608 
PROBLEMS AND SOLUTIONS. 
[383 
giving the remaining two points of intersection. If the circle be a circle of curvature, 
one of these must coincide with the point (cos 6, sin 6), that is the equation 
y sin 6 — x cos 6 — 1—4 cos 2 6 + 4m cos 6 = 0, must be satisfied by the values x = cos 6, 
y = sin 6 ; this will be the case if — 6 cos 2 6 + 4to cos 6=0, that is cos 6 = 0, giving for 
the parabola y- + 2y +1 = 0, which is not a proper Solution, or else cos0=-|, giving 
sin 6 = + (1 — |to 2 )*, so that there are two parabolas satisfying the conditions of the 
problem; if to fix the ideas we take the upper sign, the equation of the corresponding 
parabola is 
y- — 2 (1 — -i to 2 )- y + \j m 3 x +1 — ^ m 2 = 0 ; 
and it may be added that the coordinates of the focus are 
x = to — to 3 , y = (1 — A to 2 )-. 
The equation of the other parabola and the coordinates of the focus are of course 
found by merely changing the sign of the radical. The parabolas are real if to < §; 
if to = | we have a single parabola, the point of contact being in this case the vertex 
of the parabola ; and if to > f the parabolas are imaginary. 
[Professor Cayley states that he was led to the foregoing problem by the consideration 
of the curve (proposed for investigation in Quest. 1812) which is the envelope of a 
variable circle having its centre in the given circle and touching the given line. The 
required curve (which is of the sixth order) has two cusps which, it is easy to see 
geometrically, are the foci of the parabolas in the problem. Taking (cos 6, sin 6) for 
the coordinates of the centre of the variable circle, we shall have 
x = § cos 6 — to cos 26 4- \ cos 36, y = § sin 6 — to sin 26 + \ sin 36, 
for the coordinates of a point on the envelope.} 
[Vol. iv. pp. 81—83.] 
1790. (Proposed by Professor Sylvester.)—(1) If a set of six points be respectively 
represented by the six permutations of a : b : c, show that they lie in a conic, and 
write down its equation. 
(2) Hence prove that if AB, BG, G A be three real lines containing the nine 
points of inflexion of a cubic curve having an oval, the pairs of tangents drawn to 
the oval from A, B, G will meet it in six points lying in a conic. 
Solution by Professor Cayley. 
That the six points, 
1 = (a, b, c), 
4= (a, c, b), 
2 = (b, c, a), 3 = (c, a, b), 
5 =(b, a, c), 6 = (c, b, a), 
1.