PROBLEM II. In an Ilyperbola, to find any Two Conjugate Diameters, an Ordinate to one of them, and its Absciss, one from another ; viz. baving any three of them given, to find the Fourth. To find the Ordinate. diameter : That is, as d :i:: ✓ id + *) * : Y iv id + x) x; where d denotes the diameter, c its conjugate, y an ordinate to the diameter, and x its abfcifs, or its distance from the vertex of the diameter, measured upon it. Note. In the hyperbola, the less abfciss added to the diameter, gives the greater absciss. E X A M P L E. If the diaineter be 24, its conjugate 21, and the less absciss 8 ; what is the ordinate? (d+- *) * 8 By the rule, 21/32 * 7 14 the ordinate required. 7 x 16 d 24 8 CASE * The values of the several quantities, in the four cases of this problem, are easily found from the general property of the curve, viz. dd :cc:: (+1)x : yy. The demonstration of which, together with that of the preceding problem, belongs to conic Sections, To find the Abscisses. As the conjugate: Is to the diameter :: So is the root of the sum of the squares of the ordinate and semi-conjugate : To the distance between the foot of the ordinate and the center. Then, to and from this distance, add and subtract the semi-diameter, and the sum and difference will give the two abscisles. That is, + d = x the greater or less absciss, according as the upper or under sign is used. The diameter and its conjugate being 24 and 21, required the two abscisses to the ordinate 14. 21 Here dvocc + yy 2475 X 212 + 142 += + 12 4 W212 + 282 + 12 = 4V 32 + 4 + 12 = 20 7 #12 = 32 and 8, the two abscisses required. To find the Conjugate. dy That is, id + *) * : y :: 0 : ✓(+ *)* Dd 2 EX What is the conjugate to the diameter 24, the less absciss being 8, and its ordinate 147 Here dy 24 X 14 = v (d + x) * Ņ32 X 8 conjugate. 24 X 14 16 = 21, the CASE IV. To find the Diameter. To or from the root of the sum of the squares of the ordinate and semi-conjugate, add or fubtract the semi-conjugate, according as the less or greater absciss is used. Then, 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 di The less absciss being 8, and its ordinate 14, required the diameter, fuppoing the conjugate to be 21. Here Ni X 212 + 142 + 10 6X X = 21 X 8 X уу 14 X 14 W 212 + 282 + 21 = 3 X = 3 X (V 3* + 42 + 3) = 7 3 x (5 + 3) = 24, the diameter required. ÞRO PROBLEM III. Having given any Two Abscisses x, x, and their Ordinates y, y; to find the Diameter d to which they belong. Multiply each absciss by the square of the ordinate belonging to the other; multiply also the square of each absciss by the square of the other's ordinate; then divide the difference of the latter products by the difference of the former; and the quotient will be the diameter to which the ordinates belong. * If two abfcifles be 1 and 8, and their two ordinates 4 and 14; what is the diameter to which they belong? By the nature of the hyperbola, dx + xx : dx + ** :: YY : yy', or dxyy + xxyy = dxYY + **YY; hence d = хх уу 2. E. D. *YY – xyy хх үү PROBLEM IV. Given any Three Equidistant Ordinates a, b, c, with Conjugate c. From the square of the difference between A and double the said square of the middle ordinate, take 4 times the square of the product of the extreme ordinates; and call the square root of the remainder B. Then, as A is to b, fo is the given common distance, to the diameter to which the ordinates belong. And If x be the absciss to the least ordinate a, by the nature of the hyperbola we shall have these three equations, X (D + *) X * X (Dx + xx), сс CC DD DD CC CC DD CE сс D D X (D+*+ d) x (x + d) X (Dx + ** + 40x + 2D + 4dd). aa – 2bbt oc 2 dd Again, taking the first equation from the second, leaves bb-aa bb-aa D+d 2d CC CC DD DD СС DD х сс D D 2 |