290 ON THE REDUCTION OF THE GENERAL EQUATION. 



(the 'transverse' and 'conjugate' axes), a point about "which the curve 

 is symmetrical, or the 'centre.' Thus 

 x" = R cos B — K sin 6 



]__ j(a+b+m) (dcos2 B + e sin 0cos0) \ 



( a 6 — C 2~) 1 +{a+b — wi) (J mxiQ—c cos sinfl) | 



- S (a + i) J+wd (— — ) + 2»ie(--) l 



- 2 ) ( m »» J 



2 (ab — c2) 

 ce — id 



. 1 



ai - ca L (10) 



> _ cd—ae 

 V — ab—d " J 



which are the usual expressions. 



To find the points in which the curve is cut by the transverse axis, 

 whose equation is 



x — h y — k 

 cos sin ' 



substitute these values of x and y in the original equation, and we 

 obtain 



r* = e 2 (A cos + k tin 6 — r —pY 

 and the two values of r are expressed by 



+ _ (h cos + k sin — p) . 



The difference of these values is the part of the transverse axis 

 intercepted by the curve : calling this length 2 A we have 



A * = (1-/2)2 (hcond + ksinB—p)' (11) 



now (4) x cos + (5) x sin 6 gives 



h cos 6 + k sin 6 — p e 2 = — A (d cos 6 + e sin 0) . . (12) 

 and (4) 2 + (5) 2 —(6) gives 



— 2 j9€ 2 (/*cos04-£sin6)+^ 2 e 2 (l + t 2 )=A 2 (d 2 + e 2 )— a/. ..(13) 

 Hence by means of (12) 2 x c 2 + (13) x fl-e 2 ) we have 

 e 2 (7i cos (9 4- k sin — ^>) 2 



= X 2 ( e 2 (J cos + esintf) 2 + (1— e 2 ) (J 2 + e 8 — ^ ) 1 



esA 2 Jc? 2 (l-e 2 sm 2 0)+e 2 (l-e 2 cos 2 0) + 2Jee 2 sin0cos0 — ~^/ ( 



= A 2 | A IcP + \ ae* —2\cde - ^^- f X 



— A 3 | oc 2 + Id* — 2 c<fe — (ab—c*)f I 

 and, hence, 



4 * = i- iS~^A ae2+M2 - 2cde - {ah - c2 ^ \ 



