NM the difference, and мo the sum of the ordinates EQ, NS. And the differences, of the abscisses, are QR, Qs, or EK, EM. Then by cor. theor. 9, qu: as :: CK. KL NM • MO,

EK EM CK. KL NM. MO.

that is Corol. 1. The rect. CL. KL = rect, EK and the param. of rs. For the rect. CK. KL = rect. QR and the param of rs. Corol. 2. If any line CL be cut by two diameters, EK, GI ; the rectangles of the parts of the line, are as the segments of the diameters.

For EK is as the rectangle cx . KL,

and GH is as the rectangle cit. HL;

therefore EK GH: CK. KL: CH. HL.

Corol. 3. If two parallels, CL, NO, be cut by two diame. ters, EM, GI; the rectangles of the parts of the parallels will be as the segments of the respective diameters.

For
and

EK: EM:: CK. KL: NM. MO,

EK GH:: CK KL CH.

HL,

theref. by equal. EM: GH NM. MO; CH. HL.

Corol. 4. When the parallels come into the position of the tangent at P, their two extremities, or points in the curve, unite in the point of contact ; and the rectangle of the parts becomes the square of the tangent, and the same properties still follow them.

So that, EV: PV :: PV: p the param.

GW: PW:: PW:
: P,

EV: GW:: PV2: PW2,
EV: GH:: PV2: CH. UL.

THEOREM XX.

If two parallels intersect any other two parallels; the rectangles of the segments will be respectively proportional.

'For, by cor. 3 theor. 23, PK: Q! :: CK

And by the same,

PK QI: DK .KE: NI. IO;

theref. by equal. CK . KL : DK. KE ¦¦ GI. IH : NI, 10.

Corol. When one of the pairs of intersecting lines comes ⚫ into the position of their parallel tangents, meeting and limit. ing each other, the rectangles of their segments become the squares of their respective tangents. So that the constant ratio of the rectangles, is that of the square of their parallel tangents, namely,

CK. KL: DK • KE :: tang2. parallel to CL : tang2, parallel to DE.

THEOREM XXI.

If there be three tangents intersecting each other; their
segments will be in the same proportion.
That is, GI: IH: CG: GD :: DH: NE.

CG GD CK: KM,

· DH HE MN: NE.

KL, CK, MN are all equal ;

as also the 4th terms LN, KM NE.

Therefore the first and second terms, in all the lines, are proportional, namely, GI: IH:: CG: GD. DH: HE. Q. E. D.

THEOREM XXII.

The Area or Space of a Parabola, is equal to Two-Thirds of its Circumscribing Parallelogram.

Let ACB be a semi-parabola, CB the axis, F the focus, ED

the directrix; then if the line AF be supposed to revolve about F as a centre, while the line AE moves along the directrix perpendicularly to it, the area gene. rated by the motion of AE, will always be equal to double the area generated by FA; and con

Hг

E

G

D

B

sequently the whole external area AEGD = double the area

ACF.

nadi

For draw a' E' parallel, and indefinitely near, to AE; and draw the diagonals a' and A'E; then by th. 6, cor. 4, the angles E'A'A and Fa'a are equal, aa' being considered as part of the tangent at A'; and in the same manner, the angles EAA' and FAA' are also equal to each other; and since EA = AF, and E'A' = A'F; the triangles EAA' and E'A'A are each equal to the triangle AA'F; but the triangle EAA' the triangle EE'A, being on the same base and between the same parallels; therefore the sum of the two triangles EE'A and EA'A, or the quadrilateral space EAA'E' is double the trilateral space AA'F; and as this is the case in every position of FA', E'A', it follows that the whole external area EACD = double the internal area AFC.

Hence, Take DG = FB, and complete the parallelogram DGHE, which is double the triangle ABF; therefore the area ABC the area HACG, or of the rectangle AEGH, or of the rectangle ABCI, because BCBG; that is, the area of a parabola of the circumscribing rectangle.

THEOREM XXIII.

