Page images

p R



That is, if CR and cr be
Parallel to any two Lines
PHQ, pHq; then shall
CR?:cr? :: PH. HQ : pH . Hq.


[ocr errors]



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 șides, it is,

by sim. triangles, CR’ : GP? :: A CRI : A GPK, and

CR? : GH:: A CRI : A GHM; theref. by division, CR? : GP? GH* :: CRỊ: KPHM. Again, by sim. tri. CE : ch’:: ACTÉ: A CMH;

and by division, CE? : CE CH? :: A CTE: TEHM. 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? ch? :: CR2 : GP — GHor PH.HQ. In like manner CE : CE?

ch? :: cr2 : pH. Hq. Theref. by equ. CR’: cr- :: PH. HQ : PH . Hy.

Q. E. D. Corol. i. In like manner, if any other line p'u'd, parallel to cr or to pq, meet PHQ; since the rectangles PH'a, p'H'Q' are also in the same ratio of CRP to cra; therefore rect. PHQ : pHq :: PH'Q : p'H'q.

Also, if another line p'ha' be drawn parallel to PQ or CR; because the rectangles pha', phq are still in the same ratio, therefore, in general, the rect. PHQ : pHq :: P'ha': p'hq.

That is, the rectangles of the parts of two parallel lines, are to one another, as the rectàngles 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.


[merged small][ocr errors][merged small][merged small][merged small][ocr errors]

Corol. 3. And hence TE : Te ; : te : te.




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.



That is, FE + AC=CI, or FE=AI; and fe-ac=ci, or Je=BI. Where ca:CF :: CD: CI the 4th

propor. to CA, CP, CD.

f B


[ocr errors]
[ocr errors]


For, draw AG parallel and equal to ca the semi-conjugate; and join cg meeting the ordinate de produced in H. Then, by theor. 2, CAP: AG:: CD – CA?: De;

DE and, by sim. As, CAP : AG? :: CD – CA? : DH’ – AG? ; consequently DE?=DH - AG? =DH -- ca. Also FD = CF CD, and FD2 = CF - 20F.CD + CD’; but, by right angled triangles, FD? + DEʻ = FE?; therefore FE’ = CF? ca? 2CF.CD + cd? + DH”. But by theor. 4, CF2 ca? = CA’, and, by supposition, 2cF.CD = 2CA.CI; theref. Fe? = CA? 2CA . CI + CD + DH”. But, by supposition, CA: CD2 :: CF2 or CA? + AG: ci; and, by sim As, CA’ : CD2 :: CA? + AG? : CD + DH; therefore

CI2 = CD? + DH? = ch”; consequently FE’ = CA– 2CA .CI + CI. And the root or side of this square is Fe = CI In the same manner is found fe = CI + CA = BI. Corol. 1. Hence CH = ci is a 4th propor. to CA, CF, CD.


CA = AI.

Q. E. D.

[ocr errors]

Corol. 2. And ƒE + FE = 2CH or 2c1; or FE, CH, ƒE 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 - ́ca". 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.



If a Line be drawn from either Focus, Perpendicular to á
Tangent to any Point of the Curve; the Distance of their
Intersection from the Centre will be equal to the Semi-
transverse Axis.

That is, if FP, fp be perpendicular to the tangent TPP, then shall CP and cp be each equal to CA or CB.

For, through the point of contact E draw FE and ƒE, meeting FP produced in G. Then, the GEPFEP, 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 FPFG, and Fc = Ff, and the angle at F common, the side CP will be = fG or AB, that is CPCA 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, cp, CB, being equal, will be radii of the circle.

Corol: 2. CP is parallel to ƒE, 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.


[merged small][merged small][ocr errors][subsumed]
[ocr errors]


The equal Ordinates, or the Ordinates at equal Distances from the Centre, on the opposite Sides and Ends of an Hyperbola, have their Extremities connected by one Right Line passing through the Centre, and that Line is bisected by the Centre.

[merged small][merged small][ocr errors][merged small][merged small]

For, when CDCG, then also is DEGH by cor. 2 theor. 1.
But the 4D=4G, being both right angles;
therefore the third side CE = CH, and the LDCE 4 GCH,
and consequently ECH is a right line.

Coral. 1. And, conversely, if ECH be a right line passing through the centre; then shall it be bisected by the centre, or have CE = CH; also DE will be = GH, and CD = CG.

Corol. 2. Hence also, if two tangents be drawn to the two ends E, H of any diameter EH; they will be parallel to each other, and will cut the axis at equal angles, and at equal distances from the centre. For, the two CD, CA being equal to the two CG, CB, the third proportionals CT, cs will be equal also; then the two sides CE, CT being equal to the two CH, cs, and the included angle ECT equal to the included angle HCS, all the other corresponding parts are equal: and so the 4T=4s, and TE parallel to HS.

Corol. 3. And hence the four tangents, at the four extremities of any two conjugate diameters, form a parallelogram inscribed between the hyperbolas, and the pairs of opposite sides are each equal to the corresponding parallel conjugate. diameters. For, if the diameter eh be drawn parallel to the tangent TE or HS, it will be the conjugate to EH by the definition; and the tangents to eh will be parallel to each other, and to the diameter EH for the same reason.


If two Ordinates ED, ed be drawn from the Extremities E, e, of two Conjugate Diameters, and Tangents be drawn to the same Extremities, and meeting the Axis produced in T and R;


[ocr errors]

Then shall CD be a mean Proportional between cd, dr, and cd a mean Proportional between


[ocr errors]

Corol. 3.

For, by theor. 7, CD CA: CA: CT, and by the same, cd: CA: CA CR; theref. by equality, CD : cd :: CR : CT. But by sim. tri. DT: cd :: CT: CR; theref. by equality, CD: cd :: cd: DT. In like manner, cd: CD CD: dr. Corol. 1. Hence CD : cd :: CR : CT. Corol. 2. Hence also CD: cd :: de: DE. And the rect. CD. DE

cd. de, or A CDE = ▲ cde.

Also cd CD. DT, and CD2 = cd. dr. Or ed a mean proportional between CD, DT; and CD a mean proportional between cd, dr.


The same Figure being constructed as in the last Proposition, each Ordinate will divide the Axis, and the Semi-axis added to the external Part, in the same Ratio.

[See the last fig.]
That is, DA
and da

For, by theor. 7,
and by div.

and by comp.


In like manner,

DT:: DC: DB,
dR :: dc : dB.

CD: CA :: CA: CT,
CD: DB:: AD: DT,
DA: DT:: DC: DB.
da: dR:: dc: dB.

Q. E. D.

Q. E. D.

Corol. 1. Hence, and from cor. 3 to the last prop. it is,



CD. DT = AD. DB = CD2
and cd. dr = Ad. dB = CA2 cd2.

[ocr errors]

Corol. 2. Hence also CA CD2- cd2, and ca2=de2-DE2. Corol. 3. Farther, because CA2: ca2:: AD. DB or cď2: DE2. therefore CA ca: cd: DE. likewise CA

ca:: CD: de.


If fre






« PreviousContinue »