20 Prof. Cayley on a Quartic Surface. 



which, considering z as a constant, is of the form 



(# 2 +2/ 2 -a) 2 =16A(#-m) ; 



that is, the section of the surface by a plane parallel to the plane 

 of the conic is a Cartesian. 



If a and b are unequal, but if we still have /3=0, the equation 

 of the surface is 



(*« + y* + ^-«8- 7 «)* = 4fl 2 (ar-a)« + 4Ay. 



There are here two planes parallel to the plane of the conic, each 

 of them meeting the surface in a pair of circles. In fact, writing 

 as q + y' 2 =p, and therefore also y <i =p—x 2 i putting moreover 

 z*— a 2 — 7 2 =A, we have 



(p + k)* = 4 flV - 8o?ux + 4« 2 a 2 + 4b 2 (p - x*) ; 

 that is, 



or, as this may also be written, 



(l,4(6 2 -« 2 ), X; 2 -4a 2 a 2 , 4« 2 «, &-2Z/ 2 , Oyj>, so, 1) 2 =0, 



which is of the form 



(«, 5, <?,/,# 0Jj>, as, If -0 1 



and the left-hand side will break up into factors, each of the 

 form p + A# + B (so that, equating either factor to zero, we have 

 p -f A,?? + B = 0, that is, <2? 2 + ?/ 2 + A# + B = 0, the equation of a 

 circle), if only 



obc-ap-bg^O. 



Writing this under the form b{ac— g*)— af 2 = 0, and substi- 

 tuting for «, 6, c, /, # their values, we have 



Z > ==4(Z > 2 --fl 2 ),0c-/=F-4« 2 a 2 -(^~26 2 ) 2 =4(i 2 /:-Z > 4 -A 2 ), 



«/ 2 =16a 4 a 2 , 



and therefore the condition is 



(bZ-a*){b*k-b 4 -aW)-a 4 * o ~=:0 ; 

 that is, 



bZ{(b*-a*){k-b*)-aW\ = 0. 



If Z> 2 = 0, the surface is a pair of spheres; rejecting this factor, 

 we have (Z> 2 — # 2 )(£ 2 — b 2 )— # 2 « 2 =0 ; or putting for k its value, 

 the condition becomes 



(6 2 -a 2 )(^ 2 - a 2 -7 2 -6 2 )-.flV=0; 



that is, for each of the values of z given by this equation, the 

 section by a plane parallel to the plane of the conic will be a 

 pair of circles. 



