or or 2ÇK': 2IK :: AD. 2CK + AD. AG : ID3, AD. 2CK: AD. 2IK :: AD. 2CK + AD. AG: ID2; theref. by div. CK : IK :: AD. AG : ID2 and, by comp. CK: CI :: AD. AG : ID2· - AD. 2IK, AD. ID+IA, But, when the line IH, by revolving about the point 1, comes into the position of the tangent IL, 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 last proportion becomes CM CI AM2: AI2, Q. E. D. THEOREM VII. 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. That is, CA is a mean proportional between CD and CT; or CD, CA, CT, are conti nued proportionals. For, by theor. 6, CD: CT :: AD2 : AT2. that is, CD CT: (CA-CD): (CT CA), CD: DT:: CD2 +CA2: CT2 CD2, CD: DT:: CD2 +CA2 (CT + CD) DT, CD2: CD. CT :: CD2 + CA2: CD DT+CT.DT CD2: CA2:: CD ; CT. therefore (th. 78, Geom.) CD: CA :: CA: CT. Q. E. D. Corol. 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 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. THEOREM THEOREM VIII. 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. For these are as AG, DE, CH, BI, by similar triangles. THEOREM IX. 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, B FD For, draw the ordinate DE, and fe parallel to EE. By cor. 1, theor. 5, CA: CD :: CF: CA and by theor. 7, and by add. and sub. CA CD CT: CA; FE, FE; TF: Tf :: FE 2CA-FE or fe by th. 5. TF: Tf :: FE: fe; Le fee. e4 FET; Q. E. D. Corol. OF THE HYPERBOLA. THEOREM 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 another section of them perpendicular to the plane of the former; AB the axis of the hyperbolic sections; and FG, HI, ordinates perpendicular to it. Then it will be, as FG2: HI2:: AF. FB: AH.HB. For, through the ordinates FG, HI, draw the circular sections KGL, MIN, parallel to the base of M RV B D V E VA K L 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, 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 But, by the circle, KF. FL FG2, and MH. HN = Therefore the rect. AF FB : AH. HB :: FG2: HI2. . HN. HI2; But, if c be the centre, then ac CB = AC2, and ca is the Therefore AC2: AD DB:: ac2: DE2; or, by permutation, AC2: ac2 :: AD. DB : DE2; semi-conj. That is, As the transverse, So is the rectangle of the abscisses, THEOREM III. As the Square of the Conjugate Axis : The Sum of the Squares of the Semi-conjugate, and That is, Ca2: CA2 :: Ca2 + cd2 : dâ2. B For, draw the ordinate ED to the transverse AB. Then, by theor. 1, ca2: CA2 :: DE2: AD or But DB or CD2-CA2, Ca2 CA2:: cd2 : de2 CA2. theref. by compos. ca2: CA2:: ca2 + cd2 : dɛ2. In like manner, CA2: Ca2 :: CA2 + CD2: De2. Q. E. D. Corol. By the last theor. CA2; ca2 :: cn2 CA2: DE2, and by this theor. CA2 : ca2 :: CD2 + CA2 : De2, 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 foci, or burning points. THEOREM X. 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 R Let EG, eg, be two conjugate diameters parallel to the sides of the parallelogram, and dividing it into four less and equal parallelograms. Also, draw the ordinates DE, de, and CK perpendicular to PQ; and let the axis CA produced meet the sides of the parallelogram, produced if necessary, in T and t. Then, by theor. 7, and theref. by equality, CT CA: CA: CD, CT : ct :: cd : CD; but, by sim. triangles, CT : ct: theref. by equality, and the rectangle Again, by theor. 7, or, by division, and by composition, conseq. the rectangle But, by theor. 1, therefore TD cd TD DC is CD : CA: CD : CA: DA: AT, CD: DB :: AD: DT; *Corol. Because cd2 = AD DB = CA2 — CD2, therefore CA CD2 + cd?. In like manner, ca2 DE + de2. In |