138 Trans. Acad. Sci. of St. Louis. 



If the normal is to be drawn through any point in the 

 plane, P, whose co-ordinates are ?, 77, these co-ordinates 

 must satisfy equation (2), so that 



pn + Eyi = in/i + x &v ( 3 ) 



The co-ordinates x v y l also satisfy equation (2). 



If we now make x v y x variables, equation (3) becomes 



pv + £y = py + x v- W 



This equation is satisfied by the co-ordinates £ , y and also by 

 the co-ordinates x 11 y l ; and, therefore, represents a curve 

 which passes through the given point P and through the 

 intersections with the parabola of the normals through P. 



Equation (4) represents an equilateral hyperbola. The 

 equations of its asymptotes are 



x = £ — p (5) 



and 



y = 0. (6) 



Equation (5) represents a straight line parallel to the axis 

 of Y and at distance £ — p from it. 



The distance from the point P to this asymptote is evidently 



£ — (£ — p)=p; (7) 



that is, the asymptote is for all values of g at a distance p to 

 the left of P. 



The asymptote represented by equation (6) is the axis 

 of X. 



The auxiliary hyperbola can be readily constructed from 

 the well known principle of constant areas, or perhaps more 

 easily in one of the following ways: — 



Since the secant intercepts equal distances between the 

 curve and each asymptote of the hyperbola, we have only to 

 draw any secant, as QPL (Fig. 1), and make LK equal to 

 PQ; the locus of K will be the auxiliary hyperbola. 



Again, it is evident from this principle that for any secant 

 through P the triangles QRP and KML are equal ; and that 



