CONIC. SECTIONS.

stunt ratio; therefore LP 2 , is to hp 2 or QL x I-?, tig. 38, in aconstant ratio.

Prop. XXV. Fig. 43.

If two right lines QL, P p, meeting each other at any point I.,one of which is parallel to the axis, and the other - is inclined to thedirectrix in a given angle, cut a parabola in the points Q,p, andP, the rectangle under QL, and the latus rectum, will he to therectangle PLx hp in a constant ratio. Let LQ meet the direc-trix in X ; and from the centre I., at the distance LX, describe acircle, join QS, XS, S being tlie focus, Ft XS meet the circle in1', and join L.T. Draw SO perpendicular to LX, take 01 = OX,snd join SI, then SI=SX, now 1/1’= LX, and QS = QX,therefore LT is parallel to QS, and because the angle QSX =SXQ = SI\, the triangles QXS, SXI, are similar, amllXxXS: : XS : SQ : : ST : QL, therefore the rectangle lXxQL =XSx ST = DSx Sd, which bv the preceding proposition is to PLX hp in a constant ratio, and because IXxSOX, the distance ofthe focus from the directrix, therefore IX = latus rectum, therefore the rectangle under QL, and the latus rectum, is to PLx 1 -pin a given ratio.

Lemma. Fig. 43.

If a straight line be divided in two points C and D, such, thatthe rectangle CAxAD = I)BxBC, or ACxCB= BDx DA,the part AC will be e<]tial to BD. First, let CAxAD = DBxBC. Bisect CD in E; and CAx ADx EG 2 = AE 2 , alsoDBx BC-j-ED" 1 = EB 2 , but ED 2 = FX 2 , therefore BE 2 =AE 2 , and BE = A E, and therefore BD = AC. Secondly, letAC x CB = BDx DA, by bisecting AB in E, it may he shewn asabove, that ED = EC, and therefore BD = AC.

Prop. XXVI. Fig. 44, 45.,

All lines parallel to any diameter of the ellipse or hyperbola,which are terminated both" ways by the ellipse or opposite hyper-bolas, are bisected bv the conjugate diameter.

Let ACB be any diameter of an ellipse or hyperbola. Throughthe vertices A, B, draw the tangents AI., BM ; and through thecentre C draw the diameter DCK parallel to AL or BM, whichwill be the conjugate diameter. Through N any point in DKdraw LNM parallel to AB, meeting the ellipse or opposite hyper-bolas in P and Q, and the tangents AL, BM, in L and M. i’henAL being parallel to CN and BM, andl/NM parallel to ACB,AL=;BM, andLN = NM; and Cor. Prop. 34, LA 2 : PLxLQ : : BM 2 : QMxMP, therefore PLx LQ = QM X M P,therefore Lem. PL = QM, and, because LN = NM, PN =:-\Q.

Cor. 1. If the diameter DK bisect all lines parallel to AB, itwill be the conjugate diameter to AB.

Cor. 2, Fig. 44. If a straight line RDT be drawn through Dthe vertex of the conjugate diameter parallel to AB, it will touchthe curve in the point D.

Prop. XXVII.

Every diameter of a conic section bisects all its ordinates.

1st. "if the section be an ellipse, it is evident, from the lasproposition, for the ordinates of any diameter are parallel to thfconjugate diameter.

2dlv. If the section be an hyperbola, Fig. 43, of which ACB i:any diameter; in the tangent RAL take AR = AL, through Ianil R draw PLQ, l’RG, parallel to AB, meeting the opposite byuerbolas in P, Q, and F, (1, and the tangent at B in M, T. JoiiPF, cutting the diameter in V : then PF will be an ordinate whiclis bisected at V ; for bv last Prop. PI. = M Q, and FR = l Gand Cor. Prop. 24, FR x RG : HA-:: PL X LQ : LA 2 , bnLA 2 = RA», therefore, FR x RG = PL X LQ, that is RF xFT = LP x PM; therefore Lem. RF = PL, and PLRF is ;parallelogram, and therefore PF is parallel to RAL, and P\= VF.

Lastly, let the section be a parabola, Fig. 46, of which AN iany diameter, and PNQ an ordinate; through the vertex A dravthe tangent LAM ; and draw PL, QM, parallel to NA, then PI~ QM, but the rectangle tinder LP and the latus rectum is til.A- as the rectangle under MQ and the latus rectum to MA 5

7

therefore LA 2 — MA 2 , and LA = iMA, and therefore QN =

PN.

