SOLUTIONS OF HOMOGENEOUS AND CENTRAL EQUATIONS. 1049 



3. Solution of Equations consisting of two Homogeneous parts of Different 



Degrees. (Case II.) 



These equations represent a class of central curves or surfaces essentially different 

 from the preceding. In the case of equations of even degree the curve F n (x, y) = A n 

 cannot pass through the centre ; while curves which are of the form F n (x, y) = ¥ n _ p (x, y), 

 when reduced to the lowest terms, generally pass through the centre, because their 

 equations are satisfied by the values x = ; y = 0. 



Equations of the form F n (x, y) = F n _ P (x, y) are really homogeneous, or, at least, are 

 reducible to homogeneous form by division. 



Take, for example, the equation 



Ax b + Exy* = Q(H.x 3 + Ky 3 ). 



By division we have 



Ax 5 +Exy i 



H.x 3 + Ky 



= Q> 



where the left-hand side is homogeneous, and of the 2nd degree. Consequently, by the 

 results of the preceding analysis (sections 1 and 2), a solution is 



/ Q(Ha 3 +K6 3 n 

 x - a W Aa 5 +Ea¥ I 



*v 



Q(Ha 3 + K& 3 ) 

 Aa & + Ea¥ 



where a and b are any quantities whatever. This may be directly verified as follows : — 

 Calling the expression under the root-sign <u and substituting, we have 



and 



Hence 



Ax 5 + Exy* = Aa 5 ( Jwf + Ea¥( Jwf 

 = (Atf + EaV)(Ja>Y 



Q(Hx 3 +Ky 3 ) = Q(Ha 3 ( Jwf + Kb% Jw) 3 ) 



= Q(Ha 3 +K6 3 )(» 3 . 



Aaf + Exy* (Aa b +Ea¥)( Jwf 

 Q(Rx 3 + Ky 3 ) Q(Ha 3 + K& 3 )( Jw) 3 



Aa 5 + Ea¥ 

 ~Q(Ha 3 +K6 3 ) ' w 



Atf + EaV Q(Ha 3 +K6 3 ) 

 ~ Q(Ha 3 + Kb 3 ) ' Aa 5 + EaV 



— 1 , as it should be. 

 This quasi-extension of the original theorem may be formulated as follows 



