and substitutions for tan^r, cos^a from (33) in (32) give the rectangular equation to the 

 spherical hyperbola 



sin^a cos^c 



sin c -sin a 



sin y + sin 



THE POLAR EQUATION OF SPHERICAL HYPERBOLAS WITH ORIGIN AT A FOCUS 



If we choose the given point Q, (0,, A,) of the great circle base line as origin of co- 

 ordinates and a focus, then the following figure may be abstracted from Figure 5: 



(34) 



0(^0, \,) 



Q(e,\) 



The polar radius is now R = aj, /3 is the angle between R and QjQ'. k = QiQ'= S - Sj. From 

 spherical triangle Q2QQ1 we find cos ct, = cos R cos 2c - sin R sin 2 c cos jS , (35) 



and from (27) aj - R = 2a, whence 



cos (aj— R) = cos 01 cos R + sin a^ sin R = cos 2a, (36) 



sin (a, — R) = cos ai sin R + sin ffjcos R = sin 2a. 



Multiply the first of (36) by sin R, the second by cos R and add respective members to 

 solve for 



sin CTi = cos 2a sin R + sin 2a cos R. (37) 



Square and add respective members of (35) and (37) to get 



(cos R cos 2c — sin R sin 2c cosPY + (cos 2a sin R + sin 2a cos R)^ = 1. (38) 



Multiply every term of (38) by sec^R, whence it may be written 



(cos 2c - tan R sin 2c cos PY + (cos 2a tan R + sin 2a)^ = sec^R = 1 + tan^R. (39) 



Expanding (39) and writing as a quadratic in tan R find 

 tan^R (sin^2c cos^^S - sin^ 2a) + 2tan R (sin 2a cos 2a - sin 2c cos 2c cos^) (40) 



+ cos^2c-cos^ 2a = 0. 



32 



