Let any centripetal force tend to the centre C, and let it be required to find the trajectory VIKk. Let there be given the circle VR, described from the centre C with any interval CV; and from the same centre describe any other circles ID, KE cutting the trajectory in I and K, and the right line CV in D and E. Then draw the right line CNIX cutting the circles KE, VR in N and X, and the right line CKY meeting the circle VR in Y. Let the points I and K he indefinitely near; and let the body go on from V through I and K to k; and let the point A be the place from whence another body is to fall, so as in the place D to acquire a velocity equal to the velocity of the first body in I. And things remaining as in Prop. XXXIX, the lineola IK, described in the least given time[Pg 170] will be as the velocity, and therefore as the right line whose square is equal to the area ABFD, and the triangle ICK proportional to the time will be given, and therefore KN will be reciprocally as the altitude IC; that is (if there be given any quantity Q, and the altitude IC be called A), as . This quantity call Z, and suppose the magnitude of Q to be such that in some case may be to Z as IK to KN, and then in all cases will be to Z as IK to KN, and ABFD to ZZ as IK2 to KN2, and by division ABFD - ZZ to ZZ as IN2 to KN2, and therefore to Z, or as IN to KN; and therefore A × KN will be equal to . Therefore since YX × XC is to A × KN as CX2, to AA, the rectangle XY × XC will be equal to . Therefore in the perpendicular DF let there be taken continually Db, Dc equal to , respectively, and let the curve lines ab, ac, the foci of the points b and c, be described: and from the point V let the perpendicular Va be erected to the line AC, cutting off the curvilinear areas VDba, VDca, and let the ordinates Ez, Ex, be erected also. Then because the rectangle Db × IN or DbzE is equal to half the rectangle A × KN, or to the triangle ICK; and the rectangle Dc × IN or DcxE is equal to half the rectangle YX × XC, or to the triangle XCY; that is, because the nascent particles DbzE, ICK of the areas VDba, VIC are always equal; and the nascent particles DcxE, XCY of the areas VDca, VCX are always equal: therefore the generated area VDba will be equal to the generated area VIC, and therefore proportional to the time; and the generated area VDca is equal to the generated sector VCX. If, therefore, any time be given during which the body has been moving from V, there will be also given the area proportional to it VDba; and thence will be given the altitude of the body CD or CI; and the area VDca, and the sector VCX equal thereto, together with its angle VCI. But the angle VCI, and the altitude CI being given, there is also given the place I, in which the body will be found at the end of that time. Q.E.I.