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bet nyo = bb = DA'. Since NP = baz, and that MP* =****

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z at

a -bb, we have always NBZMP. Consequently the hyperbolic branch can never come in contact with its asymptote. However it perpetually approaches towards it, for as the absicissa

biz? increases, the difference between and 66 becomes less

a* sensible ; so that if we suppose z infinite, this differencc vanishes.

51. Draw MQ and AL parallel to the asymptote Cd. It is easy to see that the triangles DLA and LCA are isosceles. Let then AL=DL=m, CQ=x, QM=y. If we draw MK, parallel and equal to CQ, because of the similar triangles DLA, NQM, M'Kn, we shall have the proportions MN: DA :: QM: LA

Mn:DA :: MK : DLTherefore Mn X MN: DA? :: QMMK: AL'. But Mn * MN= DA', hence xy=mn, an equation to the hyperbola referred to its asymptotes, in which mm = + (aa+bb) is called the power of the

$ hyperbola.

52. If two parallels Ff, Gg, terminated at the asymptotes cut an hyperbola in the points m, h, p, K, we shall have Gpx pg =Fm X mf. For if we draw MmN, PpQ, perpendicular to the axis, we shall have

Fm: Mm :: Gp :Pp and mf : mN :: pg : PQ therefore Fm X mf : Mm XmN :: Gp x pg : Pp x pQ. Х

But (50) Pp XpQ=6=Mmx mN. Consequently Fm X mf=Cpx

E pg. And therefore also Kg x KG=fhxh F. 53. If we suppose that the

T
points p, K coincide in a single
point D, the line TDt will be
a tangent to the point D, and
we shall have Fm x mf=Dt X

FA
DT and fh XhF=Dix DT-
Fm X mf, therefore th (hm +mF)=

Fm (mh+hf); therefore th= Fm, and consequently TD=Dt. But if we draw DE parallel to Cl, or an ordinate to the asymptote CT, the similar triangles TDE, TiC, will give TE= EC. Consequently to draw to the hyperbola a tangent to a point D, corresponding to the ordinate DE, we must take ET=EC and draw through the points T and D, the tangent TDt.

54. Since we always have fh=Fm, in whatever manner we draw the straight line Ff, the two parts Fm, fh intercepted between the curve and its asymptotes will always be equal.

a

:

:

M

Hence we derive an easy manner of describing an hyperbola between two given asymptotes CT, Ct, and passing through a given point m.

Draw through this point the right lines Ff, MN, &c. and take fh =Fn, nN=Mm, &c, and the points n, h, &c. will be in the hyperbola.

55. According to what we have seen (58), 2 tangent TMt, terminated at the asymptotes is bisected at the point of contact M. If we draw MCM', this line is called a diameter; the tangent TMt is its

M conjugate diameter. Its ordinates are the right lines mQm' parallel to the conjugate diameter TMt or Ďcd, and the parameter of any diameter is a line, a third

С proportional to that di

D ameter and its conju

11

7 gate. The line MCM = 2CM is also called the first or principal

B

K diameter, and the line

м.

H TMt = 2TM = DCd = 2DC is the second

T

G conjugate diameter.

56. This premised, it is easy to see that a

N diameter bisects all its ordinates. For NQ: Qn ::TM : Mt, and Nm=mn. Call then CM=m, CD=TM=, CQ= ,,Qm=y; and we shall have min::Z:NQ=

But Nm X mn =TM Hence n =

na z

yy =

(28-m*), an equama

m' tion similar to that of the co-ordinates to the transverse axis. This equation gives z =

ma

(yo+n). Consequently if we make

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nz

m

yy, and

Cp=, pm=y, we shall have yy=* (zo+n*) for the equation to the co-ordinates of the second diameter CD; ít is easy to perceive the analogy which this equation has to that of the co-ordinates to the conjugate axis.

