Poats — Isogonic Transformation. 45 



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



tan ^PQO = P ° b R VT =^~' 



OQ a R i/l + sin a 



1 — tan 2 

 l + tan 2 



Hence 

 and 



therefore 



1 — tan 2 

 \| 1 + 1 + tan 2 



— tan 0. 



^.pqo = e 



POM = ^LOPM = 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 3 e 4 3' 4 -0 e D 



I 1 """ 2* "" 2 2 ^4 2 ~~ 2 2 -4 2 -6 2 ~~ ) 



i 2 3 e i 3 2 -5 e 6 

 L = 2-rra 1 — ^ — ^-ns — 



in which 



2 sin a 



e 2 = 



1 + sin a 

 and 



a — R V\ + sin a ; 

 hence 



y /. 1 sin a 



£ = %«R 1/1 + 8 in . (1 - j • (1+siDa) 



3 sin 2 a 5 sin 3 a \ 



~~ 16 ' (1+sina) 2 "" 32 ' (1 + sina) 3 ) 



When 



a = 0° Z = 2iri? Circle Maximum perimeter. 



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



Area of the ellipse 



A — Trab = 7r . RV\ + sin a . i?l/l — sin a 



— ttR 2 cos a 



