186 Mr. Morton on the Focus 



Let ^ be the vertex and FO the axis of the right cone 

 FBDE, and let AQP be any plane section of the cone, in 

 which the point P is taken (fig. 1, 2, 3.) Through VO draw 

 the plane FAA' perpendicular to the plane JQP, and let it 

 cut the latter in the line AA', and the conical surface in the 

 lines VA, FB. Describe the circle SEB touching the three 

 straight lines AA', FA, and FB in the points S, E, and B 

 respectively; and, because FO bisects the angle AVB, let the 

 point in FO be the centre of this circle, and join OB, OS. 

 Then, because the plane FAA' is perpendicular to the plane 

 AQP, and that the straight line OS which lies in the former 

 plane is perpendicular to the common section AA' of the two 

 planes, OS is perpendicular to the plane AQP. Therefore, if 

 a sphere be described from the centre with the radius OS 

 or OB, it will touch the plane AQP in the point S. Join 

 BE, and let BDE be the circular section passing through BE : 

 join FP, and let it cut the circle BDE in D, and join OD. 

 Then because the triangles FOD, FOB have two sides of the 

 one equal to two sides of the other, each to each, and the 

 included angles OFD, OFB equal to one another, the bases 

 OD, OB are likewise equal, and the angle FDO is equal to 

 the angle FBO, that is, to a right, angle. Therefore, if a sphere 

 be described from the centre O with the radius OS or OB, 

 the point D will be in the spherical surface, and every other 

 point of the line FP will be without it, and the same may 

 be shewn with regard to every other slant side of the cone : 

 therefore the sphere so described will touch the surface of the 

 cone in the circle BDE. 



Let the lines BE, A'A (produced if necessary) meet one 

 another in thfe point X, and the planes BDE, AQP in the 

 line RX. Then, because .RA' is the common section of two 

 planes, each of which is perpendicular to the plane VAA', RX 



