^ = 2 a^ shall be satisfied.
Fig. 60.
HIGHER PLANE CURVES.
191
Construct the parabola whose equation is
2/2 = 2 ace ... (1)
Let MPO be the curve. Construct also the parabola whose
equation is
x^ = ay . . . (2)
Let NPO be this curve.
Then OA (= x), the abscissa of their point of intersection
is the required edge. For eliminating y between (1) and (2),
we have
x^ = 2 a\
This problem attained to great celebrity among the ancient
geometricians. We shall point out as we proceed one of the
methods employed by them in solving it.
150. The Cissoid.
The cissoid is the locus generated by the intersection (P) of
the chord (OM') of the circle (OMM'T) with the ordinate
Fig. 61.
192 PLANE ANALYTIC GEOMETRY.
MIST (equal to the ordinate M'N' let fall from the point
M' on the diameter through 0) as the chord revolves about
the origin 0.
It may also be defined as the locus generated by the inter-
section of a tangent to the parabola ^/^ = — 8 aa; with the
perpendicular let fall on it from the origin as the point of
tangency moves around the curve.
1. To deduce the rectangular equation.
Fh^st Metliod. — Let OT = 2 a, and let P {x, y) be any point
of the curve. From the method of generation in this case
MN = WW .-. ON = N'T. From the similar triangles ONP,
ON'M', we have
NP : ON : : M'N' : ON'.
But NP = y, ON = X, M'N' = VON' . N'T = V(2 a - cc) x,
ON' = 2a -x\
.-. y : X :: V(2 a — x) x -.2 a — x.
Hence y"" = -~^ •••(!)
2 a — X
is the required equation.
Second Method. — The equation of the tangent line to the
parabola 3/^ = — 8 ax is Art. 65, (2)
, 2a
y = — sx -\-
The equation of a line passing through the origin and per-
pendicular to this line is
X
2/ = -•
s
Combining these equations so as to eliminate s, we have
x^
y =
2 a — X
for the equation of the locus.
This curve was invented by Diodes, a Greek mathematician
of the second century, B.C., and called by him the cissoid from
HIGHER PLANE CURVES. 193
a Greek word meaning '* ivy." It was employed by him in
solving the celebrated problem of inserting two mean propor-
tionals between given extremes, of which the duplication of
the cube is a particular case.
2. To deduce the polar equation.
From the figure (OP, PON) = (r, 0)
we have also r = OP = M'K = OK - OM'.
But OK = 2 a sec ^ and OM' = 2 «. cos ^ ; hence
r = 2 (z (sec 6 — cos 6),
or r = 2 a tan 6 sin 6
is the polar equation of the curve.
ScHOL. Solving (1) with respect to y, we have
V^
2/=±
An inspection of this value shows
(a) That the cissoid is symmetrical with respect to the
X-axis.
(b) That X = and x =2 a are the equations of its limits.
(c) That ic = 2 a is the equation of a rectilinear asymp-
tote (SS').
151. To duplicate the cube by the aid of the cissoid.
Let OL, Fig. 61, be the edge of the cube which we wish to
duplicate. Construct the arc BO of the cissoid, CO = a
being the radius of the base circle. Lay off CD = 2 CA =
2 a and draw DT intersecting the cissoid in B ; draw BO
and at L erect the perpendicular LE, intersecting BO in R.
Then LR is the edge of the required cube ; for the equation
of the cissoid gives
x^
y =
2 a — X
OPTS
hence HB^ = ^ (since HB =y,OB.= x, and HT =
Hi
2 a — x).
The similar triangles CDT and HBT give
CD : CT :: HB : HT.
194
PLANE ANALYTIC GEOMETRY.
But CD = 2 CT by construction ; hence HB = 2 HT
2
This value of HT in the value of HB^ above gives
2 0H3
HB2 =
HB
hence HB^ = 2 OH".
The triangles OHB and OLR are similar ; hence
HB : OH :: LR : OL
.-. HB3:OH3::LR=^::OL3
But HB^ = 2 OH^, hence LR^ = 2 OL^ ; whence the con-
struction.
152. The Witch.
Fig. 62.
V HIGHER PLANE CURVES. 195
The witch is the locus of a point P on the produced ordi-
nate DP of a circle, so that the produced ordinate DP is to
the diameter of the circle OA as the ordinate DM is to the
outer segment DA of the diameter.
It may also be defined as the locus of a point P on the
linear sine DM of an angle at a distance from its foot D equal
to twice the linear tangent of one-half the angle.
1. To deduce the rectangular equation.
First Method. — From the mode of generation, we have
DP : OA :: DM : DA
But DP = 2/. OA = 2 a, DM = VOD . DA = ^/x (2 a - x),
DA = 2 a — x\
hence ' y-2 a :: V(2 a — x) x : 2 a — x.
o ^ Cv JO /^ \
2 a — X
is the required equation.
Second Method. — Let MCO = 6 ; then by definition
o ^ ^ c, ^ la 0- — cos &)
V = 2 « tan - = 2 ft i / — ^^ '- .
2 V a (1 + cos 6)
But ft (1 — cos ^) = a, - a cos ^ = 00 — DC = OD = x, and
a (1 + cos ^) = a + ft cos ^ = 00 + DC = OD' = 2a — x;
hence
4 a^x
= 2^2-
or, squaring y
2 a — X
This curve was invented by Donna Maria Agnesi, an Italian
mathematician of the eighteenth century.
ScHOL. Solving (1) with respect to y, we have
y
= _1_ 2 a \/ ^- .
\ 2a-x
Hence (a) the witch is symmetrical with respect to the
X-axis.
(b) cc = and x = 2 a are the equations of its limits.
(c) x = 2a is the equation of the rectilinear asymptote SS'.
196
PLANE ANALYTIC GEOMETRY,
EQUATIONS OF THE FOURTH DEGREE.
153. The Conchoid.
The conchoid is the locus generated by the intersection of
a circle with a secant line passing through its centre and a
fixed point A as the centre of the circle moves along a fixed
line OX.
As the intersection of the circle and secant will give two
points P, P, one above and the other below the fixed line, it
is evident that during the motion of the circle these points
will generate a curve with two branches. The upper branch
MBM' is called the Supekior Branch ; the lower, the In-
ferior Branch. The radius of the moving circle O'P
(= OB) is called the Modulus. The fixed line OX is called
the Directrix ; the point A, the Pole.
1, To deduce the rectangidar equation.
Let P (x, y), the intersection of the circle PP'P and the
HIGHER PLANE CURVES. 197
secant AOT, be any point of the curve. Let O'P = OB = h,
and let OA = a.
The equation of the circle whose centre is at 0' (x', 0) is
(x — x'Y -\- y^ = b^-
The equation of the line AOT is
y ^= sx — a . . . (1)
Making y — in (1), we have
a
s
for the distance 00'.
But 00' = x' ; hence
^-^J+^' = ^'- • • ^^>
is the equation of the circle. If we now combine (1) and (2)
so as to eliminate s, the resulting equation will express the
relationship between the co-ordinates of the locus generated
by the intersection of the loci they represent. Substituting
the value of s drawn from (1) in (2), we have
a +y
,.xY=(b'-y^)(a + yy . . . (3)
is the required equation.
We might have deduced this equation in the following very
simple way : Draw AT || to OX, and PT || to OY. Since the
triangles ATP and O'SP are similar, we have
PS : SO' :: PT : TA :
i.e., y : -\/b^ — y^ :: a -{- y : x.
Hence xY = {b^ — y^) (a + yy.
This curve was invented by Nicomedes, a Greek mathema-
tician who flourished in the second century of our era.
It was employed by him in solving the problems of the
duplication of a cube and the trisection of an angle.
198 PLANE ANALYTIC GEOMETRY.
2. To deduce the polar equation.
From the figure we have (AY being the initial line, and A
the pole)
(AP, PAB) = (r, 6)
But AP = AO' ± OT ;
hence r = a sec 6 ^^b
is the polar equation of the curve.
ScHOL. Solving (3) with respect to x, we have
O' + y
X = ^ — ■y/b'' — if . ■
An inspection of this value shows
(a) That the conchoid is symmetrical with respect to the
Y-axis.
(5) That y ^b and y = — b are the equations of its limits.
(c) That 2/ = gives x = -I- c/d, .-. the X-axis is an asymp-
tote.
(d) If a = 0, then x = -^ V^" — y''; i.e., the conchoid be-
comes a circle.
(e) If 6 > a, the inferior branch has a loop as in the figure,
(/) li b = a, the points A' and A coincide and the loop.
disappears.
(g) li b 45° and <135° r is imaginary.
If ^ = 135°, cos 2 ^ = cos 270° = .-. r = 0.
If ^ = 180°, cos 2 ^ = cos 360° = 1 .-. r = -J- a.
An examination of these values of r shows that the curve
occupies the opposite angles formed by the asymptotes of the
hyperbola.
The curve is symmetrical with respect to both axes.
HIGHER PLANE CURVES.
203
TRANSCENDENTAL EQUATIONS.
157. The Curve of Sines.
This curve takes its name from its equation
y = sin X,
and may be defined as a curve whose ordinates are the sines
of the corresponding abscissas, the latter being considered as
rectified arcs of a circle.
Y
Fig. 67.
To construct the curve. Give values to x which differ from
each other by 30°, and find from a "Table or Natural
Sines " the values of the corresponding ordinates.
