PROPOSITION XIV.

The difference of the squares of every pair of conju gate diameters is equal to the same constant quantity, namely, the difference of the squares of the two axes.

That is, AB-ab-EG-eg', where EG, eg, are any conjugate diameters.

For, draw the ordinates ED, ed.

Then, by step 10, last dem. Cd'AD DB,

hence Cd' CD-CA",

therefore, CACD-Cd* ;

in like manner, Ca-de-DE2 ;

therefore, the diff. CA-Ca'

CD+DE-Cd-de*:

But, by right-angled triangles, CE2=CD2+DE*,

therefore, CE-Ce-CD+DE-Cd-de2.
Consequently, CA-Ca' CE'-Ce';
or, by doubling, AB-ab'EG'—eg*.

Q. E. D.

PROPOSITION

PROPOSITION XV.

All the parallelograms are equal, which are formed be tween the asymptotes and curve, by lines drawn parallel to the asymptotes.

That is, the lines GE, EK, AP, AQ, being parallel to the asymptotes CH, Cl; then the parallelogram CGEK parallelogram CPAQ.

P

H
G

CK Q

For, let A be the vertex of the curve, or extremity of the semitransverse axis AC, perpendicular to which draw AL, or Al, which will be equal to the semiconjugate.

Also, draw HEDeh parallel to Ll.

Then, by Prop. II. and, by parallels, therefore, by sub.

CA* : AL2 :: CD1-CA' : DE", CA: AL* :: CD2 : DH'; CA1: AL1 :: CA1 : DH'-DE”, or rect. HE Eh;

consequently, the square AL' the rect. HE Eh.

But, by sim. tri.

: EH,

PA : AL :: GE
GE
QA : AI :: EK : ᎬᏂ ;

and, by the same,
therefore, by comp. PA∙AQ : AL' :: GE·EK: HE Eh,
and, because AL*—HE Eh, therefore, PA AQ=GE·EK.

VOL. II.

Ccc

But,

But, the parallelograms CGEK, CPAQ2 being equiangular, are as the rectangles GE EK and PA ·AQ.

And therefore the parallelogram GK the parallelo gram PQ. That is, all the inscribed parallelograms are equal to one another.

:

COR. I. Because the rectangle GEK or CGE is constant, therefore GE is reciprocally as CG, or CG : CP :: PA GE. And hence the asymptote continually approaches toward the curve, but never meets it: for GE decreases continually as CG increases; and it is always of some magnitude, except when CG is supposed to be infinitely great, for then GE is infinitely small, or nothing. So that the asymptote CG may be considered as a tangent to the curve at a point infinitely distant from C.

COR. 2. If the abscisses CD, CE, CG, &c. taken on one asymptote, be in geometrical progression increasing; then shall the ordinates DH, EI, GK, &c. parallel to the other asymptote, be a decreasing geo

H

K

metrical progression, having the same ratio. For, all the rectangles CDH, CEI, CGK, &c. being equal, the ordinates DH, EI, GK, &c. are reciprocally as the abscisses CD, CE, CG, &c. which are geometricals. And the reciprocals of geometricals are also geometricals, and in the same ratio, but decreasing, or in converse order.

PROPOSITION

PROPOSITION XVI.

The three following spaces, between the asymptotes and the curve, are equal; namely, the sector, or trilinear space contained by an arc of the curve and two radii, or lines drawn from its extremities to the centre; and each of the two quadrilaterals, contained by the said arc, and two lines drawn from its extremities parallel to one asymptote, and the intercepted part of the other asymptote.

That is, the sector CAE PAEG QAEK, all standing on the same arc AE.

For, by Prop. XV. CPAQ=CGEK; subtract the common space CGIQ2 so shall the paral. PI the par. IK; to each add the trilineal IAE,

then is the quadrilateral PAEG-QAEK.

Again, from the quadrilateral CAEK take the equal triangles CAQ CEK, and there remains the sector CAE QAEK.

Therefore, CAE QAEK PAEG.

Q. E. D.

PARABOLA.

PARABOLA,

PROPOSITION I

The abscisses are proportional to the squares of their

ordinates.

Let AVM be a section through the axis of the cone, and AGIH a parabolic section by a plane perpendicular to the former, and parallel to the side VM of the cone; also let AFH be the common intersection of the two planes, or the axis of the parabola, and FG, HI, ordinates perpendicular to it.

For, through the ordinates FG, HI, draw the circular sections KGL, MIN, parallel to the base of the cone, having KL, MN, for their diameters, to which FG, HI, are ordinates, as well as to the axis of the parabola.

Then, by sim. tri. AF: but, because of the parallels, therefore, AF : AH

FL : HN;

AH

KF=MN;

:: KF FL : MH⚫HN.

But, by the circle, KF FLFG', and MH HN=HI';

therefore,

AF : AH :: FG:

HI'.

Q. E, D..

COR.