PROPOSITION I.

The Squares of the Ordinates of the Axis are to each other as the Rectangles of their Abſciſſes.

Let AVB be a plane paſſing through the axis of the cone; AGIH another ſection of the cone perpendicular to the plane of the former; AB the axis of this elliptic ſection; and FG, HI ordinates perpendicular to it. Then I ſay that FG² ∶ HI² ∷ AF·FB ∶ AH·HB.

For, through the ordinates FG, HI 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 ellipſe.

Now, by the ſimilar triangles AFL, AHN, and BFL, BHM, we have AF ∶ AH ∷ FL ∶ HN, and FB ∶ HB ∷ KF ∶ MH; hence, taking the rectangles of the correſponding terms, we have the rect. AF·FB ∶ AH·HB ∷ KF·FL ∶ MH·HN. But, by the nature of the circle, KF·FL = FG², and MH·HN = HI²; Therefore the rect. AF·FB ∶ AH·HB ∷ FG² ∶ HI². Q.E.D.

[9]COROL. 1. All the parallel ſections are ſimilar figures, or have their two axes in the ſame proportion; that is, AB ∶ ab ∷ DE ∶ de.

For, by ſim. triang. AB ∶ ab ∷ AQ ∶ aq, and AB ∶ ab ∷ RB ∶ rb; Theref. by comp. AB² ∶ ab² ∷ AQ·RB ∶ aq·rb. But AQ·RB = DE², and aq·rb = de²; Therefore AB² ∶ ab² ∷ DE² ∶ de², or AB ∶ ab ∷ DE ∶ de.

COROL. 2. Hence alſo, as the property is the ſame for the ordinates on both ſides of the diameter, it follows, that

1ſt. At equal diſtances from the center, or from the vertices, the ordinates on both ſides are equal, or that the double ordinates are biſected by the axis; and that the whole figure, made up of all the double ordinates, is alſo biſected by the axis.

2d. The two foci are equally diſtant from the center, or from either vertex.

COROL. 3. When the angle, which the plane of the ſection makes with the baſe of the cone, increaſes till it be⯑come equal to the angle made by the ſide of the cone and the baſe, or till the ſection be parallel to the oppoſite ſide of the baſe; then the axis becomes infinitely long, and the ellipſe degenerates into a parabola; and becauſe then the infinites FB and HB are in a ratio of equality, the general property, namely AF·FB ∶ AH·HB ∷ FG² ∶ HI², becomes AF ∶ AH ∷ FG² ∶ HI², or, in the parabola, the abſciſſes are to each other, as the ſquares of their ordinates.

PROPOSITION II.

As the Square of the Tranſverſe Axis: Is to the Square of the Conjugate ∷ So is the Rectangle of the Abſciſſes: To the Square of their Ordinate.

That is, AB² ∶ ab² or AC² ∶ ac² ∷ AD·DB ∶ DE².

For, by prop. 1. AC·CB ∶ AD·DB ∷ ca² ∶ DE²; But, if C be the center, then AC·CB = AC², and ca is the ſemi-conj. Therefore AC² ∶ AD·DB ∷ ac² ∶ DE²; or, by permutation, AC² ∶ ac² ∷ AD·DB ∶ DE²; or, by doubling, AB² ∶ ab² ∷ AD·DB ∶ DE². Q.E.D.

COROL. 1. Or, becauſe the rectangle AD·DB = CA² − CD², the ſame property is CA² ∶ ca² ∷ CA² − CD² ∶ DE², or AB² ∶ ab² ∷ CA² − CD² ∶ DE².

COROL. 2. Or, by div. AB ∶ ab²/AB ∷ CA² − CD² ∶ DE², that is, AB ∶ p ∷ AD·DB or CA² − CD² ∶ DE²; where p is the parameter ab²/AB by the definition of it.

[11]That is, As the tranſverſe, Is to its parameter, So is the rectangle of the abſciſſes, To the ſquare of their ordinate.

COROL. 3. When the axis AB is infinitely long, the curve becomes a parabola, and the infinites AB, DB are then in a ratio of equality; and then the laſt property, namely AB ∶ p ∷ AD·DB ∶ DE²; or AB·DE ∶ AD·DB ∷ p ∶ DE, becomes DE ∶ AD ∷ p ∶ DE, or AD ∶ DE ∷ DE ∶ p.

That is, in the parabola, the parameter is a third pro⯑portional to any abſciſs and its ordinate.

PROPOSITION III.

• As the Square of the Conjugate Axis∶
• Is to the Square of the Tranſverſe Axis∷
• So is the Rectangle of the Abſciſſes of the Conjugate, or the Difference of the Squares of the Semi-conjugate and Diſtance of the Center from any Ordinate of that Axis∶
• To the Square of that Ordinate.

That is, Ca² ∶ CB² ∷ ad·db or Ca² − cd² ∶ dE².

For draw the ordinate ED to the tranſverſe AB.

Then, by cor. 1. prop. 2. CA² ∶ Ca² ∷ CA²-CD² ∶ DE².

But CD² = dE², and DE² = cd², therefore CA² ∶ Ca² ∷ CA² − dE² ∶ Cd², or by alternation, CA² ∶ CA² − dE² ∷ Ca² ∶ Cd², and by diviſion, CA² ∶ dE² ∷ Ca² ∶ Ca² − Cd², and by alter. & inverſ. Ca² ∶ CA² ∷ Ca² − Cd² ∶ dE². Q.E.D.

[13]COROL. 1. If two circles be deſcribed on the two axes as diameters, the one inſcribed within the ellipſe, and the other circumſcribed about it; then an ordinate in the circle will be to the correſponding ordinate in the ellipſe, as the axis of this ordinate, is to the other axis.

That is, CA ∶ Ca ∷ DG ∶ DE, and Ca ∶ CA ∷ dg ∶ dE.

For, by the nature of the circle, AD·DB = DG²; theref. by the nature of the ellipſe, CA² ∶ Ca² ∷ AD·DB² or DG² ∶ DE², or CA ∶ Ca ∷ DG ∶ DE.

In like manner Ca ∶ CA ∷ dg ∶ dE. Moreover, by equality, DG ∶ DE or Cd ∷ dE or DC ∶ dg.

Therefore CgG is a continued ſtrait line.

COROL. 2. Hence alſo, as the ellipſe and circle are made up of the ſame number of correſponding ordinates, which are all in the ſame proportion of the two axes, it follows that the areas of the whole circle and ellipſe, as alſo of any like parts of them, are in the ſame proportion of the two axes, or as the ſquare of the diameter to the rectangle of the two axes; that is, the areas of the two circles, and of the ellipſe, are as the ſquare of each axis and the rectangle of the two, and therefore the ellipſe is a mean proportional between the two circles.

PROPOSITION IV.

The Square of the Diſtance of the Focus from the Center, is equal to the Difference of the Squares of the Semi-axis; Or, the Square of the Diſtance be⯑tween the Foci, is equal to the Difference of the Squares of the two Axes.

That is, CF² = CA² − Ca², or Ff² = AB² − ab².

For, to the focus F draw the ordinate FE; which, by the definition, will be the ſemi-parameter. Then by the nature of the curve CA² ∶ Ca² ∷ CA² − CF² ∶ FE²; and by the def. of the para. CA² ∶ Ca² ∷ Ca² ∶ FE²; therefore Ca² = CA² − CF²; and by addit. and ſubtr. CF² = CA² − Ca²; or, by doubling, Ff² = AB² − ab². Q.E.D.

COROL. 1. The two ſemi-axes, and the focal diſtance from the center, are the ſides of a right angled triangle CFa; and the diſtance Fa from the focus to the extremity of the conjugate axis, is = AC the ſemi-tranſverſe.

[15]For, as above, CA² − Ca² = CF², and by right angled triangles Fa² − Ca² = CF², therefore CA = Fa, and AB = Fa + fa.

COROL. 2. The conjugate ſemi-axis Ca is a mean proportional between AF, FB, or between Af, fB, the diſ⯑tances of either focus from the two vertices.

For Ca² = CA² − CF² = CA + CF·CA − CF = AF·FB.

COROL. 3. The ſame rectangle AF·FB of the focal diſtances from either vertex, is alſo equal to the rectangle AC·FE under the ſemi-tranſverſe and its ſemi-parameter; ſince this laſt is equal to the ſquare of the ſemi-conjugate by the definition of the parameter.

Or AF ∶ FE ∷ AC ∶ FB.

PROPOSITION V.

The Difference between the Semi-tranſverſe and a Line drawn from the Focus to any Point in the Curve, is equal to a Fourth Proportional to the Semi-tranſverſe, the Diſtance from the Center to the Focus, and the Diſtance from the Center to the Ordinate belonging to that Point of the Curve.

That is, AC − FE = CI, or FE = AI; and fE − AC = CI, or fE = BI.

Where CA ∶ CF ∷ CD ∶ CI the 4th proportional to CA, CF, CD.

For, by right angled triangles, FE² = FD² + DE².

Now draw AG parallel and equal to ca the ſemi-conjugate; and join CG meeting the ordinate DE in H.

Then, by prop. 2, CA² ∶ AG² ∷ CA² − CD² ∶ DE²; and, by ſim. tri. CA² ∶ AG² ∷ CA² − CD² ∶ AG² − DH²; conſequently DE² = AG² − DH² = Ca² − DH². Alſo FD = CF ∼ CD, and FD² = CF² − 2CF·CD + CD²; therefore FE² = CF² + Ca² − 2CF·CD + CD² − DH².

[17]But by prop. 4. Ca² + CF² = CA² and, by ſuppoſition, 2CF·CD = 2CA·CI; theref. FE² = CA² − 2CA·CI + CD² − DH².

But, by ſuppoſition CA² ∶ CD² ∷ CF² or CA² − AG² ∶ CI²; and, by ſim. tri. CA² ∶ CD² ∷ CA² − AG² ∶ CD² − DH²; therefore CI² = CD² − DH²; conſequently FE² = CA² − 2CA·CI + CI². And the root or ſide of this ſquare is FE = CA − CI = AI.

In the ſame manner is found fE = CA + CI = BI. Q.E.D.

COROL. 1. Hence CI or CA − FE is a 4th proportional to CA, CF, CD.

COROL. 2. And fE − FE = 2CI; that is, the dif⯑ference between two lines drawn from the foci, to any point in the curve, is double the 4th proportional to CA, CF, CD.

PROPOSITION VI.

The Sum of two Lines drawn from the Foci, to meet in any Point of the Curve is equal to the Tranſverſe Axis.

That is, FE + fE = AB.

For, by the laſt prop. FE = CA − CI = AI, and, by the ſame, fE = CA + CI = BI; theref. by addition, FE + fe = AB.

COROL. Hence is derived the common method of de⯑ſcribing the curve mechanically by points, or with a thread, thus.

[19]In the tranſverſe take the foci F, f, and any point I. Then with the radii AI, BI, and centers F, f, de⯑ſcribe arcs interſecting in E, which will be a point in the curve. In like manner, aſſuming other points I, as many other points will be found in the curve. Then with a ſteady hand, draw the curve line through all the points of interſection E.

Or, take a thread of the length of AB 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, keep⯑ing it always ſtretched, and its point will trace out the curve line.

PROPOSITION VII.

If from any Point I in the Axis produced, a Line IEH be drawn cutting the curve in Two Points; and from thoſe Two Points be drawn the Perpendicular Ordi⯑nates DE, GH; and if K be the Middle of DG, and C the Center or the Middle of AB: Then ſhall CK be to CI as the Rectangle of AD and AG to the Square of AI.

That is, CK ∶ CI ∷ AD·AG ∶ AI².

For, by prop. I. AD·DB ∶ AG·GB ∷ DE² ∶ GH², and by ſim. tri. ID² ∶ IG² ∷ DE² ∶ GH²; theref. by equality, AD·DB ∶ AG·GB ∷ ID² ∶ IG².