Prop. XXVIII. Pig. 47.

If a straight line cutting the hyperbola, or opposite hyperbolas,meets the asymptotes in two points; the segments between the hy-perbola or hyperbolas and asymptotes will be equal.

Let PQ cut the hyperbola, or the opposite hyperbolas in P andQ, and meet the asymptotes in R, T: the segments PR, QT, willhe equal. For ifPR he not equal to QT, let one of them, as QT,be the greater; and Q o = PR, join C o, which being produced willmeet the curve in some point r, Cor. Prop. 18. Through q drawopr parallel to QP meeting the curve in p and asymptote in r.Bisect PQ in N, and draw the diameter CNw, andPQ, pq, will beordinates tothat diameter. Because NQ := NP, and Q oz= PR,therefore N« = NR, and o N : q n :: CN : C n :: NR : nr,therefore nq — n r; but n q — up, therefore n p — u r, which isabsurd ; therefore QT is not greater than FR.

Cor. If the line TNR he supposed to move from N to A, thepoints P, Q, will coincide in A, and 1A will he equal to AG ;therefore when a line touches an hyperbola, the segments betweenthe point of contact, and the asymptotes are equal.

Prop. XXIX. Fig. 4S.

If from any point P in the hyperbola PQ two straight lines PI.,PU, be drawn to the asymptotes, and from any other point Q, inthe same or in the opposite curve, there he drawn other twostraight lines QE, QF, parallel to the two former lines PL, PH.The rectangle QE X QE will be equal to the rectangle PL x PH.

Join PQ, and let it meet the asymptotes in R and T : and be-cause the triangles TQF, TPI1, are similar, as also the trianglesRPE, PQE, QF •• PH :: TQ : TP : : RP : RQ : : PL : QF.;therefore the rectangle QF X QE isequal to the rectangle PH xPL.

Cor. Hence, if from any two points, P, Q, in the same or in theopposite hyperbolas, two straight lines, PL, QF., be drawn to thesame or to different asymptotes parallel to the other asymptote : therectangles CL x EP, CE x EQ, will be equal; also the paral-lelogram CLPI] will be equal to the parallelogram CEQF; andthe triangle CLP to the triangle CEQ.

Prop. XXX. Fig. 58.

If, from anv point K in the asymptote of a hyperbola, there hedrawn a straight line RT cutting the hyperbola, or opposite hy-perbolas in P and Q, and the other asymptote in T ; the rectanglePR • RQ is equal to the square of the semidiametev which is pa-rallel to RT.

Let AM, Bft, be the two axes, join AB meeting the asymptotein N. Draw the tangent 1A a meeting the asymptotes in I and u.Because A o is equal and parallel to BC, Prop. 17, AB is equaland parallel to a C, and 1A : In :: AN : a C, or AB ; now IA ishalf of I a, therefore AN is half of AB, and AN = NB.

Let CF be the semidiameter parallel to the fine cutting the op-posite hyperbolas, draw the tangent I.FH meeting the asymptotesin L and II, draw FE parallel to CL, meeting the asymptotes inE, and the conjugate hyperbola in D, and join CD. Because therectangle CN ■ NA,or CN • NB = CE • EF, Cor. Prop. 29, andCN • NB = CF, • ED • therefore CF. ■ EF = CE ■ ED,and EF= EI), and FE is half of ED, but because LF is half of LH,Cor. Prop. 28, and FE is parallel to CL ; FE, is also half of CE,therefore ED = CL, and they are parallel, therefore CD is equaland parallel to FII, and CD is a conjugate diameter to CF. Letthe line which cuts the hyperbola PFQ be parallel to CD or L11.'l ake any point i in the asymptote, and draw tr parallel to TR,cutting the curve or opposite curves in p and q ,'and the otherasymptote in r. Draw PY, PZ, py, pz, parallel to the asymptotes.Because the triangles PYR ,pyr, are equiangular.

PR : PY :: pr : py, and in like manner,

PT: PZ : : pt : pz, therefore

RP X PT : YP X PZ : : rp X pt : pv X pz ; but VP x PZ —f/p X pz, Cor. Prop. 29; therefore RP x P'I =xrpxpi\ orsince RP = QT, Prop. 28, PR x RQ — pr x rq\ and when'l’ is taken at F, the rectangle PR X RQ becomes FL 2 . or I- ID,

which