Tabulating the result, we have,
Value of X Corresponding Value of y
.50
.87
30°
=
IT
6 ~
.52 «
60°
=
2 TT
"6"
= 1.04 «
90°
=
37r
6
= 1.56 «
20°
47r
6
= 2.08
1.00
.87
204
PLANE ANALYTIC GEOMETRY.
Value of X
Corr
esponding
Value of y
150° -^"^ - 2.60
6
«
.50
180° = TT = 3.14
((
210° = IjL = 3.66
6
((
-.50
240° = ?Z = 4.18
6
It
-.87
270° - ^ ^ - 4.70
6
u
- 1.00
300° = ^^ '^ - 5.22
6
11
-.87
330° = ^^'^ = 5.75
6
a
-.50
360° = 2 TT = 6.28
a
Constructing these points and tracing a smooth curve
through them, we have the required locus. As x may have
any value from to -J:; oo and yet satisfy the equation of the
curve, it follows that the curve itself extends infinitely in
the direction of both the positive and negative abscissas.
158. The Cukve of Tangents.
Y
Fig.
HIGHER PLANE CURVES. 205
This curve also takes its name from its equation
y = tan x.
To construct the curve. Give x values differing from each
other by 30° and find from a Table of Natural Tangents the
corresponding values of y. Tabulating, we have,
Value of X Corresponding Value of y
"
" .57
« 1.73
30°
" 6 ~
-- .52
60°
_27r
6
= 1.04
90°
377
6
= 1.56
120°
_47r
"6"
= 2^8
150°
_57r
6
= 2.60
180°
= TT =
3.14
210°
_lir
= 3.66
« - 1.73
-.57
"
.57
6
240° =^ = 4.18 " 1.73
6
270° = ^ = 4.70 « oo
6
300° = ^^ = 5.22 " - 1.73
6
330° = 11^ == 5.75 " - .57
6
360° = 2 TT = 6.28 "
Constructing these points and tracing a smooth curve
through them, we have the locus of the equation.
This curve, together with that of the preceding article,
belong to the class of Repeating Curves, so called because
they repeat themselves infinitely along the X-axis.
206
PLANE ANALYTIC GEOMETRY.
159. The Cycloid.
This curve is the locus generated by a point on the circum-
ference of a circle as the circle rolls along a straight line.
The line OM is called the Base of the cycloid ; the
point P, the Generating Point ; the circle BPL, the Gen-
erating Circle; the line HB', perpendicular to OM at its
middle point, the Axis. The points and M are the Vertices
of the cycloid.
1. To deduce the rectangular equation, the origin being
taken at the left-hand vertex of the curve.
Let P be any point on the curve, and the angle through
which the circle has rolled, PCB = 0. Let LB, the diameter
of the circle, = 2 a.
Then OA = OB - AB and AP = CB - CK.
ButOA == a;, OB = a ^, AB = PK = a sin ^, AP = 3/, CB = a,
CK = a cos 9 ; hence, substituting, we have
X = a t) — a sm
y =z a — a cos 6
(1)
HIGHER PLANE CURVES. 207
Eliminating 6 between these equations, we have
/T If 1i
X = a cos"^ — V'2 ay — y"^ = a vers~^ -
CO (^
-^2ay-y' ... (2)
for the required, equation.
ScHOL. An inspection of (2) shows
{a) that negative values of y render x imaginary.
(b) When y = 0, x = a vers~^ = 0; but a vers"^ = 2 tt a,
or 4 TT a, or 6 TT (X, or etc. ; hence there are an infinite number
of points such as and M.
(c) When y = 2 a, x = a vers"^ 2 = tt a = OB' ; but
a vers~^ 2 = 3 tt a, or 5 tt a, ov 1 tt a, ov etc. ; hence, there are
an infinite number of points such as H.
(d) y = and y = 2 a are equations of the limits.
(e) For every value of y between the limits and 2 a there
are an infinite number of values for x.
2. To deduce the rectangular equation, the origin being at the
highest 'point H.
We have for the equations of transformation
a; = OA = OB' - PK' = tt a + cc'
2/ = AP = B'H - HK' = 2a^y'
These values in (1) above give
x' =^ a {0 — it) — a sin Q 1 .o\
y' '= — a — a cos Q f
But &, the angle through which the circle has rolled from
H, = ^ — TT ; hence
x' = aQ' -\- a sin & 1 ,^
y' = a (cos 6' -1) ] ■ ■ ' ^ ^
_ y' _ .
Hence cc' = a vers~^ — ■ -}- V— 2 ay^— y'^ . . . (5)
The invention of this curve is usually attributed to Galileo.
With the exception of the conic sections no known curve
possesses so many useful and beautiful properties. The fol-
lowing are some of the more important :
208
PLANE ANALYTIC GEOMETRY.
1. Area OPHDB'O = area HDB' = tt a".
2. Area of cycloid OHMO = 3 HDB' = 3 tt a^.
3. Perimeter OPHM = 4 HB' = 8 a.
4. If two bodies start from any two points of the curve
(the curve being inverted and friction neglected), they will
reach the lowest point H at the same time.
5. A body rolling down this curve will reach the lowest
point H in a shorter time than if it were to pursue any other
path whatever.
SPIRALS.
160. The Spiral is a transcendental curve generated by a
point revolving about some fixed point, and receding from it
in obedience to some fixed law.
The portion of the locus generated during one revolution of
the point is called a Spire.
The circle whose radius is equal to the radius-vector of the
generating point at the end of the first revolution is called
the Measuring Circle of the spiral.
161. The Spiral of Archimedes.
\^^—
r-x /
/\ "''' '
~~ ""■•^w\
/ '' \
x\\
/ / \^--
/ \
/ 1 / \
y 'a
\ [/
\ //
\ ^^ J/ \
\ / *
\ .X ^^''^>-^
^^^-^""^^
\/ '^-—
■" \
Fig. 70.
HIGHER PLANE CURVES. 209
This spiral is the locus generated by a point so moving
that the ratio of its radius-vector to its vectorial angle is
always constant.
From the definition, we have
hence r ■= cO . . . (1)
is the equation of the spiral.
To construct the spiral.
Assuming values for 6 and finding from (1) the correspond-
ing value for r, we have
Values of 6 Corresponding Values of r
"0
" . — C
4
cc 2 TT
4
,, 3 TT
" G
4
" TTC
4
" c
4
" . G
4
" 2 TTC
Constructing these points and tracing a smooth curve
through them, we have a portion of the spiral.
Since ^ = gives r =0, the spiral passes through the pole.
Since $ = cc gives r = oo, the spiral makes an infinite
number of revolutions about the pole.
Since = 2 tt gives r = 2 ir c, OA (= 2 tt c) is the radius of
the measuring circle.
45°
TT
"4
90°
_27r
4
135°
_37r
4
180°
= TT
225°
5x
4
270°
_67r
7 _
315°
/ TT
4
360°
= 277-
00
210
PLANE ANALYTIC GEOMETRY.
162. The Hyperbolic Spiral.
This curve is the locus generated by a point so moving that
the product of its radius-vector and vectorial angle is always
constant.
From the definition we have
or r = - . . . (1)
e ^ ^
for the equation of the spiral.
Fig. 71.
To construct the spiral.
Giving values to 6, finding the corresponding values of r,
we have
Values of 6 Corresponding Values of r
45° = '-
00
n
HIGHER PLANP CURVES. 211
Values of 6
Corresponding
Values of r
90° = ? TT
4
a
2c
TT
135°=^7r
4
ti
4c
3 TT
180° = TT
u
c
IT
225° = ^ TT
4
u
Ac
57r
270°=^7r
4
u
4c
315° =1,7
4
it
4c
360° = 2 TT
ii
2^
00,
a
Constructing the points we readily find the locus to be a
curve such as we have represented in the figure.
Since ^ = gives r = oo there is no point of the spiral
corresponding to a zero-vectorial angle.
Since 6 ^ oo gives r = 0, the spiral makes an infinite number
of revolutions about the pole before reaching it.
Since 6 = 2tt gives
c
2 TT
c is the circumference of the measuring circle.
ScHOL. Let P be any point on the spiral ; then
(OP, POA) = (r, 0).
With as a centre and OP as a radius describe the arc PA.
By circular measure, Arc PA = r 6, and from (1) c =r 0;
hence Arc PA = c ;
i.e., the arc of any circle between the initial line and the
spiral is equal to the circumference of the measuring circle.
212
PLANE ANALYTIC GEOMETRY.
163. The Parabolic Spiral.
This spiral is the locus generated by a point so moving
that the ratio of the square of its radius-vector to its vectorial
angle is always constant.
From the definition we have
or, r^ = c6 . . . (1)
for the equation of the spiral.
Fig. 72.
To construct the spiral.
Values of $
Corresponding
Values of r
45°='"
4
11
^ Vctt
90° _ 2 T
4
ti
i V2c7r
135° - ^7
4
ti
•1 VSCTT
180° = TT
li
Vc TT
HIGHER PLANE CURVES.
213
Values of $
225° = ^
4
270^ = —
4
315° = 1^
4
360° = 2 TT
GO
Corresponding
Values of r
*V6
Ctt
V27^
00
Constructing these points and tracing a smooth curve
through them we have the required locus.
Since ^ = gives r = 0, the spiral passes through the pole.
Since 6 = cc gives r = go, the spiral has an infinite num-
ber of spires.
164. The Lituus or Trumpet.
This curve has for its equation
or
= v/,^..w
Fig. 73.
