CONIC SECTIONS.

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34. A circle is said to touch a conic section in any point, whenthe circle and conic section have a common tangent in that point.

35. If a circle touch a conic section in any point, so that nooilier circle can he drawn between the conic section and that circle,it is said to have the same curvature with the section in the pointof contact, and is called the circle of curvature.

Proposition I. Fig. 10.

A line parallel to a side of the cone will meet the conic surfacein one point, and in one only.

Let ill, lig. 10, be parallel to AV, a side of the cone: then be-cause AV meets VD, and HI is parallel to it, III shall also meetVD ; let it meet it in E, then the part El is wholly within thecone, since it is parallel to AV ; in like maimer Eil is wholly with-out the cone, therefore Ill cannot meet the conical superficies,except in E.

Prop. II. Fig. U.

Not more than one tangent plane can he drawn through thesame side of the cone.

For if possible let each of the planes VBN, YBM, fig. 11, P 3 ')'ing through the side of the cone VB, touch the conical supci-ficies, and let BN, BM, be their common sections with the planeof the base; then because the two planes touch the cane, the com-mon sections BN, BM, are tangents to the base oi the cone, andat right angles to the radius CB, which is impossible.

Prop. III. Fig. 10.

A right line drawn through a point of a conic surface, so as nei-ther to be a tangent, nor to be parallel to a right line contained inthe conic surface, will meet either the same, or the opposite, conicsurface again in another point.

Let a plane be mawn through the vertex of the cone and theright line EE or EG, lig. 10; then that plane will cut the cone ; forif it do not, the right line FE or EG would be a tangent contraryto the hypothesis. Let VD and YA be the common sections ofthe plane and the conic surface ; then since the right line FE orEG is not parallel to V D or V H contained in the conic surface,therefore it will meet VD and VII either in the same conic sur-face as EE, or, when produced in the opposite conic surface*as EG.

Prop. IV. Fig. 12, 13, 4.

If a straight line be drawn from the vertex of a cone to a point,as B, in the plane of the base, but not in the periphery ot the base;and, through any point, as P, situated without or within the cone,another straight line, parallel to the former, be drawn to cut oitouch the conic surface or opposite surfaces; then the square ofthe line drawn from the vertex of the cone to the pomt B is to therectangle under the segments of the secant, or to the square of thetangent drawn from the point P, as the rectangle under the seg-ments of any line drawn from B to cut the base of the cone, is tothe rectangle under the segments of any line, parallel to the baseof the.cone, drawn through the point P, to cut the conic surface.

Let the point P, fig. 12, be without the base of the cone, andlet PQ, cutting the conical superficies in R and Q, be parallel toMl, and let the plane VBI’Q < ut the conic surface in the linesVRG and VQH, and the plane of the base in the line BGI1; andthrough P draw I.K parallel to G1I. Because QP is parallel toVB, and LK to Gil, therefore the triangle QPL is equi-angularto the triangle VBI1, and the triangle PRll to the triangle YG13:therefore 4. 6. E.

VB : PR :: BG : PKVB : PQ : : Bll : PL.

Consequently, VB. 4 : PR x PQ: : BG X BII: PK X PL, 23.t>. E. But the rectangle BG x BH is equal to the rectangle underthe segments of any other lme drawn from B to cut the base of thecone, 35 and 33. 3. K; and for the same reason the rectangle PKX KL is equal to the rectangle under the segments ot any otherline, parallel to the plane of the base, draw n trom P to cut theconic suriuee, because die section formed by a plane passing through

PL parallel to the base is a circle, and hence the proportion ismanifest in this case.

And if the point B, fig. 13, be within the Ktp of the cone, and astraight line, as PQR, parallel to the line V B that joins the point B

and the vertex of the cone, be drawn to cut the opposite surface*through a point P, situated without or within the cone - the pro-position may be demonstrated in this, as in the former case.

