Ad = 1 - 1 ⁄ m (supposing Bd to be taken, from B towards A ,) to AB = 1, or as m - 1 to m : so is BS = 1 ⁄ m DH , to dh , which is therefore equal to 1 / m - 1 of DH ; that is (as will appear by dividing 1, by m - 1,) to 1 ⁄ m + 1 ⁄ mm + 1 ⁄ m 3, &c. of DH .[15]
Or if Bd be taken beyond B; then as Ad = 1 + 1⁄m to AB = 1, or as m + 1 to m, so is 1⁄m DH to dh, which is therefore equal to
1 / m + 1DH; that is (as will appear by like dividing of 1 by m + 1;) = to 1⁄m - 1⁄mm + 1⁄m3 - &c. of DH.
- Let such ordinate dh, or (equal to it in the Asymptote) AF, be so divided in L, M, N, &c. (by Perpendiculars cutting the Hyperbola in l, m, n, &c.) as that FL, LM, MN, be as 1⁄m, 1⁄mm, 1⁄m3, &c. That is, so continually decreasing as that each Antecedent be to its Consequent, as 1 to 1⁄m, or as m to 1. See Fig. 5. Tab. 5.
- This is done by taking AF, AL, AN, &c. in such proportion. For, of continual Proportionals, the Differences are also continually proportional, and in the same proportion. For let A, B, C, D, &c. be such Proportionals, and their Differences a, b, c, &c. That is, A - B = a, B - C = b, C - D = c, &c.
Then, because A, B, C, D, &c. are in