IN THE HIGHER GEOMETRY. 15 



M 



^ 



+ q . AG x (A P AN) ; or, dividing by q p, 



/ m + n \ i' P \ q 



XACD = APX r- + -)XF*H X AG X 



\m -f 2nJ \q p/ q p 



m + n , 



(AP AN). Now, - x AP x PM is equal to the area 



m -\- 2n 



P 



APM; therefore the area APM together with x AP . PM, 



q -p 



q n 



and - XAG x (A P AN), or APM with - x AP . P M 

 q-p q-p 



q q 

 X A G X (A N A P), Or A P M -j X A P . P M 



q p q p 



X rect. P T, is equal to - x A c D. Now i c' is an 



q p M + N 



P 

 hyperbola of the order p + q; therefore its area is - x 



p q 



P 



rect. G H . M H. But q is greater than n therefore is 



p-q 



p X GH . HM , 



negative, and is the area M H K c : and the area 



q-p 



39 



NTKC' is equal to - x GT X TN: therefore MNTH is equal 

 q-p 



39 



to (MHKC' NTKC'), or to x (GH . MH GT . TX). 



q p 



From these equals take the common rectangle A T, and there 



P tf 



remains the area M P N, equal to - - x A p x M P 



q-p q-p 



X PT; which was before demonstrated to be, together with 



APM, equal to . A c D. Therefore M P N, together with 



M + H 



APM, that is, the area AMN, is equal to . ACD; con- 



M+N 



sequently AMN : ACD :: M:M + N; and (dividendo) AMN: 

 N M D c : : M : N. An area has therefore been found, which the 

 hyperbola ic' always cuts in a given ratio. Therefore, a 

 conic hyperbola being given, &c. Q. E. D. 



