ON THE AXES OF A CoNIC IN TKILINEARS. 391 



» 



On clearing the equation (7) of fractional forms, it will be found to 

 reduce to 



?1 -. ^(^H + K + L) + {KL %m^A + LM sm^B + MN sin^ C) 



= 0. (9) 



•wliich is probably the most simple shape it can assume. 



Cor, Hence we have at once the conditions that the curve may be 

 an ellipse, parabola, or hyperbola respectively, 



> 



KL sin ^A + anal + ... = 0. 



< 



Again, since, if r-^^r^ are the values of r, 



1 KL sin-^ + anal + ... 



p2 J, 2 pi 



the area of the curve when it is an ellipse is 



ttP 



{KL sin2^ + anal + ) ^ • 



The condition that it may represent an equilateral hyperbola is 



H -^ K + L = Q. 

 That it may represent a circle, it is plain that the roots of (7) are 

 equal when the denominators of the fractions are the same ; that is, 

 when 



_H_ _ K _ i 

 sin 2 A ~ sin 2 JS ~ sin 2 C 

 and these conditions are therefore sufficient, but it can easily be 

 proved that these conditions are also necessary, by applying the con- 

 dition of equal roots to the equation (9). Thus, this condition is 



4 {KL kM^-A + anal + ) = (If + Z H- i)^ 



©r, 



iT^ + Z* + Z2 + 2 cos 2 A. KL + 2 cos 2 B.LH+ 2 cos 2 C. HK 



= 0. 

 which is easily seen to be equivalent to 



(fl"cos25 + ircos2^ + i)2 + (a-sin2JB- J5rsin2^)2 =■ 0, 

 which requires 



H ^ K _ L 

 sin 2 A ~ sin 2 5 ~ sin 2 C ' 

 That the conic may break up into two lines not parallel, the con- 

 dition is, since the axes in this case vanish, 



P = 0. 



