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And as the common distance of the ordinates, is to the root of ĻA, so is the diameter D, to its conjugate c.

That is,
dni - aa + 4bb cc)?

and c = DNA 2bb t.cc

4 аасс

D

E X AMPLE.

Required the diameter, and its conjugate, to which belong the three ofdinates 45, 14, and 'V65, their common distance being 7.

- 2

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352

2

d N (45b – aa – cc)? — 4aacc Here the diam. Da

2bb + cc 7V (28

65 X 212 2 4 X 65 X 352 X 21
82
82

8+

65 X 21
2 X 14+
82

82
7V(322 – 51 – 65 * 3:). – 65 x 102 X 3– 7X336

24 52 – 2 X 16+ + 65 X 32

And

352

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98

od 4

d x

2

aa-2

сс

aa

DD

bb
Dtd

2bb-Laa- cc
=dx

- 1d. 2bb + cc

2bb foc

2 bb + cc Consequently, aa = x (0+ x) x x =

X

2 dd 2bb }aa – {cc;" [dX

{cc): - pp?]; and hence

2bb tec aut - aa + 466 cc)2 – 4 a acc

= the diameter D. 2bb + cc

2bb + cc Also from the equation

is deduced

2 dd
2bbtcc DVA
= the conjugate c.

2.E.D. av

d

dB

A

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Corollary. Hence the sum of the squares of the extreme ordis nates, is greater than double the square of the middle one.

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24 v 352

2 x 82

2

cc + aa

2.4.610

And

65 X 212
14? +
= 8V 5 -16*+

65 x 32 7

2 X 82 = 1V 50-322 +9X130 = *V 196 = 1X14 = 21, the conjugate c.

PROBL EM V. To find the Length of an Arc of an Hyperbola, the Arc

beginning at the Vertex.

RV L E 1. Put a = the semi-transverse axe, c = the semiconjugate, q = #^^, y = an ordinate to the axe drawn from the end of the arc required, A = the hyperbolic logarithm of 9 + vec +y3 = 2.302585093 x common log. off+Nec tuy yNcc + yg – ys

govec + vy – sccc
4
y? Ncc + yy – 7CCD

&c.
8

Then Putting * = the abfcifs of the ordinate y, we shall obtain, from the nature of the curve, * =

a v cc + yy

a, and therefore

+ ayy hence

- yy= Not ry

cy

Vi+gyy = Vect yg

Actig

Natyy * (1 + x2 92

3.594 +

58 &c), 49 2.4.6.8

But

CCA

B

2

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DEMONSTRATION.

с

сс

сс +аа

уу с с

=

co + yy

Vit

сс +аа

64

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2.

Then will the length of the arc be expressed by cx (A + ļB - 2.0 +

393

3.524

E &c). 2.4.6

2.4.6.8

D

2.4

E X A M P L E.

сс + аа

1

64

324

Required the length of the curve corresponding to the ordinate 10, the transverse and conjugate axes being 80 and 60. Here a = 40,6 = 30, and y = 10; hence 9 = 2500

; voo + yy = 10V10 = 810000

y + vec tyy 31•62277662 ; and

= 1.38742 5887; whose hyperbolic logarithm is *3274501 = A; also yn cc + yy

+ yy ---CA = 1006133 gyNcc tyy

= 641796405 us ncc + yy - 50cc

= 45698.97

c

B.

2

3cc в

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4

= D.

6

[blocks in formation]

cc + 3y -- 70CD = 3540529*3125 = E.

Then

8

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Then +

Ec=

A = '327450,

= '000764

2.4 +

3.594

E = '000012

2.4.6.8
393
+ D = '000084
2.4.6

- '000776

LB = .016607,- 3.596

+ '344141

*000776

*343365 the sum of the series, which multip. by <=

30

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a?

Using the same symbols as in the last rule, the arc

a4 + 4a?c? will be expressed by the series yx(1 + mp3

604 ao + 4a4c2 + 8a2-4 sao +242c? + 48a4c4 +64ac6

quo &c). 11 2012

4008

+

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1

1152016

or

DEMONSTRATION.

aatec
+ -99

(c For, by the demonst. of the last rule, z is = y

сс + уу =, by extracting the root of the numerator and denominator, and then dividing the one by the other, ý x a4 + 4a2c?

a + 4a4c? + 8a (4 (it

ܐq

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-y4 +

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2.40

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16

sal + 24 aoc? + 4Sa+c+ + 64ac6

y8 &c). And, by taking 4.4. the fluents, we obtain the series as in the rule.

А

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20

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a’y? a? +462 392 a4 + 4a?c? + 8c4 5y? or y X(1 +

-B + 6c4

a' + 4c? 526 + 24a4c? + 48 a2c4 + 64c6 7y?

D &c). + 4a2c? + 8c+ Putting A, B, C, &c, for the ist, 2d, 3d, &c, term.

72c4

EX AMPLE.

Taking the same example as to the last rule, in which a = 40,5 = 30, and y = 10; we shall obtain

C = '003170
B = 0'032922

59
398

*003229

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D

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+ 1'033320

*003229

the sum 1.030091 which multiplied

by y = 10 produces 10•30091 the arc, nearly as before.

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1. To 21 times the square of the conjugate, add 9 times the square of the transverse; and to the same

21 times

DEMONSTRATION.

First

avc2 + y2

- a, ufing the fame fynbols as in the

last rules, =

,

3 ago 2.4.666

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ay? a y +
+

3.5 ays
2.464

&c.

2.4.6.8c8 Then proceed as in rule 5 for the elliptic arc.

ay?

A+ (8 +1) X
A+ (B +1)x

&c) Thus y x A + B *

=yx
A + B X

202

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ay4

8c4 ау*

ay?

8c4&c)

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