Poats — Isogonic Transformation. 



49 



we have 



b 2 x 2 + 26 2 x 1 « 1 tan a -f b 2 z 2 tan 2 a 



+ b 2 z 2 se ° 2 a + a2 2/i 2 = ° 263 - ( 10 ) 



Again transforming by means of 



x x = z sin + x cos 



& = y 



z 1 = z cos — x sin 

 equation (10) becomes 



b 2 x 2 



+ cos 2 6 



— 2 tan a sin 6 cos 



+ sin 2 6 



+ 2 tan 2 a sin 2 



+ aV + 6V 



+ sin 2 6 



4- 2 tan a sin cos 



+ cos 2 



+ 2 tan 2 a cos 2 



+ 2b 2 xz 



+ sin cos 

 + tan a cos 2 



— tan a sin 2 



— sin cos 



— 2 tan 2 a sin cos 



= a 2 b 2 . 



From the coefficient of xz in (11) it is seen that 



(11) 



and 



1 — sin a 

 Equation ( 11 ) now becomes 



+ a 2 y 2 _|_ ^ hH ] _ = a 2 b 2 



1 -f- sin a 



1 — sin a 



or 



L ^ 



a 2 (l + sina) b 2 a 2 ( 1 — sin a 



= 1 



(12) 