But DB = 2CK + AG, and GB = 2CK + AD, theref. AD·2CK + AD·AG ∶ AG·2CK + AD·AG ∷ ID² ∶ IG², and, by div. DG·2CK ∶ IG² − ID² or DG²·2IK ∶ AD·2CK+AD·AG ∷ ID². or 2CK ∶ 2IK ∷ AD·2CK + AD·AG ∶ ID² or AD·2CK ∶ AD·2IK ∷ AD·2CK + AD·AG ∶ ID²; [21]theref. by div. CK ∶ IK ∷ AD·AG ∶ ID² − AD·2IK, and, by comp. CK ∶ CI ∷ AD·AG ∶ ID² − AD·ID + IA, or CK ∶ CI ∷ AD·AG ∶ AI². Q.E.D.

COROL. When the line IH, by revolving about the point I, comes into the poſition of the tangent IL, and the ordinate LM being drawn, then the points E and H meet in the point L, and the points D, K, G, coincide with the point M; and then the property in the propoſition becomes CM ∶ CI ∷ AM² ∶ AI².

PROPOSITION VIII.

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 Center.

That is, CA is a mean proportion between CD and CT; or CD, CA, CT are continued proportionals.

• For, by cor. prop. 7, CD ∶ CT ∷ AD² ∶ AT²,
• that is, [...],
• or ∷ CD²+CA² ∶ CA²+CT²,
• and ∷ CD² ∶ CA²,
• and ∷ CA² ∶ CT²;
• therefore CD ∶ CA ∷ CA ∶ CT. Q.E.D.

COROL. 1. Since CT is always a third proportional to CD, CA; if the points D, A, 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 E, of [23]every ellipſe deſcribed on the ſame axis AB, where they are cut by the common ordinate DEE drawn from the point D.

COROL. 2. Hence a tangent is eaſily drawn to the curve, from any point, either in the curve or without it.

Firſt, if the given point E be in the curve. Draw the ordinate DE of the diameter AC; and in the diameter produced take CT a third proportional to CD, CA. Then join TE for the tangent required.

But if the point T be given any where without the curve. Join CT, in which take CD a third proportional to CT, CA; and draw the ordinate DE. Then join TE as before.

PROPOSITION IX.

If there be any Tangent meeting Four Perpendiculars to the Axis drawn from theſe four Points, namely the Center, 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 prop. 8, 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 and TG, TE, TH, TI are alſo proportionals.

For theſe are as AG, DE, CH, BI, by ſimilar triangles.

PROPOSITION X.

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 = ∠ fEe.

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

By cor. 1. prop. 5, CA ∶ CD ∷ CF ∶ CA − FE, and by prop. 8, CA ∶ CD ∷ CT ∶ CA; therefore CT ∶ CF ∷ CA ∶ CA − FE; and by add. and ſub. TF ∶ Tf ∷ FE ∶ 2CA − FE or fE by prop. 6. But by ſim. tri. TF ∶ Tf ∷ FE ∶ fe; therefore fE = fe, and conſeq. ∠ e = ∠ fEe. But, becauſe FE is parallel to fe, the ∠ e = ∠ FET; therefore the ∠ FET = ∠ fEe. Q.E.D.

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

PROPOSITION XI.

If a Line be drawn from either Focus, Perpendicular to a Tangent to any Point of the Curve; the Diſtance of their Interſection from the Center will be equal to the Semi-tranſverſe Axis.

That is, if FP, fp be perpendicular to the Tangent TPp, then ſhall CP and cp be each equal to CA or CB.

For, through the point of contact E draw FE and fE meeting FP produced in G. Then, the ∠ GEP = ∠ FEP, being each equal to the ∠ fEp, and the angles at P being right, and the ſide PE being common, the two triangles GEP, FEP are equal in all reſpects, and ſo GE = FE, and GP = FP. Therefore, ſince FP = ½ FG, and FC = ½ Ff, and the angle at F common, the ſide CP will be = ½fG or ½AB, that is CP = CA or CB.

And in the ſame manner cp = CA or CB. Q.E.D.

[27]COROL. 1. A circle deſcribed on the tranſverſe axis, as a diameter, will paſs through the points P, p; becauſe all the lines CA, CP, Cp, CB, being equal, will be radii of the circle.

COROL. 2. CP is parallel to fE, and Cp parallel to FE.

COROL. 3. If at the interſections of any tangent, with the circumſcribed circle, perpendiculars to the tangent be drawn, they will meet the tranſverſe axis in the two foci. That is, the perpendiculars PF, pf give the foci F, f.

PROPOSITION XII.

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

That is, if CD = CG, or the ordinate DE = GH; then ſhall CE = CH, and ECH will be a right line.

For, when CD = CG, then alſo is DE = GH by cor. 2. prop. 1. But the ∠D = ∠G, being both right angles; therefore the third ſide CE = CH, and the ∠ DCE = ∠ GCH, and conſequently ECH is a right line.

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

[29]COROL. 2. Hence alſo, if two tangents be drawn to the two ends E, H of any diameter EH they will be pa⯑rallel to each other, and will cut the axis at equal angles, and at equal diſtances from the center. For, the two CD, CA being equal to the two CG, CB, the third pro⯑portionals CT, CS will be equal alſo; then the W 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 = ∠ S, and TE parallel to HS.

COROL. 3. And hence the four tangents, at the four extremities of any two conjugate diameters, form a pa⯑rallelogram circumſcribing the ellipſe, 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 de⯑finition; and the tangents to eh will be parallel to each other, and to the diameter EH for the ſame reaſon.

PROPOSITION XIII.

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 ſhall CD be a mean Proportional between Cd, dR, and Cd a mean Proportional between CD, DT.

For, by prop. 8, 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 CD ∶ Cd ∷ Cd ∶ DT. In like manner Cd ∶ CD ∷ CD ∶ dR. Q.E.D.

[31]COROL. 1. Hence CD ∶ Cd ∷ CR ∶ CT.

COROL. 2. Hence alſo CD ∶ Cd ∷ de ∶ DE.

And the rect. CD · DE = cd · de, or ▵ CDE = ▵ Cde.

COROL. 3. Alſo Cd² = CD·DT, and CD² = Cd·dR.

Or Cd a mean proportional between CD, DT; and CD a mean proportional between Cd, dR.

PROPOSITION XIV.

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

That is, DA ∶ DT ∷ DC ∶ DB, and dA ∶ dR ∷ dC ∶ dB.

For, by prop. 8, CD ∶ CA ∷ CA ∶ CT, and by div. CD ∶ CA ∷ AD ∶ AT, and by comp. CD ∶ DB ∷ AD ∶ DT, or DA ∶ DT ∷ DC ∶ DB.

In like manner dA ∶ dR ∷ dC ∶ dB. Q.E.D.

COROL. 1. Hence, and from cor. 3 to the laſt prop. we have Cd² = CD·DT = AD·DB = CA² − CD², CD² = Cd·dR = Ad·dB = CA² − Cd².

[33]COROL. 2. Hence alſo CA² = CD² + Cd², and Ca² = DE² + de².

COROL. 3. Farther, becauſe CA² ∶ Ca² ∷ AD·DB or Cd² ∶ DE², therefore CA ∶ Ca ∷ Cd ∶ DE. likewiſe CA ∶ Ca ∷ CD ∶ de.

PROPOSITION XV.

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;

The Diſtance on the Tranſverſe, between the Center and Ordinate, CD∶

Will be to the Diſtance PD∷

As the Square of the Tranſverſe Axis∶

To the Square of the Conjugate.

That is, CA² ∶ Ca² ∷ DC ∶ DP.

For, by prop. 2, CA² ∶ Ca² ∷ AD·DB ∶ DE², But, by rt. angled ▵ s, the rect. TD·DP = DE²; and, by cor. 1 prop. 14, CD·DT = AD·DB; therefore CA² ∶ Ca² ∷ TD·DC ∶ TD·DP, or AC² ∶ Ca² ∷ DC ∶ DP. Q.E.D.

PROPOSITION XVI.

All the Parallelograms circumſcribed about an Ellipſe are equal to one another, and each equal to the Rectangle of the two Axes.

That is, the Parallelogram PQRS = the rectangle AB·ab.

Let EG, eg be two conjugate diameters parallel to the ſides of the parallelogram, and dividing it into four leſſer and equal parallelograms. Alſo draw the ordinates DE, de, and CK perpendicular to PQ.

[36]

Then, by prop. 8, CT ∶ CA ∷ CA ∶ CD; and, by cor. 3 prop. 14, Ca ∶ de ∷ CA ∶ CD; theref. by equality, CT ∶ CA ∷ Ca ∶ de.

And by ſim. tri. CT ∶ CK ∷ Ce ∶ de, theref. by equality CK ∶ CA ∷ Ca ∶ Ce, and the rect. CK·Ce = rect. CA·Ca.

But the rect. CK·Ce = the parallelogram CEQe; theref. the rect. CA·Ca = the parallelogram CEQe; and, by doubling, the rect. AB·ab = the paral. PQRS. Q.E.D.

COROL. 1. The rectangles of every pair of conjugate diameters, are to one another reciprocally as the ſines of their included angles. For the areas of their parallelo⯑grams, which are all equal among themſelves, are equal to the rectangles of the ſides, or conjugate diameters, multiplied by the ſines of their contained angles, the radius being 1. That is, the rectangle of every two conjugate diameters, drawn into the ſine of their contained angle, is equal to the ſame conſtant quantity. And therefore the rectangle of the diameters is inverſely as the ſine of their contained angle.

[37]COROL. 2. As it is proved in this propoſition that every circumſcribing parallelogram of an ellipſe is a conſtant quantity, ſo it may hence be ſhewn that each of the ſpaces EAgP, EaeQ, GBeR, Gbgs, between the curve and the tangents, is equal to a conſtant quantity. For, ſince every diameter biſects the ellipſe, the conjugate diameters EG, eg divide the ellipſe into four equal ſectors CEAg, CEae, CGBe, CGbg; but the ſame conjugate diameters divide alſo the whole tangential parallelogram PQRs into four equal parts, or ſmall parallelograms CEPg, CEQe, CGRe, CGSg; and therefore the differences be⯑tween theſe ſmall parallelograms and the ſectors, which are the ſaid external ſpaces, muſt be all equal among themſelves.

And as the ellipſe and circumſcribing parallelogram both remain conſtant, the difference of their fourth parts will alſo be a conſtant quantity. That is, the ſaid ex⯑ternal parts are each equal to the ſame conſtant quan⯑tity.

PROPOSITION XVII.

The Sum of the Squares of every Pair of Conjugate Diameters, is equal to the ſame conſtant Quantity, namely the Sum of the Squares of the two Axes.

That is, AB² + ab² = EG² + eg², where EG, eg are any conjugate diameters.

For draw the ordinates ED, ed. Then, by cor. 2 prop. 14, CA² = CD² + cd², and ca² = DE² + de²; therefore the ſum CA² + ca² = CD² + DE² + cd² + de². But, by rt. ∠ed ▵s, CE² = CD² + DE², and ce² = cd² + de²; therefore CE² + ce² = CD² + DE² + cd² + de². conſequently CA² + ca² = CE² + ce²; Or, by doubling, AB² + ab² = EG² + eg². Q.E.D.

PROPOSITION XVIII.

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

That is, the triangle CET = the triangle CAN.

For, draw the ordinate DE. Then By ſim. triangles CD ∶ CA ∷ CE ∶ CN; but, by prop. 8, CD ∶ CA ∷ CA ∶ CT; theref. by equal. CA ∶ CT ∷ CE ∶ CN.

The two triangles CET, CAN have then the angle C common, and the ſides about that angle reciprocally pro⯑portional; therefore thoſe triangles are equal.

Namely the ▵ CET = ▵ CAN. Q.E.D.

