138 



CONIC SECTIONS. 



Ellipse. FBC, F b C, they are in all respects equal (26. 1. E.), 

 r— ' hence BC= AC. 



Cor. 2. The square of the serniconjugate axis is equal 

 to the rectangle contained by the parts into which the 

 transverse is divided at either focus. For AC l = AF.Fa 

 + FC 2 (5. 2. E.) and FB 2 s=FC 2 + CB J (47. 1. E), there- 

 fore since AC 2 =FB 2 , it follows that AF-Fa=CB 5 . 



Cor. 3. The square of the semiconjugate axis is also 

 equal to, the rectangle contained by the eccentricity 

 and the distance of either focus from its directrix. 

 For since PC : CA : : CA : CF (Def. 1.) PC.CF=CA» 

 (17. 6. E.) now FC.CF=PF.FC-}-CF' (3. 2. E.), and 

 CA 2 =BF 2 =BC 1 + CF 2 ; therefore PF.FC=BC S . 



Definition 10. 



A straight line which meets an ellipse, and being 

 produced does not cut it, is said to touch the ellipse, 

 and is called a Tangent. 



Prop. IV. Problem. 



Having given the transverse axis and a focus, and 

 consequently its directrix, to find the points in which a 

 straight line given by position perpendicular to the 

 axis, meets the ellipse. 



Fig. 5. Let Aa be the transverse axis, F the focus, PQ the 



directrix, and L I the line given by position, \i hich 

 meets the axis in G : let FM perpendicular to the axis 

 at F meet the curve in M : draw PM meeting the line 

 given by position in L : describe a circle on F as a 

 centre, with GL as a radius, meeting L I in D, and d, 

 and these will be two points in the ellipse. 



For, join FD, Fd, and draw DE, d # perpendicular 

 to the directrix. Then the triangles PGL, FFM are 

 manifestly similar, therefore GL : GP : : FM : FP; or, 

 since ED and e d are each equal to FG (:S4. 1. E.), FD : 

 • DE, also Yd : de : : FM : FP, that is in the determining 

 ratio (Cor. to Def. 1, 3, 4.); therefore D and d are points 

 in the ellipse. 



It appears that the problem can be resolved only 

 when L l has such a position that GL is not less than 

 FG. To determine the limits within which this hap- 

 pens, draw AH, a h perpendicular to the axis, meeting 

 PL in H and h ; and join FH, Fh ; then, by similar 

 triangles, FF : FM : : PA : AH : : Fa : a h, but PF : FM 

 : : PA : AF : : P o : aF (Cor. to Def. 1, 3, 4.), therefore, 

 AFIrrAF, and ah=za¥. Now, let GL meet either of 

 the lines F/i, FH in N, then GNrzGF, and G being 

 any point between A and a, GN, or GF, is less than 

 GL; but at the points A, a, GN is equal to GL, and at 

 any point in Aa produced, GN is greater than GL ; 

 therefore the problem will be possible only when the 

 point G is any where in the line A«. 



Con. 1. Every straight line perpendicular to the 

 transverse axis between its extremities, meets the curve 

 in two points, and no more ; also at the extremities it 

 meets the curve in one point only ; and beyond the 

 extremities it falls entirely without the curve. 



Cor. 2. In the ellipse, every straight line perpendi- 

 cular to the transverse axis, and terminating in the 

 eurve, is bisected by the axis. 



Cor. 3. A perpendicular to the transverse axis at ei- 

 ther of its extremities is a tangent to the curve. 



Cor. 4. The transverse divides the ellipse into parts 

 exactly alike. 



Prop. V. Problem. ■ 



The transverse axis, and a focus, and consequently 

 $9 directrix, being given, to find the points in which a 



straight line given by position' parallel to the axis meets Ellipse 

 the curve. *— -y*» mt 



Let EH, the line given by position, meet the direc- F'g* <?• 

 trix in E, and the conjugate axis B b, in H ; draw EF 

 through the focus, to meet FIG in G : On G as a cen- 

 tre, with a radius equal to a mean proportional between 

 GF and GE describe a circle, meeting EH in D and 

 d, and these will be points in the ellipse. For, join 

 DF, dF, DG, d G, and because EG : GD : : GD : GF, 

 and the angle at G, is common to the triangles EGD, 

 DGF ; these triangles are equiangular (6. 6". E.), and 

 ED : DF : : EG : GD ; therefore ED 2 : DF 2 : : EG 2 ; 

 GD 2 ('22. 6. E.) ; but EG 2 : GD 2 : : EG : GF (2. Cor. 

 20. 6. E.) : : PC : CF (2. 6. E.), and PC : CF : : AC 2 : 

 CF 2 (2. Cor. 20. 6. E.) therefore ED 2 : DF' : : AC 2 : 

 CF 2 , and ED : DF : : AC : CF ; hence D is a point in 

 the ellipse, and in the same manner it may be shewn 

 that d is a point in the ellipse. 



It appears that this problem can be resolved only 

 when GH does not exceed GD, a mean proportional 

 between EG and GF, and this circumstance restricts 

 the position of the line given by position within cer- 

 tain limits. To determine these, draw FL perpendi- 

 cular to FG, to meet GH in L. Then the triangles 

 GHE, GFL will be equiangular (32. 1. E.), there- 

 fore EG : GH : : GL : GF; hence GH.GL-EG.GF 

 (lb - . 6. E.) = GD 2 (by construction) : and as GH 2 may 

 have any magnitude that does not exceed GD 2 , there- 

 fore GPP must not exceed GH GL, that is, GH must 

 not exceed GL, or CH must not exceed CL, or CH* 

 must not exceed CH.CL ; but the triangles EPF, FCL 

 being similar, because each is similar to GCF (29- 1. and 

 8. 6. E.) LC : CF : : FP : PE or CH, therefore CH.CL 

 = FP.FC (16. 6. E.)=CB 2 (3. Cor. 3.), therefore CH* 

 must not exceed CB 2 , and CH may have any mag- 

 nitude not greater than CB, the transverse axis 



Cor. 1. If an indefinite straight line meets the con- 

 jugate axis at right angles between its extremities, it 

 cuts the curve in two points, and no more ; if it pass 

 through either extremity, it" meets the curve in one 

 point only ; and if it meets the conjugate axis produ- 7 

 ced, it falls entirely without the curve. 



Cor. 2. Every straight line perpendicular to the con- 

 jugate axis, and terminating in the ellipse, is bisected 

 by that axis. 



Cor. 3. A perpendicular to the conjugate axis at ei- 

 ther of its extremities, is a tangent to the ellipse. 



Cor. 4. The ellipse is divided into two parts ex- 

 actly alike by the conjugate axis. 



Prop. VI. 



If from any point Q in the directrix, a straight line pjg, j. 

 be drawn to meet the ellipse in D, and LF/, a parallel 

 to the directrix, passing through the focus, in H ; and 

 if another straight line be drawn to D through F, and 

 a perpendicular QK be drawn to FD ; the lines FH, 

 QK have to each other the determining ratio. 



Draw DE perpendicular to the directrix, and let DF 

 meet it in I. The triangles DHF, DQI are equiangu- 

 lar (29. 1. E.), and the triangles QIK, DIE are also - 

 equiangular, for the angles at K and E are right an- 

 gles, and the angle at I is common ; therefore FH : FD 

 (: : QI : ID) : : QK : DE (4. 6. E.), and by alternation, 

 FH : QK : : FD : DE, that is in the determining ratio. 



Cor. 1. If a straight line meet the directrix in Q, 

 and the parallel L / in H ; and if a straight line KFD 

 pass through the focus, and FH have to KQ, (the dis- 





