The Mathematical Principles of Natural Philosophy, Book III: Of the System of the World
Lemma VIII
Let ABC be a parabola, having its focus in S. By the chord AC bisected in I cut off the segment ABCI, whose diameter is and vertex. In produced take equal to one half of. Join OS, and produce it to as may be equal to 2SO. Now, supposing a comet to revolve in the arc CBA, draw, cutting AC in E; I say, the point E will cut off from the chord AC the segment AE, nearly proportional to the time.
‸For if we join EO, cutting the parabolic arc ABC in Y, and draw touching the same arc in the vertex, and meeting EO in X, the curvilinear area will be to the curvilinear area as AE to AC; and, therefore, since the triangle ASE is to the triangle ASC in the same proportion, the whole area will be to the whole area as AE to AC. But, because is to SO as 3 to 1, and EO to XO in the same proportion, SX will be parallel to EB; and, therefore, joining BX, the triangle SEB will be equal to the triangle XEB. Wherefore if to the area we add the triangle EXB, and from the sum subduct the triangle SEB, there will remain the area, equal to the area; and therefore in proportion to the area as AE to AC. But the area is nearly equal to the area; and this area is to the area as the time of description of the arc AB to the time of description of the whole arc AC; and, therefore, AE is to AC nearly in the proportion of the times. Q.E.D.
COR. When the point B falls upon the vertex of the parabola, AE is to AC accurately in the proportion of the times.
If we join cutting AC in; and in it take in proportion to as 27MI to, and draw Bn, this Bn will cut the chord AC, in the proportion of the times, more accurately than before; but the point n is to be taken beyond or on this side the point, according as the point B is more or less distant from the principal vertex of the parabola than the point.