Foats — Isogonic Transformation, 45 



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



. ^T^nn ^^ ^ R VI- sin a 



tan ^PQO — -^=r-^ = — = , r 



OQ a i? 1/1 + sin a 



1 — tan^ e 

 1 + tan2 e 



Hence 

 and 



therefore 



1 — tan^ e 

 \ ^ "^ 1 + tan2 



— tan S. 



^iPQO = e 



POM = ^OPM = 90° — (9 

 OM = MP = MQ = P'M; 



^POB = 45=* 



The vertices may consequently be determined at once as 

 well as the position and length of the axes. 



To rectify this ellipse 



3 e* 32.5 e« 



L - 27ra (^1 — 22 ~" 22T42 ~" 22.42.62 ~ J 



in which 



„ 2 sin a 



1 + sin a 

 and 



a ^ R V\ + sin a ; 

 hence 



Z=27ri2l/l + sina(l-J ^^"^ ^ 



2 (1+sina) 

 2 a 5 sin^ a 



When 



o sm' a o sm" a \ 



""16 * (l+sina)2"~32 * (l + sina)^" ) 



a = 0° L — 2'7rR Circle Maximum perimeter, 



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



Area of the ellipse 



A = irab = TT . i?l/l + sin a . P\/l — sin a 

 = 7ri22 cos a 



