Cor. 1. Hence it may be shewn, as in Art. 72. that tangents at P and Q will meet the axis produced in the same point T; that the area VQN: area VPN :: eC: EC, and that if F be any point in the axis, the area VQF: area VPF :: eC: EC.

Cor. 2. Hence, if VQ be an equilateral hyperbola, or VC=eC (Art. 92.); then since VN. NU: QN :: VC eC? (Art. 116.) VN.NU QN (prop. A.5.)

:

118. In the equilateral hyperbola, the latus rectum is equal to the minor axis, that is 2Fb=2eC.

For since (Art. 105.) VF. FU-eC, if the point N be supposed to coincide with F, the expression (cor. 2. Art. 117.) VN. NU=QN' will become VF.FU=Fb2, ·· Fb2=eC2, Fb= eC, and 2Fb=2eC. Q. E. D.

Cor. 1. Hence it again appears that the major axis, minor axis, and latus rectum of an equilateral hyperbola, are equal to

each other.

Cor. 2. Hence, because (Art. 106.) VC: EC:: EC: BF, .. (cor. 2, 20. 6.) VC: BF :: VC: EC'. But (Art. 116.) VC" : EC2:: VN. NU or CN-CV2 : PN2, ·: VN. NU or CN2 – CV: PN2 :: VC: BF.

119. If Pn be an ordinate to the minor axis EC, then will Cn+EC: PN:: EC: VC (see the following figure,)

For (34. 1.) Pn= NC and Cn= NP (cor. Art. 116.) Pn -VC: Cn2:: VC: EC2, . by adding antecedents and consequents Pn Cn2 + EC2 :: VC: EC and by inversion Cn' +EC: Pn :: EC: VC2. Q. E. D.

120. If PN any ordinate to the major axis be produced to meet the conjugate hyperbola in II, then will пN2-PN2EC2.

bП be produced to meet the hyperbola VP in the point w, wba— Пb2=2VС. Q. E. D.

121. If PT be a tangent at the point P, then will CN.CT =VC2.

Because (cor. 2. Art. 107.) ST: TF :: SP: PF, ·.· dividendo et componendo) ST-TF: ST+TF:: SP-PF: SP+ PF, that is (see Art. 112.) 2CT: SF:: 2VC: SP+ PF. But (Art. 113.) SN-NF.SN+NF=SP-PF.SP+PF, ·.· since SN-NF-SF, SP-PF-2 VC (Art. 104.), and SN+NF= 2CN, by substitution SF. 2CN=2VC.SP+PF (16.6.) SF: 2VC:: :: SP+PF: 2CN; but it has been shewn that 2CT: SF :: 2VC: SP+PF: ex æquo 2CT: 2VC :: 2VC: 2CN, that is CT: VC :: VC : CN, ·.· (17. 6.) CN.CT=VC2. Q. E. D.

Cor. 1. Because NT-CN-CT, ... CN.NT=CN.CN-CT= CN-CN.CT=CN-VC".

Cor. 2. Because in the equilateral hyperbola CN2 – VC2 = PN (because VC-EC, see the cor. to Art. 116.) ·.· CN.NT= (CN1-VC2=) PN'.

Cor. 3. Hence also, in the conjugate hyperbola Еп, if pn be an ordinate to the axis Eg, and pT a tangent at p, then will Cn.CT=EC.

122. If Pn be an ordinate to the minor axis EC, and the tangent Pt meet EC in t, then will Cn.Ct=EC2.

Because (Art. 121.) CN.CT-VC2, '.' (17. 6.) CN : VC : : VC: CT,.. (cor. 2, 20. 6.) CN: CT :: CN2: VC", . (17.5.) NT: CT :: CN2 — VC2 : VC2 : : (because by cor. Art. 116. CN2 -VC2 : PN2 :: VC2 : EC2, by alternation) PN2 : EC2. But the triangles TPN, TtC are similar, . (4. 6.) NT: CT:: PN: Ct; (from above) PN: Ct:: PN2: EC2, ·.· (16. 6.) PN.EC1 =Ct.PN, or EC2= Ct.PN; But (34.1.) PN=Cn, ·: Cn.Ct= EC. Q. E. D.

