[Barker] CORRELATION OF CURVE OF THE SECOND ORDER 31 
I add the following proof of the uniqueness of the curve of the second 
order, whatever five points on the curve be selected as the base of the 
construction and whichever of these be taken as the radiant poinis : 

First let A, B, C, D, E be the five points, and A and C the radiant 
points. Then the curve is unique, 7.e., only one curve exists with A and 
C as radiant points ; or, to put it more clearly, if a given ray from A be 
selected, say AX, to this corresponds only one ray from C, the intersec- 
tion of these rays giving a point on the curve. 
But the question arises,—if we take two other of the five points as 
radiant points shall we get the same succession of curve-points ? 
To answer this, construct first the point J’, À and C being radiant 
points and B Æ, B D base lines. Then S is the point from which the 
ranges B # and B D are in perspective. Thns A F and C F are corre- 
sponding rays. 
Next take A and B as radiant points, and C Æ and C D as base lines. 
Then S' is the point with respect to which the ranges C E and C D are 
in perspective. The same ray A Æ is taken from A, and the question is, 
will the corresponding ray, now from B, intersect it in the same point F. 
This corresponding ray is evidently B Y’'. 
Then considering the triangles XY B X!and Y Y' C, the correspond- 
ing sides BX, Y' Y intersect in S}, the sides 4° X, C Y in S, and the 
sides 4* B,C Y'in D. But S', Sand D are in the same straight line. 
Hence 41 C, B Y‘ and X Y intersect in F. 
Take now another point Æ, say, as radiant point instead of A, and 
the intersection of the ray B F' from B, with the corresponding ray from 
E, will, by what has just been proved, give the same point F. 
Thus any pair of the five points may be taken as radiant points, and 
the same points will make up the curve, since a ray in the direction of F 
will always be intersected at F by the corresponding ray, and F repre- 
sents any point on the curve. 
