T

M

For the triangles SPt, PRM, FPT are similar, because TF, PM, and tS are parallel, the angles at T, M, and t right angles, and TPF=tPS (cor. 3. Art. 57.)=PRM (29.1.); · SP : St:: PR: PM, and FP: FT :: PR : PM (4.6.), these analogies being compounded (prop. F. 5.) SP.FP: St.FT :: PR: PM2. But (Art. 78.) VC.EC= OC.PM, (VC by Art. 60.) PR: PM :: OC: EC (16. 6.); and PR2 : PM2 : : OC: EC' (22. 6.); . from above SP.FP : St.FT: OC: EC; but St.FT= EC2 (Art. 61. B.) ·.· SP.FP = OC2 (14.5.) Q. E. D.

t

E

N

R

S

80. Let OX be the conjugate and Qv an ordinate to the diameter PG, then will Pv.vG : Qu2 : : PC2 : CO3.

Draw PN, vn, QH, and Om perpendicular to the axis VU, and vr parallel to it. Then because PN is parallel to Qr, vr to TN, and Qu to PT, the triangles PTN, Qur are equiangular, and (4. 6.) Qr: (rv=by 34. 1.) Hn::

2

CN (22. 6.). But (cor. 1. Art. 67.) VC-CH: QH :: VC' -CN3: PN (being each as VC2: EC3) : ex æquo (22.5.)

CN

VCCH: Hn+Cn: VC- CN CN :: (cor. 2.

NT

:

Art. 72.) CN,NT. CN:: (15. 5.) NT: CN; (since CN:

VC :: VC: CT by cor. 1. Art. 72.; whence, by cor. 2, 20. 6,

CT

CN: CT :: CN2 : VC1= .CN1) VC2 — CH' or its equal

CN

2

CN2-Cn-Hn2 = ..Cn+· Hn (16. 6, and part 4.

CT

NT

CN

CN

CN

NT

CT

CT

actually squaring and multiplying ;) ·.· CN2

Cn'=

CN

CN

CT

NT

CN

NT

Hn' (by reduction, and from the figure); ·.· CN2 – Cn2 =

CT

. Hn2 (by dividing by CN), or NT.CN2—Cn2=CN.Hn2 ; '.' (16. 6.) CN2 —Cn2: Hn2 :: CN: NT :: (by inversion in the 7th analogy, above) CN: VC-CN2; ... (16.5.) CN2—Cn2 : CN: Hn: VC2-CN2; but (2.6.) CN: Cn :: CP: Cv, : CN2 - Cn2: CN2 :: CP2-Cv2: CP2 (part 4. Art. 69.). Also, (by similar triang. and 22. 6.) rv2=Hn2: (Cm2=by Art. 75.) VC2—CN2 :: Qv2 : CO2; ·.· (CP2 — Cv2=cor. 5.2.) Pv.vG : CP1 :: Qv2 : CO2, and (16.5.) Pv.vG : Qv2 :: PC2 : CO2. Q. E. D.

Cor. Hence it may likewise be shewn by similar reasoning, that if Qv be produced to meet the curve again in q, Pv.vG : qv2 :: PC: CX2, ·: Qu: qu:: CO: CX. But CO=CX (cor. Art. 58.), ·.· Qv=qv.

81. The parameter P to any diameter PG is a third proportional to the major axis and conjugate diameter; that is, VU : OX:: OX: P.

Let the ordinate Qv passing through the focus F meet the curve again in q; then will Qq be the parameter to the diameter PG, and (cor. Art. 80) Qu=4P. Because (Pv.vG=) PC1—Cv2 : Qv2 :: PC2 : CO2 (Art. 80.) *: Qv2: PC2 —Сv2 :: CO2 : PC2(prop. B.5.) But because Ce is parallel to tFo (Art. 60.) Pe=VC, ·: PC1—– Cv1 : (Pe-Se1) Pe2—er2 :: Pv2 : Pr2 ;; PC2 : Pe1 ·· ́ ex æquo (22.5.) Qv2

: Pe2 -er :: CO1 : (Pe2=) VC. But Pe — (Se2 =) er2 = Peter. Pe-er (cor. 5.2.)=(Art. 60.) CP. SP=(Art. 79.) CO2; : Qua : CO :: C02 : FC* and (92. 6.) Qu : C0 :: CÓ : FC, ·.: (15. 5.) 2Qv : 2C0 : : 2CO : 2VC, that is P : 0X :: 0X : VU or VU: OX:: 0X: P. Q. E. D.

82. If two ellipses RPZ, RQZ have a common diameter RZ, from any point N in which NP and NQ an ordinate to each of them be drawn, then will the tangents at P and Q meet the diameter RZ produced in the same point T.

Draw TP a tangent to the ellipse RPZ and join TQ; TQ shall be a tangent to the ellipse RQZ. For if not, let TQ meet the curve again in g and draw the ordinates nq, np and produce np, TP to meet in r. Then PN2 : pn2 :: RN.NZ : Rn.nZ :: QN3 : qn2 (cor. 3. Art. 70.), ·.· PN : pn :: QN : qn (22.6.). But the triangles PNT, rnT are similar, as are also QNT, qnT; ·.· PN : rn :: NT: nT (4.6.) :: QN : qn, ·.· PN: : pn :: PN: rn (11.5.), ':' pn=rn (14.5.), the less equal to the greater, which is absurd; meets the curve no where but in Q, consequently touches it in Q. Q. E. D.

