Produce the Tangent TD to H, and imagine the Line HGF to be drawn II, and indefinitely near DA, and LKI , and at the fame Distance from DA; and suppose the indefinitely-fhort Line IAAF. Solution. x=ADq: By the Property of a Parabola {px+pa = GFq. P The As ADT and FHT are fimilar; therefore r+a :: √ px HF; confequently (AT) è i + 2 1. +aa (TFq) :: px (A D q=p × VA.) HFq = px 2tpxa + pxáa. But HF: GF q, that is p + t t 21 pxa+pxa a tt 1 px+pa; therefore 2pxa + p t ax + pt ta, and 2 x + 2 = to Again, px-pa KIq (by the Property of a Parabola :) And fince the As TAD and TIL are fimilar; Vpx a:: IL; wherefore tttt 28 a + aa : : px ..ILq: 2 tapx + aap*; But ÌLg:IKq; that is, pr t t a 2 t p x a + p xa a ~ px-pa; wherefore - 2 t p x à + p x à d tt xa ← spa wherefore &+2. That is, by the last Step Iii 2x, and by one of the foregoing Steps, 2 x + ; confequently (by our Lemma 1.) — 2x. AP = x; then PB = b x, and the Property of an Ellipfis Given. Note, PMLA B.` Preparation. Required to find PT? Produce the Tangent TM to H, and imagine the Lines HGF and LKI to be drawn at an indefinitely little equal Distance from, and parallel to MP; and fuppofe the indefi nitely fhort Line IP PF a. Solution. By the Property of an Ellipse x x b — I ·· — · PMq :: b · p 3 The As TFH, TPM, TIL are fimilar; wherefore ! - p · (TPq :) ..P b * — * * (P Mq :) :: tt+2+a+aa(FTq) · F H q b t tpb x + 2 ta pb x + aap b œ — pxxt t 2 tabxa aad pxx And †† (TPq :) .. pbx — p** (P Mq :) : : t t − 2 ta taa tt b pbxts - pxxtt 2 tapbx +2 + dp x + 42pbx - a2px2 But FHq: that is, FGq; ttb t + b px + 2 ta p b x+a a pb x—pxxt t− 2 t ap xx―a a px x aap, which Step, after due 2tx; that is (because b ́is x, and of Confequence * abx +ttaaxx) 2 b x − 2 x x 516-28x. 3. Again, IL:IKq: t2 p b x t t − p x 3 t 2 — 2 t ap b x + 2 1 ap xx+a2 p b x — a ˆ p x x that is, pbxtt pxx + 2pxa — abp — ap; which Step, when du Now from this laft Step, and from that mark'd with an Afterick, it must follow (by our Lemma 1.) that b From the Coherence that is in, or between the Ellipfis and Hyperbola, I prefume it very easy to draw a Tangent to an Hyperbola, by the like Method with that us'd in Prop. 3. I will therefore leave it to the Reader's Practice, and here only shew, how, from the Afymptotes of an Hyperbola, to draw a Tangent to its Curve, Propofition IV. To draw fuch a Tangent as is abovemention❜d to an Hyperbola. VA=X And the Proportion of any Line as VD being L AV, to AV, Note, p being Parameter that, by the Property of an Hyperbola, pp AV x VD; But here, for fome Reafons, will (as I may) fuppose AV = x=VD Then Preparation. Produce the Tangent TD to L, and imagine the Lines LKI and HGF to be drawn at an a, or indefinitely - little equal Distance from, and I VD; then Solu |