The two triangles CET, CAN have then the angle c common, and the sides about that angle reciprocally proportional; those triangles are therefore equal, namely, the A CET = A, CAN, Corol. 1. From each of the equal tri. CET, CAN, take the common space CAPE, and there remains the external ▲ PAT = ▲ PNE. Corol. 2. Also from the equal triangles CET, CAN, CED, and there remains the ▲ TED trapez. ANED. THEOREM XIX (19). The same being supposed as in the last Proposition; then any Lines KQ, QG, drawn parallel to the two Tangents, shall also cut off equal Spaces. That is, For, draw the ordinate DE. Then The three sim. triangles CAN, CDE, CGH, are to each other as CA2, CD2, CG2; th. by div. the trap. ANED: trap. ANHG :: CA2- CD2 : CA2 — CG2. DE2: GQ2::CA2-CD2: CA2 — CG2. TED ::ANHG : GQ2. : GQ2; : KQG. But, by theor. 1, theref. by equ. trap. ANED: trap. ANHG:: DE2 But, by sim. As, tri. TED: tri. KdG :: DE2 theref. by equality, ANED : But, by cor. 2, theor. 18, the trap. ANED = ▲ ted; and therefore the trap. ANHGA KQG. In like manner the trap. ANhg = A Kqg. Q.E.D. Corol. 1. The three spaces ANHG, TEHG, KQG are all equal. Corol. 2. From the equals ANHG, KQG, take the equals Anhg, K4g, and there remains ghнG = gqQc. Corol. 3. And from the equals ghнG, gqqG, take the common space gqLHG, and there remains the ▲ LQH = A Lqk. Corol. 4. Again from the equals KQG, TEHG, take the common space KLHg, and there remains TELKA LQH. Corol. Corol. 5. And when, by the lines KQ, GH, moving with a parallel motion, KQ comes into the position IR, where CR is the conjugate to CA; then the triangle N r I A KQG becomes the triangle IRC, and the space ANHG becomes the triangle ANC; and therefore the A IRC A ANC =A TEC. Corol. 6. Also when the lines KQ and HQ, by moving with a parallel motion, come into the position ce, Me, the triangle LQH becomes the triangle ceм, and the space TELK becomes the triangle TEC; and theref. the A ceM A TECA ANCA IRC. THEOREM XX (20). Any Diameter bisects all its Double Ordinates, or the Lines drawn Parallel to the Tangent at its Vertex, or to its Conjugate Diameter. That is, if aq be parallel to the tangent TE, or to ce, then shali LQ = Lq. H T For, draw aн, qh perpendicular to the transverse. Then by cor. 3 theor. 19, the A LQH = ▲ Lgh; but these triangles are also equiangular; consequently their like sides are equal, or LQ=Lq. Corol. Any diameter divides the ellipse into two equal parts. For, the ordinates on each side being equal to each other, and equal in number; all the ordinates, or the area, on one side of the diameter, is equal to all the ordinates, or the area, on the other side of it. THEOREM THEOREM XXI (21). As the Square of any Diameter: Is to the Square of its Conjugate :: That is, CE ce2:: EL. LG or CE2 CL2: LQ2. For, draw the tangent TE, and produce the ordinate a to the transverse at K. Also draw Qн, ем perpendicular to the transverse, and meeting EG in H and M. Then, similar triangles being as the squares of their like sides, it is, or CL2. Q. E.D. Corol. 1. The squares of the ordinates to any diameter, are to one another as the rectangles of their respective abscisses, or as the difference of the squares of the semidiameter and of the distance between the ordinate and centre. For they are all in the same ratio of CE2 to ce2. Corol. 2. The above being the same property as that belonging to the two axes, all the other properties before laid down, for the axes, may be understood of any two conjugate diameters whatever, using only the oblique ordinates of these diameters, instead of the perpendicular ordinates of the axes; namely, all the properties in theorems 6, 7, 8, 14, 15, 16, 18 and 19. THEOREM XXII (22). If any Two Lines, that any where intersect each other, meet the Curve each in Two Points; then The Rectangle of the Segments of the one : That 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: GP: A CRI: A GPK, CR2 GH2:: A CRI: A GHM; and GP2 Again, by sim. tri. (E2: CH2:: A CтE: ▲ CMH; 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: CE2 'CH :: CR2: GP2- GH2 or PH.HQ. Cн2: cr2: pн. нq. In like manner CE2: CE2 Theref. by equ. CR2: cr2 :: PH. нQ: рH. нq. 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, р'H'q' are also in the same ratio of CR2 to cr2; therefore rect. PHQ: рHq: PH'Q: p'H'q. Also, if another line p'ha' be drawn parallel to rQ or CR; because the rectangles p'ha', p'hq' are still in the same ratio, therefore, in general, the rect. PHQ: рHq:: P'ha': 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. Corol. 3. And hence TE : Te :: tE: te. 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 CG meeting the ordinate DE produced in H., 2 but, by right angled triangles, Fd2 + de2 = fe2; therefore FE2 = CF2 But by theor. 4, and, by supposition, 2CF. CD = 2CA. CI; consequently And the root or side of this square is FE CI AI. In the same manner is found JE CICA BI. Q. E. D. Corol. 1. Hence CHCI is a 4th propor. to CA, CF, CD. Corol. |