CONIC SECTIONS

(Prop. 8 , Cor. 3.) but 4 AF x AP =GP* -PF 3 , 8 . '2. E ; there-fore MP J = GP- •_PL'-; ami MP 2 + PF-, or MF 2 = GI 2 , or

Fil'v 2 . Therefore MF = MK. .

Cor. OAand CG, fie. 17 ami 19 , are in the same ratio, 1 < or C t :

C A : : F A : AG*; therefore CF’ : CA : *• CA *. CG.; Hence CF,CA, and CG, me continual proportionals.

Prop. XII. Fig. 22 , 23, 24.

F t\yo straight lines DQ, D q be drawn from the point I), wherethe axis meets the directrix, through L anclT, the ordinate passingthrough the focus, which are produced both ways in the hyper-bola; and through any point P in the curve of the section a line QPpbe drawn parallel to the directrix, meeting I)L and DT in Q and91 the segment QN, which is intercepted between either of thelines and the axis, will be equal to SI’, the distance of P from thefocus.

Tiie triangles DNQ, DSL, are similar, therefore NQ : ND ::SL : SD, i. c. :: SP: ND. Prop. 11. HenceNQ=SP, and inthe same manner it may be proved that Ng — S;>.

Cor. l. If KAG be drawn tlirough the vertex parallel to thedirectrix, SA will be equal to AK or AG.

Cor. 2 . The lines DQ, Dry, touch the conic section in (hepoints L andT. For the triangle SNP being right-angled, SP orQN is always greater than PN except when r is at L, where theyemneide ; therefore DQ meets the curve only in one point L. Inlike manner it may be shewn that Dg touches the curve atT.

Prop. XIII. Fig, 22 , 23, 24.

If from the point G, where the straight line KG, which is drawnthrough the vertex parallel to the directrix, meets either of thetangents DQ, Dry, a UneGR be drawn through the focus S, andproduced both ways in the hyperbola, it will be parallel to theother tangent DQ in the parabola; it will meet it somewhere inyr in the direction GUg in the ellipse, and in the opposite directionin the hyperbola.

Let SG meet the directrix in X. The triangles SAG, SDX,are similar; now SA:=AG, Cor. 1. Prop. 12 , therefore SD=DX,but in the parabola, tig. 22 , SL=DS, therefore SL is equal andparallel to DX; and consequently XS is equal and parallel to DL.In the ellipse, tig. 23, SL is less than SD or DX, and thereforethe lines 1)L, XS, must meet when produced in the directionXGS. In the hyperbola, fig. 24, SL is greater than SD or DX ;and therefore the lines must meet when produced in the directionSGX.

Cor. 1 . Because the triangles GAS, SNR, are similar, SN willbe equal to NR.

Cor. 2 . Fig. 23, 24. Hence where Q coincides with g; in theellipse or opposite hyberbola, QN will be equal to g.M, or SM ;therefore SP will be equal to SN; and therefore SP will coincidewith SN, and the curve will meet the axis in the point M.

Cor. 3. Ilence the whole ellipse, fig. 23, is contained betweenthe lines GIv, gk, on one side of the directrix.

Cor. 4. In the parabola, lig. 22, NQ being always greaterthan NR, except at the vertex, SP is greater than SN ; thereforethe curve will meet the axis only in one point A, and it will ex-tend without limit on one side of the directrix.

Cor. 5 . In (he hyperbola, tig, 24, NQ being greater thaw NR,except at A and M, SP is greater than SN, and the two curveswill be extended without limit, on opposite sides of the directrix.

Prop. XIV. Fig. 25, 26.

The conjugate axis bisects all lines drawn parallel to the trans-verse axis, which aic terminated by the ellipse, and by the oppo-site hyperbolas. Those lines which are equally distant from thecentre are equal; and those which are nearer to the centre aregreater in the ellipse, and less in the hyperbola, than those whichare more remote.

Take CN any distance from the centre, between C and Athe ellipse, and in CA produced in the hyperbola. TalCli = CN, aud draw the. ordinates PN/>, QUq; join PQ, pmeeting R l> in n aud r. Recause P p and Q q are equal, Cor.Prop. 7, and they are bisected in N and R, the lines PQ, NR, pa e equal and parallel, and because P/i, p r, are equal to NC, arQ«, q r, equal to RC, or N C, l’Q, pg, are bisected in 11 andand they are at equal distances from the centre, because C n, Cvoi. n.—so. 56 ,

r,

are equal to PN, Np. Lastly, as Cn decreases, l’N decreases, andtherefore CN increases in the ellipse, and decreases in the hyper-bola; but Pi, is equal toCN, l’/i therefore increases iu the former,and decreases in the latter, asC », its distance from C, decreases.

