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2cK : 21K :: AD. 2cK + AD. AG : 1D, AD : 2CK :AD.21K :: AD. 2CK + AD. AG : ID?; theref. by div. CK: IK :: AD. AG : ID: AD. 2IK, and, by comp. CK :IC :: AD.AG:ID? AD. ID IA,

CK:CI :: AD. AG: Al?. But, when the line is, by revolving about the point 1, comes into the position of the tangent il, then the points E and meet in the point L, and the points D, K, G, coincide with the point m; and then the last proportion becomes CM: 0 :: AM’ : AI?.

Q. E. D.



If a Tangent and Ordinate be drawn from any Point in the

Curve, meeting the Transverse Axis; the Semi-transverse will be a Mean Proportional between the Distances of the said Two Intersections from the Centre.

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For, by theor. 6, cd: CT :: AD? : AT°. that is,

CD: CT ::(CA - CD) : (CT - CA),

CD: CT ::CD? + CA’ : CA+ ćT?, and

CD : DT :: cd’ + CA?: CT? – CD',
CD : DT :: CD? + CA:(CT + CD) DT,

cp:CD ct :: CD2 + CA?: CD . DT + CT.DT, hence

CD2: CA? :: CD DT : CT . DT, and

CD2 : CA? :: CD :CT. therefore (th. 78, Geom.) CD :CA :: CA : CT. Q. E. D



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Corol. Since ct is always a third proportional to CD, CA; if the points D, 4, remain constant, then will the point be constant also; and therefore all the tangents will meet in this point T, which are drawn from the point E, of every ellipse described on the same axis AB, where they are cut by the common crdinate DEE drawn from the point D.



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.

and by comp.

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

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

QE, D. Corol. Hence TA, TD, TC, TB and TG, TE, TH, TIS

are also proportionals. 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.

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For, draw the ordinate DE, and fe parallel to FE.

FE, By cor. 1, theor. 5, CA :CD :: GF : CA and by theor. 7, CA : CD :: CT : CA; therefore

CT : CF :: CA : CA FE; and by add. and sub. TF : tf :: FE : 2CA But by sim. tri.

TF: Tf :: fe : fe; therefore

fe = fe, and conseq; But, because therefore the

fe or fe by th: 5

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Le =
FE is parallel to fe, the Le =

FET = L 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 Ęg, 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 :ci, 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, CAP: ca:: (AD. DB or) cd : DE, therefore

CA : Ca :: cd : de;

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* Corol. Because cd? – AD . DB = CA2- 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 : de. 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. Thien, by cor. to theor. 10, CA? = CD' + cd", and

ca = DE + dez; therefore the sum CAP + ca’ = cdo + DE + cd? + de'. But, by right-angled As, CE2 = CD + DE, and

ce? = cd? + de> ; therefore the sum CEP + ce? = cd? + DE? + cd? + de?. consequently CA? + ca’ = CE? + ceR; or, by doubling, AB2 + ab2 = EGʻ + 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 Ayo be a plane passing


B through the vertex and axis of

D the opposite cones; AGIH another section of them perpendi

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

L be, as FG+: HI^:: AF . FB:AH.HB. For, through the ordinates

NI FG, HI, draw the circular sections KGL, MIN, parallel to the base of the cone, having KL, MN, for their diameters, to which FG, X1, 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 . HB :: KF , FL : MH. HN. But, by the circle, KF . FL = FGʻ, and MH . HN = HI”; Therefore the rect. AF . FB : AH . HB :: FG: H1?.


Q. E. D.

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