TQ

Cor. Hence, if RZ be bisected in C, the point C will be the centre of both ellipses, and (cor. 1. Art. 72.) CN: CR :: CR: CT.

0

Z

83. If RPZ be an ellipse, of which RZ is a diameter, and if from every point in RZ, straight lines QN be drawn, having any given ratio to the ordinates PN, and cutting the diameter RZ in any given angle, then shall the curve passing through R, Z, and all the points Q be an ellipse.

For since by hypothesis PN: QN :: OC : oC, (22.6.) PN2 : QN2 :: OC" : oC. But (Art. 80.) RN.NZ: PN2 :: CR2: OC, ex æquo (22.5.) RN.NZ: QN :: CR2: Co which (by Art. 80.) is the property of the ellipse; the curve RQOZ is an ellipse. Q. E. D.

84. If PQMG be the circle of curvature at the point P in the ellipse PVU, PG the diameter of curvature, and PH, Pt

the chords of curvature passing through the centre C, and focus Frespectively; then will

CP: CÓ :: CÓ: PH.
PK CO CO : PG.
VC : CÓ :: CO: P.

T

Join PC and produce it to M, and join Gv, HQ and QP; draw the tangent TP, and through Q and C draw Qr, OCK each parallel to TP, then will OC be the semi-conjugate diameter and Qr an ordinate to PH, and let QP be the arc in its nascent state, which may therefore be considered as common to the circle and ellipse. Then because the angle TPQ=PHQ ▼ (32.3.) PQr (29. 1.) and QPr is common to the two triangles QPr, QPH, these triangles are équiangular (32.1.),'. Fr: PQ :: PQ: PH (4. 6.); . Pr.PH =PQ'(since the arc QP is

n

indefinitely small, see Art. 35.) Qr2; ··· Pr.rM : Pr.rH :: PC2 : CO2 (Art. 80.), '.' (rM : rH, that is since r and P are indefinitely near coinciding) 2PC: PH : : PC2 : CO2; ·.· (15. 5.) PC : ¿PH :': PC' : CO2, '.' (cor. 2, 20. 6.) PC : CO : : CO : 4PH. Since CK is parallel to TP, and TP perpendicular to PG (cor. 16. 3.), CKP is a right angle (29. 1.), also PHG is a right angle (31.3.), and the angle HPG common to the triangles PKC, PHG. these triangles are equiangular, and (4.6.) PK : PC:: PH: PG :: †PH : PG. But PC: CO :: CO : PH, ex æquo (22.5.) PK PC:: CO: PH, and PC : CO :: PH: 4PG, ·: PK : CO :: CO: 4PG. Again, the triangles PnK, PvG having the angles at K and v right angles, and the angle at P common, are similar (32.1.); ·.· (Pn=by Art. 60.) VC: PK :: PG : Pv : : †PG : †Pv, and PK : CO :: CO: PG. ex æquo VC: CO: CO: Pv. Q. E. D.

:

Cor. Hence VU: 2CO :: 2CO: Pv, that is, the chord of curvature Po which passes through the focus F, is a third proportional to the major axis, and the conjugate diameter, and is consequently equal to the parameter of the diameter PM. (Art. 81.)

85. If a plane cut a cone so as neither to meet the base nor be parallel to it, the section will be an ellipse.

A

Let ABD be a cone, and let the section VEUK be perpendicular to ABC the plane of the generating triangle, VU being their common section, and the section PcQd be parallel to the base and therefore a circle, and let its common sections with ABD and VEUK be cd and PQ; let aЕKb be a section likewise parallel to the base, bisecting VU in C, having EK and ab for its common sections with the planes VEUK and ABD. Because ABD and PcQd are both perpendicular to VEUK, their common section U PQ is perpendicular to ABD (19. 11.) B and therefore perpendicular to VU and cd (conv. 4. 11.), in like manner

α

ν

C

K

D

it may be shewn that EK is perpendicular to VU and ab, ·: EK and PQ are bisected in C and N (3. 3.); and since cd and ab are parallel (16.11.), . the triangles UNc, UCa are equiangular, and UN: Ne: UC: Ca, also NV : Nd :: (CV=) UC ; Cb, ':' by compounding the terms of these analogies UN.NV: Nc. Nd :: UC :Ca. Cb. But Nc.Nd PN and Ca.Cb=EC2 (14.2.), ·: UN.NV : PN2 :: UC2 : EC which (by Art, 67.) is the property of the ellipse; therefore VEUK is an ellipse, consequently if a cone be cut by a plane which neither meets the base nor is parallel to it, the section will be an ellipse. Q. E. D.

THE HYPERBOLA.

DEFINITIONS.

86. If two straight lines FP, SP revolve about the fixed points F and S, and intersect each other in P, so that SP-FP may always equal any given straight line Z, the point P will describe the figure PVR which is called AN HYPERBOLA.

87. If two straight lines Fp, Sp revolve in like manner about F and S, so that Fp-Sp may always equal the given straight line Z, the point p will likewise describe an hyperbola pUr; this figure and the former, with respect to each other, are called

OPPOSITE HYPERBOLAS.