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foci, and a parallel to the line joining the other focus, and the point of contact from the centre, meet the tangent in the same point.

COR. 7.-If a cord pass through one of the foci, and the tangents at its extremities be produced to meet, the straight line that joins the point of concourse and the focus, will be perpendicular to the cord.

Let the tangents at the extremes of the cord DBO, passing through the focus, meet in the point T, and let a line TI be perpendicular to AD; then, because DT, TO, bisect the exterior angles of the triangle ADO, AI is equal to half its perimeter (Pl. Ge. VI. н) = AD + DB. Hence DB = DI, and triangles TDI, TDB, equal in every respect, and angle TBD = TID, or a right angle.

PROPOSITION III.

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From any point in an ellipse, a perpendicular being let fall upon the transverse, and straight lines drawn to the foci, as half the transverse is to the eccentricity, so is the distance of the centre from the perpendicular to half the difference of the straight lines drawn to the foci.

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Let D be the point in the curve, from which DA, DB, are drawn to the foci, and DK perpendicular to EF, then CF: CBCK: (AD-DB). For, let EL AD, then LF = DB, and 2CL = AD DB. But (AD + DB) (AD -DB) = AB 2CK (Pl. Ge. II. c), or EF · 2CL = AB. 2CK; therefore EF : AB = 2CK : 2CL, and CF : CB = CK : CL.

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PROPOSITION IV.

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Any two cords which pass through one of the foci of an ellipse, are to one another directly as the rectangles under their segments.

Let DO, MI, be the two cords passing through the focus B, DO: MI = DB · BO : MB · Bİ.

For, let the tangents at D, O, meet each other in R,

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through C let NQ be drawn parallel to DO, meeting the tangents in N, Q, and let RB meet NQ in P.

proved that CN or CQ

CF (II. 2, Cor. 5);

therefore NQ = EF; that CF CB = CK: CL (II. 3), therefore

CF CL

CB⚫CK, and

that DBR, or DBP, is a

right angle (II. 2, Cor. 7); therefore the triangles

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B

CLK

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DBK, BCP, are similar, and DB CP = BC · BK (Pl. Ge. VI. 4 and 16). Hence CF2

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CF CL + CF LF (Pl. Ge. II. 2) = BC. CK + DB CQ BC. CK + BC · BK +BD •PQ= BC2 + DB · PQ, and DB PQ = CF2 BC2 EB BF (Pl. Ge. II. 5, Cor).

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But DO: NQ=BO: PQ = DB BO: DB PQ.

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In like manner EF: MI

Consequently DO: MI
V. 24).

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DB BO: MB· BI (Pl. Ge.

COR. The rectangle under any cord passing through the focus and the parameter of the transverse, is equal to four times the rectangle under its segments.

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For, since EF: DO = EB · BF: DB BO, if P be the parameter of the transverse, EF P:DO P=4EB · BF: 4DB BO. But EF P= * GH2 (II. Def. 8) = 4EB⚫ BF (II. 1, Cor. 6); therefore DO P 4DB BO (Pl. Ge. V. 14).

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If a tangent to the ellipse meet either axis produced, and a perpendicular to the axis be drawn from the point of contact, then shall the semi-axis be a mean proportional between its segments intercepted from the centre by the tangent and perpendicular.

Case 1. Let the tangent DT, and the perpendicular DK, meet the transverse in T, K; then CT: CFCF: CK.

*That EF, GH, are conjugate diameters, will be proved independent of this Cor. For GH, see last figure.

For, since DT bisects the angle BDI, AT:TB=AD: DB

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(Pl. Ge. VI. A), and (AT + TB): (AT-TB) = (AD+DB): (AD-DB) (Pl. Ge. V. 15, and G).

That is

But

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CT: CB CF : CL.

CB: CF = = CL: CK (II. 3).

Therefore CT: CF = CF : CK (Pl. Ge. V. 24).

Case 2. Let the tangent DT, and the perpendicular DN, meet the conjugate in M, N; then MC: CG

CG : CN. For, let DO, perpendicular to DT, meet the transverse in O, then DO evidently bisects the angle ADB; therefore AO: OB = AD : DB (Pl. Ge. VI. A), hence AT: TB = AO: OB, and CT: CB CB: CO (Pl. Ge. V. 15, and &); therefore CT CO=CB2, but CT CK CF2, consequently CT KO CF2 CB2 CG2.

