THEOREM XVI (12).

The equal Ordinates, or the Ordinates at equal Distances from the Centre, on the opposite Sides and Ends of an Hyperbola, have their Extremities connected by one Right Line passing through the Centre, and that Line is bisected by the Centre.

That is, if CD = CG, or the ordinate DE = GH; then shall

CE CH, and ECH will be a right line.

GBS

TAD

For, when CD = CG, then also is DEGH by cor. 2 theor. 1. But the 4D=4G, being both right angles;

therefore the third side CE = CH, and the 4DCE GCH, and consequently ECH is a right line.

Coral. 1. And, conversely, if ECH be a right line passing through the centre; then shall it be bisected by the centre, or have CE = CH; also DE will be =GH, and CD = CG.

Corol. 2. Hence also, if two tangents be drawn to the two ends E, H of any diameter EH; they will be parallel to each other, and will cut the axis at equal angles, and at equal distances from the centre. For, the two CD, CA being equal to the two CG, CB, the third proportionals CT, cs will be equal also; then the two sides CE, CT being equal to the two CH, cs, and the included angle ECT equal to the included angle HCS, all the other corresponding parts are equal: and so the <T=2s, and TE parallel to HS.

Corol. 3. And hence the four tangents, at the four extremities of any two conjugate diameters, form a parallelogram inscribed between the hyperbolas, and the pairs of opposite sides are each equal to the corresponding parallel conjugate diameters. For, if the diameter eh be drawn parallel to the tangent TE or HS, it will be the conjugate to EH by the definition; and the tangents to eh will be parallel to each other, and to the diameter EH for the same reason.

THEOREM XVII (13).

If two Ordinates ED, ed be drawn from the Extremities E, e, of two Conjugate Diameters, and Tangents be drawn to the same Extremities, and meeting the Axis produced in T and R;

Then

Then shall CD be a mean Proportional between cd, dr, and cd a mean Proportional between

CD, DT.

Corol. 2.
And the rect. CD. DE

Corol. 3.

cd. de, or ▲ CDE = A cde.

Also cd CD. DT, and CD2 = cd. dr. Or cd a mean proportional between CD, DT; and CD a mean proportional between cd, dr.

THEOREM XVIII (14).

Q. E. D.

The same Figure being constructed as in the last Proposition, each Ordinate will divide the Axis, and the Semi-axis added to the external Part, in the same Ratio.

Corol. 1.

Hence, and from cor. 3 to the last prop. it is,

cd2 = CD. DT = AD. DB = CD2

and cd.dR = Ad. dB = CA2

CA2,

cd2.

Corol. 2. Hence also CA CD2-cd2, and ca2=de2 — DE2.

Corol. 3. Farther, because CA2: ca2:: AD. DB or cď2: DE2.

therefore CA

likewise CA

ca: cd: DE.
ca :: CD: de.

THEOREM

THEOREM XIX (15).

If from any Point in the Curve there be drawn an Ordinate, and a Perpendicular to the Curve, or to the Tangent at that Point: Then the

Dist, on the Trans. between the Centre and Ordinate, CD :

For, by theor. 2, CA2: ca2:: AD. DB: De2,
But, by rt. angled A s, the rect. TD. DP = DE2;
and, by cor. 1 theor. 16,

therefore

or

CD. ᎠᎢ = ᎪᎠ . ᎠᏴ ;
CA2: ca2:: TD. DC: TD. DP,
CA2: ca2:: DC

: DP.

Q. E. D.

THEOREM XX (18).

If there be Two Tangents drawn, the One to the Extremity of the Transverse, and the other to the Extremity of any other Diameter, each meeting the other's Diameter produced; the two Tangential Triangles so formed, will be equal.

The two triangles CET, CAN have then the angle c common, and the sides about that angle reciprocally proportional; those triangles are therefore equal, viz. the A CETACAN. q.e.d.

Core

Corol. 1. Take each of the equal tri. CET, CAN, from the common space CAPE,

and there remains the external ▲ PAT = A PNE.

Corol. 2. Also take the equal triangles CET, CAN, from the common triangle

CED,

and there remains the ▲ TED = trapez. ANED.

THEOREM XXI (19).

The same being supposed as in the last Proposition; then any Lines кQ, GQ, drawn parallel to the two Tangents, shall also cut off equal Spaces.

That is,

the A KQG = trapez. ANHG. and ▲ Kqg trapez. ANhg.

19

T KAS D

For, draw the ordinate DE. Then

The three sim. triangles CAN, CDE, CGH,

are to each other as

CA, CD, CG2;

th. by div. the trap. ANED: trap. ANHG :: CD2- Ca2: CG2— ca2.

But, by theor. 1, DE2:

GQ2: CD2-CA2: CG2-CA2;

theref. by equ. trap. ANED: trap. ANHG ::

But, by sim. As, tri. TED: tri. KQG :: DE2 : GQ2;

Corol. 1. The three spaces ANHG, TEHG, KQG are all

equal.

Corol. 2.

From the equals ANHG, KQG,

take the equals ANhg, Kqg,

and there remains ghHG = gqQG.

Corol. 3. And from the equals ghнG, gqaG, take the common space gqLHG,

and there remains the ▲ LOH = A Lgh.

Corol. 4. Again, from the equals KQG, TEHG,

take the common space KLHG,

and there remains

TELKA LQH.

Corol

Corol. 5. the lines Ka, GH, moving with a parallel motion, Ka comes into the position IR, where CR is the conjugate to CA; then

And when, by

the triangle KQG becomes the triangle IRC,
and the space ANHG becomes the trianglė ANC;
and therefore the AIRCA ANCA TEC.

Corol. 6. Also when the lines KQ and нQ, by moving with a parallel motion, come into the position ce, Me,

the triangle LQH becomes the triangle cem,

and the space TELK becomes the triangle TEC;
and theref. the A ceм = A TECA ANC = A IRC.

THEOREM XXII (20):

Any Diameter bisects all its Double Ordinates, or the Lines drawn Parallel to the Tangent at its Vertex, or to its Conjugate Diameter.

That is, if aq be parallel to the tangent TE, or to ce, then shall LQ=Lq.

For, draw QH, qh perpendicular to the transverse.

Then by cor. 3 theor. 21, the ▲ LOH =

A Lgh;

but these triangles are also equiangular;

conseq. their like sides are equal, or Lo = £q.

Corol. 1. 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.

Corol. 2. In like manner, if the ordinate be produced to the conjugate hyperbolas at q, q', it may be proved that VOL. III.

C

LQ'