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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 LT= <s, and TE parallel to hs.

Corol. 3. And hence the four tangents, at the four extremities of any two conjugate diameters, torm 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 us, it will be the conjugate to En by the defi. nition ; and the tangents to eh will be parallel to each other, and to the diameter En for the same reason.


If two Ordinates ed, ed be drawn from the Extremities E, C,

of two Conjugate Diameters, and Tangents be drawn to the same Extremities, and meeting the Axis produced in T

and Ri

Then shall co be a mean Proportional between cd, dr,

and cd a mean Proportional between CD, DT.

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For, by theor. 7, CD : CA :: CA : CT,
and by the same,

cd:CA :: CA : CR;
theref. by equality, CD:cd :: CR : cT.
But by sim. tri. DT: cd :: CT : CR;
theref by equality, co : cd:: cd: dt.
In like manner,

cd :D :: CD: dr. 2. E. D. Corol. 1. Hence cd : cd :: CR : CT.

Corol. 2. Hence also cd :cd:: de : DE.
And the rect. CD . DE = cd. de, or A CDE = A cde.
Corol. 3. Also cd =CD . DT, and co2 = cd. dr.

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



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.

[See the last fig.]

That is, DA: DT :: DC : DB,

and da: dr :: dc : dB. For, by theor, 7, CD :CA:: CA: CT, and by div.

CD : CA : : AD : AT,
and by comp CD: DB : : AD : DT,

DA : DT :: DC : DB.
In like manner, da : dR :: dc : dB.

Q. E. D.
Corol. I. Hence, and from cor. 3 to the last prop. it is,

cd? = CD. DT = AD . DB=CD? -CA”,

and cd . dR = Ad. dB - CAR cd.
Corol. 2. Hence also CA2 = CD? - cd>, and ca? = de?. DE2,
Corol. 3. Farther, because ca? : ca? :: AD . DB or cd? : DE.

therefore ca : ca ; : cd: DE.
likewise ca : ca : :cdide..


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:
Will be to the Dist. PA::
As Square of Trans. Axis :
To Square of the Conjugate.

That is,
CA? :ca? :: DC : DP.




For, by theor. 2, CA':ca? :: AD . DB : DE,
But, by rt. angled As, the rect. TD . DP = DE’ ;
and, by cor. theor. 16, CD. DT = AD. DB;
therefore CAP: cal :: TD.DC: TD . DP,
CAS: cal : : DC : DP

Q. E. D.



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.

That is, the triangle cer the triangle CAN


For, draw the ordinate de.

By sim triangles, CD:CA :: CE: CN;
but, by theor. 7, CD:CA :: CA:CT;

theref. by equal. Ca:ct:: CE : CN. The two triangles cer, can have then the angle c common and the sides about that angle reciprocally proportional ; those triangles are therefore equal, viz. the Acer = CAN. Q. E. D. Corol. 1. Take each of the equal tri. cer, CAN,

from the common space CAPE,

and there remains the external A PAT = A PNE. Corol. 2. Also take the equal triangles cer, CAN, from the common triangle

and there remains the A TED = trapez. ANED.


The same being supposed as in the last Proposition; theri

ány Lines KQ Ga, drawn parallel to the two Tangents, shall also cut off equal Spaces.

That is, the AKQG =

trapez. ANHG. and A KIS s

trapez. Anhg.

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For, draw the ordinate de. Then
The three sim. triangles SAN, ÇDE, CGI,
Vor, I.


are to each other as CA?, CD, CG? ; th, by div. the trap. ANED : trap ANAG:: 20'-CA? :cG-CA?. But, by theor 1, DE? : GQ? :: CDP-CA?:0G%-CA; theref. by equ. trap. ANED : trap. ANHG::


: G4?. But, by sim. As, tri. TED: tri.

KQG :: DE2

: GQ; theref. by equal. ANED: TED :: ANHG : KQG. But, by cor. 2 theor. 20, the trap. ANED = ATED ;

and therefore the trap. ANHG = A KQG.

In like manner the trap. Anhg A kig. 4. E. D. Corol. 1. The three spaces ANHG, TEHG, KQG are all equal

$ Corol. 2. From the equals ANHG, KQG,

take the equals Anhg, 699,

and there remains ghug=5996. Corol. 3. And from the equals ghug, 59QG,

take the common space gqLHG,

and there remains the A LQH = · Lgh. Corol. 4. Again, from the equals KQG, TEHG,

take the common space KLHG,
and there remains TELK = A LQI.

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the triangle KQG becomes the triangle Inc,
and the space ANHG becomes the triangle anc ;

and therefore the A IRC = - A ANC = A TEC. Corol. 6. Also when the lines KQ and hq, by moving with a parallel motion, come into the position ce. Me,

the triangle Lou becomes the triangle cem,
and the space TELK becomes the triangle TEC ;
and theref. the A COM = A TEC – A ANC = A IRC.


Any Diameter bisects all its Double Ordinates, or the Lines

drawn Parallel to the Tangent at its Vertex, or to its Conjugate Diameter.


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For, draw qh, gh perpendicular to the transverse.
Then by cor. 3 theor 21, the A LQH = A Lgh ;
but these triangles are also equiangular ;
conseq. their like sides are equal, or LQ = Lq.

Corol. 1. Any diameter divides the hyperbola into two equ l 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 a'i oʻ. it may be proved that La'=le. Or if the tangent te be produced, then ev = Ew. Also the diameter GCEn bisects all lines drawn parallel to TE or ay, and limited either by one hyperbola, or by its two conjugate hyperbolas.


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, cel : ce2 : : EL . LG or cl? CE2 : LQ?.
For, draw the tangent
TE, and produce the ordi-
nale QL 10 the transverse
at K

Also draw QH, ÈM , perpendicular to the transverse, and meeting EG in H and m. Then, similar

G triangles being as the squares of their like sides, it is,

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