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Again, from the similar triangles CTM, KDO, CT KO = CM · DK (Pl. Ge. VI. 4 and 16) CM CN; therefore

CM • CN= CG2, and CM : CGCG: CN.

COR. 1.-The rectangles under the segments of the axis, and the segments of the semi-axis produced to meet the tangent, both being divided by the perpendicular, are equal.

For EK KF CF2

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CK2 TC · CK CK2=

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CK KT. And similarly HN NGCN NM.

COR. 2.-The segments into which the axis is divided by the tangent, are directly proportional to the segments into which it is divided by the perpendicular.

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For, since CT: CF CF: CK, CT + CF : CT · CF= CF+CK: CF-CK, that is, ET: TF EK : KF.

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COR. 3.-The segments CT, CB, CO, are proportional

(proved in case 2); and, in like manner, it may be proved that CK, CL, CO, are proportional.

PROPOSITION VI.

If from any point in an ellipse, a perpendicular fall upon either axis, as the square of that axis is to the square of the other, so is the rectangle under the segments of the former, to the square of the perpendicular.

Let the perpendicular DK fall from any point in the curve upon EF, then EF2: GH2 EK KF: DK2.

For, since CT: CF CF: CK (II. 5), CF2: CK2=CT:CK (Pl. Ge. VI. 20, Cor. 2 and 3), and, by conversion, CF2: EK KFE =CT:TK. But CT: TK CM: DK or CNCG2: CN2, or DK2; therefore CF2:

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EK KFCG2: DK2, and alternately CF2: CG2 = EK · KF: DK2. Consequently EF2: GH3 = EK · KF : DK2. If the perpendicular DN fall upon GH, it may be proved in the same manner that GH2: EF2 = GN · NÍ : DN2. COR. 1.-The squares of perpendiculars, falling from points in the curve, upon either axis, are to one another as the rectangles under the segments into which they divide the axis.

For the squares of the perpendiculars are to the rectangles in the same ratio as the squares of the axes; and therefore the squares of the perpendiculars are proportional to these rectangles.

COR. 2.-If a circle be described upon the transverse, and another upon the conjugate axis, the former will contain, and the latter will be contained in, the ellipse, so that the curve lines shall touch one another only in the extremities of their common diameter.

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For DK2: EK KF GH2: EF2; but GHEF; therefore DK2 IEK KF. Now, if PF be an arc of the

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circle described on EF, PK2 = EK KF; therefore DK PK, and the point P is without the ellipse, and the circle meets the ellipse only at E and F. It may be similarly shown that the circle described on the conjugate diameter is within the ellipse.

COR. 3.-The two axes are the greatest and least diameters of the ellipse.

For, any diameter of the ellipse, except EF, is less than the diameter of the circle described on EF, and hence EF is the greatest diameter. So any diameter of the ellipse except GH, is greater than the diameter of the circle described on GH; hence GH is the least diameter of the ellipse.

COR. 4.-If a circle be described upon either axis of an ellipse, perpendiculars to the common diameter are cut proportionally by the curves.

For DK2: EK KF=GH2: EF2; but PK2 = EK · KF; therefore DK2 : PK2 = GH2 : EF2, or DK:PK=GH: EF, or the perpendiculars in the ellipse are to the corresponding ones in the circle, in the ratio of the axes, that is, in a constant ratio.

COR. 5.-Every straight line terminating in the curve, and parallel to one axis, is an ordinate to the other; and conversely.

For it was shown that EF2: GH2 = EK · KF: DK2; and it may similarly be proved that EF2: GH2-EK KF: KI2; and therefore DK = KI, and DI is an ordinate to EF. It may be similarly shown that DO' is an ordinate to GH. COR. 6. The two axes are conjugate diameters. COR. 7.-The ordinate through the focus is the

ter of the transverse.

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For, if K be the focus, EK KF = GC2, but EC2: GC2 = EK KF: DK2 = GC2 : DK2 (II. Def. 8).

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COR. 8.-Equal ordinates to either axis are equally distant from the centre; the greater ordinate is nearer the centre; and conversely.

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For EK KF = CF2 — CK2; and therefore the less CK is, the greater is EK KF, and hence the greater is DK. COR. 9.-If from any point in the curve, a straight line be drawn to the conjugate axis, equal to half the

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