( 530 ) 
the Equation is x % ■+• y i = ? n x y, in which the two 
Terms xx y -f x yy of the former Equation are wanting ; 
and its njymptote isdiftant from B by ~ B A. Again draw 
E F perpendicular to A B : let B F be called z and F E 
v ; the Equation belonging to the Curve A E B is vv 
a z z — z 3 . azz —z) 
. In the Foliate the Equation is vv — — 
a \ z a 3 z 
From thefe two laft Equations it feems that thefe Curves 
differ no more from one another than the Circle from the 
Ellipfis. I fhould be very glad to know your Opinion 
thereupon 
The Quadrature of the Curve here deferibed has fome- 
thing of Simplicity with which I was well pleafed. With 
the Radius B A and Center B deferibe a Circle AEG, let 
the Square H P ST circumfcribe it, Co that HP be pa- 
rallel to AG: prolong FE till it meat the Circumference 
of the Circle in M , and through M draw L y!/ ^parallel to 
H P. The Area B F E is equal to the Area K H L M, 
comprehended by 1(H, H L, L M and the Arc K M. 
And the Area Bfe is equal to the Area KmLH or KMPQ^ 
Therefore if B /'’and B f are equal, the two Areas B FE, 
Bfe taken together are equal to the Redangle H 
and therefore the whole Space comprehended by 
B E A X B cT G Z (fuppofing T and Z to be at an infi- 
nite Diflance) is equal to the circumfcribed Square FIS. 
N. B. This Quadrature is eajily demonfirated from the Equa- 
tion : for by it a ~Kz: a — z : : ZZ : V V, that is A F : E F :: 
M F : F B, and fo <p F the Fluxion of A F to L 1 the Fluxion 
of M F. Hence the Areola E F <p e will be always equal to 
the Areola M L I and therefore the Area A E F always 
equal to the Area M A L. 
Hence it appears that this Curve requires the Quadrature 
of the Circle to fquare it ; whereas the Foliate is exaclly qua - 
dr able, the whole Leaf thereof being but one third of the Square 
of A B, which in this is above three [evenths of the fame* Again 
in 
