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: DKCF: 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 O in the groove of EF, the extremity D will describe an ellipse.

PROPOSITION VII.

A straight line, not passing through the centre, but terminated 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.

M

D

P

For, let MN meet

CT in L, and from the

NO, be drawn perpen- E R

points M, N, let MI,

I C

K

QOF L

dicular to EF and MR, NQ, parallel to DC.
milar triangles, MI: DK = RI : CK.
And MI: DK = IL: KT; therefore (Pl. Ge.
MI2: DK2: = RI IL: CK · KT.

Then, by si

VI. 23, Cor.)

Hence EI IF: EK KF (II. 6, Cor. 1) = RI IL: CK KT.

EK KFCK KT (II. 5, Cor. 1).

EI IF RI· IL.

EI: IL

=

RI: IF.

EL: IL RF: IF.

EL: ER+FL = IL : FL.
EL: EQ + FLOL: FL.
IL: OL EQ + FL: ER+FL.
RL: QL= : EQ + FL: ER + FL.
RQ:QLRQ: ER + FL.

Consequently QLER + FL, ER FQ, CR = CQ, and MP PN.

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 dia

meter 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.

G

The triangles TDK, CQN, by reason of parallel lines, are equiangular, therefore TK: CN= DK: QN; hence TK2: CN2 DK2: QN2 = EK KF: EN NF (II. 6, Cor. 1) = CK · KT : RN NC (II. 5, Cor. 1), and TK: CN

=

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 diame

ters.

COR. 1.-If from the extremities of two conjugate semidiameters, 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= CR CN, therefore CK KT, or EK · KF =

CF2

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.

It may be similarly proved that QN2 + DK2 = CG2; 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 CK2+ KD2+ CN2+NQ2 = (CK2 +CN2) + (KD2 + NQ2) = CF2 + CG2.

PROPOSITION IX.

Any semi-diameter which meets an ordinate and tangent drawn from the same point in the ellipse, is a mean proportional between the segments intercepted from the centre.

Case 1. Let the semi-diameter CN meet a tangent and ordinate drawn from F, one extremity of either axis, in M, P. Then CM: CN=CN: CP.

M

N

P

meet the axis EF in T, K. It has already been demonstrated, with respect to either axis, that CT: CF = CF : CK; hence CM : CN = CF : CK = CT: CF = CN : CP.

Case 2. Let the semi-diameter CN meet a tangent and ordinate from any other point D in M, P, CM: CN= CN: CP.

E

O RKFT

L

Q

For, let DP, the tangent at N, DM, and NR parallel to DM, meet the axis EF in L, Q, T, R. Also, let DK, NO, be the semi-ordinates to EF from D, N. The lines DK, DT, DL, are respectively parallel to NO, NR, NQ.

Hence CO OQ: CK · KT = (II. 5, Cor. 1, and 6, Cor. 1) NO2: DK2 0Q2: KL2.

By alternation CO: 0Q=CK · KT : KL2.

But

Therefore

OQ: OR = KL : KT = KL2 : KL · KT.
CO: OR = CK: KL (by equality).

By composition CO: CR CK: CL.

CT CK CF2 = CQ • CO.

CT: CO

CQ : CK.

CT: CRCQ: CL (by equality).

And therefore CM: CN

CN : CP.

COR. 1.-Tangents, at the extremities of an ordinate, meet its diameter produced in the same point.

For if a tangent from F' meet CN produced in a point M', it may similarly be shown that CP: CN CN : CM'; therefore CM' CM, or the tangents from F and F' meet in the same point M.

COR. 2.-A straight line, drawn through the centre, from the point of intersection of two tangents, bisects the line that joins the points of contact.

This is the converse of the last corollary.

COR. 3.-If a circle be described upon any diameter of an ellipse, and through any point of it an ordinate of each curve be

if the tangent to the

circle at D meet CN produced in some point M', then, from similar triangles, CP: CD = CD: CM', or CP: CN=CN: CM'; therefore CM' CM, or the tangents meet in the same point.

=

COR. 4. The segments into which any diameter is divided by a tangent, are directly proportional to the segments into which it is divided by an ordinate from the point of contact.

For CM: CN CN : CP, and by mixing IM: MN = IP: PN; or the diameter is cut internally and externally in the same ratio; or the diameter produced is cut harmonically.

COR. 5.-If a semi-diameter be produced to meet a tangent, the rectangle under the segments into which it is divided, by an ordinate from the point of contact, is equal to the rectangle under the segments, into which the whole diameter is divided by the same ordinate. For CN2 CP CM CP2+ CP PM; therefore CP PMCN2 - CP2 = = IPPN.