the triangle

LQH becomes the triangle ceм.

and the space TELK becomes the triangle TEC;

and theref. the Acem = ATEC = ▲ANC = AIRC.

THEOREM XX.

Any diameter bisects all its double ordinates, or the lines drawn parallel to the tangeut at its vertex, or to its conjugate diameter.

For, draw aн, qh perpendicular to the transverse.
Then by cor. 3, theor. 19, the Lon= Lgh;
but these triangles are also equiangular;
consequently their like sides are equal, or LQ

1.9.

Corol. Any diameter divides the ellipse into two equal parts.

For, the ordinates on each side being equal to each other, and equal in number; all the ordinates, or the area, on one side of the diameter, is equal to all the ordinates, or the area, on the other side of it.

THEOREM XXI.

As the square of any diameter

Is to the square of its conjugate,

So is the rectangle of any two abscisses
To the square of their ordinate.

That is, ce2: ce2:: EL. LG or ce2

CL2: LQ2.

being as the squares of their like sides, it is,
by sim. triangles, ACET: ACLK :: CE2: CL2;
or, by division, ACET: trap. TELK :: CE2: CE2 — CL3.
Again, by sim. tri. ▲ cem: ▲ LQH :: Ce2: LQ2.

But, by cor. 5 theor. 19, the ACEM = ACET,
and, by cor. 4 theor. 19, the ▲1QH = trap. TELK ;
theref. by equality, ce ce2:: CE2 — CL2 : LQ3,
CE: ce2:: EL. LG: 1.Q3. Q. e. d.

or

.

Corol. 1. The squares of the ordinates to any diameter, are to one another as the rectangles of their respective abscisses, or as the difference of the squares of the semidiameter and of the distance between the ordinate and centre. For they are all in the same ratio of ce2 to ce3.

Corol. 2. The above being a similar property to that belonging to the two axes, all the other properties before laid down, for the axes, may be understood of any two conjugate diameters whatever, using only the oblique ordinates of these diameters, instead of the perpendicular ordinates of the axes; namely, all the properties in theorens 6, 7, 8, 14, 15, 16, 18, and 19.

THEOREM XXII.

If any two lines, that any where intersect each other, meet the curve each in two points; then

the rectangle of the segments of the one
is to the rectangle of the segments of the other,
as the square of the diam. parallel to the former
to the square of the diam. parallel to the latter.

For, draw the diameter CHE, and the tangent TE, and its parallels PK, RI, MH, meeting the conjugate of the diameter CR in the points T, K, I, M. Then, because similar triangles are as the squares of their like sides, it is,

by sim. triangles CR2: GP2:: ACRI: AGPK,

and

CR: GH2: A CRI: AGHM;

theref. by division, CR2: GP2

GI:: CRI: KPHM.

Again, by sim. tri. CE3: CH2: :: АСТЕ: Асми;

and by division, CE: CE3 СН2:: АСТЕ : ТЕНИ. But, by cor. 5 theor. 19, the ACTE = ▲ CIR,

and by cor. 1 theor. 19, TEHG≈ KPHG, or TEHM = KPHM; theref. by equ. ce2: ce3· CH2: CR2: GP GH2 Or PH. HQ.

::

In like manner c2: CE3 ·сн2:: cr2: ри . нq.
Theref. by equ. CR2: 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 Pн'Q, pн'q' are also in the same ratio of CR to cr2; therefore rect. PHQ: рHq:: Pи'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. PuQ: puq:: Pha' : phq'.

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 perpendi. cular 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 FG3: H2 :: AF. FB: AH.HB.

M

R

B

D

P

E

K

T

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 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 MH. HN = HI3; Therefore the rect. AF. FB: AH

HB:: FG: HI2.

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

But, if c be the centre, then AC. CB = Ac3, and ca is the

Therefore

semi-conj.

AC: AD.

DB:: ac2: DE3 ;

Q. E. D.

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

AB2: ab2:: AD. DB: De3.

or, by doubling,

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, lgk. Through c, the middle point of sa, 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.

acn, ark

AC: Cm AF: FL;

ac: cn: aF : FK.

cn: AF Fa LF
Fa: FG2.

.. AC. ca: cm.
or, ac2: ct:: AF

In like manner, for the opposite hyperbola

Ac2: ct2:: Af. fa : fg3.

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.