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2. The transverse diameter is 24, the ordinate 14, and the abscissas 8 and 32; required the conjugate.

Ans. 21.

CASE IV.

The conjugate diameter, the ordinate and two abscissas

being given, to find the transverse.

RULE.

1. Add the square of the ordinate tò the

square

of the semi-conjugate, and find the square root of their sum.

2. Take the sum or difference of the semi-conjugate and this root, according as the less or greater abscissa is used, and then say,

As the square of the ordinate,
Is to the product of the abscissa and conjugate ;
So is the sum, or difference, above found,
To the transverse required.

EXAMPLES.

1. The conjugate diameter is 72, the ordinate is 48, and the less abscissa 40; what is the transverse ?

Here 482 +36% 2304 +1296= 3600=60:
And 60+36=96.

Also 72 x 40=2880=product of the abscissa and conjugate. Whence,

cx

* This rule in algebraic terms is one x (V?c* +y=10)

=t=transverse diameter.

As 2304 : 2880 :: 96

96

17280 25920

23048276480(120=transverse required.

2304

4608 4608

2. The conjugate diameter is 21, the less abscissa 8, and is ordinate 14; required the transverse.

Ans. 24. 3. Required the transverse diameter of the hyperbola, whose conjugate is 36, the less abscissa being 20, and ordinate 24.

Ans. 60.

PROBLEM XIII.

To find the length of any arc of an hyperbola, beginning

at the vertex.

RULE.*

1. As the transverse is to the conjugate, so is the conjugate to the parameter.

* Demon. Let t=semi-transverse axe, c=semi-conjugate, a=ordinate, and y=abscissa. Then will yx(1+.

6 # +4tc2 +6+4+2+8tch -yo

-yo, &c.)=length of the

112c12 arc, as is shown by the writers on fluxions.

2c But ē Vo+y-a, and 7 =parameter=p, by the

40c8 +

t

2. To 19 times the transverse, add 21 times the parameter of the axis, and multiply the sum by the quotient of the abscissa divided by the transverse.

3. To 9 times the transverse, add 21 times the parameter, and multiply the sum by the quotient of the abscissa divided by the transverse.

4. To each of the products, thus found, add 15 times the parameter, and divide the former by the latter; then this quotient being multiplied by the ordinate will give the length of the arc nearly.

EXAMPLES.

1. In the hyperbola GAE, the transverse diameter is 80, the conjugate 60, the ordinate GH 10, and the abscissa AH 2.1637; required the length of the arc AG.

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nature of the curve. Consequently the rule becomes (15p+ 19t+21p * 7)+(15p+90+21px 7)xy=(15pt + 19tx

2x 2.2c2 +21px) :-(15pt +9tx+21px) Xy=yx :1+

&c.

3p 5p which by substituting the values of # and p, and expanding the terms, gives a series, agreeing nearly in the first three terms with the former; and therefore, the rule is an approximation.

If t=semi-transverse, c=semi-conjugate, and y=ordinate drawn from the end of the required arc; then yx (1+ +4c3y +4+4+c+8c2 5y2

-B + ca 20

t +4c of the arc.

6cf

*14c2o, &c.) = length

60 x 60 3 x 60 Here 80: 60 :: 60 :

23 x 15 =45 = 80

4 parameter.

2.1637 And (80 x 19 x 45 x 21)

1520+945 X.02704

80 =2465 X.02704=66.6536.

2.1637 Also (80 x 9+45 x 21) =720 + 945 X.02704

80 =1665 X.02704–45.0216.

Whenče 675 +66.6536-:-675 + 45.0216=741.6536; 720.0216=1.03004; and 1.03004 x 10=10.3004=length of the arc required.

2. The transverse diameter of an hyperbola is 120, the conjugate 72, the ordinate 48, and the abscissa 40 : required the length of the curve.

Ans. 62.6496. 3. Required the whole length of the curve of an hyperbola, to the ordinate 10; the transverse and conjugate axes being 80 and 60.

Ans. 20.6006.

PROBLEM XIV.

To find the area of an hyperbola, the transverse, conjugate,

and abscissa being given.

RULE.*

1. To the product of the transverse and abscissa, add 4 of the square of the abscissa, and multiply the square root of the sum by 21.

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* Demon. Let t=transverse diameter, c=conjugate, x= abscissa, y=-ordinate, and = Then it is well known 1 1

1

1 X

ed, &c.)=area of

3.5.7. 5.7.9. the byperbola.

that 4xy (3 1.3.5.

2. Add 4 times the square root of the product of the transverse and abscissa, to the product last found, and divide the sum by 75.

3. Divide 4 times the product of the conjugate and abscissa by the transverse, and this last quotient multiplied by the former will give the area required nearly.

EXAMPLES

In the hyperbola GAF, the transverse axis is 30, the conjugate 18, and the abscissa or height AH is 10; what is the

area ?

ty But

c=conjugate axis, by the nature of the ▼ tatx2

4xc hyperbola. Consequently the expression for the rule

214tc+bx+4 ta 21 toc + 9 +4 to Х

-=4xy x 75

▼ txx2 And this thrown into a series will very nearly agree with the former; which shows the rule to be an approximation.

Q. E. I.

Rule 2. If 27, 2y=bases, V, and v their distances from the centre, and the other letters as before, then will vytc

ay+cV vy-x hyp. log. of =area of the frustrum of the 4

ty+cv hyperbola. Rule 3. If t be put=transverse axis, c= = conjugate, abscissa, the area of a segment of an hyperbola,

47 tx + 4x + to cut off by a double ordinate will be

15
4cx
very nearly

.
t

and x =

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