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Corol. 1. The two semi-axes, and the focal distance from the centre, are the sides of a right-angled triangle cha; and the distance fa from the focus to the extremity of the conjugate axis, is = Ac the semi-transverse.

Corol. 2. The conjugate semi-axis ca is a mean proportional between AF, FB, or between af, fb, the distances of either focus from the two vertices.

For ca? = CAP CE? = CA + CF. CA – CF = AF.FB.


The Sum of two Lines drawn from the two Foci to meet

at any Point in the Curve, is equal to the Transverse Axis.

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For, draw AG parallel and equal to ca the semi-conjugate; and join cg meeting the ordinate de in h; also take ci a 4th proportional to CA, CF, CD. Then, by theor. 2, CA2:AG? :: CA’ – CD’: DER ; and, by sim. tri. CA?: AG? :: CA? cp2: AG consequently DE? = AGP DH? = ca? - DH. Also FD = CF 18. CD, and FDP =CF2 2CF. CD + cd?; And, by right-angled triangles, FEʻ = FD? + DE'; therefore Fe = CF? + ca? 2CF . CD + CD2 – DH'. But by theor. 4, CF + ca = CA, and by supposition, 2CF . CD = 2CA . CI; theref. FE? = CA? 2CA . CI + CD? Again, by supp. CA? : CD’:: CF* or Ca’ – AG? : ci”; and, by sim. tri. CA?: CD2 :: CA? AG? : cd? – DHR; therefore CI? cp consequently FE? = CAP 2CA . CI + ci. And the root or side of this square is Fe = CA - CI = AI. In th same manner it is found that fe = c + c = BI. Conseq. by addit. FE + fp = AI + BI = AB. Q. E. D.


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Corol. 1. Hence cı or ea – FE is a 4th proportional to CA,

ci CF, CD.

Corol. 2. And fe FE = 2cı; that is, the difference between two lines drawn from the foci, to any point in the curve, is double the 4th proportional to CA, CF, CD.

Corel. 3. Hence is derived the common method of describing this curve mechanically by points, or with a thread, thus:

In the transverse take the foci F,f, and any point 1. Then with the radii AI, B1, and centres F, f, describe arcs

B intersecting in E, which will be a 11

f point in the curve. In like manner, assuming other points i, as many other points will be found in the

Then with a steady hand, the curve line may be drawn through all the points of intersection E.

Or, take a thread of the length AB of the transverse axis, and fix its two ends in the foci F, f, by two pins. Then carry a pen or pencil round by the thread, keeping it always stretched, and its point will trace out the curve line.



If from any Point 1 in the Axis produced, a Line il be

drawn touching the Curve in one Point L; and the Ordinate lm be drawn; and if c be the Centre or Middle of AB: Then shall cm be to ci as the Square of Am to the Square of ai.


That is,



CM : CI :: AM? : AI?.


For, from the point i draw any other line IEH to cut the curve in two points E and h; from which let fall the

perpendiculars ED and HG; and bisect DG in K.

Then, by theo. 1, AD. DB : AG · GB: D E : and by sim. triangles,


IG2 : : DE : GH? ; theref. by equality, AD . DB : AG . GE :: Id? : IG?. But DB = CB+CD= AC + CD = AG + DC CG= 20K + AG, and GB = CB - CG AC CG I AD + DC -CG = 2cK + AD; theref. AD. 2CK + AD. AG:AG.2CK + AD. AG :: ID’:162, and, by div. DG . 2CK : 1G2 ID2 or DG . 21K :: AD . 2CK +






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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: A1?. But, when the line ih, by revolving about the point 1, comes into the position of the tangent it, then the points E and H meet in the point I, and the points D, K, G, coincide with the point m; and then the last proportion becomes CM:c :: AM’ : Al.

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 :: cdo + ca’ : CA+ CT, and

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

cd’: CD.CT :: CD + CA?: CD . DT + CT. IT, hence

CD2: CAP::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 t 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 ordinate 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.

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.

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

are also proportionals. and TG, TE, TH, TIS For these are as AG, DE, CH, B1, 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 FET = L fee.

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

FE; 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 = Lfee. But, because

FE is parallel to fe, the Le = L FET ; therefore the FET = 2fee.

l. 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 : cd, theref. by equality, TD: cd :: cd : CD, and the rectangle

TD. DC is - the


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

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?.


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