PROP. VI. Tangents drawn to an ellipse from the same point are to one another as the parallel semi-diameters. In fig., p. 9, let the diameter parallel to the tangent at P meet SP in k and the normal in F. Then TPL, PkF are similar right-angled triangles. Therefore TP: TL=Pk: PF =CA: PF. [Prop. XIII., p. 60. Let CD be the semi-diameter parallel to TL. TP:TQ=CD:CD, where CD' is the semi-diameter parallel to TQ. Hence TM, TL being equal, as in Prop. VI., p. 9. PROP. VII. To prove that SP.HP-CD, where CP, CD are conjugate semi-diameters. Draw the normal at P, meeting the major and minor G H axes in G, g respectively. Then, since a circle goes round SPHg (Prop. vI., p. 55), the angles PSg, PgH, in the same segment, are equal. Also LSPg=HPG. [Prop. IV., p. 54. Hence SPg, HPG are similar triangles, so that Therefore SP: PGPg: PH. SP.HP=PG.Pg = CD2. [Prop. IV., Cor. Note. Let the tangent at P meet the axes in T, t. Then by similar right-angled triangles PGT, Pgt, CP2 + CD2 = CA2 + CB3, where CP, CD are conjugate semi-diameters. Since C is the middle point of SH (fig., Prop. VII.), therefore Also SP2 + PH2 = 2 CP2 +2 CS2* 2SP.PH=2 CD". [Prop. VII. But the square on SP+ PH, or on 2CA, is equal to the squares on SP, PH together with twice the rectangle SP.PH. Therefore, from above, by addition, Hence 4 CA 2 CP2 + 2CD +2 CS2. CP+CDCA + CA2- CS [Euc. II. 4. and Or thus: =CA2+ CB2. [Prop. VIII., p. 57. Let CN, CR be the abscissæ of P, D, respectively. CN2 + CR2 = CA2, PN2 + DR2 = CB2. Todhunter's Euclid, Appendix. where CV is the abscissa of any point Q on the ellipse, measured along a diameter which meets the curve in P, and T the point in which the tangent at Q meets CP. Let the tangent at P, which is parallel to QV, meet QT in R. Complete the parallelogram QRPO. T R R Then the diagonal RO bisects PQ and is therefore a diameter. Let it be produced to the centre C. [Prop. XIII., p. 16. Then, since QV, RP are parallel, and also QT, OP, by construction, therefore PROP. X. If TPP' be any diameter and TQ the tangent at a point Q, whose abscissa is CV, then TC.TV=TP.TP'. Let the tangents at P, Q meet in R. Draw PQ. Then CR is parallel to P'Q, since it bisects PP' and also PQ. [Prop. XII., p. 16. PROP. XI. The tangent and ordinate at Q meet the diameter PCP' in T, V respectively. To prove that TP: TP'=PV: P'V. Let the tangents at P, P', which are parallel to QV, meet the tangent at Q in R, R' respectively. Then the tangents RP, RQ are as the parallel semidiameters, and therefore as the tangents R'P', R' Q. [Prop. VI. Alternando RP: R'P' = RQ: R' Q. But and Therefore RQ: R'Q=PV : P'V, [Euc. VI., 10, RP : R'P' = TP : TP', by similar triangles. TP: TP' = PV: P'V. COR. The lines TP, TV, TP' are in harmonical progression, since the first is to the third as the difference between the first and second to the difference between the second and third. Note. Any one of the last three propositions being assumed, the others follow by Euc. II. For example, let Prop. x. be assumed. Then TC2-TC.CV=TC2 – CP", or CV.CT=C.P2. PROP. XII. If CV, QV be the abscissa and ordinate of any point Q on the ellipse, then QV2: CP2 - CV2 = CD2 : CP", where CP is the semi-diameter on which CV is measured and CD that parallel to QV. Draw Qu, an ordinate of CD, and let the tangent at Q meet CD, CP produced in t, T respectively. Then and = QV: VT Ct: CT, by similar triangles, since QV, Cv are equal. But Therefore QV2: CV.VT Co. Ct: CV.CT = = CD : CP. [Prop. IX. CV.VT CV.CT- CV - CP-CV. [Prop. IX. = Therefore QV2: CP2 - CV2 = CD" : CP2. Note. Let PC meet the curve again in P'. Then Therefore or CPCV PV.VP'. [Euc. II., 5, Cor. = Again, Qu is an ordinate of the diameter CP, therefore Qv2: CD2 - Cv2=CP2: CD", CV2: CD2-QV" CP2: CD". = (i) and (ii) are different forms of Prop. XII. (i). [Prop. XII. (ii), |