214 On certain Geometrical Theorems. ["Dec. 22, 



Since two of the tangents are at right angles to each other, we 

 shall have m 1 m 2 + l = 0, and let m 1 + m 2 =/», 2m 3 =p, ^mgm^q, 

 2m 3 m 4 m 5 =r, m 3 m 4 m 5 ra 6 =s. Then substituting, we have the follow- 

 ing equations : — 



ji+p=a (1), — q + /nr + s—d . . . (4), 



— l+jLip + q = b . . . (2), — r + jus = e (5), 



-p+fiq+r=c . . . (3), —s=f (6). 



TYom these equations we obtain at once — 



fi—a—p. r=f(p — a) —e, q=l-\-b—/np = l + b — ap+p2. 

 Hence we have, substituting in (4) — 



(f+l)p 2 -(a + 2af+e)p + (l + b)+fa* + ae+f+d=0 . . (7). 

 Also substituting in (3) — 



pZ-2ap*+(a* + b-f+2)p-a(b-f+- l) + e + c=0 . . (8). 

 From (7) and (8) we easily obtain two equations of the form — 



Ap2+Bp+C=0, 



then the eliminant is at once seen to be — 



(AC - C A) 2 + (B A - AB') (B C - OB') = 0, 



the equation to the required locus. 



I have not thought it necessary to write down the values of a, b, c, 

 <fcc, as they are obtained by rules perfectly well known. 



Note by W. Spottiswoode, P.R.S. 



The second theorem in the foregoing paper follows also as an imme- 

 diate consequence of a formula given by Cayley in his " Seventh 

 Memoir on Quantics " ("Phil. Trans.," 1861, p. 286). If U repre- 

 sent the cubic and HU its Hessian, then, as is well known, HTJ passes 

 through the points of inflexion of U. Also, the function aJJ + 6/3HU 

 will represent an arbitrary curve of the third degree passing through 

 the same points ; and, on the same principle as before, its Hessian will 

 pass through its points of inflexion. Now the formula in question 

 is — 



H( a U + 6/3HU)= ^(1,0,-24S, . . )'(*,/8)*.U 

 -6S a (l, 0,-24S, . . )(«, py . HU. 