And if the point P, fig. 14, be without the cone, and the lineVB, and PS, parallel to VB, be drawn to touch the conic surface,instead of cutting it; then die plane PVB will meet the conic sur-face in a line VSM ; and BM will touch the base of the cone, andPN, parallel to BM, will touch the conic surface. And becausethe two triangles SPN and VBM are equiangular, thereforeVB : PS : : BM : PNAnd VB 2 : PS 2 :: BM 4 : PN 5

But BM 2 is equal to the rectangle under the segments of anyline draw n from B to cut the base of the cone; and PN 2 is equalto the rectangle under the segments of any line, parallel to thebase of the cone, drawn from P to cut the conic surface; andhence the proposition is manifest in this case also.

Prop.V. Fig. 15.

If a point be assumed without or within a cone, and two linesbe drawn through it to meet a conic surface, or opposite surfaces,and so as to be parallel to two straight lines given by position; thenthe rectangle under the segments of the secant, or the square ofthe tangent, parallel to one of the lines given by position, has tothe rectangle under the segments of the secant, or to the square ofthe tangent, parallel to the other line given by position, a ratiothat is constantly the same, wherever the point (from which thelines are drawn) is assumed without or within the cone.

Let VB and VC lie two straight lines, fig. 15, drawn from the vertex<n a cone to the plane of the base, and given by position (or parallelto lines given by position) ; and let PQ arnlM N be two straight linesdrawn through any assumed point, as R, to cut the conic surface,and so as to be respectively parallel to CV and VB : then as CV 4is to the rectangle CK X CL (contained by the segments of anyline draw n from C to cut the base of the cone), so let D, any as-sumed line, or magnitude, be to E ; and as VB 2 is to BG X BH(the rectangle contained by the segments of any line drawn from Bto cut the base of the coifP), so let F be to E ; and draw ST pa-rallel to the base of the cone through the point R; then, Pr. 4.

(CV 2 : CK x CL, or) L> : E :: PR x RQ : SR x RT, andBV 2 : BG x BH, or) F : E :: MR x RN : SR x RT. There-fore invertendo and ex icquo,

D: F :: PR X RQ : MR X RN.

And, as the same reasoning applies wherever the point R is as-sumed, therefore the ratio of the rectangles PR x RQ, andMR x RN is the same with, or equal to, the constant ratio ot Dto F, wherever the point R is as-umed. And, in like manner,may the proposition be demonstrated i.) all other cases, or in allpositions of the lines PQ. and MN whether they cut, or touch,the same or opposite surfaces.

i Prop. VI. Fig. 16.

If a right line, as PT, drawn through a point P in the surfaceof a cone, so as to be parallel to a right line VB contained in theconic surface, meet two parallel lines (in the points R and S) thatcut or touch the conic surface or opposite surfaces : then I’li is toPS as the rectangle under the segments of the secant, or thesquare of the tangent, drawn through the point R, is to the rec-tangle under the segments of the secant, or to the square of thetangent, drawn through the point S. Through the two parallelsPT and VB, tig. 16, draw a plane cutting the conic surface againin the line VA, and the plane of the base in the line BA; and,through R and S, draw MN-and HG parallel to AB. BecausePT is parallel to VB, and RN to SG, therefore RNGS is a paral-lelogram ; and RN = GS. It is obvious that the trianglesPMR and I’HS are equiangular: therefore PR is to PS as MR isto IIS, 4. 6. E, oras MR x RN is to IIS x SG, 1. 6. E. ButMR X RN and IIS X SG are respectively equal to the rectan-gles contained by the segments of any two lines, parallel to thebase of the cone, drawn through R and S to cut the conic surface,i nd hence the proposition is manifest, w lieu PT meets two linesparallel to the plane of the base. And if PT meet tw o parallel linesDKaiul IK, not parallel to the plane of the base; then, let thesame construction be made as before : and because DE is paral-lel toIK, and MN to Gil; therefore,

Dll X RE : M il X llN :: IS x SK : IIS X SG ;

Alternando*