CONIC SECTIONS.

8

which is equal to C D 2 ; and when P in the opposite hyperbola isat M, the rectangle PR x RQ becomes equal to CC 4 .

Prop. XXXI.

If two right lines meeting each other cut or touch a conic sec-tion, or opposite sections, the rectangles under the segments be-tween the point of concourse and the points of intersection, or thesquares of the tangents will be to each other as the squares of tiiesemidiameters to which the lines are parallel.

If the lmes be parallel to any of the diameters of the ellipse, orof the opposite hyperbolas, the proposition is evident from Prop.04, because the lines which meet each other make the same angleswith the directrix as those which pass through the centre, and thelatter are bisected in the centre. But if either of the lines PLQ,LRT, or both the lines PLQ, InLM, be parallel to some of theconjugate diameters of the hyperbola, Fig. 49, produce QLP tillit meet the asymptote in G, and through G draw FGIi parallel toI,IFF, meeting the opposite curves in F and II. Let i ll, CD,C A, be the semidiameters which are parallel to QP, MN, IFF.Then Prop. 24, PL X LQ : RL X L I :: PG X GQ : f G X GII:: CB 2 : CA 2 , Prop. 30. In like manner it mav be proved, thatRL x LT : NL X LM :: CA 2 : CD 2 ; therefore, PL x LQ:XL x LM : : CB 2 : CD-.

If the lines touch the conic section or opposite sections, thesquares of the tangents will be to each other as the rectangles un-der the segments of any two lines drawn parallel to them, whichmeet each other, and cut the section or opposite sections; andtherefore they are as the squares of the semidiameters to whichthey are parallel.

Cor. If two right lines IQ, IN, Fig. 49, 50, meeting each otherin I, touch an ellip-e or hyperbola in Q, N, anti are parallel to twoother lines VT, VN, which meet each other in V, and touch theellipse or opposite hyperbola, orhyperbolas, in T, N ; IQ : IN : :VT : VN : lor IQ 2 , IN 2 , are to each other as the squares of the se-midiameters to which they are parallel, and VT 2 , VN 2 , are in thesame ratio.

Prop. XXXIL Fig. 51, 52.

If an ordinate be drawn to anv diameter of an ellipse or an hy-perbola ; the rectangle under the abscisses will be to the square ofthe semi-ordinate as the square of the diameter is to the square < fits conjugate.

Let ACM be any diameter of an ellipse or livperhola, to whichPN Q is an ordinate, and let DCK be the conjugate diameter,which is parallel to PNQ. Then bv the preceding propositionAN x NM : PN x NQ (or PN 2 ) -.1 CA 2 : CD 2 :: AM 2 : DK 2 .

Cor. 1. Because the parameter is a third proportional to thediameter and its conjugate, the rectangle under the abscisses is tothe square of the semi-ordinate as the diameter is to the para-meter.

Cor. 2. The two conjugate diameters being constant, the rect-angle under the abscisses will vary as the square of the or-dinate.

Prop. XXXIII. Fig. 53.

If an ordinate be drawn to any diameter of a parabola; thesquare of the semi-ordinate is equal to the rectangle under the ab-scis-es and the parameter.

Let AN beany diameter of the parabola to which PNQ is anordinate. Draw the parameter TSV, cutting the diameter in F;join SA, let the diameter meet the directrix in D. Because TFis half of TV, Prop. 27, and AF is half of DF, or TF, Prop. 21,Cor. Ah’ : FT : : FT : 1 V, and AF X TV = FT 3 ; blit AF xTV : AN X TV : : TF 2 : PN 2 , Prop 25, therefore AN x TV= PNC

Cor. Because TV = 2 TF = 4 SA, 4 SA x AN = PN*,Prop. XXXIV. Fig. 54.

If two tangents be draw n at the extremities of any right linewhich is terminated by a conic section, and which does not passthrough the centre of an ellipse, they will meet each other in thediameter w hich bisects that right line.

