457 



K = xi {Xi - Xi) + y, (1/2 — y,), (15) 



we see that, as k now vanishes, the right line joining corres- 

 ponding points F and A upon these lines is perpendicular to 

 the focal line. Of the two sides of the cone which are in 

 the plane of xy, one lies between each focal and its dirigent ; 

 and it may be inferred from the equations, that the tangents 

 of the angles which the internal axis makes with a focal 

 line, with one of these sides of the cone, and with a dirigent 

 line, are in continued proportion, the proportion being that 

 of the cosine of ^ to unity. And hence it follows, that these 

 two sides of the cone, with a focal line and its dirigent, cut 

 harmonically any right line which crosses them. 



§ 5. From this discussion it appears, that the ellipsoid and 

 the hyperboloid of two sheets can be generated modularly, 

 each in one way only, the modular focal being the ellipse 

 for the former, and the hyperbola for the latter ; but that 

 the hyperboloid of one sheet can be generated in two ways, 

 each of its focals being modular, and each focal having its 

 proper modulus. The cone also admits two modes of gene- 

 ration,* in one of which, however, the focus is limited to the 

 vertex of the cone, and the directrix to its internal axis. 



* The double generation of the cone, when its vertex is the focus, may be 

 proved synthetically by the method indicated in the Examination Papers of the 

 year 1838, p. xlvi (published in the University Calendar for 1839). Supposing 

 the cone to stand on a circular base (one of its directive sections), and to be cir- 

 cumscribed by a sphere, the right lines joining its vertex with the two points 

 where a diameter perpendicular to the plane of the base intersects the sphere, 

 will be its internal and mean axes. Then if P be either of these points, Y the 

 vertex, C the point where the axis PV cuts the plane of the base, and B any 

 point in the circumference of the base, the triangles PVB and PBC will be si- 

 milar, since the angles at V and B are equal, and the angle at P is common to 

 both triangles ; therefore BV will be to BC as PV to PB, that is, in a constant 

 ratio. It is not difficult to complete the demonstration, when the focus is sup- 

 posed to be any point on one of the focal lines. 



2 R 2 



