(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: GH? = EK · KF: DK2.

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

Boere DK?; therefore CF2:


remen Y EK · KF = CG2: DK, and alternately CF2 : CG2 = EK • KF:DK. Consequently EF2 : GH = EK · KF: DKP. If the perpendicular DN fall upon GH, it may be proved in the same manner that GH2: EF = GN.NH: DN. 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. For DK2: EK · KF = GH? : EF2; but GH < EF; therefore DKP < EK KF. Now, if PF be an arc of the

circle described on EF, PK2=EK · KF; therefore DK į 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 diame

ters 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=GH? : EF2 ; but PKP = EK · KF; therefore DK2: PK? = GH: EF, 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: GH? – EK · KF:DK?; and it may similarly be proved that EF2:GH2=EK.KF:KI”; 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 parame-

ter of the transverse. For, if K be the focus, EK · KF=GC?, but EC2: GC2 = EK · KF:DK= GC2: DK2 (II. Def. 8). COR. 8.-Equal ordinates to either axis are equally dis

tant from the centre ; the greater ordinate is nearer

the centre; and conversely. For EK · KF = CF2 - CK?; 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 transverse, its segment intercepted from the same point by the transverse, will be equal to half the conjugate;

and conversely. For RD = CP = CF, and RD:OD=CP:OD=PK: DK = CF:CG, or RD:OD= CF: CG, but RD = CF, therefore OD = CG. .

Schol. On this principle, the elliptic compasses are constructed. If EF and GH represent two bars of brass with grooves, and RD a third bar equal to CF, and the part of it OD equal to CG, then two projecting pins at R and O rest in the grooves; and if RD be moved, so that the pin at R will move in the groove of GH, and that at 0 in the groove of EF, the extremity D will describe an ellipse.

PROPOSITION VII. A straight line, not passing through the centre, but termipated by the ellipse, and parallel to a tangent, is an ordinate to the diameter that passes through the point of contact.

The cord MN, parallel to the tangent at D, is an ordinate to the diameter CD.

For, let MN meet CT in L, and from the points M, N, let MI, NO, be drawn perpen- ERICK QOF LI dicular to EF and MR, NQ, parallel to DC. Then, by similar triangles, MI: DK = ŘI: CK. And MI:DK =IL:KT; therefore (Pl. Ge. VI. 23, Cor.)

MI2 : DK? = RI· IL: CK · KT. Hence EI•IF:EK KF (II. 6, Cor. 1) =RIIL:CK·KT. But

EK KF = CK · KT (II. 5, Cor. 1). Therefore EI · IF = RI· IL. And

EI : IL = RI:IF.
By composition EL:IL = RF:IF.
By alternation and

conversion EL:ER + FL = IL: FL.
In like mannerEL: EQ + FL = OL: FL.
Hence by equality IL:OL = EQ + FL: ER + FL.

By division RQ:QL = RQ: ER + FL.

Consequently QL = ER + FL, ER = FQ, CR = CQ, and MP =ÍN. Cor. 1.-Every ordinate to a diameter is parallel to the

tangent at its vertex. . For, if not, let a tangent be drawn parallel to it, then the diameter through the point of contact would bisect it; and thus the same line would be bisected in two different points, which is absurd. COR. 2.-All the ordinates to the same diameter are

parallel to one another. COR. 3.-A straight line, drawn through the vertex of a

diameter, parallel to its ordinate, is a tangent to the

curve in that point. CoR. 4.-The diameter which bisects one of two parallel

chords, will also bisect the other. COR. 5.-The straight line which bisects two parallel

chords, and terminates in the curve, is a diameter. Cor. 6.-If two tangents be at the vertices of the same

diameter, they are parallel ; and conversely. Cor. 7.-Every diameter divides the ellipse into two equal parts.

PROPOSITION VIII. Two diameters, one of which is parallel to the tangent, at the vertex of the other, are conjugate to one another.

Let the diameter PQ be parallel to the tangent DT, at the vertex of the diameter DI; then shall PQ, DI, be conjugate diameters.

For, let DK, QN, be * perpendicular to EF, and let the tangents at D, Q, meet EF in T, R. The triangles TDK, CQN, by reason of parallel lines, are equiangular, therefore TK:CN =DK:QN; hence TK?: CNP = DK2: QN2 = EK · KF: EN · NF (II. 6, Cor. 1) =CK·KT:RN:NC (II. 5, Cor. 1), and TK :CN =

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CK:RN, or DK : QN = CK:RN. Consequently the triangles DKC, RNQ, are equiangular, and DC parallel to RQ. Whence PQ, DI, are each parallel to the ordinates of the other (II. 7, Cor. 1), that is, they are conjugate diameters. Cor. 1.-If from the extremities of two conjugate semi

diameters, perpendiculars be let fall upon either axis, the rectangle under the segments, into which the axis is divided by one of the perpendiculars, is equal to the square of the segment, which the other intercepts from

the centre. For, since RQ is parallel to CD, CT: TK = CR: CN, and CT · CK:CK: KT=CR·CN: CN2. But CT-CK= CH2 = CR · CN, therefore CK · KT, or EK · KF = CN2.

By letting fall perpendiculars upon the conjugate axis from the extremities of the same conjugate diameters, the same property may be similarly proved in reference to it. Cor. 2.-Also, the sum of the squares of the segments,

intercepted from the centre, is equal to the square of half the axis upon which the perpendiculars fall; and the sum of the squares of the perpendiculars is equal

to the square of half the other axis. CN2 = EK · KF (Cor. 1) = CF CKP (Pl. Ge. II. 5, Cor.); therefore CN? + CK? = CF2.

It may be similarly proved that QN2 + DK? = CG-; for QN, DK, are equal to the segments intercepted on CG from C by perpendiculars on it from Q and D. Cor. 3.-The sum of the squares of any two conjugate

diameters is equal to the sum of the squares of the transverse and conjugate axis. For CD2 + CQ2 = CK? + KD2 + CN2 + NQ2 = (CK? +CN2) + (KD? + NQ?) = CF2 + CGP.

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