214
PLANE ANALYTIC GEOMETRY.
If ^ = 0, ?' = 00 ; if ^ = 00, r = 0. This curve has the
initial line as an asymptote to its infinite branch.
165. The Logarithmic iSph^al.
This spiral is the locus generated by a point so moving that
the ratio of its vectorial angle to the logarithm of its radius
vector is equal to unity. Hence
= 1; i.e., e = logr;
log r
or passing to equivalent numbers (a being the base), we have
r ^aO . . . (1)
for the equation of the spiral.
To construct the sjnral. Let a = 2, then
r = 29
is the particular spiral we wish to construct.
HIGHER PLANE CURVES. 215
Values of ^
Corresponding
Values of r
a
1
1 = 57.°3
u
2
2 = 114.°6
ii
4
3 = 171.°9
a
8
4 _ 229. °2
a
16
00
i(
00
- 1 = - 57.°3
(I
.5
- 2 = - 114.
°6
i(
.25
- 3 = - 171.
°9
ii
.125
_. 4 = — 229.
°2
((
.062
— 00
i( '
A smooth curve traced through these points will be the
required locus.
Since 6 = gives r = 1 whatever be the assumed value of
a, it follows that all logarithmic spirals must intersect the
initial line at a unit's distance from the pole.
Since 6 = cc gives ?• = oo , the spiral makes an infinite
number of revolutions without the circle whose radius OA = 1,
Since 6 = — oc gives r = 0, the spiral makes an infinite
number of revolutions within the circle OA before reaching
its pole.
EXAMPLES.
1. Discuss and construct the cubical parabola
_ ^^
2. What is the polar equation of the limayon, Fig, 65, the
pole being at C ?
Ans. r = 2 a cos - Q.
o
3. Let OF = OF' = a yi Fig. ^^. Show that the lemnis-
cata is the locus generated by a point so moving that the
216 PLANE ANALYTIC GEOMETRY.
product of its distances from the two fixed points F, F' is
constant and
Discuss and construct the loci of the following equations
4. a; = tan y.
5. 1/ = cos X.
G. y = sec X.
7. a; = sin y.
S. y = cot X.
2. y = cosec X.
10- y = ^
11. xV + :.,/ = !. _.. , j_^i„^-
20. Discuss and construct the locus of the equation
yi _ 96 ah/ + 100 a'^x'^ — x* = or
y = ^ Vis «2 _^ V(a; — 6 a) (x + 6 a) (x — S a) (x -{- 8 a).
21. Show that ?/ = _j- cc are the equations of the rectilinear
asymptotes of the locus represented by the equation of
Ex. 20.
12.
a^ = x^ — axy.
13.
xl + ?/§ = 1.
14.
15.
r ^= a sin 2 6.
16.
a
^ ~ sin 2 ^ '
17.
r = a sm** - .
3
18.
r2 sin2 2^ = 1.
1Q
.. _ 1 + sin 6
N
SOLID ANALYTIC GEOMETRY.
PART 11.
CHAPTER I.
CO-ORDINATES. —THE TRI-PLANAR SYSTEM..
166. The position of a point in space is determined when
we know its distance and direction from three planes which
intersect each other, these distances being measured on
lines drawn from the point parallel to the planes. Although
it is immaterial in principle what angle these planes make
with each other, yet, in practice, considerations of convenience
and simplicity have made it usual to take them at right
angles. They are so taken in what follows.
Let XOZ, ZOY, YOX be the Co-ORDiisrATE Planes inter-
secting each other at right angles. Let OX, OY, OZ be the
Co-OEDiNATE AxES and 0, their intersection, the Origin of
Co-ordinates.
Let P be any point in the right triedral angle - XYZ.
Then P is completely determined when we knoAV the lengths
and directions of the three lines PA, PB, PC let fall from
this point on the planes.
As the planes form with each other eight right triedral
angles, there are evidently se^wn other points which satisfy
the condition of being at these distances from the co-ordi-
nate planes. The ambiguity is avoided here (as in the case
217
218
SOLID ANALYTIC GEOMETRY.
of the point in a plane) by considering the directions in which
these lines are measured.
Assuming distances to the right of YOZ a,s positive, distances
to the left will be negative.
Pa
P3
p.
/j
l\
/
/\\ B/
'1 ,. P/
1 _
]/ 1
/ /
1 / / '
1 / V / '
!/ /
i
-X
/' 1 n/
'T-----. /
+x
i c
-■)
f
/
F
3
-2
p.
Fig. 75.
Assuming distances above XOY as j^ositive, distances below
will be negative.
Assuming distances in front o/XOZ as J9os^^5^^' e, distances ifo
^7ie rear will be negative.
Calling x', y', z' (= BP, AP, CP, respectively) the co-ordi-
nates of the point P in the first angle, we have the follow-
ing for the co-ordinates of the corresponding points in the
other seven :
Second Angle, above XY plane, to left YZ plane, in front
of XZ plane, (- x', y', z') V^.
Third Angle, above XY plane, to left YZ plane, in rear
of XZ plane, (- x', - y', z') P3.
CO-ORDINA TES. 219
Fourth Angle, above XY plane, to right YZ plane, in
rear of XZ plane, {x', — y', z') P4.
Fifth Angle, below XY plane, to right YZ plane, in
front of XZ plane, (x', y', - z') P5.
Sixth Angle, below XY plane, to left YZ plane, in front
of XZ plane, (— x', y', — z') Pg.
Seventh Angle, below XY plane, to left YZ plane, in
rear of XZ plane, {— x', — y', — z') P7.
Eighth Angle, below XY plane, to right YZ plane, in rear
of XZ plane, {x', — /, — z') Pg.
EXAMPLES.
1. In what angles are the following points :
(1, 2, - 3), (- 1, 3, - 2), (- 1, - 2, - 4), (3, - 2, 1).
2. State the exact position with reference to the co-ordi-
nate axes (or planes) of the following points :
(0, 0, 2), (- 2, 1, 2), (3, 1, 0), (3, - 1, 2), (2, 0, 3), (- 1, 2,
0), (0, - 1, 0), (3, 0, 1), (1, - 2, 3), (0, 0, - 2), (4, 1, 2),
(5, 1, - 1), (1, 1, - 1).
3. In which of the angles are the X-co-ordinates positive ?
In which negative ? In which of the angles are the Y-co-
ordinates positive ? In which are the Z-co-ordinates negative?
167. Projections. The projection of a point on a plane is
the foot of the perpendicular let fall from the point on the
plane. Thus A, B, and C, Fig. 75, are the projections of the
point P on the planes XZ, YZ, XY, respectively.
The projection of a line of definite length on a plane is the
line joining the projections of its extremities on that plane.
Thus OC, Fig. 75, is the projection of OP on the XY plane.
The projection of a line of definite length on another line
is that portion of the second line included between the feet of
the perpendiculars drawn from the extremities of the line of
definite length to that line.
220
SOLID ANALYTIC GEOMETRY.
Thus OM, Fig. 75, is the projection of OP on the X-axis.
KoTE. — The projections of points and lines as above de-
fined are orthogonal. Unless otherwise stated, all projections
will be so understood in what is to follow.
168. To find the length of a line joinmff two points in space.
Fig. 76.
Let P' {x', y', z') and Y' {x", y", z") be the given points.
Let L (= PT") be the required length. Draw Y'G and
P'N II to OZ; NA and CD || to OY; NB || to OX. Join N
and C and draw P'M || to NC.
We observe from the figure that L is the hypothenuse of a
right angled triangle whose sides are P'M and P''M,
Hence
L = y/p'M + P^' ;...(!)
but P'm'= NC = NB + Bc'= (OD - 0A)2 + (DC - AN)^ =
(x" - x'Y + (y" - y')\ and Pm'= (P"C - P'N)^ =
(." - zj.
•.L= V(a
r + i:y" - y'f + {^" - ^7 • • • (2)
CO-ORDINATES. 221
Cor. If x' = 0, 1/ = 0, z' = 0, then the point P' coincides
with the origin and
.•.L = Vx''2 + 2/''^' + ,^''^ ... (3)
expresses the distance of a point from the origin.
169. Given the length and the directional angles of a line
joining any i^oint tvith the origin to find the co-ordinates of the
point.
The Directional angles of a line are the angles tvhich the line
makes with the co-ordinate axes.
Let P (x, y, z), Pig. 75, be any point, then OP ^ L will be
its distance from the origin. Let POX, POY, POZ = «, ^, y,
respectively.
Since OM, OlST, OR (= x, y, z) are the projections of OP
on X, Y, Z, respectively, we have
£c = L cos « I
y = L cos y8 I ... (1)
s = L cos / J
for the required co-ordinates.
Cor. Squaring and adding equations (1), we have
^^ + Z/^ + «" = L^ (cos^ « + cos^ /3 + cos^ y) ;
but ic^ + 2/2 _j_ ^2 _ L2 ^j.^_ jg3 (^3^) .
hence cos^ « -\- cos^ /3 + cos^ y = 1 . . . (2)
That is, the sum of the squares of the directional cosines of a
space line is equal to unity.
ScHOL. The directional angles of any line, as PT", Pig. 76,
are the same as those which the line makes with three lines
drawn through P' || to X, Y, Z. The projections of PT'' on
three such lines are x" — x', if — ij , z" — z', Art. 168 ; hence
x" — x' = L cos « 1
/' - / = L cos ;8 j. ... (3)
z" — z' = Jj COS y
222 SOLID ANALYTIC GEOMETRY.
EXAMPLES.
