534] 
A “ smith’s PRIZE ” PAPER ; SOLUTIONS. 
479 
It would at first sight appear that a like question might be asked as to a 
surface ; viz. that it might be proposed to determine the angle between the normal 
and a line drawn from the point to the centre of the indicatrix conic. But this is 
not so ; in fact, taking the origin at a point on the surface, the axes of x, y being 
in the tangent plane, and the axis of z coinciding with the normal : then to the third 
order we have 
z = (A, B, G\x, y) 2 + (a, b, c, dQx, y) 3 ; 
but here, regarding z as a given constant, if we take account of the terms of the 
third order, the section is not a conic but a cubic ; and it has not in general any 
centre ; and if (as in the ordinary theory) we neglect the terms of the third order, 
thus obtaining an indicatrix conic, the centre of this conic lies on the normal, and 
there is no angle corresponding to the angle </> of the plane problem. 
The only case where there is such an angle is when the cubic terms (a, b, c, (T§x, y) 3 
contain as a factor the quadric terms (A, B, G\x, y) 2 (one relation between the 
coefficients A, B, G, a, b, c, d). For then we have 
z = (A, B, C\x, y) 2 (1 + 2lx + 2my), viz. 
z = (A, B, Gfx, y) 2 +2(l xz + myz), 
approximately to the third order; and then regarding £ as a given constant, this last 
equation represents a conic having for the coordinates of its centre, say x — clz, y = (3z, 
and there is an angle </> = tan -1 \J(a 2 + /3 2 ) ; this is, in fact, what happens in the case 
of a quadric surface, for the section by a plane parallel and indefinitely near to the 
tangent plane is then a conic, the centre of which is not on the normal ; and the 
angle <f> (in the case of a central surface) is in fact the inclination of the normal 
to the radius from the centre. 
I take the opportunity of adding a remark that the indicatrix is never a parabola, 
but in the separating case between the ellipse and the hyperbola it is a pair of 
parallel lines. The indicatrix, a parabola, is commonly obtained as follows : viz. taking 
the axes as before, but starting from an equation U = 0, the equation presents itself 
in the form 
z = (A, B, G, F, G, H\x, y, z)\ 
which, considering i as a given constant, represents a conic which, it is said, may be 
a parabola. But observe that 2 is of the order (x, yf, the terms 2Fyz + Gzx, are 
consequently of the order (x, y) 3 , but they are not all the terms of this order which 
would be obtained by the expansion of 0 as a function of (x, y); there is consequently 
no meaning in retaining them, and they ought to be rejected ; similarly the term in 
z 2 which is of the order (x, y) 4 ought to be rejected ; the equation is thus reduced to 
z = Ax 2 + 2Hxy + By 2 , 
which, when AB-H 2 = 0, represents not a parabola but a pair of parallel lines. On 
referring to Dupin’s Développements de Géométrie, éc. (see p. 49) I find that he is 
quite accurate; his expression is, “elle peut cependant être une parabole; alors elle 
se présente sous la forme de deux droites parallèles équidistantes de leur centre ” : and 
he afterwards examines in particular “ ce cas remarquable.”