PROBLEM II. In an Ilyperbola, to find any Two Conjugate Diameters, an Ordinate to one of them, and its Abfcifs, one from another; viz. having any three of them given, to find the Fourth. Is to its conjugate :: So is the mean propor. between the abfciffes : That is, as d : c :: ċ :: √(d + x) x : y ¿√(d + x) x; where d denotes the diameter, c its conjugate, y an ordinate to the diameter, and x its abfcifs, or its diftance from the vertex of the diameter, meafured upon it. Note. In the hyperbola, the lefs abfcifs added to the diameter, gives the greater abfcifs. EXAMPLE. If the diameter be 24, its conjugate 21, and the lefs abfcifs 8; what is the ordinate? The values of the feveral quantities, in the four cafes of this problem, are eafily found from the general property of the curve, viz. dd: cc :: (d+x)x : yy. The demonftration of which, together with that of the preceding problem, belongs to conic fections. CASE II. To find the Abfciffes. As the conjugate : Is to the diameter :: So is the root of the fum of the fquares of the ordinate and femi-conjugate: To the distance between the foot of the ordinate and the center. Then, to and from this diftance, add and fubtract the femi-diameter, and the fum and difference will give the two abfciffes. abfcifs, according as the upper or under fign is ufed. EXAMPLE. The diameter and its conjugate being 24 and 21, required the two abfciffes to the ordinate 14. Here ✔ce+yy 24√ × 212 + 142 12 = 4/212 +282 7 21 2 ± 12 = 4√32 + 42± 12 = 20 1232 and 8, the two abfciffes required. CASE 111. To find the Conjugate. As the mean propor. between the abfciffes: Is to the ordinate: : EXAMPLE. What is the conjugate to the diameter 24, the lefs abfcifs being 8, and its ordinate 14 To or from the root of the fum of the fquares of the ordinate and femi-conjugate, add or fubtract the femi-conjugate, according as the lefs or greater abfcifs is used: Then, as the fquare of the ordinate, is to the product of the abfcifs and conjugate; fo is the fum or difference, above found, to the di The lefs abfcifs being 8, and its ordinate 14, required the diameter, fupponing the conjugate to = 3 × √212 +282 +21= 3 × (√ 32 + 42+3)= 7 3 × (5+3)=24, the diameter required. PRO PROBLEM III. Having given any Two Abfciffes x, x, and their Ordinates x,y; to find the Diameter d to which they belong. Multiply each abfcifs by the fquare of the ordinate belonging to the other; multiply alfo the fquare of each abfcifs by the fquare 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 abfciffes be 1 and 8, and their two ordinates 4 and 14; what is the diameter to which By the nature of the hyperbola, dx + xx : dx + xx :: YY: yy, or dxyy + xxyy = dxxy + xxxy; hence d = PROBLEM IV. Given any Three Equidiftant Ordinates a, b, c, with their Common Diflance, or Common Difference d of their Abfciffes; to find the Diameter D to which they belong, together with its Conjugate c. * From the fum of the fquares of the extreme ordinates, take double the fquare of the middle one; and call the difference A. From the fquare of the difference between A and double the faid fquare of the middle ordinate, take 4 times the fquare of the product of the extreme ordinates; and call the fquare 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. If x be the abfcifs to the leaft ordinate a, by the nature of the hyperbola we shall have these three equations, CC (D x + xx + 2 dx + Dd + dd), CC=1 X (D+x+2d) x (x + 2d) = DD (Dx + xx + 4dx + 20d + 4dd). And by fubtracting the double of the fecond from the fum of the firft and third, we obtain this equation, Again, taking the first equation from the fecond, leaves bb-aa CC DD DD bb-aa D+d == × (2dx+od+ dd); hence x = X CC |