OF ARTS AND SCIENCES. 247 



r 



Cut this by a plane perpendicular to VAjU, that is, by the plane 



SlfiQ = c', 

 and we have 



SXf)S{iQ = c", 



which is an hyperbola. If now the surface be cut by planes perpendic- 

 ular to one of the other axes Q. -\- ft) or (X — jw), the planes 



our equation gives 



Sq{X + jm) = 



Sc)(;i — ^) = 



± S'Xq = SiXnQ -\- c. 



And these are both parabolas. These equations for paraboloids can be 

 verified by substituting in the cyclic forms the value of X and /x in 

 terms of c^, c^, and Cg, «p a^, and «„, given in Hamilton's Elements of 

 Quaternions, § 357, XX., and seq. 



In the general equation in 5, for finding the centre 



{g — SAju)5 -\- XSfid -\- [iSX8 = y, 



if Cj and c^ both vanish, and 



y = di -\- hk, 

 we have 



— TX^i =z SXfx ; 



whence 1 = cos < 



X is then parallel to ft, and therefore 



2XSX8 = di + hk. 



The centre must be at a finite distance in the direction of X, which is the 

 same as that of ^, at an infinite distance in the direction of k, and inde- 

 terminate in that of j ; since we may add xj to d without affecting the 

 equation. The surface must be a parabolic cylinder. Now, if h also 

 vanish, k will be in the same condition asj, and d will be anywhere in 

 a certain plane perpendicular to i. Let us study this case a moment, 

 for it is the simplest of the non-central quadrics under the cyclic form. 

 The general cyclic equation of the second degree 



(g — SXixy/ + 2 SXoSiiQ = 2 SyQ + c, 



in this case, assumes the form 



S~Xp = ^S^^ -j- c. 



