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These Proportions are the common Property of every Hyperbola, and do only differ from those of the Ellipsis in the Signs + T

and “-; as plainly appears in the following Proportions. That is, if we suppose T S the Transverse Diameter

common to both Sections (viz. both B


the Ellipsis and Hyperbola) as in the annexed Scheme: then in the Ellipfis it will be TS-Sax Sa : ab: :TSSA

XSA: 0 A B as by Sect. 1. Chap. 2. be

and in the Hyperbola it_is TS + Se x Sa: ab :: TS+SAXSA: 0

A B, as above. Therefore all, that is F

farther requir’d in the Hyperbola, may

(in a manner) be found as in the Ellisfis, due Regard being had to changing of the Sines.

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Sect. 2. To find the Latus Hirdum, or Right Parameter,

of any Hyperbola.

From the last Proportion take either of the Antecedents and its Confequent, viz, either Tax Sa: O ab. Or T'AXSA: 0 AB, to them bring in the Transverse TS for a third Term, and by those three find a fourth Proportional (as in the Ellipsis) and chat will be the Latus Re&tum.

Stax Sa: 0 ab::TS:
Thus I

Tax Sa

= the Latus 1 Rectum, which call L (as in the Parabola.) Then 2 TS: L::Tax Sa: ab.

But 3 Tax Sa:ab::TAXSA: 0 AB, therefore 2, 31 41TS:L::TAXSA: 0 AB, &c.

Consequently L is the true Latus Reflum, or right Parameter, by which all the Ordinates may be found, according to its Definition in Chap. 1. And because TS+ Sa= I a, let it be TS+

Baby TS Sa instead of Ta, then it will be

= L and in the TSXSatosa

Dabx TS
Ellipfis it would be
T'S X Sarosa


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Sect. 3. To find the focus of any Hyperbola. The Focus being that Point in the Hyperbola's Axis through which the Latus Recium must pass (as in the Ellipfis and Parabola) it may be found by this Theorem.

To the Retangle made of half the Transverse into

į half the Latus Rectum, add the Square of half the Theorem. Transverse ; the Square Root of that Sum will be the

Distance of the Focus from the Centre of the Hy(perbola.

Demonstration. Suppose the Point at F, in the annex'd Scheme, to be the Focus fought, then will FR=L. Let TC=

C S be half the Transverse; then is the Point
C call'd the Center of the Hyperbola (for a
Reason that shall be hereafter Thew'd.) A-

C gain ; let CS=d. and SF=a

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4 w

Then 1/2d:L:: 2d taxa: LL

S That is, 2 TS:L::TS+SFXFS: FR

R I 3d L=2da + aa 3+ dd 4 dd + dl=dd + 2da+aa ww? 5 vad+id=d+a= FC

BA Or 53-d16] v dutid-d=a=SF In the Ellipsis tis, 2d: L :: 2d-axa: LL. that is, įd L= 2da -- a a, &c.

The Geometrical Effection of the last Theorem is very easily perform’d, thus : make Sx = {L, viz. half the Latus Rectum; and let CS=d, as above. Upon Cx (as a Diameter) describe a Circle, and at the Vertex of the Hyperbola draw the Right Line n SN at Right Angles to Cx; then join the Points CN with a Right Line, and 'twill be CN=dta=FC.


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Now here is not only found the Distance of the Hyperbola's Facus, either from its Center C, or Vertex S, but here is also found that Right Line usually call'd its Conjugate Diameter, viz. the Line n SN, which bears the fame Proportion to the Transverse and Latus Rectum of the Hyperbola, as the Conjugate Diameter of the Ellipfis doth to its Transverse and Latus Rectum. For in the Ellipfis TS:Nn::Nn:L R. per Sect. 2. Pag. 363. Consequently TS: 1 Nn:: Nn: LR. But TS d, Nn=SN, and į LR={L. Therefore d:SN::SN: L. As at the 2d Step above.

What Use the aforesaid Line n S N is of, in Relation to the Hyperbola, will appear farther on.

1 2

Sect. 4. To describe an Hyperbola in Plano. In order to the easy describing of an Hyperbola in Plano, it will be convenient to premise the following Proposition, which differs from that of the Ellipfis in Sect. 3, Chap. 2, only in the Signs.

rif from the Foci of any Hyperbola there be draws

i two Right Lines, so as to meet each other in any Proposition. Point of the Hyperbola's Curve, the Difference of

| those Lines (in the Ellipsis 'tis their Sum) will beco

Lqual to the Transverse Diameter. That is, if F be the Focus, and it be made Cf=CF (as in the laft Scheme) then the Point f is said to be a Focus out of the Section (or rather of the opposite Section) and it will be f BFBTS.