Required the length of the lines joining the following
points :
1. (1, 2, 3), (- 2, 1, 1), _ 4. (0, 0, 0), (2, 0, 1).
Ans. V14. Ans, ^5.
2. (3, - 2, 0), (2, 3, 1). _ 5. (0, 4, 1), (- 2, - 1, - 2).
Ans. V27. Ans. V38.
3. (0, 3, 0), (3, - 1, 0). 6. (1, - 2, 3), (3, 4, 6).
Ans. 5. Ans. 7.
7. Find the distance of the point (2, 4, 3) from the origin ;
also the directional cosines of the line.
8. A line makes equal angles with the co-ordinate axes.
What are its directional cosines ?
9. Two of the directional cosines of a line are Vf and ^
What is the value of the other ?
10. If (x'', y', z') and (x", y", z") are the co-ordinates of the
extremities of a line show that
(x' + x" y' + y" z' + ^-\
\ 2 ' 2 ' 2 )
are the co-ordinates of its middle point.
THE POLAR SYSTEM.
170. The position of a space point is completely determined
when we know its distance and direction from some fixed point.
For a complete expression of the direction of the point it is
necessary that two angles should be given. The angles
usually taken are
1st, The angle which the line joining the point and the
fixed point makes with a plane passing through the fixed
point ; and 2d, The angle which the projection of the line join-
ing the points on that plane makes with a fixed line in the
plane.
CO-ORDINA TES.
223
Fig. 77.
Let O be the fixed point and P the point whose position we
wish to determine. Join and P, and let XOY be any plane
passing through 0. Leb OX be a given line of the plane
XOY. Draw PB 1 to XOY and pass the plane PBO through
PB and OP. The intersection OB of this plane with XOY
will be the projection of OP on XOY. The angles POB (6),
BOX (&'>y^'n.dic-
idar have the same equation.
YJl. If X cos a -\- y cos jB -\- z cos y ^ p be the normal equa-
tion of a plane, then x cos u -\- y cos ft -\- z cos y = p -^ d is the
equation of a parallel plane at the distance d from it.
For the directional cosines of the perpendiculars are the
same ; hence, the perpendiculars are coincident ; hence, the
planes are parallel. The distance of the planes apart is equal
to the difference of the perpendiculars drawn to them from
the origin; but this difference is^ -|- d—p\ i.e., -i- d. Hence,
the proposition.
Cor. If (x' , y', z') be a point in the plane whose distance
from the origin \q p) J^d; then
:^d ^x' cos « + y' cos ^ + z' cos y — p ■ ■ ■ (1)
232 SOLID ANALYTIC GEOMETRY.
is its distance from the parallel plane whose distance from
the origin is p. From equations {ci), Art. 174, we have
cos « = i- ^ cos 13 = ^ , cos 5' = =L ;
a b c
2 2 2
hence cos^ « + cos'-^ ft -\~ cos^ y = -2L -)- E — [_ P— — i.
a^ IP- c^
These values in (1) give
for the expression of the distance of a point from a plane
which is given in its symmetrical form.
Let the student show that the expression for d becomes
^^^AaM;E£^C£^-D _
VA--^ + B^ + C^
when the equation of the plane is given in its general form.
What is the significance of the double sign in (1), (2), and
(3)?
178. To find the equation of a i^lane which jpasses through
three given lyoints.
Let {x', tj', z'), {x", y", z"), {x"' , y"\ z"') be the given points.
Since the equation we seek is that of a plane, it must be
Aaj + B^/ + C,^ = D . . . (1)
in which A, B, C, D are to be determined by the conditions
imposed.
Since the plane is to contain the three given points, the co-
ordinates of each of these must satisfy its equation ; hence,
the following equations of condition :
Kx' + Bt/' + Cs' = D
Kx" + By'' + C;s'' = D
Aa;'" + B?/'" + C.~"' = D.
These three equations contain the four unknown quantities
A, B, C, D. If we find from the equations the values of A,
THE PLANE. 233
B, C in terms of D and the known quantities, and substitute
these vahies in (1), each term of the resulting equation will
contain D as a factor. Let
A = A'D, B = B'D, C = CD be the values found.
Substituting in (1), we have
MT>x + B'Dy + C'D« = D.
... A'cc+BV + C'a = l • • • (2)
is the required equation.
179. The preceding discussion has elicited the fact that
every equation of the first degree between three variables
represents a plane surface. It remains to be shown that every
equation between three variables represents a surface of some
kind.
Let ^=f{x, y) ■ ■ ' (1)
be any equation between the three variables (x, y, z). Since
X and y are independent, we may give them an infinite number
of values. For every pair of values thus assumed there is a
point on the XY plane. These values in (1) give the corre-
sponding value or values of z, which, laid off on the perpen-
dicular erected at the point in the XY plane, will locate one
or more points on the locus of the equation. But the number
of values of z for any assumed pair of values of x and y are
necessarily finite, while the number of pairs of values which
may be given x and y are infinite ; hence (1) must represent
a surface of some kind.
If
^=f{^, y)\ , , , (2)
z = (p{x,y)\
be the equations of two surfaces, then they will represent their
line of intersection if taken siviultaneously. For these equa-
tions can only be satisfied at the same time by the co-ordinates
of points common to both. Hence, in general, two equations
between three variables determine the position of a line in sp>ac6.
234 SOLID ANALYTIC GEOMETRY.
If z ^f{x, y) 1
z = ^{x,ij)\ . . . (3)
z-=xp{x, y) j
be the equations of three surfaces, then they will represent
their point or points of intersection when considered as simul-
taneous. Hence, in general, three equations between three
variables determine the, positions of space points.
EXAMPLES.
Find the traces and intercepts of the following planes:
1. X -2y + z =Q. 6. - - ^ + - = 1.
■^ 2 3 4
2. |.-3/+|=l. 7. |_|-| = 1.
3. .-, + 4.,= l. 8. 2_^-^f + ^ = l.
4. 2cc + 32/-4,^=0. 9. ^ + |_^ = 1.
5 ^ — ^ JL. y — ^ = 2 10?^ — ^ = ^
2^3 ■ - y ^ 4:'
11. The directional cosines of a perpendicular let fall from
2 12
the origin on a plane are - , - , -; required the equation of the
o o o
plane, the length of the perpendicular = 4.
Ans. ^ + i^ + A = 1.
6 12 6
Required the equations of the plane whose intercepts are
as follows :
12. 1, 2, 3. 14. ^, |, -2.
13. 2, - 1, 3. 15. _ 1, _ I , _ 4.
o
16. What is the equation of the plane, the equations of
whose traces are x — 3 ^z = 4 and cc -|" ^ = 4 ?
Ans. x — 3?/4-« = 4.
THE PLANE. 235
17. The co-ordinates of the projection of a point in the
plane cc — 3?/ + 2s; = 2on the XY plane are (2, 1) ; required
the distance of the point from the XY plane.
Ans. §.
Write tlie equations of the planes which contain the follow-
ing points :
18. (1, 2, 3), (- 1, 2, - 1), (3, 2, 0).
19. (4, 1, 0), (2, 0, 0), (0, 1, 2).
20. (0, 2, 0), (3, 2, 1), (- 1, 0, 2).
21. (2, 2, 2), (3, 3, 3), (- 1, - 1, - 1).
Find the point of intersection of the planes
22. 2x-\-y — z = ^. 23. 2x —ij + z = 10.
2x — 3z -\-ij = 10. x-{-y-2z = 3.
X -\- y — «=2. 2 X — 4?/-l-5s = 6.
24. 2x — y-z = 2.
2x — 3y-{-z = 10.
2x—y-{-2z = 8.
Find the distance of the point (2, 1, 3), from each of the
planes
25. x cos 60° + y cos 60° + z cos 45° = 9.
26. x-\-3y-z = 8.
27. x+^ -\-3z = 4.. 28. - - ^ + - = 1.
2 3 2 5
29. Find the equation of the plane which contains the
point (3, 2, 2) and is parallel to the plane x — 2 y -\- z = 6.
Reduce the following equations to their normal and sym-
metrical forms :
30. 2x-3y -\-z = 4. 31. 4.x -^ 2y — z = ~ .
32. ~x-\-y — ~z = ^.
3 "^ 4
33. If s, /, s" represent the sides of the triangle formed by
the traces of a plane, and a, h, c represent the intercepts,
show that s2 4. s'^ 4- s'"" ^2(a^ -^b^ + c").
236
SOLID ANALYTIC GEOMETRY.
CHAPTER III.
THE STRAIGHT LINE.
180. To deduce the equations of the straight line.
The straight line in space is determined when two planes
which intersect in that line are given. (See Art. 179.) The
equations of any two planes, therefore, may be considered as
representing a space line when taken simultaneously. Of the
infinite number of pairs of planes which intersect in and de-
termine a space line, two of its projecting planes — that is,
two planes which pass through the line and are perpendicular
to two of the co-ordinate planes — give the simplest equations.
For this reason two of these planes are usually selected.
Fig. so.
THE STRAIGHT LINE. 237
Let PBM be the plane which projects a space line on XZ,
then its equation will be of the form
X ^= sz -{- a (see Art. 176, Cor.)
in which s = tan ZBP and a = OA.
Let P'B'M' be the plane which projects the line on YZ,
then its equation will be
y=tz+b,
in which t = tan ZBT' and b = OA'.
