Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/156

 vh and ab are equal. But becaue of the imilar triangles VSH, vsh, VH is to SH as vh to sh; that is, the axe of the conic ection now decribed is to the ditance of its foci, as the axe ab to the ditance of the foci s, h; and therefore the figure now decribed is imilar to the figure aph. But, becaue the triangle PSH is imilar to the triangle psh, this figure paes through the point P, and becaue VH is equal to its axis, and VS is perpendicularly biected by the right line TR, the aid figure touches the right line TR. Q. E. F.

From three given point: to draw to a fourth point that is not given three right lines whoe differences hall be either given or none at all.

. Let the given points be A, B, C (Pl. 8. Fig. 1.) and Z the fourth point which we are to find; becaue of the given difference of the lines AZ, BZ, the locus of the point Z will be an hyperbola. whoe foci are A and B, and whoe principal axe is the given difference. Let that axe be MMN. Taking PM to MA, as MN is to AB, erect PR perpendicular to AB, and let all ZR perpendicular to PR; then, from the nature of the hyperbola, ZR will be to AZ as MN is to AB. And by the like argument, the locus of the point Z will be another hyperbola, whoe foci are A, C, and whoe principal axe is the difference between AZ and CZ; and QS a perpendicular on AC may be drawm to which (QS) if from any point Z of this hyperbola