Cor. Hence, because Cn.CT EC2 (cor. 3. Art. 121.) ··· Cn.Ct =Cn.CT and Ct CT; that is, if the perpendicular Pn cut the conjugate hyperbola in p, and tangents be drawn at P and p, the points t and T where they meet the minor axis, will be equally distant from the centre C; and conversely, if Ct=CT, the perpendicular Pn will pass through the point p.

123. The same things remaining nt: nỈ : : nP2 : np2.

For by the preceding corollary Cn. CT-EC2, '.' (17. 6.) Cn : EC :: EC: CT, ·.· (cor. 2, 20. 6.) Cn : Çİ :: Cn2 : EC2, ·¦ (componendo et dividendo) Cn + CT or nt: Cn-CT or nT:: Cn2 +EC2: Cn2 - EC2. But (Art. 119.) Cn2 + EC2 : Pn2: EC2 : V and (cor. Art. 116.) Cn2 - CE2: np2 :: EC2 : VC2 ·.·• (11.5.) CnEC: nP:: Cn2-EC Cn2+EC2 : Cn2 — EC2 :: nP2 : np2; that is, nt: nŤ :: nP2 : np. Q. E. D.

2

:

2

np2, : (alternando)

124. The normals at P and p will meet the minor axis in the same point g.

For the angles gp, gPt being right angles nP=nt.ng and np2=n'T.ng (14.2.) ·.· nP2 : np2 : : nt.ng : nÏng, `.' (Art. 123.) nt: nT::nt.ng: nỈ.ng : : ng : ng; that is, the normals at P and p cut the minor axis at equal distances from ʼn or in the same point g. Q. E. D.

Cor. In like manner it is shewn, that if NP be produced to meet the conjugate hyperbola in П, the normals from these points will meet the major axis in the same point G.

125. If CR be parallel to a tangent at P, and MPG perpendicular to it, then will the rectangle PM.PG=EC2.

Let PN be the ordinate, and draw Cm perpendicular to the tangent Pt. Because in the triangles PTG, CTt, the angles at

Produce GP, FP to M and R, then because the angles at H and M are right angles and those at P vertical, the triangles PHG, PMR are equiangular, and (4.6.) PG: PH: PR: PM, (16.6.) PH.PR=PG.PM (Art. 125.) EC2=(cor. 1. Art. 106.) L.VC. But (Art. 109.) PRVC, PH.PR= +L.PR, or PH L. Q. E. D.

127. If CR be parallel to the tangent at P, and PN, RH perpendicular to the major axis, then will CN-CH2=VC2.

Draw TR an ordinate to the minor axis, and produce it to Q, and draw the ordinate Qn. Then (cor. Art. 116.) Cn2-ČV: : Qn2:: CN3-CV: PN and Qr-VC: RH:: CN2 – CV: PN. But (Art. 120.) Qr2 — Rr2=2CV2, ·.· Qr2 — VC =VC2 + Rr2=(34.1.) VC2

t

ex æquo VC2 + CH2 : CH:: CN2-VC2: TN2:: (cor. 1. Art. 121.) CN.NT: TN' :: CN: TN, by conversion (prop. E. 5.) VC + CH2: VC :: CN: (CN-TN=) CT:: (1. 6.) CN VCCN.CT, (14.5.) VC +CH VC. Q. E. D.

CN.CT. But (Art. 121.)

CN, . CN-CH=;

Cor. Hence CH2 (=CN2—VC3) : PN2 :: VC: EC' (cor. Art. 116.), and CH : PN :: VC : EC (22. 6.)

ג

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

For (Art. 127.) VC2 + CH2 : RHa :: CN2 – VC2 : PN2 :: cor. Art. 127.) VC EC2, and VC+CH=CN, . 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.) CN2: RH2 :: VC2 : EC2 :: CN2. VC2: PN2 by subtracting antecedents and consequents VC": RH-PN :: CN2 –VC2: PN2 :: VC2 : EC2, ·.· (14.5.) RH2 —PN2-EC2. Q. E. D.

Cor. Because Cr2 — Cv2=RH2—PN2 (34.1.)=EC2, and CN2 -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—CH 2 · VC2, and (Art. 129.) RH3 —PN2 = EC3, ·.· by subtracting the latter from the former CN2 + PN2 – CH2 — RH2 — VC2-EC“. But (47.1.) CP*= CN + PN, and CR CH2 + RH2, ·. (CN2 + PN2 CH2 +RH2 =) CP2 — CR2=VC'—EC2. 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.) VCa : 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