J3S DEMONSTRATIO THEOREMATIS 



arcmfp:=zn •.p — U :f; arcus|>^i:zn ; q — U t^ 



arcus^rizr.n ; r—U: q, arcusprznri :r — U : p 



item arcusB/>:r:n ; ^ — n :p; arcus B^ — n : ^ -n : ^ 

 Denotat enim n : b arcum totius quadrantis A M B ^ 

 ideoque 4 n ; ^ totam ellipfis peripheriam. 



Problema i. 



Tabi TIL &0, Propofito in ellipfi arcu A/ in vertice A 



F'S' ^" terminato, ab alio quouis punAo p arcum abfcindere. 

 p q^ qui ab illo arcu A/ difcrepec quantitate geometrice 

 aflignabiii.» 



Solutio. 



Pofitis abrciflis , quae pundis f^pttq refpons"- 

 dent^ AF— /; k?—p\ et AQ^— ^, ex datis / et p 

 conuenienter determinari oportet q, Cum igitur prcs 

 lemmate fecundo fit 



%z:ibb\ «5^-«; kz=:b'\ Br-(«4 i)^^, et C-fS 



capiatur q ita , vt fit ; 



Ih^y {bh—ffV>i — nff) -H bbfV(bb^pplb b — npp) 



^ — 6* nffpp 



Ciitque per lemmatis conclufionem : 



fdq^y'^-fdpy'^,^ Conft. - f' 

 Ateft fdqyl^-U:qctJdpyllE:fp^n:p,ynd^. 



U: q-n-.p—Coad -~^ 



\bitantum (iipereft, vt conftans debite definiatur. Ve- 

 aumi quia pofito />— 0, fit ^— /, ad quem cafun» 



aequa.^ 



