Full text: Barrow, Isaac: Lectiones Opticæ & Geometricæ

53. APPENDICULA 3.

Præcedentia recolenti nonnulla videntur elapſa; quæ forſan ex uſu
ſit adjicere. _Demònſtrationes_ elicere poterit quiſpiam è præmiſſis; & potior inde fructus emerget.

54. Problema I.

Fig. 180.

Sit _curva_ quævis KEG, cujus _axis_ AD; & in hoc ſignatum
punctum A; curva reperiatur, puta LMB, talis, ut ſi ductâ utcun-
que rectâ PEM axi ADperpendicularis curvam KEG ſecet in E, & curvam LMB in M; nec non connectatur AE, & curvam LMB
tangat recta TM; ſit TMipſi AEparallela.

Hoc ità fiet. Per aliquodcunque punctum R, in axe AD fumptum,
protendatur recta RZad ipſam ADperpendicularis; cui occurrat re-
cta EAproducta in S; & in recta EPſumatur PY = RS; ità de-
terminetur curvæ OYY proprietas; tum ſit rectangulum ex AR, & PMæquale ſpatio AYYP(ſeu PM = {ſpat AYYP/AR}) habebit
curva LMMBconditionem propoſitam.

Adnotari poteft, ſi ſtantibus reliquis, ſit curva QXX talis, ut cum
hanc ſecet recta E Pin X, ſit PX = AS; erit ſpatium AXXP
æqualerectangulo ex AR, & curva LM, ſeu {AXXP/AR} = LM.

55. Exemp. I.

Sit ADG _circuli_ quadrans, & ductâ EPad ADutcunque per-
pendiculari, connexâque DE; deſignetur curva AMB talis, ut ſi
producta recta EPM hanc ſecet in M, ipſamque tangat recta MT,
ſit MTad DEparallela. Hocita peragetur. Ducatur AZad DG
parallela; & huic occurrat producta DEin S, & curva AYY talis
ſit, ut ſi hanc ſecet producta PEin Y, ſit PY = AS; tum capiatur
PM = {Spat. AYP/AD}; factum erit.

55.1.

Fig. 181.

Not. Quòd ſi curva QXX talis ſit, ut PX = DS (vel ſi AQ
= AD, & QXX ſit _byperbola_ angulo ADG comprehenſa) erit
curva AM x AD = ſpat. AQX P.

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