Let PQ meet the curve in P and Q ; bisect PQ in N, andthrough N draw the diameter CNT. Through P <1 raw the tan-gent PT meeting the diameter in T, and join TQ, which will

touch the conic section in Q. For draw any other line DCB pa-rallel to PNQ, meeting TP, TQ, in L and M. The triahglesTNP, TCL, are similar, as also TNQ, TCM, therefore,

NP : CL : : TN : TC : : QN : Mb', and alternatelyNP : NQ :: CL : CM ; therefore CM = CL, which is greaterthan CD or CB, therefore M is without the section, and the lineTQ meets the curve only in one point Q.

Cor. It two right lines which touch a conic section meet eachother, a right line drawn through the point of concourse bisectingthe line which joins the points of contact, will be a diameter ofthe section.

Prop. XXXV. Fig. 55.

If a tangent to any point in the parabola meet a diameter, andan ordinate be drawn to that diameter from the point of contact,the segment of the diameter between the vertex and the tangentwill be equal to the absciss.

Let TP, which touches the parabola in any point P, meet thediameter NA in T, and draw the ordinate PN. NA will beequal to AT. Draw the tangent AI meeting PT in I ; join AP,and draw the diameter IG cutting AP in F, and PN in G ; thenAF — FP, Cor. Prop. 34. T herefore Al or NG=PG; andAI is half of NP ; but Al : NP : : TA : TN, therefore TA is thehalf of TN.

Prop. XXXVI. Fig. 56, 57.

If a tangent to any point in an ellipse, or an hyperbola meet adiameter, and from the point of contact an ordinate he drawn tothat diameter, the semidiameter will be a mean proportional lie-tw een the segments of the diameter, fthich are intercepted be-tween the centre and the ordinate, and between the centre andthe tangent.

Let PT touch the ellipse or hyperbola in any point P, andmeet the diameter MA in T ; draw PNQ an ordinate to the dia-meter N1A, CN : CA : : CA : CT. Through the vertices A, M,draw the tangent Al, ML, meeting PT in 1 and L ; take CO =CN, then Cor. Prop. 31, IP: IA 1 : LP : LM, and alternatelyIP: LP : : 1A: I.M, and because Al, NP, LM, are parallel,AN : NM : : TA : TM, and by composition, (lie. 57,) or by di-vision, (tig. 56,) ON : AN : : AM : TA ; ana by taking thehalves of the antecedents, CN : AN :: CA ; AT, and by compo-sition, (tig. 57,) or by division, (tig. 56,) CA: CN : : CT: CA,and by inversion CN : CA : : CA : CT.

Prop. XXXVII. Problem. Fig. 59.

Two unequal straight lines being given, which bisect each otherat right angles; to describe an ellipse, of which the given linesshall be the axes.

Let AM, BA, be the given lines of which AM is the greater.From the centre B, with a radius equal to AC, describe a circlemeeting AM in S and IT, which will be the fori, Prop. 7, Cor. 6.Take a string equal in length to AM, and lix the extremities of itat the points S and II ; and by means of a pin at P let the stringbe stretched, and let the pin be carried round, till it return to thesame point; the point P will describe an ellipse, of which AM,B A, are the axes, as is evident from Prop. 10.

Prop. XXXVIII. Pkob. Fig. 60.

Two straight lines being given, which bisect each other at rightangles; to describe an hyperbola, of which these lines shall bethe axes.

Let AM, BA, be the given lines, bisecting each other at rightangles in C. Join Al! ; and take CS and CH, in AM producedboth ways, equal to AB. At the point II let one end of a rulerbe fixed, so that it may move round this point as a centre; andlet a string be taken, the length of which exceeds that of the rulerby a line equal to AM ; let one end of the string be fixed at L,and the other at the point $ ; apply the string, by means of a pinat P, to the side ot the ruler Lll ; ami let thornier be movedabout the centre H, while the string is constantly applied, andkept close to the ruler by the pin at P. Then the dillerence be-tween the whole length of the siring SPL and the ruler II L beingequal to AM, the difference between 11 P and PS will be equal toAM ; and the point P will describe one of the opposite hyper-bolas of which AM, B A, are the axes, as is evident from Prop, 10.

-f Prop.