PROBLEM VIII. In a hyperbola, to find the transverse, or conjugate, or ordinate, or absciss. CASE I. The transverse and conjugate diameters, and the two abscisses being given, to find the ordinate. RULE. 1. As the transverse diameter is to the conjugate, so is the square root of the product of the two abscisses to the ordinate. Note. The less absciss added to the transverse diameter, will give the greater absciss EXAMPLES. 1. In the hyperbola ABC, the transverse diameter is 30, the conjugate 18, and the absciss BD 10; what is the ordinate AD? 2. The transverse diameter is 48, the conjugate 42, and the less absciss 16; what is the ordinate? CASE II. Ans. 28. The transverse and conjugate diameters, and an ordinate being given, to find the abscisses. RULE. To the square of half the conjugate, add the square of the ordinate; and extract the square root of the sum. Then say, as the conjugate diameter is to the transverse, so is the said square root to half the sum of the abscisses, or the distance between the ordinate and the centre. Then this distance being added to half the transverse diameter, will give the greater absciss; and their dif ference will be the less absciss. EXAMPLES. 1. The transverse diameter is 30, the conjugate 18, and the ordinate 12; what are the two abscisses? Here /9+12o = √/81+144 = √√√225 = 15; then, 30 X 15 5 × 15 as 18: 30 :: 15: 5×525, half the sum of the abscisses; hence, 25 +15= 40, the greater absciss; and 25 1510, the less absciss. 2. If the transverse diameter be 48, the conjugate 42, and the ordinate 28; what are the two abscisses? Ans. 64 and 16. CASE III. The transverse diameter, the two abscisses, and the or dinate being given, to find the conjugate. RULE. As the square root of the product of the two abscisses is to the ordinate, so is the transverse diameter to the conjugate. EXAMPLES. 1. The transverse diameter is 30, the ordinate is 12 and the two abscisses are 40 and 10; required the conjugate diameter. Ff Here/40 × 10/40020, the square root of the product of the two abscisses; then, as 20:12::30: = 12 × 6 4 3 x 6 18, the conjugate required. 12 x 30 20. 2. The transverse diameter is 48, the ordinate 28, and the abscisses are 64 and 16; required the conjugate diameter. CASE IV. Ans. 42. The conjugate diameter, the ordinate, and the two abscisses being given, to find the transverse. RULE. To the square of half the conjugate add the square of the ordinate, and extract the square root of the sum. To this root add half the conjugate, when the less absciss is used, but subtract it when the greater absciss is used; and reserve the sum or difference. Then say, as the square of the ordinate is to the product of the absciss and conjugate, so is the sum or difference, above found, to the transverse required. EXAMPLES. 1. The conjugate diameter is 18, the ordinate is 12, and the less absciss 10; required the transverse dia meter. Here/92 +12o = 81+144√√√225=15; and 15+9=24. Also, 12 × 12 = 144, the square of the ordinate; and 18 x 10 180, the product of the absciss and conjugate; then, as 144: 180: 24: 15 x 24 12 180 × 24 144 = 15 × 2 = 30, the transverse required. 2. The conjugate diameter is 42, the greater absciss is 64, and the ordinate 28; what is the transverse diameter? Ans. 48. PROBLEM IX. To find the length of a hyperbolic curve. RULE. To 21 times the square of the conjugate, add 9 times the square of the transverse; and to the said 21 times the square of the conjugate, add 19 times the square of the transverse; and multiply each sum by the absciss. To each of these two products add 15 times the product of the transverse multiplied by the square of the conjugate; then say, as the less of these sums is to the greater, so is the double ordinate to the length of the curve, nearly. EXAMPLES. 1. The transverse diameter of the hyperbola ABC, is 30, the conjugate 18, the absciss BD 10, and the double ordinate AC 24; what is the length of the curve ABC? Here 182 x 21 + 302 × 9 = 324 × 21 + 900 × 9 = 68048100 = 14904 = 21 times the square of the conjugate, added to 9 times the square of the transverse ; and 6804 +900 × 19 6804+1710023904 =21 times the square of the conjugate, added to 19 times the square of the transverse; then 14904 × 10 149040, and 23904 x 10 = 239040. Again, 324 x 30 × 15 145800 = 15 times the product of the transverse multiplied by the square of the conjugate; and 149040 + 145800 294840, and 239040 + 145800 384840; then, as 294840 384840 :: 24: 31.326007, the length of the curve required. 2. The transverse diameter of a hyperbola is 45, the conjugate 27, the absciss 15, and the double ordinate 36; what is the length of the curve? PROBLEM X. Ans. 46.989. To find the area of a hyperbola, the transverse, conjugate, and absciss being given. RULE. Multiply the transverse by the absciss; to the product add of the square of the absciss; and multiply the square root of the sum by 21. Add the product, last found, to 4 times the square root of the product of the transverse and absciss; and divide the sum by 75. Divide 4 times the product of the conjugate and absciss by the transverse; then this quotient being multiplied by the former, will give the area, nearly. EXAMPLES. 1. Required the area of a hyperbola, whose transverse diameter is 30, the conjugate 18, and the absciss 10. Here 21/30 × 10+102 x=21/300 + 100 × = 21 √300 + 500 ÷ 7 = 21 √300 + 71.428571 = 21371.428571 = 21 × 19.272 = 404.712. Again, (4/30 × 10 + 404.712) ÷ 75 = (4.√√/ 300+ 404.712) 75 = (4 x 17.3205 + 404.712) ÷ 75 = (69.282404.712) ÷ 75 473.994 ÷ 75 ≈ 6.3199. 18 x 10 x 4 Lastly, 30 × 6.3199 = 18 x 4 3 × 6.3199 = 6 × 4 × 6.3199 = 24 × 6.3199 = 151.6776, the area required. 2. The transverse diameter of a hyperbola 50, the Conjugate 30, and the absciss 25; what is its area? Ans. 805.0902. |