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 required. 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. Ex. 2. The greater end of a frustum is 20, the less 10, and their distance 12, required the area. Anf. 186. Ex. 3. The greater end of a frustum is 30, the less 20, and their distance 15, required the area. Anf. 380. Ex. 4. The greater end of a fruftum is 9, theYess 6, and their distance 4, required the area. Anf. 11. PROBLEM XIV. To describe an hyperbola, 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 abfcifli PROBLEM XV. The irunfverse, conjugate, and abscissa being given, to find the or dinate, RULE. RULE. As the transverse EXAMPLE I. In the hyperbola GBH, the transverse is 60, the conjugate 36, and the abscissa AB 20, required the ordinate. Ex. 2. The transverse is 50, the conjugate 30, and the abfciffa 16, required the ordinate. Anf. 20. Ex. 3• The transverse is 45, conjugate 221, and the abscissa 15, required the ordinate. Ex. 4. The transverse diameter is 24, the conjugate 21, and the less abfciffa 8, required the ordinate. Anf. 14. Anf. 15. PROBLEM XVI. The transverse, conjugate diameters, and an ordinate, being given, to find the abfcifas. RULE. RULE. As the conjugate diameter and semi-conjugate Add to, or subtract from, the semi-transverse, this fourth proportional, according as the greater or less abscissa is required. EXAMPLE 1. The transverse diameter is 6ó, the conjugate 36, and the 07dinate 24, required the two abfciffas. 36) 1800(50 dift betw. the ordinate and centre. 180 30 semi-transverse. 80 greater abfcifla. The transverse diameter is 50, the conjugate 30, and the ordinate 20, required the abfciflas, Anf. 66 and 16 Ex. 3. The transverse diameter is 24, the conjugate 21, and the ordinate 14, required the abicillas. Anf. 32 and 3. Ex. 3. The transverse diameter is 24, the conjugate 21, and the ordinate 14, required the absciffas. Anf. 32 and 8. Ex. 4. The transverse diameter is 30, the conjugate 22, and the ordinate 15, required the abfciffas. AN 33 and 31s. PROBLEM XVII. To find the length of an arch of an hyperbolic curve, beginning at the vertex. * To 19 times the transverse add 21 times the parameter of the axis; and, to 9 times the transverse, add 21 times the pafameter, then multiply each of these sums by the quotient of the abscilla divided by the transverse. To each of the producis so 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 1. In the hyperbola GBH, the transverse is 160, the conjugate 120, the ordinate 20, and abfcifli 45, required the length of the curve GB. Fir?, Gg * From a well-known property of the hyperbola, the realangle 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 transverse and the parameter. Hence the following proportion to find the parameter :- As the transverse, is to the conjugate, |