(46 ) 



x = 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 

 r = 2,37 2,73 3,08 3,41 3,73 4,03 4,33 4,59 4,86 5,10 5,33 



6. We may now determine, where the transition represented in 

 fig. 4, takes place. (The place of the point P is also drawn in 

 figs. 1 and 2 ')). 



If we put 1 — (J^=^y in the equation {2b) of the plaitpoint curve, 

 then 



(l-2^-) + (.^- + 9^)y'f3 + (.r + g.r^-^?^) = 0. ... (a) 



V .t'(i— .f)y 



Now in the double point sought ^- must be and ^— must be 0, 



^ ox dy 



when ƒ denotes the first member of {a). This gives: 



_ 2x {\-x) + {l-2xy + 3^^ }(l-2.tO {x-\-ip) + X (l-.t-) j + 



+ 8 (.^' -f y^r 2/' (3I/-2) = , (6) 



and after division by 6y {x -\- (f) : 



x{\-x) + {x + <fyy{2y-l) = ^ (c) 



Substitution of the value of x{l — x) from (c) in {a) gives: 



(I - 2x) + {x + if) y^ U + ^^^) = 0, 

 or 



(l_2.r) + (.. + 9:)3/^i:i|^=:0 (a') 



1 — 2?/ 



So Ave have to solve ?/, a' and tp from (a'), (/;) and (c). Substitution 



of 1 — 2a; from (a'), and .t (1 — x) from (c) in {b) gives, after 



division by [x -\- (pYy : 



/I — ?>ii\ 

 -2(l-2,) + ,'(— -^j + 



+ %' !-3/^^ + (l-%)i + 3^M3.V - 2) = 0, 

 / 1 — 2?/ ) 



i. e. after multiplication by (1 — '2yY -. 



- 2 (1 - 2^)' + ^' (1 - ^yY + 



+ 33/MI-22/) j-2/(l-3^) + (l -2yrU3r(3^-2)(l-2^)=' = 0, 



from which y may be solved. The above equation gives: 



^) This point must be thought more to the left. In fig. 4 no contact but inter- 

 section takes place in the double point P. 



