And as the common distance of the ordinates, is to the root of ĻA, so is the diameter D, to its conjugate c. That is, and c = DNA 2bb t.cc 4 аасс D E X AMPLE. Required the diameter, and its conjugate, to which belong the three ofdinates 45, 14, and 'V65, their common distance being 7. - 2 352 2 d N (45b – aa – cc)? — 4aacc Here the diam. Da 2bb + cc 7V (28 65 X 212 2 4 X 65 X 352 X 21 8+ 65 X 21 82 24 52 – 2 X 16+ + 65 X 32 And 352 98 od 4 d x 2 aa-2 сс aa DD bb 2bb-Laa- cc - 1d. 2bb + cc 2bb foc 2 bb + cc Consequently, aa = x (0+ x) x x = X 2 dd 2bb – }aa – {cc;" [dX {cc): - pp?]; and hence 2bb tec aut - aa + 466 – cc)2 – 4 a acc = the diameter D. 2bb + cc 2bb + cc Also from the equation is deduced 2 dd 2.E.D. av d dB A Corollary. Hence the sum of the squares of the extreme ordis nates, is greater than double the square of the middle one. 24 v 352 2 x 82 2 cc + aa 2.4.610 And 65 X 212 65 x 32 7 2 X 82 = 1V 50-322 +9X130 = *V 196 = 1X14 = 21, the conjugate c. PROBL EM V. To find the Length of an Arc of an Hyperbola, the Arc beginning at the Vertex. RV L E 1. Put a = the semi-transverse axe, c = the semiconjugate, q = #^^, y = an ordinate to the axe drawn from the end of the arc required, A = the hyperbolic logarithm of 9 + vec +y3 = 2.302585093 x common log. off+Nec tuy yNcc + yg – ys govec + vy – sccc &c. Then Putting * = the abfcifs of the ordinate y, we shall obtain, from the nature of the curve, * = a v cc + yy a, and therefore + ayy hence - yy= Not ry cy Vi+gyy = Vect yg Actig Natyy * (1 + x2 92 3.594 + 58 &c), 49 2.4.6.8 But CCA B 2 DEMONSTRATION. с сс сс +аа уу с с = co + yy Vit сс +аа 64 2. Then will the length of the arc be expressed by cx (A + ļB - 2.0 + 393 3.524 E &c). 2.4.6 2.4.6.8 D 2.4 E X A M P L E. сс + аа 1 64 324 Required the length of the curve corresponding to the ordinate 10, the transverse and conjugate axes being 80 and 60. Here a = 40,6 = 30, and y = 10; hence 9 = 2500 ; voo + yy = 10V10 = 810000 y + vec tyy 31•62277662 ; and = 1.38742 5887; whose hyperbolic logarithm is *3274501 = A; also yn cc + yy + yy ---CA = 1006133 gyNcc tyy = 641796405 us ncc + yy - 50cc = 45698.97 c B. 2 3cc в 4 = D. 6 cc + 3y -- 70CD = 3540529*3125 = E. Then 8 Then + Ec= A = '327450, = '000764 2.4 + 3.594 E = '000012 2.4.6.8 - '000776 LB = .016607,- 3.596 + '344141 *000776 *343365 the sum of the series, which multip. by <= 30 a? Using the same symbols as in the last rule, the arc a4 + 4a?c? will be expressed by the series yx(1 + mp3 604 ao + 4a4c2 + 8a2-4 sao +242c? + 48a4c4 +64ac6 quo &c). 11 2012 4008 + 1 1152016 or DEMONSTRATION. aatec (c For, by the demonst. of the last rule, z is = y сс + уу =, by extracting the root of the numerator and denominator, and then dividing the one by the other, ý x a4 + 4a2c? a + 4a4c? + 8a (4 (it ܐq -y4 + 2.40 16 sal + 24 aoc? + 4Sa+c+ + 64ac6 y8 &c). And, by taking 4.4. the fluents, we obtain the series as in the rule. А 20 a’y? a? +462 392 a4 + 4a?c? + 8c4 5y? or y X(1 + -B + 6c4 a' + 4c? 526 + 24a4c? + 48 a2c4 + 64c6 7y? D &c). + 4a2c? + 8c+ Putting A, B, C, &c, for the ist, 2d, 3d, &c, term. 72c4 EX AMPLE. Taking the same example as to the last rule, in which a = 40,5 = 30, and y = 10; we shall obtain C = '003170 59 *003229 D + 1'033320 *003229 the sum 1.030091 which multiplied by y = 10 produces 10•30091 the arc, nearly as before. 1. To 21 times the square of the conjugate, add 9 times the square of the transverse; and to the same 21 times DEMONSTRATION. First avc2 + y2 - a, ufing the fame fynbols as in the last rules, = , 3 ago 2.4.666 ay? a y + 3.5 ays &c. 2.4.6.8c8 Then proceed as in rule 5 for the elliptic arc. ay? A+ (8 +1) X &c) Thus y x A + B * =yx 202 ay4 8c4 ау* ay? 8c4&c) |