ii 7 
of the Hyperbola. 
Moreover ; since by the notation in Art. 5 and 6, s e is = 1 , 
and V == ux , we have, by substituting these values for their 
equals in the last equation, H = zu -j- e (u x 4- E) — 2 ( 1 + ^) 
S E, (£) which is Landen's theorem in a different notation. 
8. I am now to prove, that all the labour of computing the 
eccentricity* and abscissa, and the arch itself, of this second 
ellipsis, and the subsequent operations of multiplication and 
subtraction requisite in the application of it to the rectification 
of the hyperbola, is altogether needless ; since the same end 
may be obtained by only computing and applying the fluent 
of ^ ki)’ will require no more calculation than 
must be made to find and apply the elliptic arch denoted by 
E (of which the axis, and eccentricity, and abscissa, are given). 
And the truth of this will quickly appear. For, 
9. The fractional expression, 
U y ( I EE till) 
( I — -se) U 
V ( J —uu) y(i— - uu) /(e— EE uu) j 
found above, in the equation marked (a), (see Art. 6 .) by 
reduction to a common denominator, becomes 
U-EEUUU~-U-\-'cEU ___ E£«(l— -uu) £S It y ( — •UU.) 
V(i— uu) x Vi 1 — E£ uu ) VC 1 — x y(i — ee uu) ~~ y(i~Eenu° 
Substitute this for its equal in the aforesaid equation, and we 
have V = (y)- Take the fluents, denoting 
that of by G, and there will be V = ss G 4 - — ; and 
V(i —auu) J y 11 e 5 
thence, by transposition and multiplication, H = e V— e G. (<£). 
This theorem, as far as I know, is new. 
Here then it appears, that the rectification of the hyper- 
bola is accomplished by means of the algebraic quantity e V, 
* When it is more convenient to use the conjugate semi-axis than the eccentricity, 
in the arithmetical calculation, then that must be computed instead of the eccentricity. 