Demonstration. Suppose fC, or CF=%, and S A=x, let C S, or TC=d, as before; then will fA=d+x+%, and FA=d++- 2. Again, let FB = h, and fB=b, then 2d=h-b, by the Proposition.

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From these substituted Letters it follows;
That|1dd + 2x + 2dz + 4x + 22x + zz=of A
And 2 dd + 2dx 2dz + xx —

22x + zz = 0 FA

OfA+AB=OfB, and of Ato AB=OFB 4th

} 13dd+:d L=da+da+aa =D FC=zz.

3 - dd

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= L


- dd

20 :

3 dd: 4 ZZ — - dd=dL

-dd 4 - 5

d Again 62d: L;:2d.+ * X*:0 AB, by common Properties, 5, 6 7

:: 2dx + xx: 0 AB 2dzzx+xzxx -2dddxddxx 7 8

= 0 AB dd

dd+2dx+2dz txx+2zx+zzt it8 9 2dzzxzzxx-2dx-ddxx

* =OfATO AB=hle dd

dd +2dx-zdz+xx-22x+z+ 2 +8/10 2dzzx+zzxx - 2dx-ddxx


dd 9+d11d++2d5z+2ddzx+ddzz+2dzzx+xzxx=ddhh 10 X dd 12d+--2diz-2ddzxt-ddzz+2dzzx+zzxx=adbb 13 dd +dz+zx=dh

r Altho' the Æquation at the 16th 1 2 ? 14 |dd -dz-zx=db

Step be in itself impossible, be13; d 15 dtzt b

cause z is greater than d (by the d

4th Step) yet from thence it will 14 • d 160-2

< be easy to conclude, that the Difb

ference between d and z to 16, or 172+ -d=6

will give the true Value of b; as 15-1711812d =h6

i in the 17th Step


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But because I would leave no Room for the Learner to doubt about changing the Æquation, d. ** = b into that of

+ min -d=b, it may be convenient to illustrate the whole Process in Numbers, whereby (I presume) 'twill plainly appear that b-b=TS.

In order to that, let the Transverse TS=2d=12, then d=6 suppose the Abscissa SA=*=4, and the Semi-ordinate AB=3 First | 1 | TS+SAXSA: AB::TS: L, per Seet 2.

2 12 +4X4=64:9 :: 12 : 1,6875=L Again 1 3 v ddtidL=dta=CF, per Sect. 3. 3, viz. | 4 | 36+5,0625 = 6,408 =CF=% Then 5 d +x+x=6+4 + 6,408 = 16,408 =ÍA And 60+*-%=6+4-6,408 = 3,592 = FA

i, viz.

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But 9

269,2224 = Of A
8 12,9024 = OFA

9 = 0 AB, for A B = 3 by Supposition,
+ 9 10 278,2224 = Of A to A B = OfB
+911 21,9024 = OFAAB= DFB

12 16,68 - fB
4,08 = FB

12-13/14) 12,00 = f B-FB=TS. Which was to be prov'd.



Io *
II w?

If this proposition be truly understood, it must needs be eaty to conceive how to describe the Curve of any Hyperbola very readily by Points, when the Transverse Diameter and the Focus are givea (or any other Data by which they may be found, as in the precedent Rules) thus :

Draw any streight Line at Pleafure, and on it set off the Length of the given Transverse IS, and from its extream Points or Limits, viz. TS, fet off Tf=SF, the Distance of the given Foaus (viz. the Pointf without, and F within the Seation, as before); that done, upon the Point f (as a Center) with any affum'd Radius greater than TS, defcribe an Arch of a Circle ; then from that Radius take the Transverse TS, making their Diffe. rence a second Radius, with which, upon the Point F within the Section, describe another Arch to cut or cross the first Arch, as at B; then will that Point B be in the Curve of the Hyperisla, by the last Proposition. And therefore 'tis plain, that, proceeding on in this Manner, you may find as many Points (like B) as may be thought convenient (the more there are, and nearer they are together, the better) which being all join'd together with an even Hand (as in the Parabola) will form the Hyperbola requir’d.

There are several other Ways of delineating an Hyperbola in Plano: One Way is, by finding a competent Number of Ordi. nates, as by Section 1, &c. but I think none so eafy and expeditious as this mechanical Way: I fhall therefore, for Brevity's Sake, pafs over the reft, and leave them to the Learner's Practice, as being easily deduced from what hath been already laid,

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