Q. E. D.

*

The Solid Content of a Paraboloid (or Solid generated by the Rotation of a Parabola about its Axis), is equal to Half its Circumscribing Cylinder.

Let GHBD be a cylinder, in which two equal paraboloids are inscribed; one BAD having its base BCD equal to the lower extremity of the cylinder; the other Geit in. verted with respect to the former, but of equal base

and altitude. Let the plane

LR parallel to each end of the cylinder, cut all the three solids, while a vertical plane may be supposed to cut them so as to define the parabolas shown in the figure. Then, in the semi-parabola ACB, p. AP = PM3, also, in the semi-parabola ACG, p. cr = PN3,

consequently, by addition, p. (AP + CP)=p. AC = PM2 + PN3. But, p. ACCB2 = PL2.

Therefore PL = PM2 + PN2:

That is, since circles are as the squares of their radii, the

This demonstration was given by Lieut. Drummond of the Royal Engineers, when he was a gentleman Cadet at the Royal Military Acadeiny.

circular section of the cylinder, is equal to the sum of the corresponding sections of the two paraboloids.

The same property evidently obtains for any sections what. ever parallel to BD; it therefore holds for the two parabo. loids. In other words, the cylinder is equal to the two pa. raboloids taken together: wherefore, since the two parabo loids, having equal bases and equal altitudes, are equal to one another, it follows that each paraboloid is half of its circumscribing cylinder.

THEOREM XXIV.

Q. E. D.

The Solidity of the Frustum BEGC of the Paraboloid, is equal to a Cylinder whose Height is DF, and its Base Half the Sum of the two Circular Bases EG, BC.

Then, by the last theor.

pc X AD = the solid arc,

CX AF the solid AEG,

pc X (AD-AF) = the frust. begc.

AF = DFX (AD + AF),

theref. 1pc X DFX (AD+AF) = the frust. BEGC. But, by th. 1, p × AD = Dc3, and p X AF = FG2; theref. c X DF X (Dc2 + FG2) = the frust. BEGC.

Q. E. D.

PROBLEMS, &C. FOR EXERCISE IN CONIC SECTIONS.

1. Demonstrate that if a cylinder be cut obliquely the section will be an ellipse.

2. Show how to draw a tangent to an ellipse whose foci are F, f, from a given point r.

3. Show how to draw a tangent to a given parabola from a given point P.

4. The diameters of an ellipse are 16 and 12. Required the parameter and the area.

5. The base and altitude of a parabola are 12 and 9. Required the parameter, and the semi-ordinates corresponding to the abscissæ 2, 3, and 4.

6. In the actual formation of arches, the voussoirs or arch. stones are so cut as to have their faces always perpendicular

to the respective points of the curve upon which they stand. By what constructions may this be effected for the parabola and the ellipse?

7. Construct accurately on paper, a parabolt whose base shall be 12 and altitude 9.

8. A cone, the diameter of whose base is 10 inches, and whose altitude is 12, is cut obliquely by a plane, which enters at 3 inches from the vertex on one slant slide, and comes out at 3 inches from the base on the opposite slant side. Requir ed the dimensions of the section?

9. Suppose the same cone to be cut by a plane parallel to one of the slant sides, entering the other slant side at 4 inches from the vertex, what will be the dimensions of the section?

10. Let any straight line EFR be drawn through F, one of the foci, of an ellipse, and terminated by the curve in E and R; then it is to be demonstrated that EF. FR=ER.¦ pa

rameter.

11. Demonstrate that, in any conic section, a straight line drawn from a focus to the intersection of two tangents makes equal angles with straight lines drawn from the same focus to the points of contact.

12. In every conic section the radius of curvature at any point is to half the parameter, in the triplicate ratio of the distance of the focus from that point to its distance from the tangent.

Also, in every conic section the radius of curvature is proportional to the cube of the normal.

Also, let re be the radius of curvature at any point, r, in an ellipse or hyperbola whose tranverse axis is AB, conju.

gate ab, and foci F and ƒ: then is rc =

(PF Pf)

rf) }

‡AB . ab

Required demonstrations of these properties.