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If there be any Tangent meeting Four Perpendiculars to the

Axis drawn from these four Points, namely, the Centre, the two Extremities of the Axis, and the Point of Contact; those Four Perpendiculars will be Proportionals.



That is,
AG : DE :: CH : BI.


For, by theor. 7, TC : AC :: AC : DC, theref. by div. TA : AD :: TC : AC or CB, and by comp.

TA : TD :: TC : TB, and by sim. tri. AG : DE :: CH : BI.

& E. D. Corol. Hence TA, TD, TC, TB

are also proportionals. and TG, TE, TH, TI For these are as AG, DE, CH, BI, by similar triangles.


If there be any Tangent, and two Lines drawn from the

Foci to the point of Contact; these two Lines will make equal Angles with the Tangent.

That is,
the L FET = L fee.




For, draw the ordinate DE, and fe parallel to FE. By cor. 1, theor. 5, CÁ : CD :: GF : CA and by theor. 7, CA : CD :: CT : CA; therefore

CT : CF :: CA : CA and by add. and sub. TF : Tf :: Fe : 2CA FE or fe by th. 5. But by sim. tri. TF: Tf :: Fe : fe ; therefore

fe = fe, and conseq. Le = fee. But, because FE is parallel to fe, the Le= LFET; therefore the L FET = 2 fee.

Q. E. D.


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


All the Parallelograms circumscribed about an Ellipse are

equal to one another, and each equal to the Rectangle of the two Axes.

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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 CK perpendicular to PQ; and let the axis CA produced meet the sides of the parallelogram, produced if necessary, in T and t. Then, by theor 7, CT : CA :: CA : CD, and

ct :CA::CA : cd; theref. by equality, CT : ct :: cd : CD; but, by sim. triangles, ct:ct :: TD: cl, theref. by equality,

TD: cd :: cd : CD, and the rectangle TD. DC is = the square cd?. Again, by theor. 7, CD; CA :: CA : CT, or, by division, CD: CA :: DA : AT, and by composition, CD : DB :: AD : DT; conseq.

the rectangle CD DT = cd = AD. DB*. But, by theor. 1, CA? : ca:: (AD . DB or) cd : DE', therefore

CĄ : Ca :: cd : DE;


* Corol. Because cd' =ĄD . DB=CA – CD',

therefore ca' = CD? + cd'. In like manner, ca’ - DE+ de'.


In like manner,

CA : ca :: CD :de,

ca : de :: CA : CD. But, by theor. 7, CT: CA :: CA : CD; theref. by equality, CT:CA :: ca : sle. But, by sim. tri. CT : CK ::ce: de; theref. by equality, CK : CA :: ca : ce, and the rectangle ck . ce = ca . ca. But the rect. CK. ce = the parallelogram cepe, theref.the rect. CA . ca = the parallelogram CEPE, conseq. the rect. AB . ab = the parallelogram PQRS. Q. E. D.


The Sum of the Squares of every Pair of Conjugate Dia

meters, is equal to the same constant Quantity, namely, the Sum of the Squares of the two Axes.

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For, draw the ordinates ED, ed. Then, by cor. to theor. 10, CA? = cd? + cd', and

ca DE + de; therefore the sum CA? + ca? = cp + DE+ cd? + de'. But, by right-angled As, CE2 = CD? + DE, and

ce = cd + de; therefore the sum ÇE? + ce = CD + DE + cd? + de . consequently

CA? + ca? = ce? + ce?; or, by doubling, AB+ ab = EG+ eg.

Q. E. D. Note. All these theorems in the Ellipse, and their demonstrations, are the very same, word for word, as the corresponding number of those in the Hyperbola, next following, having only sometimes the word sum changed for the word difference.




The Squares of the Ordinates of the Axis are to each other

as the Rectangles of their Abscisses. LET AVB be a plane passing


B through the vertex and axis of

D the opposite cones; AGIH another section of them perpendi

P cular to the plane of the former; AB the axis of the hyperbolic sections; and FG, HI, ordinates perpendicular to it. Then it will


I be, as FGʻ: HI:: AF. FB:AH .HB. For, through the ordinates


11 FG,H1, 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 EFK, 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 . AB :: KF . FL : NH. HN. But, by the circle, KF. FL = FG%, and mh . HN = HI”; Therefore the rect. AF . FB : AH . HB :: FG? : Hi'.

Q. E. D.

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For, by theor. I, AC.CB: AD. DB :: ca® : ;
But, if c be the centre, then AC CB = AC?, and ca is the

semi-conj. Therefore

AC? : AD . DB : : acé : DE2; or, by permutation, Ac : ac :: AD . DB : De’; or, by doubling, AB? : ab? :; AD . DB : DE?.

Q. E. De ab? Corol. Or, by div. AB : :: AD, DB or cda CA?: DE, that is, AB:p :: AD . DB or CD

CA? : DE2;

ab? where p is the parameter

by the definition of it. AB


That is, As the transverse,

Is to its parameter,
So is the rectangle of the abscisses,
To the square of their ordinate.



As the Square of the Conjugate Axis
To the Square of the Transverse Axis ::
The Sum of the Squares of the Semi-conjugate, and
Distance of the Centre from any Ordinate of the Axis :
The Square of their Ordinate.

That is,
ca: CA :: ca* + cd: de.


For, draw the ordinate ed to the transverse AB. Then, by theor. 1. ca? : CA? :: DE : AD . DB or CD? - CA,

ca3 : CA? :: cdo : de? CA”. But

ca’ : CAP :: ca: CA?. theref. by compos. ca’ : CA:: ca+ cd? : de?. In like manner,

CA? : cao :: CA + CD2 : De?. Q. E, D. Corol. By the last theor. CA?: ca*::cp- CAP: DE,

and by this theor. CA’: ca? :: CD? + CAP: De’,

therefore De: De2 :: CD -CA: CD + Ca. In like manner,

de:de? :: cd? - ca:cd+ ca.


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