::

Cor. Hence CH' (CNVC): PN :: VC EC (cor. Art. 116.) and CH : PN :: VC : EC (22. 6.)

128. The same things remaining CN: RH:: VC: EC,

For (Art. 127.) VC +CH: RH :: CN - VC2 : PN2 :: (cor. Art. 127.) VC: EC, and VC2+CH2=CN2, ·: CN2 : RH2 :: VC2 : EC2 and (22.6.) CN : RH :: VC : EC. Q. E. D. 129. If CR be parallel to the tangent PT and PN, RH ordinates to the major axis, then will RH-PN2=EC2.

:

Because (Art. 128) CN RH2 :: VC2 : EC2 :: CN2VC2 : PN2 by subtracting antecedents and consequents VC2 : RH-PN2 :: CN2 –VC2 : PN2 :: VC2 : EC2, ·.· (14.5.) RHo -PN-EC2. Q. E. D.

Cor. Because Cr2 — Cv2 —RH2—PNo (34.1.)=EC2, and CN? —CH2=VC2 (Art. 127.), ·.· if CP be conjugate to CR, CR is also conjugate to CP.

130. If CP and CR be semi-conjugate diameters, then will CP2-CR2=VC2—EC2.

Because (Art. 127.) CN2-CH2=VC2, and (Art. 129.) RH-PN=EC", · by subtracting the latter from the former CN2 + PN2 – CH2 — RH2=VC2-EC. But (47.1.) CP2 = CN+PN, and CR CH2 + RH2, '.: (CN2 + PN2 — CH'+RH) CP-CR-VC-EC'. Q. E. D.

-

=

131. The same things remaining, if PL be drawn perpendicular to CR, then will CR. PL=VC.EC.

Draw Cm parallel to PL, then because (Art. 128.) CN : RH :: VC : EC, ·.· (16 5.) CN : VC :: RH: EC. But the triangles CTm, RCH (having the alternate angles RCH, CTm equal (29. 1.), and the angles at H and m right angles) are similar, and (4.6.) CT: Cm :: CR: RH, ·. (compounding the two latter proportions,) CN.CT (=by Art. 121.) VC2 : VC.Cm :: RH.CR: RH.EC :: CR : EC,' '.' (15. 5.) VC : Cm : : CR : EC, ·.· (16.6.)=CR.Cm=VC.EC; but Cm=PL (34. 1.), ·.· CR.PL =VC.EC. Q. E. D.

Cor. 1. Hence (16.6.) VC: PL :: CR: EC, and (22.6.) VC PL:: CR2 : EC2.

:

Cor. 2. Let VC=a, EC=b, CP=x, and PL=y; then because

Cor. 3. Hence, if tangents be drawn at the extremities of any two conjugate diameters (cor. 2. Art, 108.) a parallelogram will be formed, and all the parallelograms that can be formed by the tangents in this manner are equal to each other, as appears from the foregoing demonstration, being each equal to 2VC.2EC= VU.EK; see the figure to Art. 133.

132. If CA be a semi-conjugate to CP, then will FP.PS =CA3.

Let FP and

CA be produced to meet in R, and draw FY, SZ perpendicular to the tangent at P. Then the triangles FPY, PRL, and SPZ being equiangular, (4. 6.) FP: FY:: PR: PL and SP : SZ :: PR: PL, ·.· compounding

these proportions FP.SP FY.SZ

:

:: PR2: PL ::

(Art. 109.) VC2: PL2: (cor. 1. Art. 131.) CA: EC2. But (Art. 111.) FY. SZ=EC', . (14.5.) FP.SP=CAo. Q. E. D.

133. If through the vertex V the straight line ek be drawn equal and parallel to the minor axis EK, and from the centre C straight lines CM, Cm be drawn through e and k meeting any ordinate (PN) to the major axis, produced in M and m; then will PM.Pm=Ve". See the following figure.

Because (cor. Art. 116.) CN2 –VC2 : PN2 :: VC2 : EC and (4. and 22. 6.) CN2: NM2 :: VC2 : (Ve2=) EC2, (19.5.) VC: NM2-PN2 :: VC2 : EC2, ·. (14.5.) NM3 — PN2 EC2 = Ve2. But (cor. 5.2.) NM- PN2 = NM+PN.NM-PN=PM.Pm; PM.Pm Ve2. Q. E. D.

Cor. 1. Hence, in like manner pm.pM may be shewn to be equal to Vk2=Ve2, ··· PM.Pm=pm.pM; and if any other line

any finite distance it does not entirely vanish. For the same reason as pM increases, pm decreases; and at an infinite distance from C becomes infinitely small, but does not vanish; . CM and Cm continually approach the curve, but do not meet it at any finite distance, they are therefore asymptotes.

Cor. 1. Hence it appears that CM, Cm are likewise asymptotes to the conjugate hyperbolas; for Ve, Vk being respectively equal and parallel to EC, CK, :.' (33. 1.) Ee, Kk will each be equal and parallel to VC; and by the same reasoning it is plain that CM, Cm continually approach the conjugate hyperbolas, but do not meet them at any finite distance from the centre.