COROL. 1. From each of the equal tri. CET, CAN, take the common ſpace CAPE, and there remains the external ▵ PAT = ▵ PNE.

COROL. 2. Alſo from the equal triangles CET, CAN, take the common triangle CED, and there remains the ▵ TED = trapez. ANED.

PROPOSITION XIX.

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

That is, the Triangle KQG = Trapez. ANHG. and the Triangle Kqg = Trapez. ANhg.

For draw the ordinate DE. Then The three ſim. triangles CAN, CDE, CGH, are to each other as CA², CD², CG²; theref. by div. the trap. ANED ∶ trap. ANHG ∷ CA² − CD² ∶ CA² − CG². But, by prop. 1, DE²: GQ² ∷ CA² − CD² ∶ CA² − CG². theref. by equ. trap. ANED ∶ trap. ANHG ∷ DE² ∶ GQ². But, by ſim. ▵s, tri. TED ∶ tri. KQG ∷ DE² ∶ GQ²; theref. by equal. ANED ∶ TED ∷ ANHG ∶ KQG. But, by cor. 2 prop. 18, the trap. ANED = ▵ TED; and therefore the trap. ANHG = ▵ KQG. In like manner the trap. ANhg = ▵ Kqg. Q.E.D.

COROL. 1. The three ſpaces ANHG, TEHG, KQG are all equal.

[41]COROL. 2. From the equals ANHG, KQG, take the equals ANhg, Kqg, and there remains ghHG = gqQG.

COROL. 3. And from the equals ghHG, gqQG, take the common ſpace gqLHG, and there remains the ▵ LQH = ▵ Lqh.

COROL. 4. Again from the equals KQG, TEHG, take the common ſpace KLHG, and there remains TELK = ▵ LQH.

COROL. 5. And when, by the lines KQ, GH, moving with a parallel motion, KQ comes into the poſition IR, where CR is the conjugate to CA; then the triangle KQG becomes the triangle IRC, and the ſpace ANHG becomes the triangle ANC; and therefore the ▵ IRC = ▵ ANC = ▵ TEC.

COROL. 6. Alſo when the lines KQ and HQ, 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 ▵ CeM = ▵ TEC = ▵ ANC = ▵ IRC,

PROPOSITION XX.

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 Qq 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. prop. 18, 19, the ▵ LQH = ▵ Lqh; but theſe triangles are alſo equiangular; conſequently their like ſides are equal, and therefore LQ = Lq. Q.E.D.

[43]COROL. 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.

PROPOSITION XXI.

• 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 CE² ∶ Ce² ∷ EL·LG or CE² − CL² ∶ LQ².

For draw the tangent TE, and produce the ordinate QL to the tranſverſe at K. Alſo draw QH, eM perpen⯑dicular to the tranſverſe, and meeting EG in H and M.

Then ſimilar triangles being as the ſquares of their like ſides, we ſhall have, by ſim. triangles, ▵ CET ∶ ▵ CLK ∷ CE² ∶ CL²; or, by diviſion, ▵ CET ∶ trap. TELK ∷ CE² ∶ CE² − CL². Again, by ſim. tri. ▵ ceM ∶ ▵ LQH ∷ ce² ∶ LQ².

But, by cor. 5 prop. 19, the ▵ ceM = ▵ CET, and, by cor. 4 prop. 19, the ▵ LQH = trap. TELK; theref. by equality, CE² ∶ ce² ∷ CE² − CL² ∶ LQ², or CE² ∶ Ce² ∷ EL·LG ∶ LQ². Q.E.D.

[45]COROL. 1. The ſquares of the ordinates to any diameter, are to one another as the rectangles of their reſpective abſciſſes, or as the difference of the ſquares of the ſemi-diameter and of the diſtance between the ordi⯑nate and center. For they are all in the ſame ratio of CE² to ce².

COROL. 2. The above being the ſame property as that belonging to the two axes, all the other properties before laid down, for the axes, may be underſtood of any two conjugate diameters whatever, uſing only the oblique ordinates of theſe diameters inſtead of the perpendicular ordinates of the axes; namely, all the properties in pro⯑poſitions 7, 8, 9, 12, 13, 14, 18 and 19.

PROPOSITION XXII.

If any Two Lines, that any where interſect each other, meet the Curve each in Two Points; then

• The Rectangle of the Segments of the one∶
• Is to the Rectangle of the Segments of the other ∷
• As the Square of the Diam. Parallel to the former∶
• To the Square of the Diam. Parallel to the latter.

That is, if CR and Cr be Parallel to any two Lines PHQ, pHq; then ſhall CR² ∶ Cr² ∷ PH·HQ ∶ PH·Hq.

For draw the diameter CHE, and the tangent TE and its parallels PK, RI, MH, meeting the conjugate of the diameter CR in the points T, K, I, M. Then, becauſe ſimilar triangles are as the ſquares of their like ſides, we have, by ſim. triangles, CR² ∶ GP² ∷ ▵ CRI ∶ ▵ GPK, and CR² ∶ GH² ∷ ▵ CRI ∶ ▵ GHM; theref. by diviſion, CR² ∶ GP² − GH² ∷ CRI ∶ KPHM. Again, by ſim. tri. CE² ∶ CH² ∷ ▵ CTE ∶ ▵ CMH; and by diviſion, CE² ∶ CE² − CH² ∷ ▵ CTE ∶ TEHM. [47]But, by cor. 5 prop. 19, the ▵ CTE = ▵ CIR, and by cor. 1 prop. 19, TEHG = KPHG, or TEHM = KPHM; theref. by equ. CE² ∶ CE² − CH² ∷ CR² ∶ GP² − GH² or PH·HQ. In like manner CE² ∶ CE² − CH² ∷ Cr² ∶ pH·Hq. Theref. by equal CR² ∶ Cr² ∷ PH·HQ ∶ pH·Hq. Q.E.D.

COROL. 1. In like manner, if any other line p′H′q′, parallel to Cr or to pq, meet PHQ; ſince the rectangles PH′Q, p′H′q′ are alſo in the ſame ratio of CR² to Cr²; theref. the rect. PHQ ∶ pHq ∷ PH′Q ∶ p′H′q′.

Alſo, if another line P′hQ′ be drawn parallel to PQ or CR; becauſe the rectangles P′hQ′, p′hq′ are ſtill in the ſame ratio, therefore, in general, the rectangle PHQ ∶ pHq ∷ P′hQ′ ∶ p′hq′.

That is, the rectangles of the parts of two parallel lines, are to one another, as the rectangles of the parts of two other parallel lines, any where interſecting the former.

COROL. 2. And when any of the lines only touch the curve, inſtead of cutting it, the rectangles of ſuch become ſquares, and the general property ſtill attends them.

That is, CR² ∶ Cr² ∷ TE² ∶ Te², or CR ∶ Cr ∷ TE ∶ Te. and CR ∶ Cr ∷ tE ∶ te.

COROL. 3. And hence TE ∶ Te ∷ tE ∶ te.

PROPOSITION I.

The Squares of the Ordinates of the Axis are to each other as the Rectangles of their Abſciſſes.

Let AVB be a plane paſſing through the vertex and axis of the oppoſite cones; AGIH an⯑other ſection of them perpen⯑dicular to the plane of the for⯑mer; AB the axis of the hy⯑perbolic ſections; and FG, HI ordinates perpendicular to it. Then FG² ∶ HI² ∷ AF·FB ∶ AH·HB.

For, through the ordinates FG, HI 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 BFK, BHM, we have AF ∶ AH ∷ FL ∶ HN, and FB ∶ HB ∷ KF ∶ MH; hence, taking the rectangles of the correſponding terms, we have the rect. AF·FB ∶ AH·HB ∷ KF·FL ∶ MH·HN. But, by the nature of the circle, KF·FL = FG², and MH·HN = HI²; Therefore the rect. AF·FB ∶ AH·HB ∷ FG² ∶ HI². Q.E.D.

[49]COROL. 1. All the parallel ſections are ſimilar figures, or have their two axes in the ſame proportion; that is, AB ∶ ab ∷ DE ∶ de.

For, by ſim. triang. AB ∶ ab ∷ AQ ∶ aq, and AB ∶ ab ∷ RB ∶ rb; Theref. by comp. AB² ∶ ab² ∷ AQ·RB ∶ aq·rb. But AQ·RB = DE², and aq·rb = de²; Therefore AB² ∶ ab² ∷ DE² ∶ de², or AB ∶ ab ∷ DE ∶ de.

COROL. 2. Hence alſo, as the property is the ſame for the ordinates on both ſides of the diameter, it follows, that

1ſt. At equal diſtances from the center, or from the vertices, the ordinates on both ſides are equal, or that the double ordinates are biſected by the axis; and that the whole figure, made up of all the double ordinates, is alſo biſected by the axis.

2d. The two foci are equally diſtant from the center, or from either vertex.

COROL. 3. When the angle, which the plane of the ſection makes with the baſe of the cone, decreaſes till it be⯑come equal to the angle made by the ſide of the cone and the baſe, or till the ſection be parallel to the oppoſite ſide of the baſe; then the axis becomes infinitely long, and the hyperbola degenerates into a parabola; and becauſe then the infinites FB and HB are in a ratio of equality, the general property, namely AF·FB ∶ AH·HB ∷ FG² ∶ HI², becomes AF ∶ AH ∷ FG² ∶ HI², or, in the parabola, the abſciſſes are to each other, as the ſquares of their ordinates.

PROPOSITION II.

As the Square of the Tranſverſe Axis: Is to the Square of the Conjugate ∷ So is the Rectangle of the Abſciſſes: To the Square of their Ordinaté.

That is, AB² ∶ ab² or AC² ∶ ac² ∷ AD·DB ∶ DE².

For, by prop. 1. AC·CB ∶ AD·DB ∷ ca² ∶ DE²; But, if C be the center, then AC·CB = AC², and ca is the ſemi-conj. Therefore AC² ∶ AD·DB ∷ ac² ∶ DE²; or, by permutation, AC² ∶ ac² ∷ AD·DB ∶ DE²; or, by doubling, AB² ∶ ab² ∷ AD·DB ∶ DE². Q.E.D.

COROL. 1. Or, becauſe the rectangle AD·DB = CA² − CD², the ſame property is CA² ∶ Ca² ∷ CD² − CA² ∶ DE², or AB² ∶ ab² ∷ CD² − CA² ∶ DE².

COROL. 2. Or, by div. AB ∶ ab²/AB ∷ CD² − CA² ∶ DE², that is, AB ∶ p ∷ AD·DB or CD² − CA² ∶ DE²; where p is the parameter ab²/AB by the definition of it.

[51]That is, As the tranſverſe, Is to its parameter, So is the rectangle of the abſciſſes, To the ſquare of their ordinate.

COROL. 3. When the axis AB is infinitely long, the curve becomes a parabola, and the inſinites AB, DB are then in a ratio of equality; and then the laſt property, namely AB ∶ p ∷ AD·DB ∶ DE²; or AB·DE ∶ AD·DB ∷ p ∶ DE, becomes DE ∶ AD ∷ p ∶ DE, or AD ∶ DE ∷ DE ∶ p.

That is, in the parabola, the parameter is a third pro⯑portional to any abſciſs and its ordinate.

PROPOSITION III.

As the Square of the Conjugate Axis ∶

To the Square of the Tranſverſe Axis ∷

So is the Sum of the Squares of the Semi-conjugate, and Diſtance of the Center from any Ordinate of this Axis ∶

To the Square of that Ordinate.

That is, ca² ∶ CA² ∷ ca² + cd² ∶ dE².

For draw the ordinate ED to the tranſverſe AB.

Then, by prop. 1. CA² ∶ ca² ∷ CD² − CA² ∶ DE².