But the tAvo planes determine the line ; hence
X = sz -\- a )^ ,-.s
y = tz + b\ ' ' ' ^^
are the required equations.
Cob. 1. If a = and 5 = 0, then
X = sz^ _ _ _ ^2')
y = tz)
are the equations of a line which pass through the origin.
Cob. 2. If s = and ^^ = 0, we have
^ = ''1 ... (3)
y = bS
for the equation of a line || to the Z-axis.
CoR. 3. Since equations (1) express the relation existing
between the co-ordinates of every point on the space line, if
we eliminate z from these equations we obtain the immediate
relation existing between x and y for points of the line. But
this relation is evidently the same for all points in the pro-
jecting plane of the line which is L to XY and therefore for
its trace on XY. But the trace is the projection of the line
on XY ; hence, eliminating, we have
sy — tx = bs — at . . . (4)
for the equation of the projection of the line on XY.
181. We have found, Art. 169, SchoL, for the length of a
line joining two points the expression
L ^ x" -x' ^ if - if ^ z" - z'
cos « cos ^ cos y
238 SOLID ANALYTIC GEOMETRY.
Eliminating L and letting x", y", z" {= x, y, z) be the co-
ordinates of any point on the line, we have
X — X y — y z — z
... (1)
cos « COS ^ COS 7
for the Symmetrical Equation of a straight line.
182. To find ivhere a line given by the equations of its
projections pierces the co-ordinate planes.
CT — ^^ I ft f
Let ^"^ T 7 (he the equations of the line.
y = tz -\-b ^ -L
1. To find where the li7ie jnerces the XY-plane.
The equation of the XY-plane is
z = Q.
Since the jDoint of intersection is common to both the line
and the plane, its co-ordinates must satisfy their equations.
Hence
X ^= sz -\- a
y = tz -\-b
z =
are simultaneous equations. So treating them we find
{a, b, 0)
to be the required point.
2. To find tvhere the line p)ieTces the XZ-plane.
The equation of the XZ-plane is
Combining this with the equations of the line, we have
at — ■^^ A ^
t t
for the required point.
3. To find xohere the line pierces the YZ-plane.
X ^ sz -\- a')
y = tz -\- b > are simultaneous ;
x = )
hence [0, ^^ ~ ^^ ^
s s
is the required point.
THE STRAIGHT LINE. 239
183. To find the equations of a line passing through a given
point.
Let (x\ y', z') be the given point.
Since the line is straight its equations are
X = sz + a I .-|^^
y = tz + b^
in which the constants are unknown.
Since it is to pass tlirougii the point (x', y',z') its equations
must be satisfied for the co-ordinates of this point ; hence
the equations of condition :
X = sz -\- a I /9^
y' = tz' +h \ ' ' ' ^'^^
As the three conditions imposed by these four equations
cannot, in general, be fulfilled by a straight line, we must
eliminate one of them. Subtracting the first equation in
group (2) from the first in group (1) and the second in group
(2) from the second in group (1), we have
iC CC = S (Z ^ / ) f'X\
y-y' = t{z-z')S ' - ' ^^
for the general equations of a straight line p)Cissing through a
point.
184. To find the equations of a line passing through two
given p)oi7its.
Let (x', y', z'), {x'\ y", z") be the given points.
As the line is straight its equations are
y = tz-{-b\' ' ' ' ^ ^
in which the constants are to be determined.
As it is to pass through (x', y' , z'), we must have
x' = sz' -\- a
(2)
y' = tz' -^b ' ^ ^
240
SOLID ANALYTIC GEOMETRY.
As it is to pass through (x", y", z"), we must have also
x" = sz" + a ~> /ON
y'^ = tz" -\-l\
As these six equations impose four conditions on the line,
we must eliminate two of them. The conditions of the
proposition, however, require the line to pass through the
two points ; hence we must eliminate the other two.
Elimiting a and h from groups (1) and (2), by subtraction,
we have
X — x' = s {z — z'")^
y-y'^t{z-z')\ ■ • • ^^)
Now, eliminating a and b from (2) and (3), we have
x' — x" = S {z' — «") ) /f-N.
^J -y" = t {z' - z") i
Eliminating s and t between (4) and (5), we have
x = (z — z)
y-y'-^-f^,i?-^')
for the required equations.
(6)
EXAMPLES.
1. Given the line ^, ^^ 4 ^ _ 3 [• required the equation of
the projection on XY.
Ans. 2 X — y = 5.
2. How are the following lines situated with reference to
the axes ?
x=2} y=Ol 2/ = 01
=S}
Find the co-ordinates of the points in which the following
lines pierce the co-ordinate planes :
X = 3z —
y = 2z +
1}
A X = — Z —1\
*• 2/ = 2^ + 3 )
THE STRAIGHT LINE. 241
6. Given (2, 1, - 2), (3, 0, 2) ; required
(a) The length of the line joining the points.
(h) The equation of the line.
(c) The points in which the line pierces the co-ordinate
planes.
Find the equations of the lines which pass through the
points :
7. (2, 1, 3), (3, - 1, - 1). 9. (2, - 1, 0), (3, 0, 0).
8. ( - 1, 2, 3), ( - 1, 0, 2). 10. (1, - 1, - 2), (- 1, -2, - 3).
11. The projections of a line on XZ and YZ make angles
of 45° and 30° respectively with the Z-axis, and the line in
space contains the point (1, 2, 3) ; required the equations of
the line.
X = z — 2.
12. The vertices of a triangle are (2, 1, 3), (3, 0, — 1),
(—2, 4, 3) ; required the equations of its sides.
13. Is the point (2, — 1, 3) on the line which passes through
(- 1, 3, 2), (3, 2, - 2) ?
14. Write the equations of a line which lies in the plane
x-2y + 3z = l.
Note. — Assume two points in the plane; the line joining
them will be a line of the plane.
15. Find the equation of a line through (1, — 2, 2) which
is parallel to the plane x — y -\- z = 4:.
T J- 2 s; = 3 ^
16. Find the point in which the line _ ^ t o _ a r
pierces the plane Sx-{-2y — z = 4:.
17. Required the equation of the plane which contains the
^ T x-2z-l=0} . x-z-5 = I
t^^°l^^^^ 2/-2^-2 = 0| ^^'^ 2/-4^ + 6 = j
242 SOLID ANALYTIC GEOMETRY.
18. Find the point of intersection of the planes
19. Find the equations of the projecting planes of the line
2x ^3y — z= Q\-
20. Which angles do the following planes cross ?
X— y-\-z = 4:, 2x-\-y — 3z = 2, X— 2y — z = l.
185. To find the intersection of two lines given hy their
ec[uations.
_ . X ^ sz -\- a^ :, X = s'z -\- a'
Let ^ ; 7 ^ and ., ,,
y =^tz -\-b '^ y = tz -{-
be the given equations. Since the point of intersection is com-
mon to both lines, its co-ordinates must satisfy their equations.
Hence these equations are simultaneous. But we observe that
there are four equations and only three unknown quantities ;
hence, in order that these equations may consist (and the lines
intersect), a certain relationship must exist between the con-
stants which enter into them. To find this relationship, we
eliminate x between the fii^st and third, y between the second
and fourth, and z between the two equations which result.
We thus obtain
(s - s') (b - b') - (t- t') (a — a') =
for the required equation of condition that the two lines shall
intersect. If this condition is satisfied for any pair of as-
sumed lines the lines will intersect, and we obtain the
co-ordinates of this point by treating any three of the four
equations which represent them as simultaneous. So treating
the first, second, and third we obtain
s«' — s'a .a' — a , -, a' — a
— — :^ ' ^ z — J + ^' - — V
for the co-ordinates of the required point.
ISToTE. — We were prepared to expect that our analysis
would lead to some conditional equation, for in assuming the
equations of two space lines it would be an exceptional case
THE STRAIGHT LINE.
243
if we so assumed them that the lines which they represent
intersected. Lines may cross each other under any angle in
space without intersecting. In a plane, however, all lines
except parallel lines must intersect. Hence, no conditional
equation arose in their discussion,
186. To find the angle between two lines, given by their
equations, in terms of functions of the angles which the lines
make with the axes.
Let
sz -\- a
tz^b
, a; = s'z 4- »'
and ,, ,,
y = tz^b
''}
be the equations of the two lines. The angle under which
two space lines cross each other is measured by the angle
formed by two lines drawn through some point parallel to
their directions.
Let OB and OC be two lines drawn through the origin
parallel to the given lines. Then
^ = f| and^=^.
y = tz\ y = tz
will be their equations. The angle between these lines is the
angle sought. Let go (= BOC) be this angle.
244 SOLID ANALYTIC GEOMETRY.
Let a', p', y' represent the angles which the line BO makes
with X, Y, Z, respectively; and u", (3'', y" the angle which
CO makes with the same axes. Take any point P' (x', y' , z')
on OB and any point V (x", y", z") on CO and join them by
a right line forming the triangle P'OP".
Let OP' = V, OV"^ V, and PT" = L.
