128 Proceedings of the Royal Society of Edinburgh. [Sess. 
and the locus of Q is given by 
y* = (aiy-p)(x-a) (3) 
(V.) Let P lie on xy —p. 
If 0 1 is the point at infinity on the y- axis, O x PQ parallel to OY, and 
lpo 2 q = 7 t/ 2, the locus of Q is given by 
y = — x{a = x) 2 /p. 
Cor. 9. In this corollary Maclaurin proves the generality of this con- 
struction for a singular cubic by using it to describe a cubic having a 
double point at 0 1? and passing through other six points 0 2 P 1 P 2 . . . P 5 . 
Construct AP 1 0 1 0 2 , and let 0 ± = a, 0 2 = /3. Take a quadrilateral P0 1 Q0 2? 
in which 0 1 = a, 0 2 — /3, and place P in coincidence with P 2 , P 3 , P 4 , P 6 , when 
Q will take up positions Q 2 , Q 3 , Q 4 , Q 5 . Construct the conic through O l5 
Q 2 , Q 3 , Q 4i Q 5 , and restrict the vertex Q to lie on this conic, when P traces 
out a cubic having a double point at O x and passing through OgPjPg . . . P 5 . 
There cannot be two such cubics ; for, if there were, they would require to 
be considered as intersecting in ten points, whereas they cannot cut in 
more than nine points. 
§ 40. Prop. II. 
When the locus conic , as in Prop. I, passes through neither 0 1 nor 0 2 , 
. then the same method of proof shows that the curve traced by Q is a curve 
of the fourth order, having double points at 0 1 and 0 2 , and also at a 
third point. 
For let OjOg cut the conic in and A 2 . To A 4 and A 2 corresponds a 
common point B on the quartic, which is thus a third double point. 
Cor. 4. If O x Q and 0 2 Q coincide simultaneously with C^Og, the locus 
is only of the third degree, with ordinary points at 0 4 and 0 2 and a double 
point at B. 
[Maclaurin might have shown how to use Prop. II to construct a 
quartic having three double points O p 0 2 , 0 3 , and through five other 
points P x . . . P 5 . 
Construct AO^Og, and let 0 1 = a, 0 2 = /3, 0 3 = y. Use the quadri- 
lateral POjQOg in which 0 1 = a and 0 2 = /3, and place P in coincidence 
with P l5 P 2 , . . . P 6 , when Q takes up the positions Q x , Q 2 , . . . Q 5 . Let 
the five points Q determine the conic C. Now restrict Q to lie on C, and 
P will trace out a quartic having double points at ChC^Og and through 
p x p 2 . . . p 5 . 
There cannot be two such quartics, for, if so, they would require to be 
considered as intersecting in seventeen points, which is impossible.] 