But CD² = dE², and DE² = cd², therefore CA² ∶ ca² ∷ dE² − CA² ∶ cd², or by alternation, CA² ∶ dE² − CA² ∷ ca² ∶ cd², and by compoſition, CA² ∶ dE² ∷ ca² ∶ ca² + cd², and by alter. & inverſ. ca² ∶ CA² ∷ ca² + cd² ∶ dE². In like manner CA² ∶ ca² ∷ CA² + CD² ∶ De². Q.E.D.

[53]COROL. By the laſt prop. CA² ∶ ca² ∷ CD² − CA² ∶ DE², and by this prop. CA² ∶ ca² ∷ CD² + CA² ∶ De², therefore DE² ∶ De² ∷ CD² − CA² ∶ CD² + CA². In like manner de² ∶ dE² ∷ cd² − ca² ∶ cd² + ca².

PROPOSITION IV.

The Square of the Diſtance of the Focus from the Center, is equal to the Sum of the Squares of the Semi-axes.

Or, the Square of the Diſtance between the Foci, is equal to the Sum of the Squares of the two Axes.

That is, CF² = CA² + ca², or Ff² = AB² + ab².

For, to the focus F draw the ordinate FE; which, by the definition, will be the ſemi-parameter. Then by the nature of the curve CA² ∶ ca² ∷ CF² − CA² ∶ FE²; and by the def. of the para. CA² ∶ ca² ∷ ca² ∶ FE²; therefore ca² = CF² − CA²; and by addit. CF² = CA² + ca²; or, by doubling, Ff² = AB² + ab². Q.E.D.

COROL. 1. The two ſemi-axes, and the focal diſtance from the center, are the ſides of a right angled triangle CAa; and the diſtance Aa is = CF the focal diſtance.

[55]For, as above, CA² + ca² = CF², and by right angled ▵s, ca² + ca² = Aa², therefore CF = Aa, and Ff = Aa + Ba.

COROL. 2. The conjugate ſemi-axis ca is a mean proportional between AF, FB, or between Af, fB, the diſ⯑tances of either focus from the two vertices.

For ca² = CF² − CA² = CF + CA·CF − CA = AF·FB.

COROL. 3. The ſame rectangle AF·FB of the focal diſtances from either vertex, is alſo equal to the rectangle AC·FE under the ſemi-tranſverſe and its ſemi-parameter; ſince this laſt is equal to the ſquare of the ſemi-conjugate by the definition of the parameter.

Or AF ∶ FE ∷ AC ∶ FB.

PROPOSITION V.

The Difference between the Semi-tranſverſe and a Line drawn from the Focus to any Point in the Curve, is equal to a Fourth Proportional to the Semi-tranſverſe, the Diſtance from the Center to the Focus, and the Diſtance from the Center to the Ordinate belonging to that Point of the Curve.

That is, AC − FE = CI, or FE = AI; and fE − AC = CI, or fE = BI.

Where CA ∶ CF ∷ CD ∶ CI the 4th proportional to CA, CF, CD.

For, by right angled triangles, FE² = FD² + DE².

Now draw AG parallel and equal to ca the ſemi-conjugate; and join CG meeting the ordinate DE produced in H.

Then, by prop. 2, CA² ∶ AG² ∷ CD² − CA² ∶ DE²; and, by ſim. Δs, CA² ∶ AG² ∷ CD² − CA² ∶ DH² − AG²; conſequently DE² = DH² − AG² = DH² − ca². Alſo FD = CF ∼ CD, and FD² = CF² − 2CF·CD + CD²; therefore FE² = CF² − ca² − 2CF·CD + CD² + DH².

[57]But by prop. 4. CF² + Ca² = CA² and, by ſuppoſition, 2CF·CD = 2CA·CI; theref. FE² = CA² − 2CA·CI + CD² + DH².

But, by ſuppoſition CA² ∶ CD² ∷ CF² or CA² + AG² ∶ CI²; and, by ſim. ▵s, CA² ∶ CD² ∷ CA + AG² ∶ CD² + DH²; therefore CI² = CD² + DH² = CH²; conſequently FE² = CA² − 2CA·CI + CI². And the root or ſide of this ſquare is FE = CI − CA = AI.

In the ſame manner is found fE = CI + CA = BI. Q.E.D.

COROL. 1. Hence CH = CI is a 4th proportional to CA, CF, CD.

COROL. 2. And fE + FE = 2CH or 2CI; or FE, CH, fE are in continued arithmetical progreſſion, the common difference being CA the ſemi-tranſverſe.

COROL. 3. From the demonſtration it appears that DE² = DH² − AG² = DH² − Ca². Conſequently DH is every where greater than DE; and ſo the aſymptote CGH never meets the curve, though they be ever ſo far pro⯑duced: but DH and DE approach nearer and nearer to a ratio of equality as they recede farther from the vertex, till at an infinite diſtance they become equal, and the aſymptote is a tangent to the curve at an infinite diſtance from the vertex.

PROPOSITION VI.

The Difference of two Lines drawn from the Foci, to meet in any Point of the Curve, is equal to the Tranſverſe Axis.

That is, fE − FE = AB.

For, by the laſt prop. FE = CI − CA = AI, and, by the ſame, fE = CI + CA = BI; theref. by ſubtraction, fE − FE = AB.

COROL. Hence is derived the common method of de⯑ſcribing the curve mechanically by points, thus.

[59]In the tranſverſe AB, produced, take the foci F, f, and any point 1. Then with the radii AI, BI, and centers F, f, deſcribe arcs interſecting in E, which will be a point in the curve. In like manner, aſſuming other points 1, as many other points will be found in the curve.

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

In the ſame manner are conſtructed the other two hyper⯑bolas, uſing the axis ab inſtead of AB.

PROPOSITION VII.

If from any Point 1 in the Axis produced, a Line IEH be drawn cutting the curve in Two Points; and from thoſe Two Points be drawn the Perpendicular Ordi⯑nates DE, GH; and if K be the Middle of DG, and C the Center or the Middle of AB: Then ſhall CK be to CI as the Rectangle of AD and AG to the Square of AI.

That is, CK ∶ CI ∷ AD·AG ∶ AI².

For, by prop. 1. AD·DB ∶ AG·GB ∷ DE² ∶ GH², and by ſim. ▵s, ID² ∶ IG² ∷ DE² ∶ GH²; theref. by equal. AD·DB ∶ AG·GB ∷ ID² ∶ IG².

But DB = 2CK − AG, and GB = 2CK − AD, theref. AD·2CK − AD·AG ∶ AG·2CK − AD·AG ∷ ID² ∶ IG², and, by div. DG·2CK ∶ IG² − ID² or DG·2IK ∷ AD·2CK − AD·AG ∶ ID². or 2CK ∶ 2IK ∷ AD·2CK − AD·AG ∶ ID², or AD·2CK ∶ AD·2IK ∷ AD·2CK − AD·AG ∶ ID²; [61]theref. by div. CK ∶ IK ∷ AD·AG ∶ AD·2IK − ID², and, by div. CK ∶ CI ∷ AD·AG ∶ ID² − AD·ID + IA, or CK ∶ CI ∷ AD·AG ∶ AI². Q.E.D.

COROL. When the line IH, by revolving about the point 1, comes into the poſition of the tangent IL, and the ordinate LM being drawn, then the points E and H meet in the point L, and the points D, K, G, coincide with the point M; and then the property in the propoſition becomes CM ∶ CI ∷ AM² ∶ AI².

PROPOSITION VIII.

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 Center.

That is, CA is a mean proportion between CD and CT; or CD, CA, CT are continued proportionals.

• For, by cor. prop. 7, CD ∶ CT ∷ AD² ∶ AT²,
• that is, [...],
• or ∷ CD² + CA² ∶ CA² + CT²,
• and ∷ CD² ∶ CA²,
• and ∷ CA² ∶ CT²;
• therefore CD ∶ CA ∷ CA ∶ CT. Q.E.D.

COROL. 1. Since CT is always a third proportional to CD, CA; if the points D, A, 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 E, of [63]every hyperbola deſcribed on the ſame axis AB, where they are cut by the common ordinate DEE drawn from the point D.

COROL. 2. Hence a tangent is eaſily drawn to the curve, from any point, either in the curve or without it.

Firſt, if the given point E be in the curve. Draw the ordinate DE of the diameter AC; and in the diameter produced take CT a third proportional to CD, CA. Then join TE for the tangent required.

But if the point T be given any where without the curve. Join CT, in which take CD a third proportional to CT, CA; and draw the ordinate DE. Then join TE as before.

PROPOSITION IX.

If there be any Tangent meeting Four Perpendiculars to the Axis drawn from theſe four Points, namely the Center, 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 prop. 8, 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 and TG, TE, TH, TI are alſo proportionals.

For theſe are as AG, DE, CH, BI, by ſimilar triangles.

PROPOSITION X.

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 = ∠ fEe.

For draw the ordinate DE, and fe parallel to FE. By cor. 1. prop. 5, CA ∶ CD ∷ CF ∶ CA + FE, and by prop. 8, CA ∶ CD ∷ CT ∶ CA; therefore CT ∶ CF ∷ CA ∶ CA + FE; and by add. and ſub. TF ∶ Tf ∷ FE ∶ 2CA + FE or fE byprop. 6. But by ſim. tri. TF ∶ Tf ∷ FE ∶ fe; therefore fE = fe, and conſeq. ∠ e = ∠ fEe. But, becauſe FE is parallel to fe, the ∠ e = ∠ FET; therefore the ∠ FET = ∠ fEe. Q.E.D.

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

PROPOSITION XI.

If a Line be drawn from either Focus, Perpendicular to a Tangent to any Point of the Curve; the Diſtance of their Interſection from the Center will be equal to the Semi-tranſverſe Axis.

That is, if FP, fp be perpendicular to the Tangent TPp, then ſhall CP and Cp be each equal to CA or CB.

For, through the point of contact E draw FE and fE meeting FP produced in G. Then, the ∠ GEP = ∠ FEP, being each equal to the ∠ fEp, and the angles at P being right, and the ſide PE being common, the two triangles GEP, FEP are equal in all reſpects, and ſo GE = FE, and GP = FP. Therefore, ſince FP = ½ FG, and FC = ½Ff, and the angle at F common, the ſide CP will be = ½fG or ½AB, that is CP = CA or CB.

And in the ſame manner Cp = CA or CB. Q.E.D.

[67]COROL. 1. A circle deſcribed on the tranſverſe axis, as a diameter, will paſs through the points P, p; becauſe all the lines CA, CP, Cp, CB, being equal, will be radii of the circle.

COROL. 2. CP is parallel to fE, and cp parallel to FE.

COROL. 3. If at the interſections of any tangent, with the circumſcribed circle, perpendiculars to the tangent be drawn, they will meet the tranſverſe axis in the two foci. That is, the perpendiculars PF, pf give the foci F, f.

PROPOSITION XII.

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

That is, if CD = CG, or the ordinate DE = GH; then ſhall CE = CH, and ECH will be a right line.

For, when CD = CG, then alſo is DE = GH by cor. 2. prop. 1. But the ∠D = ∠G, being both right angles; therefore the third ſide CE = [...] and the ∠ DCE = ∠ GCH, and conſequently ECH is a right line.

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

[69]COROL. 2. Hence alſo, if two tangents be drawn to the two ends E, H of any diameter EH; they will be pa⯑rallel to each other, and will cut the axis at equal angles, and at equal diſtances from the center. For, the two CD, CA being equal to the two CG, CB, the third pro portionals 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 = ∠ S, and TE parallel to HS.

COROL. 3. And hence the four tangents, at the four extremities of any two conjugate diameters, form a pa⯑rallelogram circumſcribing the ellipſe, 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 de⯑finition; and the tangents to eh will be parallel to each other, and to the diameter EH for the ſame reaſon.

PROPOSITION XIII.

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 ſhall CD be a mean Proportional between cd, dR, and cd a mean Proportional between CD, DT.

For, by prop. 8, 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 CD ∶ cd ∷ cd ∶ DT. In like manner cd ∶ CD ∷ CD ∶ dR.

