142 



Ellipse, less than DL'+CL 1 , that is BC 

 v -- V — and BC less than DC. 



Prop. XVIII. 



If an ordinate be drawn to any diameter of an ellipse, 

 the rectangle contained by the abscissae is to the square 

 of the semiordinate as the diameter to its parameter. 



tig. co. Eet DE be a semiordinate to the diameter Vp, let 



PG be the parameter of the diameter, and Qq the con- 

 jugate diameter : By Def. 15. Vp: Qq : : Qq : PG > there- 

 fore, Vp :VG::Vp l : Qq 2 (2 Cor. 20. 6". E.), but Vn 1 : 

 Qq 1 : : VE.Ep : DE 2 (16.) ; therefore, VE.Ep : DE' : : 

 Vp : PG. 



Con. 1. Let the parameter PG be perpendicular to 

 the diameter Vp ; join PG, and from E draw EM pa- 

 rallel to PG, meeting PG in M. The square of DE, 

 the semiordinate, is equal to the rectangle contained 

 by PE and EM. 



For VE.Ep : DE* : : P ( > : PG, and 

 Vp : PG : : Ep : EM : PE.Ep : PE.EM, 

 Therefore DE : = PE.EM. 



Scholium, 



If the rectangles VGEp, KGKM be completed, it 

 will appear that ED ! is equal to. the rectangle MP, 

 which rectangle is less than the rectangle KP con- 

 tained by the abscissa PE, and parameter PG, by a rect- 

 angle KH similar, and similarly situated to LP, the 

 rectangle contained by the diameter and its parameter. 

 It was on account of the deficiency of the square of the 

 ordinate from the rectangle contained by the abscissa 

 " and parameter, that Apollonius called the curve, to 

 which the property belonged, an Ellipse. 



Prop. XIX. 



If from the vertices of two conjugate diameters of 

 an ellipse, there be drawn ordinates to any third dia- 

 meter, the square of the segment of that diameter in- 

 tercepted between either ordinate and the centre, is 

 equal to the rectangle contained by the segments be- 

 tween the other ordinate and the vertices of the same 

 diameter. 



Fig. 21. Let Vp, Qq be two conjugate diameters, and PE, 



QG semiordinates to any third diameter Aa, then 

 CG z z=AE.Ea; and CE 2 zrAG.Ga. Draw the tangents 

 PH, QK meeting Aa in H and K. The rectangles 

 HC.CE and KC.CG are equal; for each is equal CA 2 

 (14.), therefore HC : CK : : CG : CE. But the triangles 

 HPC, CQK are evidently similar (Cor. Def. 14..), and 

 PE being parallel to QG, their bases CH, KC are si- 

 milarly divided at E and G, therefore HC : CK : : HE : 

 CG; wherefore CG : CE : : HE : CG, consequently CG 2 

 = CE.EH=(Cor. 1. 11.) AE.Ea. In like maimer it 

 may be shewn, that CE 2 =AG.Ga. 



Cor. 1. Let lib be the diameter that 'is conjugate 

 to Aa ; then Aa is to BA, as CG to PE, or as CE to QG. 



For Aa 2 : B6 2 : : AE.Ea, or CG 2 : PE 2 , therefore Aa : 

 Bb : : CG : PE. In like manner A« : B b : : CE : QG. 



Cor. 2. The sum of the squares of CE, CG, the seg- 

 ments of the diameter to which the semiordinates PE, 

 QG are drawn, is equal to the square of CA, the semi- 

 diameter. 



For CE 2 + CG 2 =CE 2 +AE.E«=CA 2 . 



Cor. 3. The sum of the squares of any two conju- 

 gate diameters, is equal to the sum of the squares of 

 fciie axes. 



Let Aa, Bb be the axes, and Vp. Qq any two conju- 



CONIC SECTIONS. 



is less than DC 1 , 



gate diameters; draw PE, QG perpendicular to Aa, ZiIipiB 

 and PL, QM perpendicular to B6 ; then CE 2 -f-CG 2 = N — -v* 

 CA 2 , andCM 2 + CL 2 . or GQ 2 -fPE 2 = CB 2 ; therefore 

 CE 2 -f-PE 2 -fCG 2 -fGQ 2 = CA 2 - r .CB 2 ; that is (4-7. I. 