Prom the triangle P'OP'', we have
L'2 + L''2_L2 ,,.
cos cp = y^TjTT • ■ • (1)
But Art. 168, equation (3) and (2)
L'2 = x''- + y'- + z'%
L"2 = x"^^ y"' + z"%
U = (x" - x'f + {y" - yj + iz!' - z'f
= x"^ + 2/"2 ^ ^//2 ^ ^'2 ^ ^/2 + ~'2 _ 2 (a;V + yY + «'0-
Substituting these values in (1), we have
x'x" + yY + «'^'' /o\
cos q> = ^£,£,,^ • • • (2)
But Art. 169, (1)
x' = L' cos a', ?/' = L' cos f3', z' = L' cos y'
x" = V cos a", y" = L" cos (3'', z" = V cos y"
Substituting in (2), we have
cos qo = cos a' cos a" + COS /3' COS yS" + cos "/ cos /" . . . (3)
for the required relation.
Cor. If (p = 90°
cos a' COS a" -\- COS /5' COS ^" + COS y' COS y" = . . . (4)
187. To find the angle which two space lines Tnake with each
other in terms of functions of the angles which the projections
of the lines 'make with the co-ordinate axes.
T ^i_ X ^^ SZ f -I X — s z
Let , y and .,
y = tz\, y = tz
be, as in the preceding article, the equations of the lines
THE STRAIGHT LINE. 245
drawn through the origin parallel to the given lines. Since
P' (x', y', z'), Fig. 81, is a point on the first line, we have
x' = sz'
'2 _l_ V2
and, Art. 168, L"-^ = x"^ + tj'^ + s'
Eliminating, we find
sh' , tV , \!
y
•^ - Vi + s--^ + ^^ VI +5' + ^- Vi + s^ + ^^'
and since P" {%" , y", z") is a point on the second line, we
have
x" = s'z"
y" = t'z",
and, Art. 168, L"^ = x"'- + y"^ + s"^.
Hence,
y
Vl + s'2 + ^'2 V 1 + s'-' + t"' Vl + s'" + t'^
But, Art. 169,
f 0(1 S f/ 00 s
cos « = =: ^^=^z=^ , cos (X = — - = z=z3^^=-
L' Vl+6'-' + ^' ^ Vl + s'2 + ^''
o, y t or, 4: ac.
Case 1. ^ > go. We find this supposition in (2) gives a >
and c > ; hence, ^^ < 4 ac, i.e., the intersection is an
ellipse.
If ^ > gp and c = 0, the equation resulting from introducing
this supposition in (1) can only be satisfied by the point
(0, 0) ; hence it is the equation of two imaginary lines inter-
secting at the origin.
THE STRAIGHT LINE. 253
If (jp = 0, equation (1) becomes
y'^ tan ^ -\- x' tan ^ =^ c^,
that is, the intersection is a circle.
Case 2. 6 = (p. This supposition in (2) gives a > and
G = .-. b" = 4ac. Hence the intersection is a parabola.
If ^ = qo and c = 0. From (1), we have
7/'^ tan '^6 = ; i.e., y =
which is the equation of the X-axis — a straight line.
If ^ = go = 90° andc = gc, then the cone becomes a cylinder,
and the cutting plane is perpendicular to its base. The inter-
section is therefore two parallel lines.
Case 3. 6 0, c <
.*. 5^ > 4 ac. Hence the intersection is an hyperbola.
It . 4 ac, is the equation of an hyperbola for all
values of n.
(c) Plane \\ to XZ-plane. Let y = 2^^Q the equation of such
a plane. Combining with (2), Art. 192, we have after reduc-
tion
xHan'-e -z^ + 2 cz +^/ tan ^ ^ - c^ = . . . (5)
254
SOLID ANALYTIC GEOMETRY.
which, since 6^ > 4 ac, is the equation of an hyperbola for all
values of p.
Hence, in all possible positions of the cutting plane, the
intersection is an ellijyse, a parabola, an hyperbola, or one of
their liviiti7ig cases.
JSToTE. — Equations (3), (4), (5) of case 4 are the equa-
tions of the projections of the curves of intersection on the
planes to which they are parallel. But the projection of any
plane curve on a parallel plane is a curve equal to the given
curve ; hence the , conclusions of case 4 are true for the
curves themselves.
194. We have defined the conies, Art. 191, as the curves
cut from the surface of a right circular cone by a plane, and
assuming this definition we have found and discussed their
general equation. Art. 193.
A conic, hoivever, tnay be otherwise defined as the locus
generated by a point so moving in a plane that the ratio of its
distance from a fixed point and a fixed line is always constant.
195- To deduce the general equation of a conic.
Y
Fig.
Let us assume the definition of Art. 194 as the basis of the
operation. Let F be the fixed point and OY the fixed line.
Let P be the generating point in any position of its path.
THE STRAIGHT LINE. 255
Draw FO J_ to OY, and take OY and OX as co-ordinate axes.
Draw PL i| to OY, PD J_ to OY, and join P and F. Let OF
FP
By definition = e = a constant.
From triangle FPL, FP^ = FL^ + PL^ ; . . . (1)
but FL2 = (OL - OYf = (x - j^f, LP^ = y'- ■ and FP^ =
e^DP^ = eV.
These values in (1) give
e- x^ = (x — iSf + xf'
or, after reduction,
7/2 + (1 - (?) x" - Ipx +^2 ^ ... (2)
for the required equation.
CoR. Comparing (2) with (1), Art. 138, we find
a = 1, 5 = 0, and c = (1 — e^),
hence V - ^ac = - ^ (Y - (?) = ^ {e" - V) . . . (3)
Case 1. The fixed point not on the fixed line ; i.e., p not
zero.
If e < 1, &2 < 4 ac ; hence equation (2) is the equation of
an ellipse.
If e = 1, 6^ = 4 ac ; hence equation (2) is the equation of a
parabola.
If e > 1, 5^ > 4 ac; hence equation (2) is the equation of
an hyperbola.
Case 2. The fixed point is on the fixed line, i.e., ^ = 0.
In this case (2) becomes
y'i j^ (I - e}) x"" = . . . (4)
If 6 < 1, equation (4) represents two imaginary lines inter-
secting at origin.
If 6 = 1, equation (4) represents one straight line (the
X-axis).
If e > 1, equation (4) represents two straight lines inter-
secting at the origin.
Hence, equation (2) represents the conies or one of their lim-
iting cases.
= 2) X -2z-\- 3 = 0}
-If' 2j-z = 2 \
256 SOLID ANALYTIC GEOMETRY.
GENERAL EXAMPLES.
1. Find the point of intersection of the lines
x = 2z -\-ll x=z -{-2
y = 3z+2S^ y = 4.z + l
and the cosine of the angle between them.
5
Ans. (3, 5, 1) ; cos go = — 1|
2. Eequired the equation of the line which passes through
(1, — 2, 3) and is parallel to
- = 2- + n. Ans. - = 2.-5 1
y = 2-z^ y = -« + l|
3. What is the angle between the lines
X -\-z = 2
y
Ans. (p = 90°.
4. What is the distance of the point ( — 3, 2, — 1) from
the line
ic + 3« + 1 == 0| ^
2/ = 4. + 3 r
5. A line makes equal angles with the co-ordinate axes ;
required the angles which it makes with the co-ordinate
planes.
6. The equation of a surface is x^ -\- y"^ -\- z^ — 2x — Ay —
6z =2; what does the equation become when the surface is
referred to a parallel system of axes, the origin being at
(1, 2, 3) ? Ans. x^-\-y^ + z^ = 16.
7. Given the line _ ^ ^"T [■ , required the projection of
the line on XY and the point in which the line pierces the
co-ordinate planes. Ans. in part, 2 3/ + cc = 4.
8. Eequired the distance cut off on the Z and Y axes by
the projections of the line '^ i o ^^ o [- on YZ.
z= -6
Ans. 3
2/ = o •
THE STRAIGHT LINE. 257
9. How are the following pair of lines related ?
x^2z^2 \ x = 2z-l \
2/__^_4| y=-z-\-2\
10. What are the equations of the line which passes through
the origin and the point of intersection of the lines
x=2z-^l\ x = z + 2 \^ . x = -Sz\
11. What is the distance of the point (3, 2, — 4) from the
origin ? What angle does this line make with its projection
on XY ?
12. A straight line makes an angle of 60° with the X-axis
and an angle of 45° with the Y-axis ; what angle does it make
with the Z-axis ? Ans. 60°.
13. What are the cosines of the angles which the line
X = 3 z — 1
, o r makes with the co-ordinate axes ?
y= — z-^2
14. A line passes through the point (1, 2, 3) and makes
V2 1 1
angles with X, Y, Z whose cosines are — — > o ' 2 ' respect-
ively ; required
(a) the equation of the line,
(b) the equation of the plane J_ to the line at the point,
(c) to show that the projections of the line are J_ to the
traces of the plane.
2 12
15. The directional cosines of two lines are - , -- , ~ and
o o o
^^, -, -. What is the cosine of the angle which they
2 2 2
make with each other ? _
3 -f 2 V2
Ans. Cos 'x'y' + ^x'z' + Yy'z' -f GV + Wy'
+ IV + K = . . . (2).
260 SOLID ANALYTIC GEOMETRY.
Since the original axes were supposed rectangular the nine
angles a', /3; y' etc., are connected by the three relations
cos^ «' + cos^ /3' + cos^ y' = 1.
cos2 a" 4- cos^ ^" + cos^ y" = 1.
COS^ a'" + C0S2 /3'" + COS^ f" = 1.
If we take the new axes also rectangular, which is desir-
able, the nine angles will be connected by the three additional
relations
cos a' COS a"' -{- cos ^' COS /3" + COS y' COS y" = 0.
cos «' COS a'" -|- COS (3' COS (3'" + COS y' cos y"' = 0.