Q.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 ▵ CDE = ▵ cde.

COROL. 3. Alſo cd² = CD·DT, and CD² = cd·dR.

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

PROPOSITION XIV.

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

That is, DA ∶ DT ∷ DC ∶ DB, and dA ∶ dR ∷ dC ∶ dB.

[See the laſt fig.]

For, by prop. 8, CD ∶ CA ∷ CA ∶ CT, and by div. CD ∶ CA ∷ AD ∶ AT, and by comp. CD ∶ DB ∷ AD ∶ DT, or DA ∶ DT ∷ DC ∶ DB.

In like manner dA ∶ dR ∷ dC ∶ dB.

Q.E.D.

COROL. 1. Hence, and from cor. 3 to the laſt prop. we have

cd² = CD·DT = AD·DB = CD² − CA², CD² = cd·dR = Ad·dB = CA² − cd².

COROL. 2. Hence alſo CA² = CD² − cd², and ca² = de² − DE².

COROL. 3. Farther, becauſe CA² ∶ ca² ∷ AD·DB or cd² ∶ DE², therefore CA ∶ ca ∷ cd ∶ DE. likewiſe CA ∶ ca ∷ CD ∶ de.

PROPOSITION XV.

• 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;
• The Diſtance on the Tranſverſe, between the Center and Ordinate, CD∶
• Will be to the Diſtance PD ∷
• As the Square of the Tranſverſe Axis ∶
• To the Square of the Conjugate.

That is, CA² ∶ Ca² ∷ DC ∶ DP.

For, by prop. 2, CA² ∶ Ca² ∷ AD·DB ∶ DE², But, by rt. angled ▵ s, the rect. TD·DP = DE²; and, by cor. 1 prop. 14, CD·DT = AD·DB; therefore CA² ∶ Ca² ∷ TD·DC ∶ TD·DP, or CA² ∶ Ca² ∷ DC ∶ DP. Q.E.D.

PROPOSITION XVI.

All the Parallelograms inſcribed between the four Con⯑jugate Hyperbolas are equal to one another, and each equal to the Rectangle of the two Axes.

That is, the Parallelogram PQRS = the rectangle AB·ab.

Let EG, eg be two conjugate diameters parallel to the ſides of the parallelogram, and dividing it into four leſſer and equal parallelograms. Alſo draw the ordinates DE, de, and CK perpendicular to PQ.

[74]

Then, by prop. 8, CT ∶ CA ∷ CA ∶ CD; and, by cor. 3 prop. 14, ca ∶ de ∷ CA ∶ CD; theref. by equality, CT ∶ CA ∷ Ca ∶ de. And by ſim. tri. CT ∶ CK ∷ Ce ∶ de, theref. by equality CK ∶ CA ∷ Ca ∶ Ce, and the rect. CK·Ce = rect. CA·Ca. But the rect. CK·Ce = the parallelogram CEQE; theref. the rect. CA·Ca = the parallelogram CEQE; and, by doubling, the rect. AB·ab = the paral. PQRS. Q.E.D.

COROL. 1. The rectangles of every pair of conjugate diameters, are to one another reciprocally as the ſines of their included angles. For the areas of their parallelo⯑grams, which are all equal among themſelves, are equal to the rectangles of the ſides, or conjugate diameters, multiplied by the ſines of their contained angles, the radius being 1. That is, the rectangle of every two conjugate diameters, drawn into the ſine of their contained angle, is equal to the ſame conſtant quantity. And therefore the rectangle of the diameters is inverſely as the ſine of their contained angle.

PROPOSITION XVII.

The Difference of the Squares of every Pair of Con⯑jugate Diameters, is equal to the ſame conſtant Quantity, namely the Difference of the Squares of the two Axes.

That is, AB² − ab² = EG² − eg², where EG, eg are any conjugate diameters.

For draw the ordinates ED, ed. Then, by cor. 8 prop. 9, CA² = CD² − cd², and ca² = de² − DE²; theref. the difference CA² − Ca² = CD² + DE² − Cd² − de². But, by rt. ∠ed ▵s, CE² = CD² + DE², and Ce² = Cd² + de²; therefore CE² − Ce² = CD² + DE² − Cd² − de². conſequently CA² − Ca² = CE² − Ce²; Or, by doubling, AB² − ab² = EG² − eg². Q.E.D.

PROPOSITION XVIII.

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

That is, the triangle CET = the triangle CAN.

For, draw the ordinate DE. Then By ſim. triangles CD ∶ CA ∷ CE ∶ CN; but, by prop. 8, CD ∶ CA ∷ CA ∶ CT; theref. by equal. CA ∶ CT ∷ CE ∶ CN.

The two triangles CET, CAN have then the angle C common, and the ſides about that angle reciprocally pro⯑portional; therefore thoſe triangles are equal.

Namely the ▵ CET = ▵ CAN. Q.E.D.

[77]COROL. 1. Take each of the equal tri. CET, CAN, from the common ſpace CAPE, and there remains the external ▵ PAT = ▵ PNE.

COROL. 2. Alſo take the equal triangles CET, CAN, from the common triangle CED, and there remains the ▵ TED = trapez. ANED.

PROPOSITION XIX.

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

That is, the Triangle KQG = Trapez. ANHG. and the Triangle Kqg = Trapez. ANhg.

For draw the ordinate DE. Then The three ſim. triangles CAN, CDE, CGH, are to each other as CA², CD², CG²; theref. by div. the trap. ANED: trap. ANHG ∷ CD² − CA² ∶ CG − CA². But, by prop. 1, DE²: GQ² ∷ CD² − CA² ∶ CG² − CA². theref. by equ. trap. ANED ∶ trap. ANHG ∷ DE² ∶ GQ². But, by ſim. ▵s, tri. TED ∶ tri. KQG ∷ DE² ∶ GQ²; theref. by equal. ANED: TED ∷ ANHG ∶ KQG. But, by cor. 2 prop. 18, the trap. ANED = ▵ TED; and therefore the trap. ANHG = ▵ KQG. In like manner the trap. ANhg = ▵ Kqg. Q.E.D.

COROL. 1. The three ſpaces ANHG, TEHG, KQG are all equal.

[79]COROL. 2. From the equals ANHG, KQG, take the equals ANhg, Kqg, and there remains ghHG = gqQG.

COROL. 3. And from the equals ghHG, gqQG, take the common ſpace gqLHG, and there remains the ▵LQH = ▵Lqh.

COROL. 4. Again from the equals KQG, TEHG, take the common ſpace KLHG, and there remains TELK = ▵LQH.

COROL. 5. And when, by the lines KQ, GH, moving with a parallel motion, KQ comes into the poſition IR, where CR is the conjugate to CA; then the triangle KQG becomes the triangle IRC, and the ſpace ANHG becomes the triangle ANC; and therefore the ▵IRC = ▵ANC = ▵TEC.

COROL. 6. Alſo when the lines KQ and HQ, 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 ▵CeM = ▵TEC = ▵ANC = ▵IRC,

PROPOSITION XX.

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 Qq 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 prop. 19, the ▵ LQH = ▵ Lqh; but theſe triangles are alſo equiangular; conſequently their like ſides are equal, and therefore LQ = Lq. Q.E.D.

[81]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 pro⯑duced to the conjugate hyperbolas at Q′, q′, it may be proved that LQ′ = Lq′. Or if the tangent TE be pro⯑duced, then EV = EW. And the diameter GCEH biſects all lines drawn parallel to TE or Qq, and limited either by one hyperbola, or by its two conjugate hyperbolas.

PROPOSITION XXI.

• 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 CE² ∶ Ce² ∷ EL·LG or CL² − CE² ∶ LQ².

For draw the tangent TE, and produce the ordinate QL to the tranſverſe at K. Alſo draw QH, eM perpen⯑dicular to the tranſverſe, and meeting EG in H and M.

Then ſimilar triangles being as the ſquares of their like ſides, we ſhall have, by ſim. triangles, ▵ CET ∶ ▵ CLK ∷ CE² ∶ CL²; or, by diviſion, ▵ CET ∶ trap. TELK ∷ CE² ∶ CL² − CE². Again, by ſim. tri. ▵ ceM ∶ ▵ LQH ∷ ce² ∶ LQ². But, by cor. 5 prop. 19, the ▵ ceM = ▵ CET, and, by cor. 4 prop. 19, the ▵ LQH = trap. TELK; theref. by equality, CE² ∶ Ce² ∷ CL² − CE² ∶ LQ², OF CE² ∶ ce² ∷ EL·LG ∶ LQ². Q.E.D.

[83]COROL. 1. The ſquares of the ordinates to any diameter, are to one another as the rectangles of their reſpective abſciſſes, or as the difference of the ſquares of the ſemi-diameter and of the diſtance between the ordi⯑nate and center. For they are all in the ſame ratio of CE² to Ce².

COROL. 2. The above being the ſame property as that belonging to the two axes, all the other properties before laid down, for the axes, may be underſtood of any two conjugate diameters whatever, uſing only the oblique ordinates of theſe diameters inſtead of the perpendicular ordinates of the axes; namely, all the properties in pro⯑poſitions 7, 8, 9, 12, 13, 14, 18 and 19.

COROL. 3. Likewiſe, when the ordinates are con⯑tinued to the conjugate hyperbolas at Q′, q′, the ſame pro⯑perties ſtill obtain, ſubſtituting only the ſum for the dif⯑ference of the ſquares of CE and CL, That is, CE² ∶ Ce² ∷ CL² + CE² ∶ CQ′². And ſo LQ² ∶ LQ′² ∷ CL² − CE² ∶ CL² + CE².

COROL. 4. When, by the motion of LQ parallel to itſelf, that line coincides with EV, the laſt corollary be⯑comes CE² ∶ Ce² ∷ 2CE² ∶ EV², or ce² ∶ EV² ∷ 1 ∶ 2, or ce ∶ EV ∷ 1 ∶ √2, or as the ſide of a ſquare to its diagonal.

That is, in all conjugate hyperbolas; and all their diameters, any diameter is to its parallel tangent, in the conſtant ratio of the ſide of a ſquare to its diagonal.

PROPOSITION XXII.

If any Two Lines, that any where interſect each other, meet the Curve each in Two Points; then

• The Rectangle of the Segments of the one∶
• Is to the Rectangle of the Segments of the other∷
• As the Square of the Diam. Parallel to the former∶
• To the Square of the Diam. Parallel to the latter.

That is, if CR and cr be Parallel to any two Lines PHQ, pHq; then ſhall CR² ∶ Cr² ∷ PH·HQ ∶ pH·Hq.

For draw the diameter CHE, and the tangent TE and its parallels PK, RI, MH, meeting the conjugate of the diameter CR in the points T, K, I, M. Then, becauſe ſimilar triangles are as the ſquares of their like ſides, we have, by ſim. triangles, CR² ∶ GP² ∷ ▵ CRI ∶ ▵ GPK, and CR² ∶ GH² ∷ ▵ CRI ∶ ▵ GHM; theref. by diviſion, CR² ∶ GP² − GH² ∷ CRI ∶ KPHM. Again, by ſim. tri. CE² ∶ CH² ∷ ▵ CTE ∶ ▵ CMH; and by diviſion, CE² ∶ CH² − CE² ∷ ▵ CTE ∶ TEHM. [85]But, by cor. 5 prop. 19, the ▵ CTE = ▵ CIR, and by cor. 1 prop. 19, TEHG = KPHG, or TEHM = KPHM; theref. by equ. CE² ∶ CH² − CE² ∷ CR² ∶ GP² − GH² or PH·HQ. In like manner CE² ∶ CH² − CE² ∷ Cr² ∶ pH·Hq. Theref. by equal CR² ∶ Cr² ∷ PH·HQ ∶ pH·Hq. Q.E.D.

