450 Prof. Cay ley on Differential Equations and Umbilici. 

 that is 



z*-2zx*/¥=F- (a 2 -b 2 )y 2 ~z{x 2 + y 2 ) = > 

 so that there is a tangent cone the equation whereof is 



z 2 —2zx\/a T ^¥-~ (d l - b*)y* = 0, 

 or, as it might be written, 



(z-xy/d^ 2 ) 2 -{a 2 --b 2 ){x 2 ±y 2 ) = 0. 

 The equation is that of a cone of the second order, meeting 

 the plane of zx in the lines z = 0, z=z2x\Za 2 —b 2 (and therefore 

 such that its sections parallel to the plane of xy are parabolas), 

 and meeting the plane of yz in the lines z — + y\/a 2 — b 2 ( the 

 origin being at the vertex of the cone or conical point of the 

 surface). 



Returning to the original origin, and to the equation of the 

 surface written in the form 



z 2 + z{a 2 + b 2 -x 2 -y 2 ) + a 2 b 2 -b 2 x 2 -a 2 y 2 =0, 



calling this for a moment z 2 + 2Bz + C = 0, the differential 

 equation is C' 2 -4BB'C' + 4CB' 2 = Q; or, substituting, this is 



{b 2 x + a 2 yp) 2 -(a 2 + b 2 -x 2 -y 2 ){x + yp)(b 2 x + a 2 yp) 

 + (d 2 b 2 -b 2 x 2 -a 2 y 2 ){x + yp) 2 =z0 ; 

 or, reducing, this is 



(a 2 -b 2 )xy{xy{p 2 -l)-(a 2 -b 2 — a. + y 2 )p}=0, 



or say 



xy{xy(p 2 -l)-{a 2 -b 2 -x 2 + y 2 )p}=0, 



where the factor xy arises from the level lines (z + b 2 = 0, y = 0) 

 and (z + a 2 =0, x=0). Throwing out this factor, the equation 

 becomes 



xy(p 2 -l)-(a 2 -b 2 -x 2 +y 2 )p=0 ) 



which is satisfied identically by z + b 2 = 0, y=0, a? 2 =« 2 — b 2 . 

 The first derived equation is 



(xp+y)(p 2 -l)+2(x-yp)p=0, 



which for the values in question gives 



P(P 2 +i)=o, 



where the factor p=0 corresponds to the section y=0 by the 

 plane z + b 2 =0: and taking the conical point for origin, and 

 observing that the polar of the line a?=0, y = in regard to the 

 tangent cone is z—x\/a 2 —b 2 = 0j then writing the equation of 

 the tangent cone in the form 



(z-x\/a 2 -b 2 ) 2 -(a 2 -b 2 )(x 2 +y 2 )=0 } 



