Again, by sim. tri. ce3: cn2:: ACTE: ▲ CMH;
CE: CE2 CH3 : : АСТЕ : ТЕНМ.

and by division,

But, by cor. 5 theor. 19, the ACTE = ▲CIR,

and by cor. I theor. 19, тEHG =≈ KPHG, or TEHM = KPHM ; theref. by equ. ce2: ce2· :: CH2: CR2: GP2- GH2 or PH. HQ.

In like manner cÊ2: CE2 CH2: cr2: pu. иq.

Theref. by equ. cr3: cr3 :: PH. HQ: рн. нq..

Q. E. D.

Corol. 1. In like manner, if any other line p'n'q', parallel to cr or to pq, meet PHQ; since the rectangles PH'Q, p'H'q' are also in the same ratio of CR to cr2; therefore rect. PHQ: рHq: Pu'q : p'u'q'.

Also, if another line p hq' be drawn parallel to PQ or CR; because the rectangles, r ho', p hq' are still in the same ratio, therefore, in general, the rect. PHQ: puq:: Pho': p hq'.

That is, the rectangles of the parts of two parallel lines, are to one another, as the rectangles of the parts of two other parallel lines, any where intersecting the former.

Corol. 2. And when any of the lines only touch the cure, instead of cutting it, the rectangles of such become squares, and the general property still attends them.

OF THE HYPERBOLA.

THEOREM I.

The Squares of the Ordinates of the Axis are to each other as the Rectangles of their Abscisses.

LET AVB be a plane passing through the vertex and axis of the opposite cones; AGIH another section of them perpendicular to the plane of the former; AB the axis of the hyperbolic sections; and FG, HI, ordinates perpendicular to it. Then it will be, as FG2: HI :: AF. FB: AH.HB.

For, through the ordinates FG, HI, draw the circular sections KGL, MIN, parallel to the base of

M

R

D

P

E

K

the cone, having KL, MN, for their diameters, to which FG, HI, are ordinates, as well as to the axis of the hyperbola. Now, by the similar triangles AFL, AHN, and BFK, BHM, it is AF AH :: FL: HN,

and FB: HB :: KF : MH;

hence, taking the rectangles of the corresponding terms, it is, the rect. AF. FB: AH. HB :: KF. FL: MH. HN. But, by the circle, KF. FL = FG2, and мH . HN = Therefore the rect. AF. FB: AH. HB :: FG2: HI'.

= HI2;

For, by theor. 1, ca . CB : AD. DB:: ca3: DE3;

But, if c be the centre, then AC. CB = Ac2, and ca is the

semi-conj.

Therefore or, by permutation, ac2 : ac2:: AD. DB : DE2;

AC: AD. DB:: ac2 : De2;

or, by doubling,

AB: ab:: AD. DB : DE2.

Q. E. D.

:: AD • DB or CD3-CA2 : de3,

DB or CD2—CA2 : DE2;

ab2

AB

That is, As the transverse,
Is to its parameter,

by the definition of it.

So is the rectangle of the abscisses,
To the square of their ordinate.

Otherwise, thus:

Let a continued plane, cut from the two opposite cones, the two mutually connected opposite hyperbolas HAG, hag, whose vertices are a, a, and bases нG, hg, parallel to each other, fall. ing in the planes of the two parallel circles LGK, Igk. Through c, the middle point of aa, let a plane be drawn parallel to that of LGK, it will cut in the cone LVK a circular section whose diameter is mn; to which circu lar section, let ct be a tangent at t. Then, by sim. tri.

ACM, AFL

and, by sim. tri.

асп, aFk

}

H
L

AC cm: AF: TL;

ac: cn: aF: FK.

In like manner, for the opposite hyperbola

AC: ct: Af. fa: fg.

::

Here ct is what is usually denominated the semi-conjugate to the opposite hyperbolas HAK, hak: but it is evidently not in the same plane with them.

THEOREM III.

As the Square of the Conjugate Axis

Is to the Square of the Transverse Axis,

So is the Sum of the Squares of the Semi-conjugate, and distance of the Centre from any Ordinate of the Axis, To the Square of the Ordinate.

For, draw the ordinate ED to the transverse AB.

Then, by theor. 1, ca2 CA: DE2: AD

or

But

DB or CD2

ca: CA :: cd2 dɛ2 - CA3.

:

theref. by compos. ca2: ca2:: c^2 + cd2 : dɛ3.

In like manner,

CA ca: CA + CD2: De3.

and by this theor.

Q. E. D.

Corol. By the last theor. CA2: ca2:: CD2-CA: DE3,
cA2: ca2:: CD2+CA2 : De3,
DE: De :: CD2 — C3a : CD3 + CA2.
De2 de

therefore

In like manner,

:

THEOREM IV.

cd2-ca2: CD2 + ca3.

The Square of the Distance of the Focus from the Centre, is equal to the Sum of the Squares of the Semi-axes.

Or, the Square of the Distance between the Foci, is equal to the Sum of the Squares of the two Axes.

For, to the focus F draw the ordinate FE; which, by the definition, will be the semi-parameter. Then, by the nature CA2: ca3: CF3 CA: FE2;

Corol. 1. The two semi-axes, and the focal-distance from the centre, are the sides of a right-angled triangle caa; and the distance sa is CF the focal distance.

Corol. 2. The conjugate semi-axis ca is a mean proportional between AF, FB, or between af,ƒB, the distances of either focus from the two vertices.

For ca CF2 CA2 = (CFCA). (CF — CA) = AF. Fb.

THEOREM V.

The Difference of two Lines drawn from the two Foci to meet at any Point in the Curve, is equal to the Transverse Axis.

That is,

- FE AB.

fe

a

FDI

For, draw AG parallel and equal to ca the semi-conjugate; and join cc, meeting the ordinate DE produced in H; also take ci a 4th proportional to CA, CF, cd.

Then, by th. 2, CA

and, by sin. consequently

: AG :: CD2 CA2: DE2; AS, CA: AG2 :: CD2

CA2: DH2

-

DE3 DH3 AG2 = DH2

--

Also, FD CF CD, and FD2 = CF3 2CF. CD + CD2; and, by right-angled triangles, FE2 = FD2 + DE3. therefore FE2 = CF3

But, by theor. 4, CF3 ca2 CA2,

and, by supposition, 2cr. CD = 2ca. ci;

theref. FECA2 - 2CA. CI+ CD + DH';

And the root or side of this square is FE CI
In the same manner, it is found that fe
Conseq. by subtract. fe

CD.

Corol. 1. Hence CHCI is a 4th proportional to CA, CF,

Corol. 2. And fe + FE 2CH or 2c1; or FE, CH, fe, are in continued arithmetical progression, the common difference being ca the semi-transverse.