For, draw the diameter CHE, and the tangent TE, and its parallels PK, RI, MH, meeting the conjugate of the diameter CR in the points T, K, I, M. Then, because similar triangles are as the squares of their like sides, it is, by sim. triangles, CR: GP2 :: A CRI: A GPK, CR2: GH2:: A CRI: A GHM; and CR2: GP2 GH CRI: KPHM. theref. by division, Again, by sim. tri. and by division, But, by cor. 5 theor. and by cor. I theor. theref. by equ. CE: CE In like manner CE2: CE2 CH2: cr2: рн. нq. Theref. by equ. CR2: cr2:: PH. HQ: рH. нy. CE2: CE2 CH2: A CTE: TEHM. 19, the A CTE = Á CIR, 19, TEHG = KPHG, or TEHM = KPHM; CH2; CR2: GP2- GH2 or PH.HQ. 2 Q. E. D. Corol. 1. In like manner, if any other line p'H'q, parallel to cr or to pq, meet PHQ; since the rectangles PH'Q, p'H'q' are also in the same ratio of CR to cr2; therefore rect. PHQ: рHq:: PHQ: p ́H'q. Also, if another line pha' be drawn parallel to ra or CR; because the rectangles p'ho', p'hq' are still in the same ratio, therefore, in general, the rect. PHQ: pнq:: Pha: p'hq'. That is, the rectangles of the parts of two parallel lines, are to one another, as the rectangles of the parts of two other parallel lines, any where intersecting the former. Corol. 2. And when any of the lines only touch the curve, instead of cutting it, the rectangles of such become squares, and the general property still attends them. The Sum or Difference of the Semi-Transvers and the drawn from the Focus to any Your n de Curve, i ena to a Fourth Proportional to the beni-travery, the De tance from the Centre to the Focus, and the Distance from the Centre to the Grimace belongingu na Forte Curve. That is FE+AC=C1, or FE = £J; and E-AC=CL QE=BL. Where CA: CF :: CD: HE 4th propor. to CA, OF, DD. # For, draw a paralel and equal to IL the semi-conjugate; and join cc meeting the ordinate Then, by theor. 2, and, by sim. A 5, consequently Also FD CF CD, and F = OF ~ vg. ID but, by right angled triangles, F2 4 DE = 18 therefore FE = CF2 But by theor. 4, CF2 20F. CD + CD + DE2. DE = CÉ, and, by supposition, 2c.c = 204. C15 theref. FE CA2 - 20A. CI + 12 + DE2. But, by supposition, and, by sim A52 therefore consequently FE2 = CA2 204.01 + CR Corol. 1. Hence CHCI is a 4th propor. CI — CA = AL = BL. QE. D. to CA, CF, CD. Corol For, draw the diameter CHE, and the tangent TE, and its parallels PK, RI, MH, meeting the conjugate of the diameter CR in the points T, K, I, M. Then, because similar triangles are as the squares of their like sides, it is, by sim. triangles, CR2: GP :: A CRI : A GPK, Again, by sim. tri. and by division, CE2: CH2 :: ACTE: ACMH; CH2 :: A CТЕ: ТЕНМ. But, by cor. 5 theor. 19, the ACTE = A CIR, and by cor. I theor. 19, TEHG = KPHG, or TEHM = KPHM; theref. by equ. CE: CE In like manner CE: CE 2 CH2 :: CR2: GP2- GH2 or PH.HQ. CH2 :: cr2: pH. нд. Theref. by equ. CR2: Cr2 :: PH. HQ: pH. Hy. Q. E. D. Corol. 1. In like manner, if any other line p'н'д', parallel to cr or to pq, meet PHQ; since the rectangles PH'Q, p'H'q' are also in the same ratio of CR to cr2; therefore rect. PHQ: PHq :: PHQ: píq'. Also, if another line p'ha' be drawn parallel to PQ or CR; because the rectangles p'ha', p'hq are still in the same ratio, therefore, in general, the rect. PHQ: PHq :: Phá: phg. That is, the rectangles of the parts of two parallel lines, are to one another, as the rectangles of the parts of two other parallel lines, any where intersecting the former. Corol. 2. And when any of the lines only touch the curve, instead of cutting it, the rectangles of such become squares, and the general property still attends them. Corol. 3. And hence TE : Te :: tE : te. e SECTION SECTION II. OF THE HYPERBOLA. THEOREM XIV (5). The Sum or Difference of the Semi-transverse and a Line drawn from the Focus to any Point in the Curve, is equal to a Fourth Proportional to the Semi-transverse, the Distance from the Centre to the Focus, and the Distance from the Centre to the Ordinate belonging to that Point of the Curve. For, draw AG parallel and equal to ca the semi-conjugate; and join co meeting the ordinate DE produced in H. Then, by theor. 2, CA2: AG2 : : CD2 and, by sim. As, CA: AG2:: CD2 CA2: DE2; CA2: DH2 - AG2; 2CF.CD + CD2; DE2=DH2 – AG2 = DH2 - са2. consequently Also FD = CFS CD, and FD2 = CF2 but, by right angled triangles, FD2 + DE2 = FE2; 2 therefore FE2 = CF2 - ca2 2CF.CD + CD2 + DH2. But by theor. 4, CF2 - ca2 = CA2, and, by supposition, 2CF.CD = 2CA.CI; theref. FE = CA2 - 2CA.CI + CD2 + DH2. But, by supposition, CA2: CD2 :: CF2 or ca2 + AG2 : CI2; and, by sim As, CA2: CD2 :: CA2 + AG2: CD2 + DH2; therefore CI2 = CD + DH2 = CH2; FE2 = CA 2CA. CI + CI2. consequently And the root or side of this square is FE = CI - CA = A In the same manner is found fE = CI + CA = BI. Q. E. D. Corol. 1. Hence CH = CI is a 4th propor. to CA, CF, CD. Corol. Corol. 2. And fE + FE = 2CH or 2c1; or FE, CH, fe are in continued arithmetical progression, the common difference being ca the semi-transverse. Corol. 3. From the demonstration it appears, that DE2= DH2 AG2 = DH2 ca2. Consequently DH is every where greater than DE; and so the asymptote CGH never meets the curve, though they be ever so far produced: but DH and DE approach nearer and nearer to a ratio of equality as they recede farther from the vertex, till at an infinite distance they become equal, and the asymptote is a tangent to the curve at an infinite distance from the vertex. THEOREM XV (11). If a Line be drawn from either Focus, l'erpendicular to a Tangent to any Point of the Curve; the Distance of their Intersection from the Centre will be equal to the Semitransverse Axis. That is, if FP, fp be perpendicular to the tangent TPр, then shall cp and cp be each equal to ca or cв. G B TA M For, through the point of contact e draw Fe and fe, meeting FP produced in G. Then, the GEP= ∠FEP, being each equal to the fep, and the angles at P being right, and the side PE being common, the two triangles GEP, FEP are equal in all respects, and so GE = FE, and GP = FP. Therefore, since FP = FG, and FC = +Ff, and the angle at F common, the side CP will be = f or AB, that is CP = CA or CB. And in the same manner cp = CA or CB. Q. E. D.. Corol. 1. A circle described on the transverse axis, as a diameter, will pass through the points P, p; because all the lines CA, CP, cpр, св, being equal, will be radii of the circle. Corol: 2. CP is parallel to fe, and cp parallel to FE. Corol. 3. If at the intersections of any tangent, with the circumscribed circle, perpendiculars to the tangent be drawn, they will meet the transverse axis in the two foci. That is, the perpendiculars PF, pf give the foci F, f. THEOREM 1 |