CONIC SECTIONS. 



161 



Sections 

 of a 

 Cone. 



ig. 92. 



rated two superfices which meet at V, and include with- 

 in them a solid called a Cone. 



2. The point V is called the Vertex of Ike Cone. 



3. The circle ADB is called its Base. 



4. Any straight line drawn from the vertex of the 

 cone to a point in the circumference of the base is call- 

 ed a Side of the Cone. 



5. A straight line VC drawn from the vertex to the 

 centre of the base, is called the Axis of the Cone. 



Cor. Any section of a cone through the axis is a tri- 

 angle. 



Prop. I. 



If a cone be cut by a plane parallel to its base, the 

 section is a circle. 



Let EFGM be the section, and the triangles VAB, 

 VCD two other sections made by any two planes 

 through the axis ; let EG, OF be the common sections 

 of these and the section EFG. Then, EO, FO, GO 

 are parallel to AC, DC, BC respectively : therefore 

 AC : EO ( : : CV: OV) : : DC : FO; but AC = DC ; 

 therefore EO = FO ; and in like manner it appears 

 that GOrrFO ; therefore EFGM is a circle of which 

 EG is the diameter. 



Prop. II. 



If a cone be cut by a plane which neither passes 

 through the vertex, nor is parallel to the base, the 

 section will be one or other of these figures, viz. a 

 circle, an ellipse, an hyperbola, or a parabola. 



Let PX be the common section of the base, and a 

 plane passing through V, the vertex of the cone ; and 

 let FKM be a section of the cone parallel to the plane 

 VPX. From C, the centre of the base, draw CN per- 



Eendicular to PX; meeting the circumference in A and 

 i ; and let a plane pass tlu'ough V, A, B, and meet the 

 plane XPV in the line NV, the surface of the cone in 

 VA, and VB, and the plane of the section FKM in KL ; 

 then because the planes XVP, MKF are parallel, KL 

 is parallel to VN. 



Let EFGM be a section of the cone parallel to the 

 base, meeting the plane VAB in EG, and the plane 

 FKM in FM ; and let L be the intersection of EG and 

 FM; then EG will be parallel to BN ; and FM is pa- 

 rallel to PX, and therefore will make the same angle 

 with LK, wherever the lines FM, LK cut each other ; 

 and since BN is perpendicular to PX ; EG is perpen- 

 dicular to FM. Now the section EFGM is a circle, of 

 which EG is the diameter (1.) therefore FM is bisect- 

 ed at L, and FL'=EL.LG. 



Now there will be four cases, resulting from the dif- 

 ferent positions of the line VN in respect of the cone. 

 Cases 1. and 2. Let the line PNX fall without the 

 base of the cone (Fig. 92.) Then KH will meet the 

 opposite sides in K and H, below the vertex. Draw 

 KR, HQ parallel to AB ; then, the triangles KLG, 

 KHQ are similar, as also HLE, HKR ; therefore, 

 KL : LG : : KH : HQ, 

 and HL : LE : : KH : KR ; 

 therefore taking the rectangles of the corresponding 

 terms, 



KL.LH : GL.LE or LF J : : KH J : HQ.KR. 

 Now if the angles VKH, VAB are equal, then also 

 VKH=VRK, and HKQ=HRK; and as the alternate 

 angles RKH, HKQ are equal, the triangles RKH, 

 HKQ are similar, so that RK : KH : : KH : HQ, and 

 HK^HQ.KR, therefore in this case KL.LH=LF% 



VOL. Til. TART I. 



Fig. 93. 



and consequently HFK is the circumference of a circle, 

 of which KH is the diameter (35. 3. E.) 



Note. This particular section of a cone is called a 

 Suhcontrary Section. 



If, however, the angles VKH, VAB be unequal, 

 then HK ! will not be equal to HQ.KR, but will have 

 to it the same ratio wherever the point L is situated in 

 the line HK ; therefore, in tins case KL.LH has a con- 

 stant ratio to LF 1 , and consequently the section HFK 

 is an ellipse, of which HK is a diameter, and LF an 

 ordinate (1. Cor. 16. Sect. II.) 



Case 3. Next let the line PNX fall within the base 

 of the cone, (Fig. 93.) then the plane FKM will meet 

 the base in a line SID, and the line KL will meet AV, 

 the side of the cone, at H a point above the vertex. 

 Draw VT perpendicular to IH, and by similar tri- 

 angles 



HT : TV : : LH : LE, 

 and KT:TV::KL:LG; 

 therefore, HT.KT : TV* : : HL.LK : LE.LG or LF'. 

 Hence it appears that in this case, HL.LK has to LF* 

 a constant ratio; therefore the section DFKMS is 

 an hyperbola, of which KH is a transverse diameter, 

 and FM an ordinate to that diameter. (2. Cor. 24. Sect. 

 III.) 



Case 4. Lastly, let the line PNX touch the cir- Fig. 94 

 cumference of the base in A, ( Fig. 94.) In this case, 

 DIS, the common section of the base and the plane 

 FKM, is evidently parallel to FLM, and perpendi- 

 cular to AB; therefore DPzrAI.IB, and DP : FL l : : 

 AI.IB : EL.LG : : IB : LG. But since EG is parallel to 

 AB, and IK parallel to AV, we have IB : LG : : KI : 

 KL ; therefore DP : FL= : : KI : KL ; hence it follows 

 that the section DFKMS is a parabola, of which KLI 

 is a diameter, and DIS, FLM ordinates to that diame- 

 ter. (Cor. to 12. Sect. IV.) 



SECTION VII. 

 Of the Areas of the Conic Sections. 



Prop. I. 



If an ordinate AB, and abscissa PQ of a parabola be Fig. 05. 

 completed into a parallelogram ABCD ; the area of the 

 parabola included between the ordinate and the curve 

 is two thirds of the parallelogram. 



Draw the chords AP, PB; also draw the diameters 

 Ff, I i, to bisect the semiordinates AQ, QB inland i, 

 and draw the chords AF, FP, PI, IB. Again, draw 

 the diameters E e, G g, H h, K k to bisect the distan- 

 ces between the points A,f, Q, i f B, and draw the 

 chords AE, EF, FG, GP, PH, HI, IK, KB; and sup- 

 pose this operation of drawing diameters to be repeat- 

 ed any number of times. By this process there will be 

 formed successive series of triangles ; the first consist- 

 ing of one term, viz. APB ; the second of two, viz. the 

 triangles AFP, PIB ; the third of four, viz. AEF, FGP, 

 PHI, 1KB, and so on. 



Let F/meet PD in M, and the chord PA in N ; and 

 because PA is manifestly an ordinate to the diameter 

 Ff, (Def. 8. Sect. IV.) and PM a tangent, at the ex- 

 tremity of the ordinate; FN=J MN; but because PM 

 =A/, it follows that MN=N f s=| PQ ; therefore FN 

 = \ PQ. Now the triangles AFP, AQP, which have 

 the same ba*e AP, being to each other as their alti- 



