Poats — Isogonic Transformation. 45 



To determine the vertices of the ellipse we have (Fig. 2) 



* /P nn P0 b - R VX - sin a 

 tan ^.PQO = -^ryr = — = — — / , . " 



OQ a R l/l + sin a 



1 — tan 2 

 1 + tan 2 



4 



1 — tan 2 

 1 + 1 + tan 2 



Hence 

 and 



therefore 



= tan 0. 



^LPQO = 



^.POM = ^.OPM = 90° — 

 OM = MP = MQ = P'M; 



^LP'OB = 45° 



The vertices may consequently be determined at once as 

 well as the position and length of the axes. 



To rectify this ellipse 



/ e 2 3e« 3 2 -5 e 8 \ 



L — 2-ira (1 — g 2 — 22742"" 2 2 -4 2 -6 2 / 



in which 



„ 2 sin a 



e 2 = 



1 + sin a 



and 



a — R Vl + sin a ; 

 hence 



/ 1 sin a 



X = 2ttR 1/1 + sin a (1 - g • (1+sina 



) 

 sin 2 a 5 sin 3 a 



_....) 



16 ' (1+sina) 2 32'(l + sina) 3 

 When 

 a = 0° Z = 2tt.R Circle Maximum perimeter, 



a = 90° i = 4 i?l/2 Straight line Minimum perimeter. 



Area of the ellipse 



A = irab = it . i?l/l+ sin a . i?l/l — sin a 

 = rrrR 2 cos a 