COROL. 1. In like manner, if any other line p′H′q′, parallel to Cr or to pq, meet PHQ; ſince the rectangles PH′Q, p′H′q′ are alſo in the ſame ratio of CR² to cr²; theref. the rect. PHQ ∶ pHq ∷ PH′Q ∶ p′H′q′.

Alſo, if another line P′hQ′ be drawn parallel to PQ or CR; becauſe the rectangles P′hQ′, p′hq′ are ſtill in the ſame ratio, therefore, in general, the rectangle PHQ ∶ pHq ∷ p′hQ′ ∶ p′hq′.

That is, the rectangles of the parts of two parallel lines, are to one another, as the rectangles of the parts of two other parallel lines, any where interſecting the former.

COROL. 2. And when any of the lines only touch the curve, inſtead of cutting it, the rectangles of ſuch become ſquares, and the general property ſtill attends them.

That is, CR² ∶ Cr² ∷ TE² ∶ Te², or CR ∶ Cr ∷ TE ∶ Te. and CR ∶ Cr ∷ tE ∶ te.

COROL. 3. And hence TE ∶ Te ∷ tE ∶ te.

PROPOSITION XXIII.

If a Line be drawn through any Point of the Curves, Parallel to either of the Axes, and terminated at the Aſymptotes; the Rectangle of its Segments, meaſured from that Point, will be equal to the Square of the Semi-axis to which it is parallel.

That is, the Rect. HEK or HeK = Ca², and the Rect. hEK or hek = CA².

For draw AL parallel to ca, and aL to CA. Then by the parallels, CA² ∶ Ca² or AL² ∷ CD² ∶ DH²; and, by prop. 2, CA² ∶ Ca² ∷ CD² − CA² ∶ DE²; theref. by ſubtr. CA² ∶ Ca² ∷ CA² ∶ DH² − DE² or HEK. But the antecedents CA², CA² are equal, theref. the conſequents Ca², HEK muſt alſo be equal.

[87]In like manner we have, again by the parallels, CA² ∶ ca² or AL² ∷ CD² ∶ DH²; and, by prop. 3, CA² ∶ ca² ∷ CD² + CA² ∶ De²; theref. by ſubtr. CA² ∶ ca² ∷ CA² ∶ De² − DH² or HeK. But the antecedents CA², CA² are the ſame, theref. the conſequents ca², HeK muſt be equal. In like manner, by changing the axes, is hEk or hek = CA². Q.E.D.

COROL. 1. Becauſe the rect. HEK = the rect. HeK. therefore EH ∶ eH ∷ eK ∶ EK. And conſequently HE is always greater than He.

COROL. 2. The rectangle hEK = the rect. HEk. For, by ſim. tri. Eh ∶ EH ∷ Ek ∶ EK.

PROPOSITION XXIV.

All the Rectangles are equal which are made of the Segments of any Parallel Lines cut by the Curve, and limited by the Aſymptotes.

That is, the Rect. HEK = Rect. HeK. and the Rect. hEk = hek.

For each of the rectangles HEK or HeK is equal to the ſquare of the parallel ſemi-diameter CS; and each of the rectangles hEk or hek is equal to the ſquare of the parallel ſemi-diameter CI. And therefore the rectangles of the ſegments of all parallel lines are equal to one another. Q.E.D.

[89]COROL. 1. The rectangle HEK being conſtantly the ſame, whether the point E is taken on the one ſide or the other of the point of contact 1 of the tangent parallel to HK, it follows that the parts HE, KE, of any line HK, are equal.

And becauſe the rectangle HeK is conſtant, whether the point e is taken in the one or the other of the oppoſite hy⯑perbolas, it follows that the parts He, Ke are alſo equal.

COROL. 2. And when HK comes into the poſition of the tangent DIL, the laſt corollary becomes IL = ID, and IM = IN, and LM = DN.

Hence alſo the diameter CIR biſects all the parallels to DL which are terminated by the aſymptote, namely RH = RK.

COROL. 3. From the propoſition, and the laſt corollary, it follows that the conſtant rectangle HEK or EHE is = IL². And the equal conſtant rect. HeK or eHe = MLN or IM² − IL².

COROL. 4. And hence IL = the parallel ſemi-diameter CS.

For the rect. EHE = IL², and the equal rect. eHe = IM² − IL², theref. IL² = IM² − IL², or IM² = 2IL²; but, by cor. 4 prop. 21, IM² = 2CS², and therefore IL = CS.

And ſo the aſymptotes paſſes through the oppoſite angles of all the inſcribed parallelograms.

PROPOSITION XXV.

The Rectangle of any two Lines drawn from any Point in the Curve, Parallel to two given Lines, and Limited by the Aſymptotes, is a Conſtant Quantity.

That is, if AP, EG, DI be Parallels, as alſo AQ, EK, DM Parallels, then ſhall the rect. PAQ = rect. GEK = rect IDM.

For, produce KE, MD to the other aſymptote at H, L. Then, by the parallels, HE ∶ GE ∷ LD ∶ ID; but, EK ∶ EK ∷ DM ∶ DM; theref. the rectangle HEK ∶ GEK ∷ LDM ∶ IDM. But, by the laſt prop. the rect. HEK = LDM; and therefore the rect. GEK = IDM = PAQ. Q.E.D.

PROPOSITION XXVI.

All the Parallelograms are equal which are formed be⯑tween the Aſymptotes and Curve, by Lines Parallel to the Aſymptotes.

That is, the paral. CGEK = CPAQ.

For, by prop. 25, the rect. GEK or CGE = rect. PAQ or CPA. But the parallelograms CGEK, CPAQ being equiangular, are as the rectangles CGE, CPA. And therefore the parallelogram GK = PQ. That is, all the inſcribed parallelograms are equal to one another. Q.E.D.

[92]COROL. 1. Becauſe the rectangle GEK or CGE is conſtant, therefore GE is reciprocally as CG, or CG ∶ CP ∷ PA ∶ GE. And hence the aſymptote conti⯑nually approaches towards the curve, but never meets it ∶ for GE decreaſes continually as CG increaſes; and [...]t is always of ſome magnitude, except when CG is ſuppoſed to be infinitely great, for then GE is infinitely ſmall, or no⯑thing. So that the aſymptote CG may be conſidered as a tangent to the curve at a point infinitely diſtant from C.

COROL. 2. If the abſciſſes CD, CE, CG, &c, taken on the one aſymptote, be in geometrical progreſſion in⯑creaſing; then ſhall the ordinates DH, EI, GK, &c, pa⯑rallel to the other aſymptote, be a decreaſing geometri⯑cal progreſſion, having the ſame ratio. For, all the rectangles CDH, CEI, CGK, &c, being equal, the ordi⯑nates DH, EI, GK, &c, are reciprocally as the ab⯑ſciſſes CD, CE, CG, &c, which are geometricals. And the reciprocals of geometricals are alſo geometricals, and in the ſame ratio.

COROL. 3. But if the abſciſſes CD, CE, CG, &c, be arithmeticals, or in arithmetical progreſſion, then ſhall the ordinates DH, EI, GK, &c, be harmonicals, or in harmonical progreſſion. For the reciprocals of arith⯑meticals, are harmonicals.

PROPOSITION XXVII.

Every Inſcribed Triangle, formed by Any Tangent and the two Intercepted Parts of the Aſymptotes, is equal to a Conſtant Quantity; namely Double the Inſcribed Parallelogram.

That is, the triangle CTS = 2 paral. GK.

For, ſince the tangent TS is biſected by the point of contact E, and EK is parallel to TC, and GE to CK; therefore CK, KS, GE are all equal, as are alſo CG, GT, KE. Conſequently the triangle GTE = the triangle KES, and each equal to half the conſtant inſcribed parallelogram GK. And therefore the whole triangle CTS, which is compoſed of the two ſmaller triangles and the parallelo⯑gram, is equal to double the conſtant inſcribed parallelo⯑gram GK. Q.E.D.

PROPOSITION XXVIII.

The three following Spaces, between the Aſymptotes and the Curve, are equal; namely, the Sector or Trilinear Space contained by an Arc of the Curve and two Radii, or Lines drawn from its Extremities to the Center, and each of the two Quadrilaterals, contained by the ſaid Arc, and two Lines drawn from its Extremities parallel to one Aſymptote, and the intercepted Part of the other Aſymptote.

That is, the ſector CAE = PAEG = QAEK, all ſtanding upon the ſame arc AE.

For, by prop. 26, CPAQ = CGEK; ſubtract the common ſpace CGIQ, ſo ſhall the paral. PI = the par. IK; to each add the trilineal IAE, then is the quadr. PAEG = QAEK. Again, from the quadrilateral CAEK take the equal triangles CAQ, CEK, and there remains the ſector CAE = QAEK. Therefore CAE = QAEK = PAEG. Q.E.D.

PROPOSITION XXIX.

If from the Point of Contact of any Tangent, and the two Interſections of the Curve with a Line parallel to the Tangent, three parallel Lines be drawn in any Direction, and terminated by either Aſymptote; thoſe three Lines ſhall be in continued Proportion.

That is, if HKM and the tangent IL be parallel, then are the parallels DH, EI, GK in continued proportion.

For, by the parallels, EI ∶ IL ∷ DH ∶ HM; and, by the ſame, EI ∶ IL ∷ GK ∶ KM; theref. by compoſ. EI² ∶ IL² ∷ DH·GK ∶ HMK; but, by prop. 24, the rect. HMK = IL²; and theref. the rect. DH·GK = EI², or DH ∶ EI ∷ EI ∶ GK. Q.E.D.

PROPOSITION XXX.

Draw the ſemi-diameters CH, CIN, CK; Then ſhall the ſector CHI = the ſector CIK.

For, becauſe HK and all its parallels are biſected by CIN, therefore the triangle CNH = tri. CNK, and the ſegment INH = ſeg. INK; conſequently the ſector CIH = ſec. CIK.

COROL. 2. If the geometricals DH, EI, GK be parallel to the other aſymptote, the ſpaces DHIE, EIKG will be equal; for they are equal to the equal ſectors CHI, CIK.

So that by taking any geometricals CD, CE, CG, &c, and drawing DH, EI, GK, &c, parallel to the other aſymp⯑tote, as alſo the radii CH, CI, CK. [97]then the ſectors CHI, CIK, &c, or the ſpaces DHIE, EIKG, &c, will be all equal among themſelves. Or the ſectors CHI, CHK, &c, or the ſpaces DHIE, DHKG, &c, will be in arithmetical progreſſion.

And therefore theſe ſectors, or ſpaces, will be analogous to the logarithms of the lines or baſes CD, CE, CG, namely CHI or DHIE the log. of the ratio of CD to CE, or of CE to CG, or of EI to DH, or of GK to EI, and CHK or DHKG the log. of the ratio of CD to CG, &c, or of GK to DH, &c.

PROPOSITION I.

The Abſciſſes are Proportional to the Squares of their Ordinates.

Let AVM be a ſection through the axis of the cone, and AGIH a parabolic ſection by a plane perpendicular to the former, and parallel to the ſide VN of the cone; alſo let AFH be the common interſection of the two planes, or the axis of the parabola, and FG, HI ordi⯑nates perpendicular to it.

Then I ſay that AF ∶ AH ∷ FG² ∶ HI².

For, through the ordinates FG, HI draw the circular ſections KGL, MIN parallel to the baſe of the cone, having KL, MN for their diameters, to which FG, HI are ordi⯑nates, as well as to the axis of the parabola.

Then by ſimilar triangles AF ∶ AH ∷ FL ∶ HN; but, becauſe of the parallels, KF = MH; therefore AF ∶ AH ∷ KF·FL ∶ MH·HN. But, by the circle, KF·FL = FG², and MH·HN = HI²; Therefore AF ∶ AH ∷ FG² ∶ HI². Q.E.D.

