diameter to PCG. From any point Q, draw Qv parallel
to a tangent at P; then Qu is called the ordinate to the
diameter PCG, and Pv, v G the abscissas.

=SC+AC× SC-AC.

=SC+CM× SC-AC (for CM=AC.)
=MS × SA.

COR. For same reason, AH x HM=BC'.

PROPERTY 2. (Prop. 3. of Ellipse.)

(75.) The Latus-rectum LST is a third proportional
to the major and minor axis.

For HL-SL=2AC (by construction);

(Fig. in page 44.)

Again,

.. HL=2AC-SL,

and HL=4AC2+4AC× SL+SL.

HL2=SL+SH (Euc. 47. 1.) =SL2+4SC2 (SH=2SC), =SL2+4AB3,

=SL2+4AC2+BC".

Hence 4AC2+4AC × SL+SL=SL2+4AC2+4BC';

.. 44CX SL=4 BC",

and 2AC 2BC: 2BC: 2SL,

or

:

AM : BO :: BỎ : LT.

PROPERTY

PROPERTY 3. (Prop. 4. of Ellipse.)

(76.) If PYT bisects the angle HPS, it will be a tangent to the Hyperbola in P.

If PT be not a tangent, let it cut the hyperbola in Q. Draw Sys at right angles to PT, meeting HP in s; and

join HQ, QS, $Q. In the right-angled triangles Sy P, sy P, Py is common, and 4 SPy= L sPy; .. Sy=sy, and Ps=PS. Again, in the right-angled A'QSy, Qsy, there is Sysy and Qy common; .. Qs=QS.

Since Qs=QS, HQ-Qs=HQ-QS=AM; and since Ps=PS, HP-Ps (or Hs)=HP-PS=AM; hence HQ-QS=Hs; .. HQ=Hs+Qs, i. e. the side HQ of the triangle HsQ is equal to the sum of the two sides Hs, sQ, which is impossible (Euc. 20. 1.) Hence TP does not cut the curve; in the same manner it be proved that it does not cut it in any other point, and consequently it is a tangent at P.

may

(77.) COR.

(77.) COR. When P comes to A, the HPS= two right angles; therefore a tangent at A is perpendicular to the axis AM.

(78.) If tangents ef, gh be drawn at the extremities of any diameter PCG, they will be parallel to each other.

Taps & 011 3dart
Mike Or H

Compleat the parallelogram SPHG, of which HP, PS are two sides; then since its opposite sides are equal HGGS will be equal to HP-PS, and by a process similar to

that

that in the Ellipse (Art. 43.), it may be proved that G is a point in the opposite hyperbola, C the center of the hyperbola, and PG a diameter.

=

HGS

Now in the parallelogram SP HG, the opposite
HPS; but by Art. 76. PT bisects the HPS,

and for the same reason Gg bisects the
the g GP is equal to
parallel to gh.

HGS; hence

GPT, and therefore ef is

(79.) COR. In the same manner it might be proved, that if tangents be drawn at the extremities of any other diameter DCK, they will be parallel; hence (as in the Ellipse) if tangents be drawn at the extremities of any two diameters PCG, DCK, they will form, by their intersection, a parallelogram eghf.

(80.) If SP and CD be produced till they intersect each other in E, then PE=AC.

Draw HI parallel to CDE or ef, and produce SE to meet it in I. Since HI is parallel to Pf, the exterior LSPf interior PIH, and HPƒ = alternate L PHI; but 4 SPf= L HPf, because PT bisects LHPS; .. L PHI= L PIH, and consequently PI = PH. Again, because CE is parallel to HI, and SC= CH, SE must be equal to EI.

(81.) If the ordinate PN be drawn to the major axis, then ANxNM: PN :: AC: BC. (Fig. in page 51.)

Let AC or CM=a, Then, by Art. 9.

SC or CH=b,
CN=x,
PN=y,,

ba2xx-axx+a=a1y",

or SC-ACX AN× NM=AC' × PN'. But, by construction, SC-AC2=BC' ;

.. BCX ANX NM AC2x PN, and ANX NM: PN:: AC: BC'.(*)

COR. 1. Since ANx NM-CN-ACX CN+AC= CN-AC'; .. CN-AC: PN: AC: BC'.

(82.) COR. 2. Produce NP to p, and draw any ordinate pm at right angles to Cm, then (since the conjugate hyperbola Bp is described with BC as major and AC minor axis) Bm x m0: pm2 :: BC: AC3, or (since Bmxm0=Cm-BCx Cm+BC) Cm-BC' : pm :: BC: AC'.

(83.) Cor.

(*) The general property of the Hyperbola analogous to the 10th Property of the Ellipse, viz. Pv xvG: Qu:: PC': CD, will be found at the end of the Properties of the Hyperbola derived from its relation to the Asymptote.