258 



Geometrical Theorems on the 



the hyperbolic arc A'T' above its tangent T'P', by twice the 

 elliptic arc A"L. 



Take the point L so that when the ordinate MLH is drawn 

 to meet in H the semicircle described on the axis, the angle 

 HC"N may be equal to 

 half the angle PCF (or 

 PC'F', for the triangles 

 PCF and PC'F' have 

 all their sides and angles 

 equal) ; draw C"D per- 

 pendicular to C"H, and 

 DE to A" a", meeting the 

 ellipse in E\ then, by km. 2, C"E will be conjugate to C"L, 

 and by km. 4 it will be equal to FP, since the angle B"C"D is 

 equal to HC"M, and is therefore half of PCF, whilst the semi- 

 axes C"A", C"B", are the sum and difference of PC and CF. 

 Hence the point L thus found is that required by the enuncia- 

 tion. Take p, p', h, indefinitely near to P, P', H, and similarly 

 related to each other ; let Fp and Fp f intersect TP and T'P' in 

 q and q', and with the centre C" and a radius equal to C"E de- 

 scribe the evanescent arc Jk. Then FP, F'P' are always per- 

 pendicular to TP, T'P', and therefore Pq is ultimately the 

 increment of the difference between the arc AT and the tangent 

 TP (km. 1.) and P'q' the increment of the difference between 

 the arc A'T' and the tangent T'P'; also by km. 3. Kk is ulti- 

 mately equal to LI, the increment of the arc A"L. Now the 

 angles CFP and CFp are equal to C'PF and C'p'F* ; therefore 

 PFp is equal to the sum of P'F'p' and PC'p, or to PTfy' with 

 twice HG"h : but FP, F'P', and C"Kv& all equal, and there- 

 fore P7 is equal to P'q' and twice Kk, or to P'<?' and twice LI. 

 Whence the proposition is manifest. 



Schol The angle ff'C"H is half the angle fCP oif'C'P', 

 and therefore by km. 4, the semidiameter C"L is equal to the 

 straight line fP or/'P'. This gives another and easier method 

 of finding the point L. 



Hence every arc of a hyperbola may be found by means of 