Cor. 2. If VE be joined, the right angled triangles VEC, VeC having CE=Ve and VC common, are equal in all respects (4.1.) ·.· VE=eC, and the angle CVE=VCe. In like manner it follows that VK=Ck, and since EC=CK (Art. 108.) ·.· the right angled parallelograms CEeV, CKkV are equal (36. 1.) and consequently similar, and the four diameters Ce, EV, Ck, KV are equal, . (cor. Art. 241. Part 8.) CD, De, ED, DV, CZ, Zk, KZ, ZV are equal to each other; and because Vk=CK = EC·.· (33.1.) EV and Ck are parallel; in like manner it is plain that KV and Ce are parallel.

135. The position of any diameter with respect to the axis being given, that of its conjugate may be determined. For (Art. 133.) NM-PN2=EC2, and (Art. 129.) RH2-PN2= EC2. NM=RH, ·.· if CP be a semi-diameter, PN an ordinate at P to the major axis produced to the point M in the asymptote, and MR be drawn parallel to the major axis, then if RC be joined, RC will be conjugate to CP by cor. to Art. 129. And in the same manner the position of the conjugate to any other diameter is known. Q. E. D.

136. If a straight line Xr be drawn in any position cutting the curve in Q9, and the tangent TPt be parallel to it, then will QX.Qx=PT.Pt. See the figure to Art. 141.

Through Q and P draw Ww, Zz perpendicular to the axis ; then the triangles XQW, TPZ, wQx, and zPt being similar QW : QX :: PZ: PT (4. 6.) and Qw : Qx :: Pz : Pt; these proportions being compounded QW.Qw : QX.Qx :: PZ.Pz : PT.Pt. But (cor. Art. 133.) QW.Qw=PZ.Pz·.· (14.5.) QX.Qx=PT.Pt. Q. E. D.

Cor. By similar reasoning qx.qX-PT.Pt, ; QX.Qx=qx.qX. 137. The same construction remaining QX=qx.

For QX.Qx=QX.Qq+qx=QX.Qq+QX.qx. And qx.qX= qx.qQ+QX=qx.qQ+qx.QX; ··· (since QX.Qx=qx.qX_by_the preceding corollary) QX.Qq+QX.qx==qx.qQ+gx.QX, from these equals take away QX.qx, and the remainders are equal, viz. QX.Qq=qx.Qq, divide both sides by Qq, and QX=qx. Q. E. D.'

Cor. Hence, if Xx move parallel to itself so as to coincide with Tt, the points Q and q will each coincide with P, and Qq will vanish; also QX and qr will coincide with, and be equal to TP and tP respectively; ·.· (since QX=qx) TP=tP, ·.· QX.Qx =TP.

138. The same construction remaining if through P, the diameter Go be drawn, Qu=qv.

Because the triangles XvC, TPC are similar, and also xvC, tPC; ·. (4. 6. and 16.5.) vX : PT :: vC : PC :: vx: Pt. But PT Pt by the preceding cor. (14.5.) vX=va. But (Art. 137.) QX=qx, .• vX—QX=vx—qx or Qv=qv. `Q. E. D.

Cor. Hence vX2 —vQ2 = PT1. For (cor. 5. 2.) vX3 —vQ2: vX―vQ.vX+vQ=QX.Qx=(cor. Art. 137.) TP2.

139. If PH, VD be parallel to an asymptote Cx, then will PH.CH=VD.CD See the figure to Art. 134.

Through the points V and P draw the straight lines ek, Mm each perpendicular to the axis CN, and Pd, Vv parallel to CX. Because the triangles PMH, VeD, Pdm, and Vek are similar, ·.· (4. 6.) PH : PM :: VD: Ve, and (Pd=) CH : Pm :: (Vv=) CD: Vk and by compounding PH.CH: PM.Pm :: VD.CD: Ve.Vk. But (Art. 133.) PM.Pm (Ve2=) Ve‚Vk, ́‚ (14.5.) PH.CH=VD.CD. Q. E. D.

Cor. 1. Hence, because (cor. 2. Art. 134.) CD=VD, ·: PH.CH CD2=VD2.

Cor. 2. Hence also, if PH be produced to meet the conjugate hyperbola in R, RH.CHED.CD=VD.CD= CD2 or VD2. Cor. 3. Hence, because PH.CH=(CD2=) RH.CH, by dividing these equals by CH, PH=RH.

140. If PT be a tangent at P meeting the asymptotes in 7' and X, and CR be joined, then will CR and TX be parallel and CR TP=PX.

For PH being parallel to CT one side of the triangle CXT, (2.6) PX: PT:: XH: HC. But (cor. Art. 137.) PX=PT, (prop. A. 5.) XH-HC;. In the triangles PXH, RCH there are the two sides XH, HP=CH, HR respectively, and the vertical angles at H equal (15.1.) ·.· PX=(PT=) CR; also the angle HRC=HPX (4,1.) ·.· CR and TPX are parallel (27.1.) Q. E. D.

141. If PG and DO be conjugate diameters, and Qu an ordinate to PG, then will Pv.vG : Qua:: CP: CD2.

At the point P draw the tangent PT, and produce the ordinate vQ to meet the asymptote in X.