[99]COROL. 1. Hence the third proportional EG²/AE or HI²/AH is a conſtant quantity, and is equal to the parameter of the axis by cor. 5 prop. 1 of the ellipſe.

Or AE ∶ EG ∷ EG ∶ P the parameter.

Or the rectangle P·AE = EG²

COROL. 2. Hence alſo as the property is the ſame for the ordinates on both ſides of the diameter, it follows that the ordinates DH, HI, on both ſides of the axis are equal, or that the double ordinates are biſected by the axis; and that the whole figure, made up of all the double ordinates, is alſo biſected by the axis.

PROPOSITION II.

As the Parameter of the Axis∶ Is to the Sum of any Two Ordinates∷ So is the Difference of thoſe Ordinates∶ To the Difference of their Abſciſſes∶

That is, P ∶ GH + DE ∷ GH − DE ∶ DG, Or, P ∶ KI ∷ IH ∶ EI.

For, by cor. 1 prop. 1, P·AG = GH², and P·AD = DE²; theref. by ſubtrac. P·DG = GH² − DE². But [...], theref. the rectangle P·DG = KI·IH, or P ∶ KI ∷ IH ∶ DG or EI. Q.E.D.

COROL. Hence becauſe P·EI = KI·IH, and, by cor. 1 prop. 1, P·AG = GH², therefore AG ∶ EI ∷ GH² ∶ KI·IH.

And ſo any diameter EI is as the rectangle of the ſeg⯑ments KI, IH of the double ordinate KH.

PROPOSITION III.

The Diſtance from the Vertex to the Focus is equal to ¼ of the Parameter, or to Half the Ordinate at the Focus.

That is, AF = ½FE = ¼P.

For, the general property is AF ∶ FE ∷ FE ∶ P. But, by the definition, FE = ½P; therefore alſo AF = ½FE = ¼P. Q.E.D.

PROPOSITION IV.

A Line drawn from the Focus to any Point in the Curve, is equal to the Sum of the Focal Diſtance and the Abſciſs of the Ordinate to that Point.

That is, FE = FA + AD = GD, taking AG = AF.

For ſince FD = AD ∼ AF, theref. by ſquaring, FD² = AF² − 2AF·AD + AD², But, by cor. 1 prop. 1, DE² = P·AD = 4AF·AD; theref. by addition, FD² + DE² = AF² + 2AF·AD + AD², But, by rt. angled ▵s, FD² + DE² = FE²; therefore FE² = AF² + 2AF·AD + AD², and the root or ſide is FE = AF + AD, or FE = GD, by taking AG = AF. Q.E.D.

[103]COROL. 1. The difference FE − FE is always equal to DD the difference of the abſciſſes.

COROL. 2. Hence alſo the curve is eaſily deſcribed

by points. Namely, in the axis produced take AG = AF the focal diſtance, and draw a number of lines EE per⯑pendicular to the axis AD; then with the diſtances GD, GD, GD, &c, as radii, and the centre F, draw arcs croſſing the parallel ordinates in E, E, E, &c. Then draw the curve through all the points E, E, E.

PROPOSITION V.

If a Line be drawn from any Point in the Axis produced, to cut the Curve in two Points; then ſhall the ex⯑ternal Part of the Axis be a Mean Proportional between the two Abſciſſes of the Ordinates to the two Points of Interſection.

That is, AI a mean proportional between AD, AG, or, AD ∶ AI ∷ AI ∶ AG.

For, by prop. 1, AD ∶ AG ∷ DE² ∶ GH²; and, by ſim. tri. ID² ∶ IG² ∷ DE² ∶ GH²;* theref. by equal. AD ∶ AG ∷ ID² ∶ IG²; and, by diviſion, [...] or AD ∶ ID ∷ ID ∶ ID + IG; and by diviſion AD ∶ AI ∷ ID ∶ IG*, and again by div. AD ∶ AI ∷ AI ∶ AG. Q.E.D.

COROL. 1. * Hence we have AD ∶ AI ∷ DE ∶ GH.

[105]COROL. 2. If the line be ſuppoſed to revolve about the point 1; then as it recedes farther from the axis, the points E and H approach towards each other, the point E deſcending, and the point H aſcending, till at laſt they meet in the point C when the line becomes a tangent to the curve at C. And then the points D and G meet in the point M, and the ordinates DE, GH in the ordinate CM. Conſequently AD, AG, becoming each equal to AM, their mean proportional AI will be equal to the abſciſs AM. That is, the external part of the axis, cut off by a tangent, is equal to the abſciſs of the ordinate to the point of con⯑tact.

COROL. 3. Hence the tangents to all parabolas, which have the ſame abſciſſes, meet the axis produced in the ſame point. For if the abſciſs AM be the ſame in all, the ex⯑ternal axis AI, which is equal to it, will be the ſame alſo.

COROL. 4. And hence alſo a tangent is eaſily drawn to the curve.

For if the point of contact C be given; draw the ordi⯑nate CM, and produce MA till AI be = AM; then join IC the tangent.

Or if the point 1 be given; take AM = AI, and draw the ordinate MC, which will give the point of contact C, to which draw IC the tangent.

PROPOSITION VI.

If a Tangent to the Curve meet the Axis produced; then the Line drawn from the Focus to the Point of Contact, will be equal to the Diſtance of the Focus from the Interſection of the Tangent and Axis.

That is, FC = FT.

For draw the ordinate DC to the point of contact.

Then, by cor. 1 prop. 5, AT = AD; therefore FT = AF + AD. But, by prop. 4, FC = AF + AD; therefore by equality, FC = FT. Q.E.D.

COROL. 1. If CG be drawn perpendicular to the curve, or to the tangent, at C; then ſhall FG = FC = FT.

For draw FH perpendicular to TC, which will alſo biſect TC, becauſe FT = FC; and therefore, by the na⯑ture of the parallels, FH alſo biſects TG in F. And con⯑ſequently FG = FT = FC.

So that F is the center of a circle paſſing through T, C, G.

[107]COROL. 2. The ſubnormal DG is every where equal to the conſtant quantity 2FA, or FT, or ½ P the ſemi-para⯑meter.

For draw the tangent AH parallel to DC, making the triangle FHA ſimilar to GCD.

Then DC = 2AH, becauſe DT = 2DA ∶ conſequently DG = 2FA = ½P.

COROL. 3. The tangent at the vertex AH, is a mean proportional between AF and AD.

For becauſe FHT is a right angle, therefore AH is a mean between AF, AT, or between AF, AD, becauſe AD = AT. Likewiſe FH is a mean between FA, FT, or between FA, FC.

COROL. 4. The tangent TC makes equal angles with FC and the axis FT, or with the line ICK drawn parallel to the axis.

For becauſe FT = FC, therefore the ∠ FCT = ∠ FTC = its altern. ∠ ICT. Alſo the angle GCF = the angle GCK.

COROL. 5. And becauſe the angle of incidence GCK is = the angle of reflection GCF; therefore a ray of light falling upon the curve in the direction KC, will be reflected to the focus F. That is, all rays parallel to the axis, are reflected to the focus, or burning point.

PROPOSITION VII.

If an Ordinate be drawn to the Point of Contact of any Tangent, and another Ordinate produced to cut the Tangent; It will be as the Difference of the Ordinates∶ Is to the Difference added to the external Part∷ So is Double the firſt Ordinate∶ To the Sum of the Ordinates.

That is, KH ∶ KI ∷ KL ∶ KG.

For, by cor. 1 prop. 1, P ∶ DC ∷ DC ∶ DA, and P ∶ 2DC ∷ DC ∶ DT or 2DA. But, by ſim. triangles, KI ∶ KC ∷ DC ∶ DT; therefore by equality, P ∶ 2DC ∷ KI ∶ KC, or, P ∶ KI ∷ KL ∶ KC. Again, by prop. 2, P ∶ KH ∷ KG ∶ KC; therefore by equality, KH ∶ KI ∷ KL ∶ KG. Q.E.D.

COROL. 1. Hence, by compoſition and diviſion, we have KH ∶ KI ∷ GK ∶ GI, and HI ∶ HK ∷ HK ∶ KL, alſo IH ∶ IK ∷ IK ∶ IG; that is, IK is a mean proportional between IG and IH.

COROL. 2. And from this laſt property we can eaſily draw a tangent to the curve from any given point 1. Namely, draw IHG perpendicular to the axis, and take IK a mean proportional between IH, IG; then draw KC parallel to the axis, and C will be the point of contact, through which and the given point 1 the tangent IC is to be drawn.

PROPOSITION VIII.

If there be any Tangent, and a Double Ordinate drawn from the Point of Contact, and alſo any Line parallel to the Axis, limited by the Tangent and Double Ordi⯑nate∶ then ſhall the Curve divide that Line in the ſame Ratio, as the Line divides the Double Ordinate.

That is, IE ∶ EK ∷ CK ∶ KL.

For, by ſim. triangles, CK ∶ KI ∷ CD ∶ DT or 2DA; but, by the def. the param. P ∶ CL ∷ CD ∶ 2DA; therefore by equality, P ∶ CK ∷ CL ∶ KI. But, by prop. 2, P ∶ CK ∷ KL ∶ KE; therefore by equality, CL ∶ KL ∷ KI ∶ KE; and, by diviſion, CK ∶ KL ∷ IE ∶ EK. Q.E.D.

PROPOSITION IX.

The ſame being ſuppoſed as in Prop. VIII: then ſhall the External Part of the Line between the Curve and Tan⯑gent, be proportional to the Square of the intercepted Part of the Tangent, or to the Square of the inter⯑cepted Part of the Double Ordinate.

That is, IE is as CI² or as CK². and IE, TA, ON, PL, &c, are as CI², CT², CO², CP², &c, or as CK², CD², CM², CL², &c.

For, by prop. 8, IE ∶ EK ∷ CK ∶ KL, or, by equality, IE ∶ EK ∷ CK² ∶ CK·KL. But, by cor. prop. 2, EK is as the rect. CK·KL, and therefore IE is as CK², or as CI². Q.E.D.

[111]COROL. 1. As this property is common to every po⯑ſition of the tangent, if the lines IE, TA, ON, &c, be ap⯑pended to the points I, T, O, &c, and movable about them, and of ſuch lengths as that their extremities E, A, N, &c, be in the curve of a parabola in ſome one poſition of the tangent; then making the tangent re⯑volve about the point C, the extremities E, A, N, &c, will always form the curve of ſome parabola, in every poſition of the tangent.

COROL. 2. The parameter of the axis is alſo a third proportional to IE and CK.

For, by this prop. EK ∶ KL ∷ IE ∶ CK; and, by prop. 2, EK ∶ KL ∷ CK ∶ P; therefore by equality, IE ∶ CK ∷ CK ∶ P.

PROPOSITION X.

The Abſciſſes of any Diameter, are as the Squares of their Ordinates.

That is, CQ, CR, CS, &c, are as QE², RA², SN², &c. Or CQ ∶ CR ∷ QE² ∶ RA², &c.

For, draw the tangent CT, and the externals EI, AT, NO, &c, parallel to the axis, or to the diameter CS.

Then, becauſe the ordinates QE, RA, SN, &c, are pa⯑rallel to the tangent CT, by the definition of them, there⯑fore all the figures IQ, TR, OS, &c are parallelograms, whoſe oppoſite ſides are equal, namely IE, TA, ON, &c, are equal to CQ, CR, CS, &c. Therefore by prop. 9, CQ, CR, CS, &c, are as CI², CT², CO², &c, or as their equals QE², RA², SN², &c. Q.E.D.

COROL. Here, like as in prop. 2, the difference of the abſciſſes, are as the difference of the ſquares of their ordinates, or as the rectangles under the ſum and dif⯑ference of the ordinates, the rectangle under the ſum and difference of the ordinates being equal to the rectangle under the difference of the abſciſſes and the parameter of that diameter, or a third proportional to any abſciſs and its ordinate.

PROPOSITION XI.

