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Therefore in the hyperbola, the squares of the ordinates of the transverse acis are to the rectangles of their aðscisso, (that is of the distances to the two vertices,) as the square of the second or conjugats axis is to the square of first or transverse axis. If we take the centre Č for the origin of the abscissæ, then cal
66 ling CP=z, we shall have yy=(22-aa,) this equation is some
what more simple than the preceding one. It gives zz= (66+yy);
სს therefore if we draw MQ perpendicular on the conjugate axis CB, prolonged if necessary; and if we call the co-ordinates CQ, MQ,
66 for the equation to the co-ordinates of the second axis.
42. If a=b, the hyperbola is said to be equilateral, and its equations are yy=20x+xz, and yy=x2-aa, according as the origin of the abscissæ is at the vertex or at the centre; and the equation to its second axis becomes yy=aa+22.
43. If, having drawn BA, we take on both sides of the centre C, CF =CJ=BA, the points F, f, are the foci of the hyperbola. The double ordinate Dd passing through one of the foci, is called the parameter, and a line FM, or f M drawn from one of these points to any point of the curve is called the Radius vector. This premised, the distance CF=v(aa+bb). Hence FA X Fa= ) x
Conseqnently the semi-conjugate axis of ihe hyperbola is a mean proportional between the two distance from one of the foci to the two verlices.
bb The ordinate DF=- ✓(a + b2-a) = therefore the para
2bb_466 meter p=Dd=.
It is therefore a third proportional
2a to the transverse and conjugate axes. The parameter of the second axes is a line q, a third proportional to the conjugate and transverse 44. If we introduce the expression of the parameter into the
bb equations of the hyperbola yy= (2ax +xx); and yy=(zz--aa) we shall have yy=... (2ax +ex) and yy= (zz-aa) similarly
the equation yy= (66 +zz) which belongs to the second or cona
2a jugate axis becomes yy =
(66+22) = 24
45. Call CF=Cf=c, we shall have FM=v(yy+zz--2cz + cc)
-a and fM=<z +a; hence f M-FM=2a. Therefore in the hyperbola the difference of the lines drawn from the two foci to any point of the curve is every where equal to the transverse aris.
From this we may deduce an easy method of describing an hy. pobola, whose axis are 2a and 26. Take an interval Ff=2 vaa +66), and make use of a rule fMO, longer, in proportion as you desire to have a greater portion of the hyperbola : fix one extremity of it at one of the foci, to the point f, for example, so that it may revolve freely about that point. Take then a thread FMO equal in length to f MO——2a. Fasten one of the ends of this thread to the point o of the rule, and the other to the focus F. This done, draw the rule away from the axis as far as the thread FMO will allow, and then bring it again gradually towards the axis, taking care to keep the thread always extended by means of a point or style M which glides along the rule /MO. The curve described during this motion by the point M, will be an hyperbolical branch AM, since the difference of the radius-vectors will be every where equal to the transverse axis.
46. This same property will enable us to draw a tangent MT, to any point m of the hyperbola. For if we conceive they arc Mm to be infinitely little, and draw the lines fM, fin, FM, Fm, we may prove, much in the same manner as in the
TAF ellipse, that the angles fmM, Mm F are equal, and consequently that if we bisect the angle f MF by the line MT, this line will be the required tangent
This being premised, in the triangle f MF, we have fM: MF:: fT: FT and therefore by composition f M+FM :JM :: fT+FT
aa+cz :fT or
aa+cz :: 2c :
+c, therefore fT-C or CT = , which gives this proportions CP : CA :: CA : CT, by which it is easy to find the point T, and consequently to draw the tangent MT. 47. We may observe that CT being equal to
it is always positive as long as z is so. Therefore every tangent to the hyperbola cuts the axis in some point T, between A and C. But as the abscissa increases, the line CT diminishes, so that it is infinitely little or nothing when the abscissa is infinitely great.
Hence we see that
through the centre C we may draw two right lines cX, Cx which will be the limits of the tangents to the hyperbola, these lines, whose position we shall soon determine, are called the asymptotes of the hyperbola. 48. The subtangent PT =%
and the tangent bb
32 - bb
209+). If we draw the normal MN we shall have the subnore
da mal PN =
rallel to MP, we shall have PT : PM :: AT : AS, or
Now if we suppose z infinite, the %+a
%+a quantity will not differ from unity. Consequently we shalı
z+a then have AS=b. Hence it follows that if we draw AD and Ad perpendicular to CA, and each equal to the semiconjugate axis b, the lines CD, Cd, drawn so as to pass through the points D, d, and the centre C, will be the asymptotes' to the hyperbola MAM', and if prolonged in a contrary direction, they will be those of the opposite hyperbola.
If the hyperbola is equilateral, the angle DCd, made by the asymptotes, is a right angle, for then DA=Ad=CA.
The hyperbola referred to its asymptotes has many properties; here follow the principle of them:
50. If though any point N of the asymptote we draw the straight line Nn parallel to the line Dd, we shall have CA : DA ::CP: NP
M bz or a:b::2: there.
bz fore NM y,
back and Mn="+y. Con
la sequently NM x Mn=
boz-yo = bb = DA'. Since NP =
bon", and that MP• =6*: 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 increases, the difference between bé zand bo za
- bb becomes less
ao 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, C&=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? :: QM MK : AL'. But Mnx MN= DA', hence xy=mn, an equation to the hyperbola referred to ils asymptotes, in which mm = (aa +66) 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. But (50) PpXpQ=66=MmX mN. Consequently Fm X mf=Cpx pg. And therefore also Kg x KG=fhxh F.
53. If we suppose that the 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 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.
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 no h, &c. will be in the hyperbola.
55. According to what we have seen (58), a 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 conjugate diameter. Its ordinates are the right lines mQm parallel to the conjugate diameter
d TMt or Ďcd, and the parameter of any diameter is a line, a third proportional to that di
D ameter and its conju- PI gate. The line MCM = 2CM is also called the first or principal
В. diameter, and the line
H TMt = 2TM = DC4
T = 2DC is the second
G conjugate diameter.
56. This premised, it is easy to see that a
N. 111' diameter bisects all its ordinates. For NQ:Qn ::TM:Mt, and Nm=mn. Call then CM=m, CD=TM=n, CQ= ,,Qm=y; and we shall have min::Z: NQ=
But Nm X mn
=TM Hence no = yy, and yy = (2*—m*), an equation similar to that of the co-ordinates to the transverse axis.
ma This equation gives z = (yo+m*). Consequently if we make
na Cp=z, 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