440 



NATURE 



[March 13, 1890 



— which is also the centre of the conic — and producing the line 

 yO to y, so that Oy' may be equal to Oy, he determines a 

 second point of contact y' on the conic, by which means the 

 problem is reduced to the case dealt with in the preceding pro- 

 position, showing how to construct the curve when three tangents 

 and two points are given. Having in this way found five points 

 on the transformed conic, Newton next proceeds to retransform 

 the whole of the figure to its original shape, in order to apply 

 his well-known method of constructing a conic of which five 

 points are known. 



Fig. I. 



Now all these transformations and retransformations of lines 

 and quadrangles involve very tedious and laborious operations, 

 which can be avoided by borrowing a few simple principles of 

 modern geometry. The following two original solutions of the 

 above problem will serve to illustrate this statement. 



Solution, — Casel. When the given point of contact x lies 

 on one of the given four tangents. — Assume the given point of 

 contact X and the neighbouring aisex B of the quadrangle as 

 centres of projection, and the given tangent lines EA and C'T> 

 as punctuated lines. The meaning of the term "punctuated 

 line," familiar to students of modern geometry, will appear in 

 the sequel. 



It will be seen that the fourth tangent AB cuts the first punc- 

 tuated line EA in A and the second punctuated line CD in A'. 

 Now, according to a proposition of modern geometry, if the 

 points A and A', in which the tangent AB intersects the two 

 punctuated tangents EA and CD, be projected by rays -rA and 

 BA' issuing from their respective centres of projection x and B, 

 those rays will meet in a point A, situate on what is termed the 

 perspective line of the pencils x and B. 



Next imagine the tangent AB to revolve upon the curve 

 so as gradually to approach the limiting position BC. In that 

 case A will approach C, B will fall upon C, and the inter- 

 section of the projecting rays xC and BC will coincide with C, 

 which is therefore a second point on AC, the required perspec- 

 tive line of the pencils .r and B. Wherefore, in order to find a 

 fifth or any number of tangents to the curve, choose any point 

 E on the punctuated line EA, and project this point from x, the 

 corresponding centre of projection, upon the perspective line 

 AC in e ; and then project e from the second centre of projec- 

 tion B upon the corresponding punctuated line CD in D. The 

 line ED is a fifth tangent to the conic, and any number of 

 tangents can be drawn in precisely the same way. Then, let F 

 be any other point on EA. Join and produce ¥x, intersecting 

 the perspective line AC in/; and from the centre B project / 

 upon the punctuated tangent C'T> in F'. Then the line FF' 

 will be a sixth tangent to the conic. 



Cor. I. — Since the lines AC, BD, and xE all meet in the 

 same point e, it follows that, in any pentagon ABCDE circum- 

 scribed to a conic, the opposite diagonals AC and BD and the line 

 joining the fifth point E to the opposite point of contact x all 

 meet in the same point. 



Case II. When the given point of contact z lies otitside of the 

 four tangeitts AEDC'B. — By the corollary. Case I., if AB be 

 the fifth tangent, it must pass through the given point of con- 

 tact z in such a direction that the diagonals CA and EB may 

 intersect in a point I situate on a given line T>z. 



Now let AB revolve about the fixed point of contact s as a 

 fulcrum, whilst A and B describe the lines EC and CC (Figs. 

 I and 2). Then, necessarily, s will be the centre of perspectivity 

 of the punctuated lines EC and CC, whose centres of projection 

 are respectively C and E. But, by a well-known proposition of 

 geometry of position, when the points of two converging punc- 

 tuated lines, such as EC and CC, are projected from opposite 

 centres in this fashion, the locus of the successive intersections 

 of the rays CA and EB, or in other words the variable position 

 of the point I, will describe a conic, which in the present 

 instance is a hyperbola. But the problem is how to find the 

 point I on the transversal Ls without constructing the hyperbola, 

 four points on which are already known. For it will be 

 observed that, when A coincides with E, the point B will lie 

 on the prolongation of Ez, and the corresponding projecting 

 rays Y.z and CE will meet in E, a point on the hyperbola. 

 Similarly C is a second point on the hyperbola. Again, as AB 

 continues to revolve about the fixed centre of perspectivity z, its 

 intersections A and B with the punctuated lines EC and CC 

 will ultimately coalesce in the point C, common to both those 

 lines. Hence, since in that case the rays projecting the double 

 point C from the centres E and C meet in C, this point must lie 

 on the hyperbola. 



Fourthly, if the line Co be produced tp intersect the line EC 

 in N, it can be easily shown that i, the third point in the 

 harmonic ratio Gjsj'N, is a fourth point on the hyperbola. A fifth 

 point can be found by simply drawing AB in any direction 

 traversing z and intersecting EC in A' and CC in B', and then 

 projecting A' and B' from the centres C and E respectively by 

 rays CA' and EB' which will meet in a fifth point upon the 

 hyperbola. 



Thus, given these or in fact any five points EDi'TH (Fig. 2} 



Fig. 



on the hyperbola, it is possible to find the point of intersection I 

 of the given transversal Ls with the hyperbola without con- 

 structing the curve. First describe any circle in the plane of 

 the five points, choosing two of these, such as E and i, as 

 centres of projection from which to project the remaining three 

 points DHT upon the given transversal L: in the points dht 



