LAW OP THE UNIVERSE. 255 



6D 2 fx a\ 3D 2 /2 1 



2 ' 2 



3 D 2 2 1 3D 2 



~^"* Therefore 3 x y - 2 ( x ~ a ) y = ~^s~' 



2 211 



and -, AM X PM = - xy; and - (a: - a)y = - SM . PM = 



o 2 2i 



3 D 2 



S M P ; therefore the sector A S P = ;- : so that the radius 



m 2 



from the focus S cuts off the given area, and therefore P is 



g 



the point where the comet or other body will be found in - 



parts of the time. 



If the point is to be found by computation, we can easily 



1 8 a 2 I) 2 

 find the value of y by a cubic equation, y 3 -j- 3 a*y = - - , 



Ttl 



and making B L = ?/, L P parallel to A M, cuts A P in the 

 point P required. Sir Isaac Newton gives a very elegant 

 solution geometrically by bisecting A S in G, and taking the 

 perpendicular G E to the given area as 3 to 4 A S, or to S B, 

 and then describing a circle with the radius E S ; it cuts the 

 parabola in P, the point required. This solution is infinitely 

 preferable to ours by the hyperbola, except that the demon- 

 stration is not so easy, and the algebraical demonstration far 

 from simple. 



It is further to be observed, that the place being given, 

 either of these solutions enables us to find the time. Thus, 



3 D 2 



in the cubic equation, we have only to find . It is equal 



m 2 



y 3 _i_ 3 a 8 u 

 to - t ; and as D 2 is the given integer, or period of e.g. 



Ct 



half a year, the body comes to the point P in a time which 



bears to D 2 the proportion of unity to -- - _ . 



J 3 2 



