1915-16.] The “Geometria Organica” of Colin Maclaurin. 93 
Cor. 6. When a + /3 = ir the curve is a hyperbola in general, but a para- 
bola when l is parallel to OO'. 
Cor. 7. If l passes through the centre of the circle OKO', the angle AOB 
is a right angle and the hyperbola is equilateral. 
§ 5. Prop . III. 
But if for any position of P on l, OQ and O'Q coincide simultaneously 
with 00', the conic degenerates into a straight line (along with 00'). 
The proof given is analytical. 
Cor. 2. This may happen, for example, when a + /3 = 7r, and l is inclined 
at angle a to 00' (viz. when P is at infinity on l). 
Cor. 3. In particular this is so when a = /3 = 7r/2, and l perpendicular 
to 00', when Q is on another line perpendicular to 00', which is the 
image of l in the mid-point of 00'. 
Cor. 4 contains an important statement. 
Find I on l such that L OO'I — /3, and let 100' = a'. 
Let P trace out l , and let Q' be taken in quadrilateral POQ'O' for 
angles a and /3. Then Q' traces out a straight line. The angle QOQ' = 
a— a and is therefore constant, and O', Q, Q' are collinear. Hence we 
may trace the conic locus of Q by making Q' lie in V and taking QOQ' = 
a —a) i.e. in the preceding constructions we may replace an angle by a 
straight line rotating round one of the fixed points. 
§ 6. Prop. IV 
proves the converse of Prop. I by solving the problem : — To describe a 
conic through Jive given points. 
Let the points be A, B, C, D, E. Form ACAB, and let ZCAB = ct, 
L CBA = f3. Rotate angles a and /3 round A and B respectively, and let 
the intersection of two arms be in D and then in E, while the intersection 
of the other arms comes to be at D' and E'. 
Let the line D'E' be taken for l ; then if P traces out l, Q generates a 
