As the tranfverfe Is to the conjugate, RULE. So is the fquare root of the product of the two abfciffas EXAMPLE I. In the hyperbola GBH, the tranfverfe is 60, the conjugate 36, and the abfciffa AB 20, required the ordinate. Ex. 2. The tranfverfe is 50, the conjugate 30, and the abfciffa 16, required the ordinate. Anf. 20. Ex. 3. The tranfverfe is 45, conjugate 221, and the abfciffa 15, required the ordinate. Anf. 15. Ex. 4. The transverse diameter is 24, the conjugate 21, and the lefs abfciffa 8, required the ordinate. Anf. 14. PROBLEM XVI. The tranfverfe, conjugate diameters, and an ordinate, being gi ven, to find the abfciffas. RULE. As the conjugate diameter Is to the tranfverfe, RULE. So is the fquare root of the fum of the fquares of the ordinate and femi-conjugate To the distance between the ordinate and centre. Add to, or fubtract from, the femi-tranfverfe, this fourth proportional, according as the greater or lefs abfciffa is required. EXAMPLE I. The transverse diameter is 66, the conjugate 36, and the ozdinate 24, required the two abfciffas. 18=324 fquare of the femi-conjugate. 900(30 36:60:30 66 36)1800(50 dift betw. the ordinate and centre. 180 30 femi-tranfverfe. The tranfverfe diameter is 50, the conjugate 30, and the ordinate 20, required the abfciffas, Anf. 663 and 16}. Ex. 3. The tranfverfe diameter is 24, the conjugate 21, and the ordinate 14, required the abfcifas. Anf. 32 and 8. Ex. 3. The tranfverfe diameter is 24, the conjugate 21, and the ordinate 14, required the abfciffas. Anf. 32 and 8. Ex. 4. The tranfverfe diameter is 30, the conjugate 224, and the ordinate 15, required the abfciffas. PROBLEM XVII. Anf. 33 and 34. To find the length of an arch of an hyperbolic curve, beginning at the vertex. RULE. To 19 times the tranfverfe add 21 times the parameter * of the axis; and, to 9 times the tranfverfe, add 21 times the parameter, then multiply each of thefe fums by the quotient of the abfciffa 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 tranfverfe is 160, the conjugate 120, the ordinate 20, and abfciffa 4, required the length of the curve GB. * From a well-known property of the hyperbola, the redangle contained by the tranfverfe and the parameter is equal to the fquare 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 tranfverfe, is to the conjugate, So is the conjugate, to the parameter. Ex. 2. Let the tranfverfe be 80, the conjugate 60, the ordinate 10, and the abfciffa 2.1637, required the length of the arch GB. Anf. 10.3 Ex. 3. The tranfverfe is 120, the conjugate 72, the ordinate 48, and the abfciffa 49, required the length of the arch. Anf. 62.6496. PROBLEM XVIII. To find the area of an hyperbola, the tranfverfe, conjugate, and abfciffa being given. RULE. To the product of the tranfverfe and abfciffa add of the fquare of the abfciffa, and multiply the fquare root of the fum by 21. |