CT: CF CA: CA + FE;

TF: Tf :: FE :2CA + FE or fɛ by th. 5.
TF: Tf :: FE: fE;

Corol. As opticians find that the angle of incidence is equal to the angle of reflexion, it appears, from this proposition, that rays of light issuing from the one focus, and meeting the curve in every point, will be reflected into lines drawn from the other focus. So the ray fe is reflected into FE. And this is the reason why the points F, f, are called foci, or burning points.

THEOREM X.

All the Parallelograms inscribed between the four Conjugate Hyperbolas are equal to one another, and each equal to the Rectangle of the two Axes.

That is,

the parallelogram PQRS =
the rectangle AB. ab.

E

R

D

Let EG, eg be two conjugate diameters parallel to the sides of the parallelogram, and dividing it into four less and equal parallelograms. Also, draw the ordinates DE, de, and cê perpendicular to PQ; and let the axis produced meet the sides of the parallelograms, produced, if necessary, in T and t.

ct

Then, by theor. 7, CT: CA :: CA: CD,
: CA :: CA: cd;
CT ct : cd: CD;

and

theref. by equality,

but, by sim. triangles, CT: ct :: TD : cd,

theref. by equality,

and the rectangle

Again, by theor. 7, or, by division,

TD: cd :: cd: CD,

TD DC is = the square

and, by composition, CD: DB :: DA : DT;

conseq. the rectangle

CD

AD. DB ➡ CD2 — CA2,
CD2- cd2.

de2- 'DE".

But,

The Difference of the Squares of every Pair of Conjugate Diameters, is equal to the same constant Quantity, namely the Difference of the Squares of the two Axes.

All the Parallelograms are equal which are formed between the Asymptotes and Curve, by Lines drawn Parallel to the Asymptotes.

For, let A be the vertex of the curve, or extremity of the semi-transverse axis AC, perp. to which draw AL or Al, which will be equal to the semi-conjugate, by definition 19. Also, draw HEDeh parallel to Ll,

Then, by theor. 2, CA: AL2 :: CD2 and, by parallels,

DE2 or

CA: AL:: CD: DH2;
CA2: AL2:: CA2: DH2

rect. HE. Eh;

the rect. HE. Eh.

theref. by subtract.

conseq. the square AL

But, by sim. tri. and, by the same, theref. by comp. and, because

PA. AQ : AL2 :: GE . EK: HE. Eh;

AL2=HE. Eh, theref. PA. AQ= GE, EK.

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

Therefore the parallelogram GK = the paral. PQ. That is, all the inscribed parallelograms are equal to one another.

Q. E. D.

Corol. 1. 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 towards 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.

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

THEOREM.

THEOREM XIII.

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 Asymp

=

FOR, by theor. 12, CPAQ CGEK;
subtract the common space CGIQ,

there remains the paral. PI= the par. IK;
to each add the trilineal IAE, then

the sum is the quadr. PAEG QAEK.
Again, from the quadrilateral CAEK
take the equal triangles CAQ, CEK,
and there remains the sector CAE QAEK.
Therefore CAE QAEK PAÉG.

Q. E. D.

OF THE PARABOLA.

THEOREM 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 also let AFH be the comcone; 'mon intersection of the two planes, or the axis of the parabola, and FG, HI ordinates perpendicular to it.

12

K

D

M

Then

Then it will be, as AF: AH :: FG2: HI2.

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 similar triangles, AF: AH :: FL: HN; but, because of the parallels,

KF = MH;

KF. FL: MH. HN.
HN = HI2;

AF AH:: FG2: HI2.

Q. E. D.

Corol. Hence the third proportional or

FG2 HI2

is a con

AF

AH

stant quantity, and is equal to the parameter of the axis by

As the Parameter of the Axis :
Is to the Sum of any Two Ordinates ::
So is the Difference of those Ordinates:
To the Difference of their Abscisses :

So that any diameter EI is as the rectangle of the segments KI, IH of the double ordinate Kн.

THEOREM