SOLUTIONS OF HOMOGENEOUS AND CENTRAL EQUATIONS. 1067 



the y of equation (2) = (x + y).sin 6, and expand in lines and columns. The expansion 

 of the first term of (2) forms the first line, that of the second term is the second line, 



and so on. 



x i x 3 y x 2 y 2 xy % y* 



Term 1 {1 -4 +6 -4 +1}Dcos0 



Term 1 {1-2 + 2 -1 } E.cos 3 sin 6 



Term 3 {1 -2 +1 } F. cos 2 sin 2 



Term 4 {1+2 -2 -1 } G. cos 6 sin 3 



Term 5 {1 +4 +6 +4 +l}H.sin0. 



The coefficients of the new equation are of course the numerical quantities multiplied by 

 the quantities outside the brackets. The two uneven terms in the new equations are 



accordingly 



{ - 4D cos 6 - 2E cos 3 6 sin 6 + 2G cos 6 sin 3 6 + 4H sin B}x*y , 

 and 



{ - 4D cos + 2E cos 3 sin 6- 2G cos 6 sin 3 + 4H sin 6}xy s . 



The coefficient of x 3 y cannot be changed into that of xy* by interchanging the signs 

 + and — . Hence a value of 6 which makes the term x*y disappear will not in general 

 make the term xy 3 disappear. In order that both terms may disappear, we must have 

 D = H; E = G; cos = sin 0. Hence the condition of the existence of a pair of conjugate 

 axes equally inclined to the axes of (2) is that the equation (2) be symmetrical. The 

 conjugate diameters thus found are evidently the principal axes of symmetry of the 

 curve. 



If in equation (2) the second and fourth terms are supposed to be wanting, so that 

 the curve is already referred to its principal axes of symmetry, then in the new equation 

 the two uneven terms will be 



{-4Dcos0 + 4Hsin0}:z 3 2/ and { -4Dcos0 + 4Hsin#}a;;i/ 3 . 



Their coefficients are identical, and the value, tan 6 = D/H, will reduce both terms to zero, 

 leaving an equation consisting of even powers, and therefore referred to conjugate 

 diameters. The diameters thus found are the secondary axes of the curve ; and the 

 relation, tan 6 = D/H shows that they are the diagonals of the circumscribing parallelogram 

 whose sides are parallel to the principal axes. When the highest power of y is negative, 

 a real solution is impossible, and the diagonals in question are the asymptotes of the 

 hyperbolic curve of the fourth degree. These results might have been found directly by 

 considering that every quartic homogeneous equation of even powers is necessarily a 

 projection of an equiaxial form. Because the given equation, when referred to conjugate 

 axes, is of the form 



S+^y+U=i. 



a 



We have then only to take P =pa 2 b 2 , or P/a 2 6 2 =p, in order to obtain the equation in 

 projection form. 



VOL. XXXV. PART IV. (NO. 23). 7 Z 