 E.), CP 2 -}-CQ 2 =CA 1 -f-CB 2 , therefore Vp*+Qq>- 

 A« 2 -f-RZ> 2 . 



Prop. XX. 



If four straight lines be drawn touching an ellipse at 

 the vertices of any two conjugate diameters, the paralle- 

 logram formed by these lines is equal to the rectangle 

 contained by the transverse and conjugate axes. 



Let Vp, Qq be any two conjugate diameters, the pa- Fig. -2.1i 

 rallelogramVXYZ formed by tangents to the curve at 

 their vertices, is equal to the rectangle contained by 

 Aa, Bb the two axes. Produce Aa, one of the axes, 

 to meet the tangent pY in N, join QN, and draw p\ c 

 QG perpendicular to Aa. Because CN : CA : : CA : CI 

 (14.), and CA : CB : : CI : QG (1 Cor. lfj.) ex asq. CN : 

 CB : : CA : QG, therefore CN.QG = CB.CA. But 

 CN.QG=twice trian. CNQ=paral. CpYQ, therefore 

 the parallelogram C/jYQ= CB.CA, and taking the qua- 

 druple of these, the parallelogram VXYZ is equal t» 

 the rectangle contained by Aa and Bb. 



Prop. XXI. 



If two tangents, at the vertices of any diameter of aa 

 ellipse, meet a third tangent, the rectangle contained 

 by their segments, between the points of contact and 

 the points of intersection, is equal to the square of the 

 semi-diameter to which they are parallel. And the 

 rectangle contained by the segments of the third tan- 

 gent, between its point of contact and the parallel tan- 

 gents, is equal to the square of the semi-diameter to 

 which it is parallel. 



Let PH, p h, tangents at the vertices of a diameter Fig. 26. 

 Vp, meet HDa, a tangent to the crave, at any point D 

 in H and h. Let CQ be the semi-diameter, to which 

 the tangents PH, p h are parallel, and CR that to which 

 Hh is parallel, then VH.pk = CQ\ and DH.D/«=CR\ 

 If the tangent HDA be parallel to Vp, the proposition 

 is manifest. If it is not parallel, let it nfeet the semi- 

 diameters CP, CQ in L and K. Draw DE, RM paral- 

 lel to CQ, and DG parallel to CP. Because LP.L/;= 

 LE.LC (2 Coy. 14.) LP: LE : : LC:Lp, and because 

 PH, ED, CK,ph, are parallel, PH : ED : : CK :ph, 

 wherefore PH.p/(=ED.CK ; but ED.CKz=CG.CK= 

 CQ 2 (14.), therefore PLT.p«=CQ 2 . Again, the tri- 

 angles LED, CMR are evidently similar; and LE, LD 

 similarly divided at P and H, also at p and h, therefore 

 PE : HD : : (LE : LD : :) CM : CR, also pE : h D : : 

 (LE : LD) CM : CR, hence taking the rectangles of 

 the corresponding terms, PE.j;E : HD.AD : : CM 2 : CR" . 

 But if CD be joined, the points D and R are evidently 

 the vertices of two conjugate diameters (Cor. Def. 14.) '- 

 and therefore PE.;;E=CM 2 (19.), therefore RD.kD= 

 CB*. 



Cor. The rectangle contained by LD and DK, the 

 segments of a tangent intercepted between D the point 

 of contact and Vp, Qq, any two conjugate diameters is 

 equal to the square of CR, the semi-diameter to which 

 the tangent is parallel. Let the parallel tangents PH, 

 p h meet LK in H and h, and draw DE a semi-ordi- 

 nate to Vp. Because of the parallels PH, ED, CK, 

 ph, LE : LD : : EP : DH, and EC : DK : : Ep : DA, 

 therefore LE.EC : LD.DK : : EV.Ep : DH.DA. But 

 LE.EC = EV.Ep (1. Cor 14.), therefore LD.DK = 

 DH.D/; = (by this Prop.) CR'. 



