CRX RD=CR x RP, and consequently RD= RP, i. e. PD is bisected by the asymptote.

(108.) COR. 2. Since SA=AB, SA is a constant quantity; hence RP varies inversely as CR; when CR

therefore is infinite, RP will become equal to 0; which coincides with what has already been said as to the asymptotes touching the curve at an infinite distance.

PROPERTY 19.

Join CD, and produce it to K; draw the diameter PCG; then will DCK be the conjugate diameter to PCG. (109.) Draw ef touching the curve in the point P, and

meeting

meeting the asymptotes XCZ, UCY in the points e and f; then, by Art. 103, Pf will be equal to Pe; and since PR is parallel to Cf, CR will also be equal to Re. Hence, in the triangles CRD, PRe, we have CR=Re, RD=RP,

and

CRD

Pe, and the

Le RP, .. (Euc. 1. 4.) CD is equal to
DCR equal to the ReP; consequently

DCK is parallel to the tangent ef, and is therefore the conjugate diameter to PCG (73.)

(110.) COR. Join De, then eDCP will be a paral

lelogram,

lelogram, whose diagonal is Ce; and as De is parallel to the diameter PCG, it touches the conjugate hyperbola in D. Complete the parallelogram eghf, as in Art. 79. then, in the same manner as it has been proved that Ce is the diagonal of the parallelogram e DCP, it might also be proved that the point h would be found in the asymptote XCZ, and the points g, f in the asymptote UCY; these asymptotes are therefore the prolongation not only of the diagonals of the parallelogram described about the major and minor axes, but also of the parallelogram described about any two conjugate diameters.

PROPERTY 20.

(111.) Draw the ordinate Qv, then PyxvG : Qv2 PC: CD.

Produce vQ to X, then, by sim. A3, CvX, CPe,

Cv

CP :: v X' : Pe';

:: vX'-Pe: Pe',

and Cv-CP: vX-Pe2 :: CP2 : Pe. (A)

But Cv-CP=Cv¬CP× Cv+CP=PvxvG. By Art. 105, v X-vQ'=Pe', :.vX'—Pe'=vQ'. Now (109.) Pe'=CD'.

Hence, by substitution in Proportion (A), we have

PvxvG Qv2 :: CP1 : CD.

This property of the hyperbola is analogous to the tenth property of the Ellipse,

x. Отѣ

(") For CD being equal and parallel to Pe, De must be equal and parallel to CP (Euc. 33. 1.)

X.

On the Properties of the EQUILATERAL Hyperbola.
In Art. 72, it was observed, that if the axes of the
hyperbola becomes equal, it is then said to be equilateral:
in this case the figure possesses some peculiar properties,
which it may be worth while to investigate.

(112.) Let the annexed figure represent an equilateral

* then since the axes ACM, BCO are equal, it is evident that if a circle be described upon the axis ACM it will pass through the extremities of the axis BCO, and that the rectangular figure abdc which circumscribes those axes will be a square. Draw the diagonals ad, cb, and produce them each way to X, V, U,Z; then XCZ, UCV will be the asymptotes to the four hyperbolas; and as the angles a CB, c C B are each of them half a right angle, the angle a Cc will be a right angle. Since the asymptote XCZ cuts the asymptote UCV at right angles in the center C, it will also cut all other lines BA, DP, pQ, &c. (drawn parallel to UCV) at right angles. Now," by Art. 106, CRX RP=sA'; and for the same reason Cex eQ=sA'; .. CR× RP=CexeQ, or CR: Ce :: eQ ; RP; hence if any points R, e, &c. are taken in the asymptote, and from them ordinates RP, eQ, &c. are drawn at right angles to it, then the abscissas CR, Ce, &c. will be to each other inversely as the ordinates RP, eQ, &c.b

(113.)

(*) In the Figure, the letter R should have been placed over the point where PD intersects the asymptote CX.

(") Let Cs or sA=a, CR=x, RP=y, then (since CR ×

a2x log. x+Cor.; suppose therefore the area ASRP to begin from s, it would vanish when x=C's or a; hence a2x log. a+ Cor.=0, and Cor.=-a2x log. a, the area ASRP is therefore equal to a3× log. x — a3× log. a = axlog.. Suppose now

a

that Cs=a=1, then a' and a would each be equal to 1, and we should have area As RP=log, x; and thus if the abscissas

CR,