1J.S7 



of u and v from the equations 



Oil Ov 



as appears e.g. at once from the example : 



,,.= _|_ y--! __ /.,2 _ 2^.,. _2by = R^ — 2a' — 2//^ 



representing- a system of hyperboloids of revolution with \ertieal axes. 



df df df 



If ;^ , :;^ , :f- are to be =0, then .v = a, i/ = //, c = ()an(l 



O.v 0// Oz 



if this point will lie on f=:0, then a' -\- Jf must be :^ R'. 

 The equation then assumes the form : 



{.v—ay + (>f-by — kz':=o, 



and consequently represents now a cone of revolution with its vertex 

 in -r. = a, y = h, z ^=iO, and the locus of those vertices is the circle 

 x' +/ = R'. 



On the other hand we find easily that j& = is represented here by 

 .^.2 + v/^ — 2kz' = 2K'; 

 it appears at once therefore that the circle x" -\- if = R"^ does not 

 lie on E here. This is easy to understand. If we consider x as 

 independent variable, then y, z, ii, v become in consequence of the 

 nodal locus functions of x (cf. the analogous reasoning in § 2) and 

 as these functions must satisfy the equation /'=0, we can write: 



0/ 0/ dy df dz ^ df du df dv 



d.t? dy da: ' dz dx du dx dv dx 



df df df 

 Now for a point of the curve ^ , ^^ , ^ =; 0, so that remains: 

 ^ dx dy dz 



df du df dv 



^ H / — = 0, 



Oil dx Ov dx 



0/ df 

 from which of course it need not ensue that -- = ^ = 0, thoui>h 



du dv " 



on the other hand this is not impossible. 



That, however, ~ =: ~ = is a particular, and not the general 



du Ov ° 



case, appears as follows (cf. § 2). 

 By elimination of x, y, z from 



df df df 



dx dy dz 



we find a function y {ii, v) of a and v, which becomes zero for 

 these systems of values of ii. and v, which determine an integral 

 surface with a node. Conversely an infinite number of functions 



