THE CURVE ON ONE OF THE COORDINATE PLANES. 

 These coefficients A , A 1} &c, may now be deduced from the expressions, 



+ {t-2¥)(^^[b\a 2 +u 2 ) + tf . 



B'=t(L 2 +ty 



+ 4a 2 & 2 [6 2 (J 1 - u 2 ) - t](L 2 + 1) 

 + 8a\t-2b i )(b 2 -u 2 )(L + L x + 1) 



C'-jV(^-^ 2 )-^ + (^)(L + 2 



B' = (L. 2 + tf-8a 2 (b 2 -u 2 )(L + L 1 + t 2 ) , 



L = b 2 (a 2 +v 2 ) 

 L!=u 2 (6u 2 -4y 



L. 2 = 2a 2 b 2 + L +L 1 



= 3a 2 6 2 - u 2 (4a 2 + 76 2 ) + 6u 4 . 



481 



where 



The factor 



su 6 _ y 2 z 2 u® 



is now of dimension zero. 



Considering s = y 2 z 2 in itself we may transform it in referring p to two axes /, ¥, of 



which / is the bissecting line of the angle between j, h 



Having then 



p=jy+1cz=j'y'+k'z\ 

 we get 



y'~ z ' 1 , , . >\ 



y =J2~> Z =T2 iy+Z) ' 

 and 



We put 



y'=r cosS 



z'=r sin (£ 