Cor. l. The conjugate axis divides the ellipse into two equaland similar parts, and the'two opposite hyperbolas are equal and si-milar. *

Cor. 2 . Suppose I’Q, fig..2(1, which is always parallel to AM,to move from the centre towards, 13, the points P, Q, will coincideat B, and the line EPQwill become a tangent to the ellipse;therefore the ordinates to the conjugate axis are parallel to thetangent at its vertex. A line parailei to I 3 /;, and passing tliroughA, is a tangent.

Prop. XV. Fig. 26.

The square of the semi-ordinate lo.the conjugate axis, in theellipse, is to the rectangle under the abscissa; as the square of thetransverse, axis to the square of the conjugate axis.

For lb;Q being parallel to AM, it is perpendicular to B 6 , andit is bisected in n, by the last Prop, it is therefore au ordinate to134, and by Prop. 7, PN : , or Cm* : 13C 2 : : ANxNM ; AC 5 ,and by division B«X nb: BC 2 : : CN 2 or 1’;;-: AC 2 , and by alter-ation and inversion Pn 3 : B«x nb :: AC 2 : BC 2 : : AM" : 13A ; .

Prop. XVI. Fig. 20 , 21.

All the diameters of an ellipse or hyperbola are bisected in thecentre.

From any point P in the curve draw PC to the centre, and PNperpendicular to the axis. Take C« = CN; and draw »G paral-lel to NP, but on the other side of the axis, let it meet the cure cin G, and join CG. Then because Cn = CN, the semi-ordinatesGn, PN, will be equal, Cor. S, Prop. 7, and the angles at N andn are right angles, therefore the triangles CNP, C«G, are equal,therefore CG = CP, and the angle «CG is equal to NCP ; henceit follows that GGP is a straight line, which is bisected at C.

Prop. XVII. Fig. 27.

If lines SH, SA, be drawn from the focus S to the points II, h,in which the asymptotes cut the directrix, they will be perpendi-cular to the asymptotes; and these lines, as also AG, A a, the seg-ments of the tangent at the vertex, which is intercepted betweenthe asymptotes, are each equal to half the conjugate axis.

For’Cor. Prop. 11 , CIS : CA : : CA : CD, that is, CS : CII: :Cli : CD; now the angle I1CS Is common to the two trianglesCHD, CSH, therefore these triangles are similar, and the angleCHS = CDH = a right angle. In the same way it may be prov-ed that CAS is a right angle.

Again, SID = SC* - CII* = SC 5 - CA 2 = CI3 2 by Prop. 7,Cor. 6 ; therefore SII = CB. In the same way it may be provedthat S/i=CB: and because CH = CA, and CHS, CAu, areright angles, and the angle HCS is common to the triangles SIIC,a AC, these triangles are equal, and Aa = SII =.RC.

Cor. 1 . Radius is to the sine of the angle contained by theasymptote and directrix in the constant ratio, mentioned Prop. 11 ,for CA or CII is to CD in the same ratio, aud CH : CD :: ra-dius : sine CHD.

Cor. 2 . If a line PG, fig. 28, be drawn from any point P inthe hyperbola, or in the opposite curve, parallel to the asymptote,meeting the directrix in G, PG will be equal to I’S. Draw l’Eperpendicular to the directrix; and because the. angle PGE —CHD, PG'istb PE in the above-mentioned ratio, or as SP to PE,therefore PG = PS.

Prop. XVIII. Fig. 28.

The asymptotes never meet the curve, but any other linsdrawn parallel to an asymptote will meet one of the hyperbolas.

For, if it be possible, let the asymptote meet the curve in thepoint R. Join RS, and draw RN perpendicular to the directrix.Then by Cor. 1 , last Prop. HR is to RN, as SR to RN, there-fore RS’e=RH, and the angle 11SI1 = RHS, which is impossi-ble ; for by last Prop. RHS is a right angle. In the same way itmay be proved that it cannot meet the opposite curve. Let anyother line GP be drawn parallel to the asymptote; and first let itbe nearer to the focus. Join SG, and produce it to meet tneasymptote GII in I: then the angle SGP-nSlII, which is lessC than