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By Geom. (160).... FM2=PM2 + FP2, R2 = f + cV + 2cax + x*,

and, by theorem 4, cor. 3, ..... yi = a*—c2a> + <?x2 x%.

Wherefore, eliminating y, by adding these equations together, we have

the equations R* - a2 + 2cax + cV.

Then, extracting the roots of each side of this equation, we have R = a + ex. In the same manner will be found r = a—cx\ therefore R + r = 2a.

272. Corollary 1.—Hence R = a + -x, and r = a—-x, for e=ca or c = '-.

273. Corollary 2.—Because R = a + g, we shall have aR = a2 + ex; and, by transposition, a(R- a)=s£e, hence R — a : x :: e: a.

274. Corollary 3.—Since the ratio of e to a, or *that CF to CA is constant, if we produce a A to R, and find the point R by making CF : CA :: CA : CR, and draw RX perpendicular to aR, and MX parallel to «R, and let CR=c?; then will R—a : x :: a: d; wherefore Rd—ad=ax, or, by transposition, Rd=ad+ax = a{d+x). Therefore d+x : R C." d : a; wherefore MX is to MF always in the constant ratio of CR to CA; and hence we have another method of constructing the ellipse.

N.B. The line RX is called the directrix. Several writers make this the fundamental principle from which all the other properties emanate.

275. Corollary 4.—Hence CR : CA Cl aR : aF; because, when the point M comes to a, the line MX or PR will become aft, and the radius vector, FM, will become Fa.

276. Corollary 5.—Hence CA : CF :: aR : aF.

ELLIPSE.—THEOREM 6.

277. The tangents from the corresponding points in the curves of a circle and ellipse, made by the prolongation of an ordinate of the ellipse, will meet the axis major produced in the same point. T.

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Let ANa be a circle described upon the axis major, and let PN be a tangent to the circle in N. Join TM; then, if TM does not touch the ellipse, let it cut it in M, I; and, through I, draw the ordinate HO, meeting the ellipse in I, the circle in L, and the tangent in O.

rTPM,THI TPxHI = PMxTH

By similar triangles .... ] ___ _TT _.„ TT _

(TPN.THO PNxTH = TPxHO

and, by theorem 3, cor. 4, .. PN : PM :: HL : HI .\ PM x HL = PN x HI.

Therefore, by multiplication, we shall find HL = HO, which is impossible; therefore TM does not cut the ellipse, and it must, in consequence, be a tangent at M.

ELLIPSE.—THEOREM 7.

278. In the straight line, Ta, of the axis major, the semi-axis, CA, is a mean proportional between the abscissa, CP, and the distance, CT, from the centre to the intersection of the tangent.

For the triangle CNT (see the preceding diagram) is right-angled at N, since PN meets the circle at N; and, because PN is perpendicular to TC, we shall have,

by the similar triangles PCN, NCT, .... CP x CT = CN*

and by the circle CN* = CA2.

Therefore, by multiplication, CP x CT = CAI

If CT=«, this conclusion, analytically expressed, is vx—d; or, if * be the subtangent, then will u=s + x; and, therefore, (s + x) x=a*, that is, x + ar^a2.

279. Corollary 1.—Whence, if z be any other abscissa, and S its subtangent, and if s + z=v, then vzsstf; and, since vx—c^, therefore ux = vz.

280.—Corollary 2.—When the abscissa becomes equal to the excentricity, x becomes e; and, consequently, vx-ct becomes ue=4p, or Se+^=za\ 11. z

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ELLIPSE.—THEOREM 8.

281. The line bisecting the angle at any point in the curve, formed by one of the lines drawn from one of the focii, and the prolongation of the line drawn from the other focus, will be a tangent to the curve.

Let MT bisect the angle /MK; then, if MT is not a tangent, it will cut the ellipse; let p be the point where it meets the curve, and, making MK=M/, join fK,fp, Fp, Kp, and let MT cut K/, in *.

Then, in the triangles fMz, KM*, fM being = KM, Ms common, and the included angles fMz, KM as, equal; the base J% is = Ka;, and the angle fzM=KzM.

Again, in the triangles Jim, Kpz, fz being = Ksr, pz common, and the angles fzp, Kzp, equal, the based is =Kp. Therefore Fp+pK=Fp+pf= FM+M/=FM+MK=FK; wherefore the sum of the two sides, Fp,pK, of the triangle FpK, is equal to the third FK; which is impossible; therefore as MT cannot cut the ellipse: it must be a tangent.

282. Corollary.—Hence, because the angle KM/ is bisected, the angle FMf will also be bisected by the normal MN. For since the angle TMf is = TMK, and TMK equal to the opposite angle FML; there- fore TM/ is = FML: and, because the angles TMN and LMN are each a right angle; therefore, taking away the angles TMf, LMF, which are equal, from the right angles TMN, LMN, there will remain the angle NM/equal to the angle NMF.

ELLIPSE.—THEOREM 9.

283. If there be any tangent meeting the four perpendiculars to the transverse axis, the one at the extremity of the transverse, next to the point of intersection, will be to that which terminates in the point of contact, as that which passes through the centre is to the remaining one at the other extremity of the transverse.

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