COS «" cos «'" + cos (3" cos ^'" + cos y" cos r'" = 0.
This will leave three of the nine angles to be assumed arbi-
trarily. Let us give to them such values as to render the co-
efficients D', E', and F' each equal to zero in equation (2).
The general equation will thus be reduced to the form
A'x'^ + By^ + G'^'' + G'x' + Hy -f IV + K = 0,
or, omitting accents,
Ax^ -\- By' -\- Gz^ + Gx -\- mj -\- Iz + K = . . . (3)
In order to make a further reduction in the form of the
equation let us endeavor to move the origin without changing
the direction of the axes. The formulae of transformation
will be (Art. 189)
X = a -\- x', 1/ ^ b -{- y', z = c -\- z' .
Equation (3) will become
A (a + x'Y + B (^< + y'Y + C (c + z!y -f G (a + a?') +
H (Z» + y') + I (c + ^0 + K = 0.
Developing, omitting accents, and placing Ao? + BIP- -\- Cc^
+ Ga + HS + Ic + K = L, the equation takes the form
Aa;2 -j- B^/^ + Gz" + (2 Aa + G) cc + (2 Bh +H) y + (2 Cc +1 )z
+ L = 0.
In order now to give definite values to the quantities a, 5,
and c, which were entirely arbitrary, let us assume
G (,= _ H , I „.
2A ' 2B ' 20
SURFACES OF THE SECOND ORDER. 261
2 Aa + G = 0, 2 B6 + H = 0, 2 Cc + I = . . . (4)
If these values of a, b, and c be finite, the general equation
reduces to the form
A^2 + By^ + C^^ + L = . . . [A],
a form which will be set aside for further examination.
It may be remarked that equations (4) are of the first
degree, and will give only one value to each of the quantities
a, b, and c, and there is therefore only one position for the
new origin.
If, however, either A, B, or C be zero, then a, b, or c will
become infinite, and the origin will be removed to an infinite
distance. This must be avoided.
Let us suppose A = 0, while B and C are finite. We may
then assume 2 B6 + H = 0, and 2 Cc + I = 0, but we cannot
assume 2 A.a -|- G = 0.
Having assumed the values of b and c as indicated, let us
assume the entire constant term equal to zero. This will give
B^»2 4_ Cc2 + Ga + H6 + Ic + K = 0,
B&2 + Cc2 + H5 + Ic + K
or a = ! ! ! 1
G
and the general equation will be reduced to the form
By'' + Cz^ + Gx = . . . (B),
a second form set aside for examination.
We must observe that this last proposed transformation
will also fail when G = 0, that is, when the first power of x,
as well as the second power of x, is wanting in the general
equation.
And without making the second transformation we have a
third form for examination, viz. :
By2 + c^^ + Hy + I^ + K = . . . (C)
Lastly, two of the terms involving the second powers of
the variables may be wanting, and the equation (1) then
becomes
C^2 + Gx + Hy + I;s + K = . . . (D)
262 SOLID ANALYTIC GEOMETRY.
It is apparent, therefore, that every equation of the second
degree involving three variables can be reduced to one or
another of the four forms
Kx" + B?/2 + C«- + L = . . . (A)
B?/2 + C,^^ + G^ = . . . (B)
B2/2 + C«2 + Hy + I« + K = . . . (C)
C«2 + Gx + Hy + I^ + K = . . . (D)
We will examine each of these forms in order, beginning
with the first form :
Aa;2 + B7/2 + C^2 _^ L = . . . (A)
This equation admits of several varieties of form according
to the signs of the coefficients.
1. A, B, and C positive, and L negative in the first member.
2. A, B, C, and L positive.
3. Two of the coefficients as A and B positive, C and L
negative.
4. Two of the coefficients as A and B positive, C negative,
and L positive.
No other cases will occur.
Case 1. Kx" + By'- + Cs^ = L,
in which form all of the coefficients are positive.
In order to determine the nature of the surface represented
by this equation, let it be intersected by systems of planes
parallel respectively to the co-ordinate planes. The equations
of these intersecting planes will he x = a, y ^= b, z = c. Com-
bining the equations of these planes with that of the surface,
we find the equations of the projections on the co-ordinate
planes of the curves of intersection.
When X = a,
■By^ + C«2 = L - Aa^
an ellipse.
" y = h,
Ax2 + C«2 = L - BP
an ellipse.
" z = c,
Ax^ + By' =L- Cc2
an ellipse.
Thus we see that the sections parallel to each of the co-
ordinate planes are ellipses.
SURFACES OF THE SECOND ORDER.
263
The section made by the plane x = a is real when
L — A(x- > or a < -j- y— r-j and imaginary in the contrary
case.
The section made by the plane y = h is real when
ft < J^ t/ — , and imaginary when ^ > J^ V/^~ •
The section made by the plane 2: = c is real when
c < -t y -j;- , and imaginary when c > i W — - .
Thus we see that the surface is enclosed within a rectangu-
lar parallelepiped whose dimensions are
When a
B
and
ovb = ^
\ — or c =
V B
±
the
A'' ^VB -^VC
sections become points.
When a = 0,h = 0, and c = 0, we find the sections made,
by the co-ordinate planes to be
B^/' + Cz^ = L.
Kx^ + C«2 = L.
Xx- + B?/- = L.
E
264 SOLID ANALYTIC GEOMETRY.
These are called the principal sections of the surface. The
principal sections are larger ellipses than the sections parallel
to them, as is indicated by the magnitude of the absolute
term.
The surface is called the Ellipsoid.
It may be generated by the motion of an ellipse of variable
dimensions whose centre remains on a fixed line, and whose
plane remains always perpendicular to that line, and whose
semi-axes are the ordinates of two ellipses Avhich have the
same transverse axis, but unequal conjugate axes placed at
right angles to each other. The axes of the principal sections
are called the axes of the ellipsoid.
If we represent the semi-axes of the ellipsoid by a, J, and
c, we shall have
and the equation of the surface
kx' + B?/- + Cs- = L becomes
r^ 7/2 z^
-^ + i^ + -^ = 1, or
a2 ^ ^2 ^ ^2
Tr(?x^ -f- o^c^'if' -\- a-lrz' = a}lrc^.
These are the forms in which the equation of the ellipsoid
is usually given.
If we suppose B = A, then b = a, and the equation becomes
^•' + 2/" I «' _ 1
cr c-
and the surface is the ElUi^soid of Bevohdion about the axis
of Z.
If A = B = C, then a = h = c, and the equation becomes
^2 _|_ yi j^ ^ji ^ (.^2^ and the surface is a sphere.
If L = 0, the axes 2 sj^, 2 y/-^, 2 sj ^
reduce to zero, and the ellipsoid becomes a point.
SURFACES OF THE SECOND ORDER. 265
Case 2. If L be negative in the second member, the equa-
tion A.x^ + B?/^ + Cs;^ = — L will represent an imaginary
surface, and there will be no geometrical locus.
Hence the varieties of the elUjosoid are
(1) The ellipsoid proper with three unequal axes.
(2) The ellipsoid of revolution with two equal axes.
(3) The sphere.
(4) The point.
(5) The imaginary surface.
Case 3. In this case the equation takes the form
Ax^ + B2/2 _ C^2 ^ L,
in which A, B, C, and L are essentially positive.
Cutting the surface by planes as before, the sections will be,
when cc = a, By^ — Cz^ = L — Ao-^, a hyperbola, having its
transverse axis parallel to the Y-axis when «,<_[_ t / — , but
parallel to the Z-axis when a > -[- i / And when a = J-
— - , the intersection becomes two straight lines whose pro-
jections on the plane of YZ pass through the origin.
When y = b, Kx"^ — Gz^ = L — B5^, a hyperbola, with simi-
lar conditions as above.
When z = c, Kx"^ + Bi/ = L + Cc^, an ellipse real for all
values of c.
Since the elliptical sections are all real, the surface is con-
tinuous, or it consists of a single sheet.
The principal sections are found by making successively
a = 0, which gives B?/^ — Cz^ = L, a hyperbola.
5 = 0, " " Ax'' - Cz^ = L, «
c = 0, " " Aa;2 -f By^ = L, an ellipse.
The surface is called the elliptical hyperholoid of one sheet.
The equation may be reduced to the form
x^ . y'^ z'^ _ -,
a?- y c^
/
266
SOLID ANALYTIC GEOMETRY.
This surface may be generated by an ellipse of variable
dimensions whose centre remains constantly on the Z-axis,
and whose plane is perpendicular to that axis, and whose
semi-axes are the ordinates of two hyperbolas having the
same conjugate axis coinciding with the Z-axis, but having
different transverse axes placed at right angles to each other.
Fig. B.
If we suppose A = B, then will a = h, and the equation of
the surface becomes
9 1 ^
- — = 1,
the hyperboloid of revolution of one sheet.
If A = B = C, we have x"" -^ y"^ — z" = a^, the equilateral
hyperboloid of revolution of one sheet.
If L = 0, the equation represents a right cone having an
elliptical base ; and if A = B this base becomes a circle.
SURFACES OF THE SECOND ORDER. 267
Hence we have the following varieties of the hyxjerholoid
of one sheet.
1. The hyperboloid proper, with three unequal axes.
2. The hyperboloid of revolution.
3. The equilateral hyperboloid of revolution.
4. The cone.
Case 4. A.x'' + Bif - Gz'' = - L, where A, B, C, and L
are essentially positive.
