Page:Encyclopædia Britannica, Ninth Edition, v. 6.djvu/310

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ABC—XYZ

CONIC SECTIONS 40. PROP. XIX. If POP , pOp be any two chords, and CD, Cd the semi-diameters parallel to them, then PO . OP : pO . 0/ = CD 2 : Cd* . From the last proposition we have PO . OP : RO . OR = CD 2 : CR 2 , and also pO . Op : RO . OR = Cd* : CR* . Therefore PO . OP : CD 2 = RO . OR : CR 2 =pO . Op ; Cd*, or PO . OP : pQ . Op = CD 2 : Cd 2 . PROP. XX. If from a point Q on one asymptote (fig. 40) ordinates QPM, QDN be drawn to two conjugate hyperbolas in P, D, PD will be parallel to the other asymptote. QM 2 : QN 2 = QM 2 : CM 2 = BC 2 : AC 2 and QM 2 - PM 2 : QN 2 - DN 2 = BC 2 : AC 2 . . . PM 2 : DN 2 = BC- : AC 2 = QM 2 : QN 2 or PM :DN=QM : QN . Therefore DP is parallel to NM (Eucl. vi. 2), and NM is parallel to BA, and therefore to the other asymptote. COROLLARY. It follows therefore that CP, CD are conjugate (Prop, xvi.) PROP. XXI. If CP, CD be conjugate semi-diameters, CP 2 - CD 2 = CA 2 - CB 2 . Let ordinates PM, DN (fig. 41) in the two hyperbolas be produced, they will meet in a point Q on the asymptote (Prop, xx., Cor.) Then CP 2 - CD 2 == CM 2 + MP 2 - CN 2 - ND 2 = QN 2 + MP 2 -QM 2 -DN 2 = QN 2 -DN 2 -(QM 2 -PM 2 ) = CA 2 -CB 2 (Prop. xiv.) It follows that if the tangent at P meets the axes in T, T , then PT.PT = CD 2 . For PT:CD = PM:CN and PT :CD = CM:DN .-. PT.PT :CD 2 = PM.CM : CN . DN =PM.QN :QM.DN = 1 : 1 .-. PT.PT = CD 2 . Fig. 4L PUOP. XXII. If CP, CD be conjugate semi-diameters, the area of the triangle CPQ is constant. Produce QP (fig. 42) to meet the other asymptote in ; and join MON, PLD. They are parallel to CQ . QO : LO = QM : PM (Eucl. vi. 2.) QO 2 :L0 2 = QM 2 : PM 2 .-. Q0 2 -L0 2 : Q0 2 = QM 2 -PM 2 : QM 2 or QL . LC : Q0 2 = BC 2 : QM 2 N 4CL.LQ :CQ 2 = BC 2 : QM 2 QM 2

CB 2

CB 2 or 4CL.LQ : BC 2 = CQ 2 = CA = CS 2 .-. 4CL.LQ = CS 2 . Now in the right angled triangle PQD, L is the middle point of the hypotenuse ; therefore . . 4CL.LP = CS 2 . If PL be drawn parallel to CL to meet the other asymptote, 4PL.PL =-CS 2 , and the area of the quadrilateral CLPL is constant. It follows that, if the tangent at P meets the asymptotes in K, K , the area of the triangle CKK is constant ; also that the area of the quadrilateral formed by the tangents at the extremities of two conjugate diameters is constant. PART IV. -THE CONE AND ITS SECTIONS. DEFINITIONS. If through the point V, without the plane of the circle ADB (fig. 43), a straight line AV be drawn, and produced indefinitely both ways, and if the point V remain fixed while the straight line AV is moved round the whole cir cumference of the circle, a superficies of two sheets, which is called a cone, will be generated by its motion. K The fixed point V is called the vertex of the cone. The circle ADB is called the base of the cone. Any straight line drawn from the vertex to the circumference of the base is called a side of the cone. A straight line VC drawn through the ver tex of the cone, and the centre of the base, is called the axis of the cone. If the axis of the cone be perpendicular to the base, it is called a right cone. If the axis of the cone be not perpendicular to the base, it is called a scalene cone. PROP. I. If a cone be cut by a plane passing through the vertex, the section will be a triangle. Let ADVB be a cone of which VC is the axis ; let AD be the common section of the base of the cone and the cutting plane ; join VA, VD. When the generating line comes to the points A and D, it is evident that it will coincide with the straight lines VA, VD ; they are therefore in the surface of the cone, and they are in the plane which passes through the points V, A, D, therefore the tri angle VAD is the common section of the cone and the plane which passes through its vertex, PROP. II. If a cone be cut by a plane parallel to its base, the section will be a circle, the centre of which is in the axis. Let EFG be the section made by a plane parallel to the base of the cone, and VAB, VCD two sections of the cone made by any two planes passing through the axis VC ; let EG, HF be the common sections of the plane EFG and the planes VAB, VCD. Because the planes EFG, ADB are parallel, HE, HF will be parallel to CA, CD, and AC:EH = (VC:VH = )CD:HF; but AC = CD, therefore EH =.HF. For the same reason GH = HF; therefore EFG is a circle of which H is the centre and EG the diameter. PROP. III. If a scalene cone ADBV (fig. 44) be cut through the axis by a plane perpendicular to the base, making the triangle VAB, and from any point H in the straight line A V a straight line II K be drawn in the plane of the triangle VAB, so that the angle VHK may be equal to the angle VBA, and the cone be cut by another plane passing through HK perpendicular to the plane of the triangle ABV, the common section HFKN of this plane and the cone will be a circle. Take any point L in the straight line HK, and through L draw EG parallel to AB, and let EFGN be a section parallel to the base, passing through EG ; then the two planes HFKN, EFGN being perpendicular to the plane VAB, their common section FLN perpendicular to ELG, and since EFGN is a circle (by last Prop.), and EG its diameter, the square of FL is equal to the rectangle contained by EL and LG (Eucl. iii. 35) ; but since the angle VHK is equal to VBA or VGE, the angles EHK,EGK are equal, therefore the points E, H, G, K, are ii the circumference of a circle (Eucl. iii. 21), and HL . LK = EL . LG (Eucl. iii. 35) =FL 2 , therefore the section HFKN is a circle of which HLK is a diameter (Eucl. iii. 35).

This section is called a Subcontrary Section.