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PROBLEM XIII.

To find the area of the fruftum of a parabola.

RULE.

Divide the difference of the cubes of the two ends of the frustum by the difference of their squares, multiply this quotient by the altitude, and the product will be the area requi

red.

EXAMPLE I.

In the parabolic frustum DABE, the two parallel ends DE, AB, are 12 and 2c, and the altitude FC 6, required the area.

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Ex. 2. The greater end of a frustum is 20, the less 10, and

their diftance 12, required the area.

Ans. 186.

Ex. 3. The greater end of a frustum is 30, the less 20, and Anf. 380.

their distance 15, required the area.

Ex. 4. The greater end of a frustum is 9, the less 6, and their

distance 4, required the area.

Ans. 113.

PROBLEM XIV.

To defcribe an byperbola, the transverse and conjugate diameters being given.

RULE,

Draw AB the transverse diameter, and BC the conjugate at right angles to it; bisect AB in c, and and with the centre c, and radius cE, describe the circle EFDf, cutting AB produced in the points F, f, and these points will be the foci.

In AB produced take any convenient number of points x, x, &c. and from F and f as centres, and radii Bx, Ax, describe arches interfecting in the points m, m, &c. Join these points, and it will form the hyperbolic curve required.

Note. If through the points E and D straight lines be drawn from c, they will be the asymptotes of the hyperbola.

Any three of the four following particulars being given, to find a fourth, viz. the transverse, conjugate, ordinate, and its abfciffa.

PROBLEM XV.

The tranfverfe, conjugate, and abfcifssa being given, to find the or

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RULE.

As the transverse

Is to the conjugate,

So is the square root of the product of the two abscissas
To the ordinate.

EXAMPLE I.

In the hyperbola GBH, the transverse is 60, the conjugate 36, and the abscissa AB 20, required the ordinate.

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Ex. 2. The transverse is 50, the conjugate 30, and the ab

fciffa 16, required the ordinate.

Anf. 20.

Ex. 3. The transverse is 45, conjugate 224, and the abfciffa

15, required the ordinate.

Anf. 15.

Ex. 4. The transverse diameter is 24, the conjugate 21, and

the less abfciffa 8, required the ordinate.

PROBLEM XVI.

Anf. 14.

The transverse, conjugate diameters, and an ordinate, being gi

ven, to find the abfciffas.

RULE. RULE.

As the conjugate diameter

Is to the transverse,

So is the square root of the fum of the squares of the ordinate and femi-conjugate

To the distance between the ordinate and centre.

Add to, or fubtract from, the femi-transverse, this fourth proportional, according as the greater or less abscissa is requi

red.

1

EXAMPLE I.

The transverse diameter is 60, the conjugate 36, and the ordinate 24, required the two abscissas.

182=324 square of the femi-conjugate.
242=576 square of the ordinate.

900(30

9

00

36:60::30

60

36)1800(50 dist betw. the ordinate and centre. 180 30 femi-transverse.

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The transverse diameter is 50, the conjugate 30, and the or

Ans. 66 and 16.

Ex. 3. The transverse diameter is 24, the conjugate 21, and

dinate 20, required the abscissas,

the ordinate 14, required the abfciflas.

Anf. 32 and 8.

Ex. 3. The transverse diameter is 24, the conjugate 21, and

the ordinate 14, required the abfciffas.

Anf. 32 and 8.

Ex. 4. The transverse diameter is 30, the conjugate 224, and

the ordinate 15, required the abfciffas.

Anf. 33 and 34.

PROBLEM XVII.

To find the length of an arch of an hyperbolic curve, beginning at

the vertex.

RULE.

To 19 times the transverse add 21 times the parameter * of the axis; and, to 9 times the transverse, add 21 times the parameter, then multiply each of these sums by the quotient of the abfcifla divided by the transverse.

To each of the products fo found add 15 times the parameter, and divide the former by the latter, and multiply this quotient by the ordinate, the product will be the length of the arch nearly.

EXAMPLE I.

In the hyperbola GBH, the transverse is 160, the conjugate 120, the ordinate 20, and abscissa 4+, required the length of the curve GB.

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* From a well-known property of the hyperbola, the recaangle contained by the transverse and the parameter is equal to the square of the conjugate; that is, the conjugate is a mean proportional between the tranfverfe and the parameter. Hence

the following proportion to find the parameter :-

As the transverse, is to the conjugate,

So is the conjugate, to the parameter.

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