322 Prof. G. H. Darwin. On Jacobis Figure of [Nov. 25, 



And since when a. vanishes, /3 also vanishes, the equation 



L -M+ N cos 2 7 tan 2 /3=0 



is satisfied by a = 0, fi=Q. 



That is to say there is a solution of the problem which makes 

 a=b. 



Thus there is solution which gives us an ellipsoid of revolution. 

 When sina=0, we have also /3=0, A = l, and 



L= 



2-n- cos 7 

 sin 3 7 J 



sin 2 7 cfry— 



tt cos 7 , . N 

 ■ (sm 7 cos 7 — 7) , 

 sm d 7 



2V= 



27TCOS 7 



sin 3 7 j 



tan 2 7 ^7= 7rC °f ^ (27—2 tan 7). 



sm3 7 



Therefore 



TT 



— [27 — 2 tan 7— (1 + tan 2 7)(sin 7 cos 7 — 7)]; 



tan°7 • 



= W7 [7(3 + tan27) " 3tan7] ' 



(14)* 



and the eccentricity of the ellipsoid of revolution is sin 7. 



To find the other solution when a. is not zero, we have by com- 

 parison between (10) and (11), 



L . 



M . 



sin 2 a sm 3 rv ^ „ 



— = E — F 



27r cos j3 cos 7 



1 



sin 2 a sin 3 7 sin 2 a sin 7 cos 7 • _ 



— ^F-Esec 2 *, )> . (15) 



27r cos (3 cos 7 cos 2 a cos /3 



sin 2 asin 3 7 _ 

 N . - — — =tan 2 *(E —tan 7 cos /3). 



Z7r cos j3 cos 7 

 Hence the first of (12) gives 

 -(2F-E)+Esec 2 * 



j. o n ^ sin 2 a sin7 cos 7 ' 



+ tan 2 a tan 2 /3 cos 2 7 (E — tan 7 cos 0) = — 1 = 0, 



v • cos 2 «cos/3 



* Compare with Thomson and Tait's ' Nat. Phil.,' § 771, (3) ; or any other work 

 which gives a solution of the problem. 