Intersecting the surface as before we have, when x = a,
B,r/2 _ Cs;2 =^ — L — Ka^, a hyperbola having its transverse
axis parallel to the axis of Z.
When y = h, A-x" — C«' = - L -B^^^ A hyperbola hav-
ing its transverse axis parallel to the axis of Z.
When ^_= c, Ax^ + By^ = — L + Cc^, an ellipse real when
c > -i; i/ — ., and imaginary when c < -^ y ~7r' Since the
sections between the limits « = J- y — are imaginary, but
real beyond those limits, it follows that there are two distinct
sheets entirely separated from each other.
The surface is called the hyperboloid of two sheets.
The principal sections are found by making successively
a = 0, which gives By^ — Cs;^ = — L, a hyperbola with its
transverse axis coinciding with the Z-axis.
b =0, which gives Ax^ — Cz^ = —Jj, a, hyperbola with its
transverse axis coinciding with the Z-axis.
c = 0, which gives Ax^ -f- By^ = — L, an imaginary ellipse.
The semi-axes of the first section are i/jz_andt/_.
V B V C
Those of the second section are 1/ and i/— . And those
A V C
of the imaginary section are y — . (— 1) and i / (— !)•
The distances 2 i/ , 2 v/-:^ , and 2 1/-^- are called the axes
268
SOLID ANALYTIC GEOMETRY.
Fig. C.
of the surface. Eepresenting the semi-axes by a, b, and c,
the equation of the surface may be reduced to the form
„9 "1" 70 o
If we suppose A = B, then a = b, and the equation re-
duces to
x^ + f _ ^' _ 1
the hyperboloid of revolution of two sheets.
SURFACES OF THE SECOND ORDER. 269
If A = B = C, the equation becomes
x^ -\- if — z^ = — a-,
which represents tlie surface generated by the revokition of
an equilateral hyperbola about its transverse axis.
Finally, if L = 0, the surface becomes a cone having an
elliptical base, and the base becomes a circle when A = B.
We have, therefore, the following varieties of the liyper-
boloid of two sheets :
1. The hyperboloid proper having three unequal axes.
2. The hyperboloid of revolution.
3. The equilateral hyperboloid of revolution.
4. The cone.
We will now examine the second form,
B2/2 + C.~2 + G a; = . . . (B)
Three cases apparently different present themselves for
examination.
(1). B and C positive and G negative in the first mem-
ber.
(2). B, C, and G positive.
(3). B positive and C and G negative.
Case 1. The equation may be written
B7/2 + C«2 = Ga;
in which B, C, and G are essentially positive.
Let the surface be intersected as usual by planes parallel
respectively to the co-ordinate planes.
When X = a, B?/^ + Cs^ = Ga, an ellipse real when a > 0,
and imaginary when a < 0.
When y = b, Cz'^ = Gx — B6^, a parabola with its axis
parallel to the axis of X.
When z = c, B?/^ = Gx — Cc^ a parabola with its axis
parallel to the axis of X.
The principal sections are found by making a = 0,b = 0,
and c = 0.
When a =0, B?/^ + Gz^ = 0, a point, the origin.
When b = 0, Gz'^ = Ga:;, a parabola with its vertex at the
origin.
270
SOLID ANALYTIC GEOMETRY.
When c, = 0, Bt/^ = Gx, a parabola with vertex at the origin.
Since every positive value of x gives a real section, and
every negative value of x an imaginary section, the surface
consists of a single sheet extending indefinitely and contin-
uously in the direction of positive abscissas, but having no
points in the opposite direction from the origin.
The surface is called the elliptical paraboloid. It may be
generated by the motion of an ellipse of variable dimensions
whose centre remains constantly on the same straight line,
and whose plane continues perpendicular to that line, and
whose semi-axes are the ordinates of two parabolas having a
common axis and the same vertex, but different parameters
placed with their planes perpendicular to each other.
Fig. D.
Case 2. If we suppose G to be positive in the first member
so that the equation will take the form
B7/2 4- C^- = - Gx,
SURFACES OF THE SECOND ORDER. 271
the sections perpendicular to the X-axis will become imaginary
when X > 0, and real when a; < 0.
In other respects the results are similar to those deduced in
case 1.
Thus the equation will represent a surface of the same form
as in case 1, but turned in the opposite direction from the
co-ordinate plane of YZ.
If B = C, the surface becomes the paraboloid of revolution.
Case 3. By^ _ C^^ = Go?.
Intersect the surface by planes as before.
When X ^= a, B?/^ — Gz^ = Ga, a hyperbola with transverse
axis in the direction of the Y-axis when a > 0, and in the
direction of the Z-axis when a. < 0.
When y = b, Gz^ = — Qx -\- B6^, a parabola having its axis
in the direction of the X-axis and extending to the left.
When z = c, By^ = Gx + Cc^, a parabola having its axis in
the direction of the X-axis and extending to the right.
Since every value of x, either positive or negative, gives a
real section, the surface consists of a single sheet extending
indefinitely to the right and left of the plane of YZ. This
surface is called the Hyperbolic Paraboloid. To find its princi-
pal sections make x, y, and z alternately equal to zero.
When X = 0, By'^ = Cz^, two straight lines.
When y = 0, Cz~ = — Gx, a parabola with axis to the left.
When z = 0, By^ = Gx, a parabola with axis to the right.
The hyperbolic paraboloid admits of no variety.
Now taking up form (C), By"" + Cs;^ + H?/ + I« + K ^ 0,
we see that it is the equation of a cylinder whose elements
are perpendicular to the plane of YZ, and whose base in the
plane of YZ will be an ellipse or hyperbola according to
the signs of B and C.
The fourth form (D), Cz" ^ Gx -{- B.y + Iz -\- K = repre-
sents a cylinder having its bases in the planes XZ and YZ
272
SOLID ANALYTIC GEOMETRY,
Fig. E.
parabolas, and having its right-lined elements parallel to the
plane XY and to each other, but oblique to the axes of X and Y.
The preceding discussion shows that every equation of the
second degree between three variables represents one or an-
other of the following surfaces :
1. The ellipsoid with its varieties, viz. ; the ellipsoid
proper, the ellipsoid of revolution, the sphere, the point,
and the imaginary surface.
SURFACES OF THE SECOND ORDER.
273
2. The hyperboloid of one or two sheets, witli their
varieties, viz. : the hyperboloid proper of one or two sheets,
the hyperboloid of revolution of one or two sheets, the
equilateral hyperboloid of revolution of one or two sheets,
the cone with an elliptical or circular base.
3. The paraboloid, either elliptical or hyperbolic, with the
variety, the paraboloid of revolution.
4. The cylinder, having its base either an ellipse, hyper-
bola, or parabola.
Surfaces of Revolution. — The general equatioii of surfaces
of revolution may be deduced by a direct method, as follows :
Fig. F.
Let the Z-axis be the axis of revolution, and let the equa-
tion of AB, the generating curve in the plane of XZ, be
x^ = fz.
Let P be the point in this curve which generates the circle
274 SOLID ANALYTIC GEOMETRY.
PQR, and let r be the radius of the circle. We will have
r^ z= x^ -\- y^.
The value of r may also be expressed in terms of z from
the equation of the generatrix in the plane of XZ as follows :
r2^CF^= Od' =fz.
Equating these two values of r we have
x^ + y"" =fa
as the general equation of surfaces of revolution.
It will be observed that the second value of r'^ is the value
of x^ in the equation of the generatrix. Hence, to find the
equation of the surface of revolution we have only to substi-
tute x^ + 'if of the surface for x^ in the generatrix.
Surface of a Sjihere. — Equation of generatrix x^ -\- z"^ = W:
Hence the equation of the surface of the sphere is
x^J^ifJ^z'' = ^K
Ellipsoid of Revolution. —
Generatrix — + — = 1.
a" c^
Surface ^!J:J^ + ^ = i.
a^ &
Similarly, the equation of the hyperboloid of revolution is
x^ -\- y- _ z' _ ^
a^ c-
Paraholoid of Revolution. —
x"^ = 'ipz, the generatrix.
x^ -{- y- = 4,2jz, the surface of revolution.
Cone of revolution, z = mx -\- /3 the generatrix,
7Ji ' m?
Hence x"^ + y-
_(z-Jl
or m^ (x^ + y-) = (z — 0)-
SURFACES OF THE SECOND ORDER. 275
EXAMPLES.
1. What is the locus in space of 4 a;^ + 9 y^ = 35 ? Qf
9 ^2 — 16 ?/- = 144 ? Of a;' + 2/' = r^ ? Of y' + -' = r^ ? Of
y2 + 8 a; = ?
2. Determine the nature of the surfaces a;^ + 2/^ + 4 «^ = 25,
3. Find the equation of the surface of revolution about the
axis of Z whose generatrix is s = 3 a; + 5.
4. Find the equation of the cone of revolution Avhose inter-
section with the plane of XY is a;^ + y^ = 9, and whose vertex
is (0, 0, 5.)
5. Determine the surfaces represented by
ic2 ^ 4 2/' + 9 ^2 ^ 36.
a;2 _|. 4 ^2 _ 9 ^2 ^ 3e_
a;2 + 4 ?/' = 9 «2 _ 36.
4 3/2 _|_ 9 ^2 _ 36 ^_
4 ?/2 - 9 «2 = 36 a;.
J
Date Due
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