L x Pv will be to Gv x Pv as L to Gv; and (by
the properties of the conic ſections) the rectangle GvP
is to as to ; and (by cor. 2. lem. 7.)
to , the points Q and P coinciding, becomes
a ratio of equality; and Qx or is to
as to , that is, as to ,
or (by lem. 12.) aſ to : and, compounding
all thoſe ratio's together, we ſhall have L x QR to
as or to , or as 2PC to Gv.
But the points P and Q coinciding, 2PC and Gv
are equal. And therefore the quantities L x QR
and , proportional to them, will be alſo equal. Q. E. I.
Let thoſe equalſbe drawn into, and we ſhall
have to . And therefore (by cor. 1 & 5. prop. 6.) the centripetal force is reciprocally
as , that is, reciprocally in the
duplicate ratio of the diſtance SP. Q. E. I.
Find out the force tending from the centre C of the
hyperbola. This will be proportional to the diſtance
CP. But from thence (by cor. 3. prop. 7.) the
force tending to the focus S will be as , that is,
becauſe PE is given, reciprocally as. Q. E. I.
And the ſame way it may be demonſŧrated, that the
body having its centripetal changed into a centrifugal
force, will move in the conjugate hyperbola.