8t 



♦ KNOWLEDGE 



[Jan. 30, 1865. 



enabled to lay down upon them with far greater accuracy 

 than on the present "one-inch" maps, the facts which he 

 ascertains about the strata. 



CHATS ON 



GEOMETRICAL MEASUREMENT. 



Bv Richard A. Pkoctor. 



THE HYPERBOLA AND PARABOLA. 



(Cont'iiiieil fruni i). 15.) 



A. In dealing with the ellipse and parabola, you deter- 

 mined (or rather, you showed me how to determine) the 

 area of a segment cut cti' by any chord. Can you do the 

 f?ame for the hyperbola? 



M. Let us see. Note first that our proof for the 



Fig. 3. 



area between an asymptote, the curve, and two ordinates 

 parall -1 to the other asymptote, applies equally whether 

 the points from which ordinates are drawn are on the same 

 side or on different sides of the axis. Thus if we draw 

 (Fig. .3) two crdinates K E, PL parallel to C N' to meet 

 ON, 



area F L E K = nar D D' log ^ 

 ' " CL 



Hence it' we draw ihe chord PK we get at once what you 



require ; for we have 



Area P Q K = trap P L E K -par D D', log _ 



= i (K E -f PL) L E sin K C N'-par 



D D' log 2J?. 

 " CL 



A. You u.ie here the trigonometrical ratio, the sine; is 

 that legitimate in geometrical work 1 



M. When we have come to use logarithms we need not 

 stickle at the trigonometrical fanction.s. Howbeit, the 

 trigonometrical ratios are purely geometrical in origin and 

 meaning. You no more take from their geometrical cha- 

 racter by making them the subject of analytical processes, 

 than you de/irive a square of its geometrical cliaracter by 

 showing that if the side of the square be a unit of length, 

 the diagonal can only be represented by a non-terminating 

 decimal. 



A. In passing, I notice that the area between the hyper- 

 l<oIa m\i it.s asymptotes is infinite. 



M Yes ; f j[ we can take C E as large as we please, so 



that wherever L be taken, the ratio CE : C L, and there- 



fore log—-!, may be made as large as we please. 

 OL 



A. What relation holds between such areas asAPK'cD 

 and A P K' E' D', obtained by drawing ordinates A D, 

 KV to one asymptote, and A I)', K' E', to the other'! 



M. A very simple one. These areas are equal. 



A. Can this be easily proved'? 



J/. Very easily, either from what we already know or 

 independently. Thus we have learned that \ 



Area A P K' e D=D D' log ^- ■ 



and 



Ce 

 CE' 



■ Area APK'E'D' = DD'log 



"C D- 



but, since par e E'= par D D', and they have a common angle, 



we have CE':OD::CD:C<^ 



log = log ; and area A P K' e D 



''CD » ' ■ ' 



wherefore 



C« 



=area A P K' E' D'. 

 A. Is the independent [iroof simple ? 

 j1/. Very simple indeed. From two adjacent points, P,^:>, 

 draw the ordinates P L,^>/, P M, p m. Then we have 

 par L M = par I m, 

 Wherefore par Jjp =z par m P ; 



And these are corresponding elements of the two areas we 

 are dealing with, A P K' « D and A P K' E' !>'. Therefore, 

 these areas are equal. 



A. Might you not have obtained the area A Q K L more 

 easily from what you have just established, than by your 

 former method 1 



M. More symmetrically, perhaps, but not more simply. 

 Suppose we want the area K A K', where K h K' is perp. 

 to the axis C A h. We have at once 



Area KA K' = trap KEeK' ■^^' '--^^ 



= ACNN' 



D D' . log 



° Ce 



parCK'-2ANEK 



-DD'log/'2^V 

 iCD 



;d; 



=^2 (AC AN 

 - D I)' log 



ANEK 



ACD A- 

 C_E^ 



c; 



as before. 



A. Shall we now leave the conic sections ■? 



M. First we may note another way of dealing with the 

 parabola, suggested by a property of the tangent to the 

 curve. 



Fig. t. 



Let P Q, Fig. 4, be adjacent points on the arc A PQ, so 

 that Q P Y is the tangent, ultimately, at P ; A the vertex ; 

 S the focus ; B A S M axial : K M perp. to axis ; K a A-, 

 Q h n, P c 7)1 perp. to directrix, B k ; Ac ha the tangent at 

 A. Join S P, S Q, S K, S m (cutting A a in Y), and draw 

 mi parallel to PT. Then, since SY = Y?n, by well- 

 known property, — 



A «PQ = J par QPjrtj; 



