162 



CONIC SECTIONS. 



tucles ; they WJH al«o be to one another as FN to QP, 

 which making equal angles with the common base, 

 have the same ratio as the altitudes ; therefore the tri- 

 angle AFP is ^ of the triangle APQ; and as in like 

 manner it may be shewn that the triangle BIP is I of 

 the triangle BPQ ; the triangles AFP, BIP are to- 

 gether ^ of the triangle APB. 



If a tangent be drawn at F, and produced to meet 

 the two diameters E e, Gg, it may be proved in the same 

 manner, that the triangles AEF, PGF are the fourths 

 of the triangles AFN, PFN, and therefore, that their 

 sum is \ of the triangle AFP ; and as it may also be 

 shewn that the triangles PHI, BKI are together equal 

 to £ of the triangle PIB, the four triangles AEF, FGP, 

 PHI, 1KB are together I of the sum of the triangles 

 AFP, PIB. Proceeding in this way, it appears that at 

 every succeeding bisection, a new set of triangles is 

 produced, which together are I of the area of the pre- 

 ceding set. 



Put S for the area of the parallelogram ADCB, then 

 we have 

 Triangle APB= |S, 

 Polygon AFPIB=(i+|.)S, 

 Polygon AEFGPHIKB=(A-K+A)S- 

 From the law according to which the terms in these 

 numerical series are formed, it is evident that whatever 

 be the number of sides of the polygon inscribed in the 

 parabola, its area will be expressed by the geometrical 

 series (^ + 4-+T2+^r+ & c -) S, the number of terms 

 depending on the number of sides of the polygon. 

 Suppose now the number of sides to be continually in- 

 creased, then its area will manifestly approach continu- 

 ally to the area of the parabola, which is its limit, and 

 will at last differ from it by less than any assignable 

 space. Therefore the area of the parabola is exactly 

 equal to the product of S, by the fraction which ex- 

 presses the sum of the geometrical series continued in- 

 definitely; now, observing that the first term is ^, and 

 common ratio \, the sum is readily found by a known 

 formula to be \, (See Algebra, Art. 207.) therefore 

 the area of the parabola is j- of the parallelogram 

 ABCD. 



Definition. 



If the two axes of an hyperbola be equal, it is called 

 an Equilateral Hyperbola. 



Prop. II. 



F! j. 96, 97. Let C be the centre of a circle, or of an equilateral 

 hyperbola, and CA the semitransverse axis ; let ACB, 

 BCD be two equal sectors of the circle, or hyperbola, 

 and let AHK, a tangent at the vertex A, meet the se- 

 midiameters CB, CD in H and K; then, 



in the circle, 2 CA : AK:: CA S — AH* : CA.AH, 

 and ift the hyperbola, 2 CA : AK : : CA 2 -)- AH 2 : CA.AH. 



For draw DA, D a to the extremities of the axis; let 

 the chord DA meet the semidiameter CB in O, and 

 draw DE perpendicular to the axis; and because 

 AO=:OD, (Geometry, and Prop. 34. Part III.) and 

 AC=Ca, the line CO, a D are parallel (2.6. E.) there- 

 fore the triangles CAH, a ED are similar; but because 

 ED*=AE.Ea (Prop. 24. of Part III.) AE:ED:: 

 ED: Ert (16. 6. E.) hence the triangles DEA, a ED 

 are similar, (6. 6. E.) and each is similar to CAH ; 

 therefore 



Ea : ED : : AC : AH : : AC : AC.AH, 

 EA ; ED : : AH: AC : : AH a : AC.AH, 



Area 1 ! of 

 the Conic 

 Sections. 



Hence in the circle, 



Efl-EA or 2 CE : ED : : AC*— AH* : AC.AH; 

 and in the hyperbola, (24. 5. E.) 



Ea + EA or 2CE.ED::AC 2 + AH 2 :AC.AH. ' «"—■ ' 



But 2 CE : ED : : 2 CA : AK 

 therefore 2 CA : AK : : AC 2 =p AH 2 : AC.AH. 



Prop. III. 



If two ellipses, or two hyperbolas, have a common Fi". 93,99, 

 axis A a, and a semiordinate DD'E be drawn from the 

 same point E in the axis, to each of the curves; the 

 areas ADE, AD'E included between the common ab- 

 scissa, the ordinates, and the two curves are to one ano- 

 ther as the conjugate axes BC, B'C. 



Let the common abscissa AE be divided into any 

 number of equal parts Ap, pq, qYL ; draw the semior- 

 dinates pm'm, qn'n, and the chords Am, mn, «D, 

 also A m' , m! n', n' D'. And because BC* : DE 2 (: : 

 AC.Crt : AE.En) : : B'C : D'E* (1. Cor. 16. Part II, and 

 2. Cor. 24. Part III.), therefore BC : DE : : B'C : D'E, 

 and BC : B'C : : DE : D'E ; in like manner it appears 

 that BC : B'C : : nq: n'q : : mp : m' p. Now, the tri- 

 angle Apm is to the triangle Apm! as mp to m' p, 

 that is, as BC to B'C ; and the trapezoid p in n q is evi- 

 dently to the trapezoid p m' n' q, also the trapezoid 

 <7»DE to the trapezoid qn'D'JL in the same ratio; 

 therefore the polygon A m n DE is to the polygon 

 Awj'w'D'E also as BC to B'C : and this will be true, 

 whatever be the number of equal parts into which the 

 common abscissa AE is divided. Suppose, now, that 

 number to be indefinitely increased, then it is evident 

 that the polygons will approach continually to the el- 

 liptic, or hyperbolic areas in which they are- inscribed, 

 so as at'last to differ from them by less than any as* 

 signable space; therefore the elliptic, or hyperbolic 

 spaces themselves, which are the limits of the polygons, 

 are also to one another as BC to B'C. 



Cor. 1. Hence it appears, that the area of an ellipse 

 is to that of its circumscribing circle, as the conjugate 

 axis to the transverse axis. For a circle may be con-- 

 sidered as an ellipse having its axes equal. 



Cor. 2. It also appears, that the area of any segment 

 of an ellipse may be found from that of a correspond- 

 ing segment of a circle : And the area of any hyperbo- 

 lic segment from the corresponding segment of an 

 equilateral hyperbola, having the same transverse axis. 



Prop. IV. Problem. 



To investigate a formula that shall express the area 

 of a circle, or equilateral hyperbola. 



Let ACD be a sector of a circle or hyperbola, and p;_ jo^ 

 AC the radius of the circle, or semitransverse axis of 101. 

 the hyperbola. Let the sector be bisected by the se- 

 midiameter CB, and let each of the sectors ACB, BCD 

 be bisected by the semidiameters CP, CQ; and pro- 

 ceed in this way, by repeated bisections, to divide the 

 whole sector first into two equal parts, then into four, 

 then into eight, and so on. Draw tangents to the 

 curves at the alternate points A, B, D ; and because, 

 from the nature of the curves, the chords which join 

 these points are ordinates to the semidiameters CP, 

 CQ, which pass between them, the tangents will inter- 

 sect each other at G, L, points in the semidiameters 

 (1. Cor. Prop. 13. Sect. III.) Draw the chord AB join- 

 ing any two contiguous points of contact, and let it 

 meet CP in I ; and because the triangles ACI, BCI are 

 equal (38. 1, E.), as also the triangles AG I, BGI, by 