If a Line be drawn parallel to any Tangent, and cut the Curve in two Points; then if two Ordinates be drawn to the Interſections, and a third to the Point of Con⯑tact, theſe three Ordinates will be in Arithmetical Pro⯑greſſion, or the Sum of the Extremes will be equal to Double the Mean.

That is EG + HI = 2CD.

For draw EK parallel to the axis, and produce HI to L. Then, by ſim. triangles, EK ∶ HK ∷ TD or 2AD ∶ CD; but, by prop. 2, EK ∶ HK ∷ KL ∶ P the param. theref. by equality, 2AD ∶ KL ∷ CD ∶ P. But, by the defin. 2AD ∶ 2CD ∷ CD ∶ P; theref. the 2d terms are equal, KL = 2CD, that is EG + HI = 2CD. Q.E.D.

PROPOSITION XII.

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

That is, ME = MH.

For to the axis AI draw the ordinates EG, CD, HI, and MN parallel to them, which is equal to CD.

Then, by prop. XI, 2MN or 2CD = EG + HI, therefore M is the middle of EH.

And for the ſame reaſon all its parallels are biſected. Q.E.D.

PROPOSITION XIII.

The Parameter of any Diameter is equal to four Times the Line drawn from the Focus to the Vertex of that Diameter.

That is, 4FC = p the param. of the diam. CM.

For draw the ordinate MA parallel to the tangent CT; as alſo CD, MN perpendicular to the axis AN, and FH perpendicular to the tangent CT.

Then the abſciſſes AD, CM or AT being equal by cor. 2 to prop. 5, the parameters will be as the ſquares of the ordinates CD, MA or CT, by the definition; that is, P ∶ p ∷ CD² ∶ CT², But, by ſim. tri. FH ∶ FT ∷ CD ∶ CT; therefore P ∶ p ∷ FH² ∶ FT².

But, by cor. 3 prop. 6, FH² = FA·FT; therefore P ∶ p ∷ FA·FT ∶ FT², or, by equal. P ∶ p ∷ FA ∶ FT or FC. But, by prop 3, P = 4FA, and therefore p = 4FT or 4FC. Q.E.D.

COROL. Hence the parameter p of the diameter CM is equal to 4FA + 4AD, or to P + 4AD, that is, the pa⯑rameter of the axis added to 4AD.

PROPOSITION XIV.

If an Ordinate to any Diameter paſs through the Focus, it will be equal to Half its Parameter; and its Abſciſs equal to One Fourth of the ſame Parameter.

That is, CM = ¼p, and ME = ½p.

For join FC, and draw the tangent CT. By the parallels, CM = FT; and, by prop. 6, FC = FT; alſo, by the laſt prop. FC = ¼p; therefore CM = ¼p. Again, by the defin. CM or ¼p ∶ ME ∷ ME ∶ p, and conſequently ME = ½p = 2CM.

COROL. Hence, of any diameter, the double ordinate which paſſeth through the focus, is equal to the parameter, or to quadruple its abſciſs.

PROPOSITION XV.

If there be a Tangent, and any Line drawn from the Point of Contact and meeting the Curve in ſome other Point, as alſo another Line parallel to the Axis, and limited by the Firſt Line and the Tangent∶ then ſhall the Curve divide this Second Line in the ſame Ration, as the Second Line divides the Firſt Line.

That is, IE ∶ EK ∷ CK ∶ KL.

For draw LP parallel to IK or to the axis. Then, by prop. 9, IE ∶ PL ∷ CI² ∶ CP², or by ſim. tri. IE ∶ PL ∷ CK² ∶ CL². Alſo, by ſim. tri. IK ∶ PL ∷ CK ∶ CL, or IK ∶ PL ∷ CK² ∶ CK · CL; therefore by equality, IE ∶ IK ∷ CK · CL ∶ CL², or IE ∶ IK ∷ CK ∶ CL; and, by diviſion, IE ∶ EK ∷ CK ∶ KL. Q.E.D.

PROPOSITION XVI.

If a Tangent cut any Diameter produced, and if an Ordinate to that Diameter be drawn from the Point of Contact; then the Diſtance in the Diameter pro⯑duced, between the Vertex and the Interſection of the Tangent, will be equal to the Abſciſs of that Ordi⯑nate.

That is, IE = EK.

For, by the laſt prop. IE ∶ EK ∷ CK ∶ KL. But, by prop. 12, CK = KL, and therefore IE = EK. Q.E.D.

COROL. 1. The two tangents CI, LI, at the extremi⯑ties of any double ordinate CL, meet in the ſame point of the diameter of that double ordinate produced. And the diameter drawn through the interſection of two tangents, biſects the line connecting the points of contact.

COROL. 2. Hence we have another method of draw⯑ing a tangent from any given point 1 without the curve. Namely, from 1 draw the diameter IK, in which take EK = EI, and through K draw CL parallel to the tangent at E; then C and L are the points to which the tangents muſt be drawn from 1.

PROPOSITION XVII.

If from any Point of the Curve there be drawn a Tan⯑gent, and alſo Two Right Lines to cut the Curve; and Diameters be drawn through the Points of In⯑terſection E and L, meeting thoſe Two Right Lines in two other Points G and K: Then will the Line KG joining theſe laſt Two Points be parallel to the Tangent.

For, by prop. 15, CK ∶ KL ∷ EI ∶ EK; and by comp. CK ∶ CL ∷ EI ∶ KI ∷ GH ∶ LH by parallels But, by ſim. tri. CK ∶ CL ∷ KI ∶ LH; theref. by equal. KI ∶ LH ∷ GH ∶ LH; conſequently KI = GH, and therefore KG is parallel and equal to IH. Q.E.D.

OTHERWISE.

By prop. 15, EI ∶ EK ∷ CK ∶ KL ∷ CE ∶ EG by the parallels. So that the two triangles CEI, KEG have their angles at E equal, and the ſides about thoſe angles pro⯑portional; they are therefore ſimilar; conſequently the angle at C is equal to the angle at G, and KG parallel to CH. Q.E.D.

COROL. 1. Hence we have KI = GH.

COROL. 2. Alſo KI or GH is a mean proportional between EI and LH.

For, by the parallels, EI ∶ KI ∷ GH or KI ∶ LH.

PROPOSITION XVIII.

If a Line be drawn from the Vertex of any Diameter to cut the Curve in ſome other Point, and an Ordinate of that Diameter be drawn to that Point, as alſo another Ordinate any where cutting the Line, both produced if neceſſary: The Three will be continual Proportionals, namely, the two Ordinates and the Part of the Latter limited by the ſaid Line drawn from the Vertex.

That is, DE, GH, GI are continual Proportionals, Or DE ∶ GH ∷ GH ∶ GI.

For, by prop. 10, DE² ∶ GH² ∷ AD ∶ AG; and, by ſim. tri. DE ∶ GI ∷ AD ∶ AG; theref by equality DE ∶ GI ∷ DE² ∶ GH², that is, of the three DE, GH, GI, 1ſt ∶ 3d ∷ 1ſt² ∶ 2d²; therefore 1ſt ∶ 2d ∷ 2d ∶ 3d, that is DE ∶ GH ∷ GH ∶ GI. Q.E.D.

COROL. 1. Or their equals GK, GH, GI are pro⯑portionals; where EK is parallel to the diameter AD.

[121]COROL. 2. Hence we have DE ∶ AG ∷ p ∶ GI, where p is the parameter, or AG ∶ GI ∷ DE ∶ p. For, by the defin. AG ∶ GH ∷ GH ∶ p.

COROL. 3. Hence alſo the three MN, MI, MO are pro⯑portionals, where MO is parallel to the diameter, and AM parallel to the ordinates.

For, by prop. 10, MN, MI, MO, or their equals AP, AG, AD, are as the ſquares of PN, GH, DE, or of their equals GI, GH, GK, which are proportionals by cor. 1.

PROPOSITION XIX.

If a Diameter cut any Parallel Lines terminated by the Curve; the Segments of the Diameter will be as the Rectangle of the Segments of thoſe Lines.

That is, EK ∶ EM ∷ CK·KL ∶ NM·MO.

Or, EK is as the rectangle CK·KL.

For draw the diameter PS to which the parallels CL, NO are ordinates, and the ordinate EQ parallel to them.

Then CK is the difference, and KL the ſum of the ordinates EQ, CR; alſo NM the difference, and MO the ſum of the ordinates EQ, NS. And the differences of the abſciſſes, are QR, QS, or EK, EM.

Then by cor. prop. IO, QR ∶ QS ∷ CK·KL ∶ NM·MO, that is EK ∶ EM ∷ CK·KL ∶ NM·MO.

COROL. 1. The rect. CK·KL = rect. EK and the param. of PS.

For the rect. CK·KL = rect. QR and the param. of PS.

[123]COROL. 2. If any line CL be cut by two diameters EK, GH; the rectangles of the parts of the line, are as the ſegments of the diameters.

For EK is as the rectangle EK·KL, and GH is as the rectangle CH·HL; therefore EK ∶ GH ∷ CK·KL ∶ CH·HL.

COROL. 3. If two parallels CL, NO be cut by two diameters EM, GI; the rectangles of the parts of the parallels will be as the ſegments of the reſpective diame⯑ters.

For EK ∶ EM ∷ CK·KL ∶ NM·MO, and EK ∶ GH ∷ CK·KL ∶ CH·HL, theref. by equal. EM ∶ GH ∷ NM·MO ∶ CH·HL.

COROL. 4. When the parallels come into the poſition of the tangent at P, their two extremities, or points in the curve, unite in the point of contact P; and the rectangle of the parts becomes the ſquare of the tan⯑gent, and the ſame properties ſtill follow them.

So that, EV ∶ PV ∷ PV ∶ p the param.

GW ∶ PW ∷ PW ∶ p, EV ∶ GW ∷ PV² ∶ PW², EV ∶ GH ∷ PV² ∶ CH·HL.

PROPOSITION XX.

If two Parallels interſect any other two Parallels; the Rectangles of the Segments will be reſpectively Proportional.

That is, CK·KL ∶ DK·KE ∷ GI·IH ∶ NI·IO.

For, by cor. 3 prop. 19, PK ∶ QI ∷ CK·KL ∶ GI·IH; and by the ſame, PK ∶ QI ∷ DK·KE ∶ NI·IO; theref. by equal. CK·KL ∶ DK·KE ∷ GI·IH ∶ NI·IO.

COROL. When one of the pairs of interſecting lines comes into the poſition of their parallel tangents meeting and limiting each other, the rectangles of their ſegments become the ſquares of their reſpective tangents. So that the conſtant ratio of the rectangles, is that of the ſquare of their parallel tangents, namely, CK·KL ∶ DK·KE ∷ ſq. tang. parallel to CL ∶ ſq. tang. parallel to DE.

PROPOSITION XXI.

If there be Three Tangents interſecting each other; their Segments will be in the ſame Proportion.

That is, GI ∶ IH ∷ CG ∶ GD ∷ DH ∶ HE.

For through the points G, I, D, H, draw the diameters GK, IL, DM, HN; as alſo the lines CI, EI, which are double ordinates to the diameters GK, HN, by cor. 1 prop. 16; therefore the diameters GK, DM, HN, biſect the lines CL, CE, LE; hence KM = CM − CK = ½ CE − ½ CL = ½ LE = LN or NE, and MN = ME − NE = ½ CE − ½ LE = ½ CL = CK or KL.

But, by parallels, GI ∶ IH ∷ KL ∶ LN, and CG ∶ GD ∷ CK ∶ KM, alſo DH ∶ HE ∷ MN ∶ NE.

But the 3d terms KL, CK, MN are all equal; as alſo the 4th terms LN, KM, NE.

Therefore the firſt and ſecond terms, in all the lines, are proportional, namely GI ∶ IH ∷ CG ∶ GD ∷ DH ∶ HE. Q.E.D.