[■ 28 s ] 
Hence = ”i±i‘ From 
•Xg Xg, • I 
which equation, by taking the fluxions, we have, 
. tt 
XX- — + - 
'^S xg X rn — «"P — %Xm^ -\-n - xt^->ct‘^ 
tt 
m + n^Xtt — ft T > 4. ■ • 7 >r — fj 
-T — - ■■■ . But js beme = ^ 
i2 Hi " — •*■ 
: — n* ■ — t Xm4-n \ — t i 
7/i — gx‘ 
X '• iV n 
zg X m — n* - — f X m-\-nf — fi' 
as obferved in the preceding article, it appears that 
^ X .vi is = z. It is obvious, therefore, that ^ 
z IS ^ + 4 X .. 
W* + « X:t ft 
m — nf — f X m -f nf — f f" ' 
m — »)* X i — ft , 1 X i — ft 
-:.y ^ 4 X 
rx + 4 ^ 
m — nf — t xm-i-n 
I 
#2J2 
m — — f X — f'^ 
_ I -A I I 
= T ^ T X . 
m — n\ — t 
m-fra 
1 , 1 m-\r‘nf — f 
X f + X X 
}H «• 1 
x/. From whence 
taking the fluents by the theorems in art. i. and 2 . we 
havea = ae(fig. 3.) =4x+ °’’ ^^° (fig- i-) +7 (%• .2-) 
confequently the hyperbolic arc ADis=DP + ae+2/~4ae. . 
Thus, beyond my expe6tation, I And, that the hyperbola . 
may in general be re6lified by means of two ellipfes. .. 
Writing e and f for the quadrantal arcs ad, ad, (fig» 
2. and 3.) refpedlively, and l for the limit of the differ-^ 
ence dp- ad, whilft the point of conta61 (d) is fuppofed 
to be carried to an infinite diftance from the vertex a of 
the hyperbola (fig. i .), we find.* 
2 F-E = L, 
the value of. ae being =yF when t \ : 
6 . that:x 