57. Let now aCA be the transverse axis of the hyperbola, and suppose that BA represents the semiconjugate axis ; if we draw

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DE, TG, MPK, perpendicular to this axis, and ML, tK parallel to it, the triangles MTL, MK, CDE, will be equal and similar. If we call CP=u, PM=, CE=K=ML=r, MK=DE=TL=s, and as before CM=m, TM=u, CA=a, AB = b, we shall have TG2+s, CG = u+r, and z + siu+r::b: a, and consequently as +as=bu + br; besides TL : ML:; MP: PS, or s:r::%: and u? _a?

aasz PS=

Therefore r = -; substituting this

62 u value in the equation az +as=bu+br we have (bu-as) (64-az)

Now bumaz cannot be equal to zero; therefore bu-as=o. Hence bu=as, and consequently az=br, Therefore we have CP: DE::a:6::CE: MP.

58. Consequently Io the triangles CED, CMP, are equal in surface II°. If we draw DM, we shall have DMC, or s CDTM=to the

811 tuz-S1-z trapezium DEMP=(s+2)( ) =

2

2 b bbuu-aazz a z? 62

= 1 ab; therefore the parallelo2a 26

2ab

2ab gram TT constructed on the conjugale diameters is equal to the rece tangle of the axes.

62 IIIO. DE=CCP

bce
Hence DEP = CP=6*+PM', and

a DE'— PM=6.

IV°. CE = MP, and CE:= MP=CP-a. Consequently CP_CE'=a'.

Vo. a'_62=CP + PMS-DE-CE:=CM_CD". Therefore the difference of the squares of any two conjugate diameters is equal to the difference of the squares of the two ares. Hence also in the equilateral hyperbola the conjugale diameters are equal.

59. Call P the angle DCM comprized between the two conjugate diameters; we shall have the two equations mn sin P=ab, and m_no=a_62; by means of these two expressions it will be easy to resolve the two following problems.

Problem 1. Given the two axes a and b of an hyperbola, to find two conjugate diameters making with each other a given angle P.

The preceding equations give m =N

+ 4

sin :P a'-6")* a' b*

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=

a ba

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+

sin P

There now only remains to find the direction of one these diameters, or the angle MCP which I shall call C. In the triangle CMP

am

am sin C we have 1 :m::sin C: PM=m sin C. Therefore CE

6 and in the triangle DCE we have 1:n:: cos (P+C):

am sin C

6 Consequently sin C = cos P cos C-sin P sin C, and tang C=

bn bn cos P 62

cot P, because a=

mn sin P am + bn sin P-6*+ma

6 Problem 11. Given two conjugate diameters m and n of a hyperbola, and the angle P which they make with each other, to find the two axes and their direction?

We have first

a

=v{}(m=») + } v (+*+2mn* +n*Amo n° cos "P) } °=v

=v{1{m?n") + } v ((mo+n*)?—4 m n cos *P)} of=v{}(n=m*

) + v ((m*+n) 44m*n* cos
(* v op

*P)}
=v{ } (no—m) + } v ((mon") + 4 mono sin “P)}

and then

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Lastly, we find the direction of the axes, or the angle C, as in the preceding problems.

But it is better to employ the asymptotes. Through the extremity M of the principal diameter CM, draw TMt making with MQ, the angle ÎMQ=P, and having taken TM=M=n, draw CT, Ct. Then if we bisect the angle TCt by the line CA, we shall have the direction of the transverse axis.

OF SEVERAL OTHER CURVES.

Among the curves most frequently used in geometry the conic sections undoubtedly occupy the first rank : but there are several other curves of which it will be necessary to give some account.

I. THE CONCHOID OF NICOMEDES.

a

60. If through a point B taken at pleasure without a straight line GH, we draw the lines BQM, BAD, &c. such that their parts QM, AD, &c. be equal, the curve MDM which passes through the points M, D, &c. is called a conchoid. The point B is its pole, the line

M D
GH is the directrix ; and if below

M
GH we take the equal parts Qm,
Ad, &c., the curve which passes H

P Q A
through the points m, d, &c. deter-
mined in this manner, will be the
inferior conchoid, or rather the

ld

mil lower part of the same conchoid.

61. From its construction it follows, I', that H is the asymp

B tote of the curve; II° that dD measure its greatest breadth, when BA is perpendicular to GH. But as BA may be greater, less, or equal to dA, let us investigate the figure of the curve in these three

In the first case it will be such as represented in the preceding figure.

M D Á In the second case, it will have a node Bndn'.

H A

G

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cases.

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