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 http://www.archive.org/details/cu31924001105448 
 
Cornell University Library 
 QA 485.H86 
 
 Elements of the conic sections, 
 
 3 1924 001 105 448 
 
ELEMENTS 
 
 CONIC SECTIONS. 
 
 BY 
 RUFUS B. HOWLAND, B.C.E., 
 
 Fbofessob op Mathematics in Wyoming Seminabt. 
 
 LEACH, SHEWELL & SANBORN, 
 
 BOSTON AND NEW YOEK. 
 
coptbight, 1887, 
 By RUFUS B. howl and. 
 
 Electbottped bt J. S. CusHiNG & Co., Boston. 
 
PEEFAOE. 
 
 For two years I have been giving my advanced classes 
 in ■ Geometry work in conic sections by means of notes 
 copied bj' the hektograph. The amount of work has grown 
 on my hands till it has become too large for convenient 
 use in its present form ; so it is now put in the hands of 
 the printer. 
 
 To show the connection of the curves by their definition, 
 and not to follow too closely the line of reasoning in other 
 American works on this subject, I used Boscovitch's defini- 
 tions, and find by so doing many theorems can be demon- 
 strated with less work than by using the ordinary definitions. 
 
 The demonstrations are necessarily longer than, those 
 relating to the straight lice and circle, but there is a cor- 
 responding advantage in discipline to the student. 
 
 Although more details are given than in English works, 
 it is hoped that enough has been omitted to require the 
 independent thought of the student ; and the author trusts 
 it may be of some use to those desiring a knowledge of 
 this extensive subject. 
 
 Kingston, Pa., 
 
 Jan. 28, 1887. 
 
ELEMENTS 
 
 COI^IC SECTIOI^S. 
 
 o>0<o 
 
 DEFINITIONS. 
 
 A curved line is a line no part of which is straight, and 
 points on the curve are so situated that they answer the 
 requirements of some definition. To define the curved lines 
 called conic sections, we assume a fixed straight line, called 
 a directrix, and a fixed point without this line, called a, focus. 
 
 A conic section is a curved line the distance from any 
 point in v/hich to the directrix is in a constant ratio to its 
 distance from the focus. 
 
 There are three classes of curves coming under this defi- 
 nition. 
 
 The parabola, iu which any point is the same distance from 
 the directrix as from the focus; i.e., the constant ratio of 
 the above definition equals unity. 
 
 Take BD as the directrix, and F as the focus of the 
 parabola PCIf; then BP=PF and DE^EF. 
 
2 ELEMENTS OF THE CONIC SECTIONS. 
 
 The ellipse, in which any point is farther from the direc- 
 trix than from the focus ; i.e., the constant ratio is greater 
 than unity. 
 
 • B 
 
 I 
 
 d- 
 
 ^ 
 
 ^ 
 
 ^ 
 
 A^ 
 
 ^V 
 
 K^ i 
 
 ■>£ 
 
 jTake DG as the directrix, and F. as the focus ; then 
 
 PF ~ HF KF LF' 
 
 and DP>PF. 
 
 The hyperbola, in which any point is nearer the directrix 
 than the focus ; i.e., the constant ratio is less than unity. 
 
 7^ 
 
 
 K 
 
 D P/ 
 
 
 V 
 
 
 /I 
 
 
 B' JM 
 
 A 
 
 g\f\i 
 
 1 
 
 K 
 
 Take DA as the directrix, and F the focus of the hyper- 
 
 bola PiffiO; then 
 
 and 
 
 DP^DH 
 PF HF' 
 
 DP<PF. 
 
 Pboblem I. 
 
 Having given a point on a line, to find a second point on 
 the line such that the ratio of its distances from a fixed line 
 
PROBLEM I. 
 
 (the directrix) and from a fixed point (the focus) sliall equal 
 the ratio of the correspondiug distances from the given point. 
 
 Let DB be the dii-ectrix, and F the focus. Take P as the 
 given point on the line DH, and suppose the problem solved, 
 and H to be the required point ; then 
 
 AP:PF::BH:HF; (1) 
 
 or, by alternation, 
 
 AP:BH::PF:HF; (2) 
 
 AP and BH being perpendicular to DB by construction, 
 
 AP:BH::DP:DH; (3) 
 
 comparing (2) and (3), 
 
 PF:HF::DP:DH; 
 
 therefore DF bisects the angle PFE. 
 
 Hence, to find the required point, join the focus F with 
 the point D in which the line intersects the directrix, and 
 draw a line making an angle DFE equal to DFP: the point 
 in which this line EF meets the given line is the required 
 point. 
 
 CoKOLLAKY I. As but one line can be drawn from F, 
 making with DF an angle equal to DFP, there are but two 
 points on a line from which a given ratio, formed as in the 
 problem, can be made. 
 
 Corollary II. If the point P be moved along the line 
 DH so that DFP approaches a right angle, its equal DFE 
 
4 ELEMENTS OF THE CONIC SECTIONS. 
 
 also approaches a right angle, and when each becomes a right 
 angle, the line EFH coincides with FP, and H with P; 
 hence there will- be but one point on a line. from which a 
 given ratio can be formed when the line joining this point to 
 the focus is perpendicular to the line joining the focus to 
 the intersection of the given line and the directrix. 
 
 Exercises. Find a second point on the line DQ oi each 
 of the following figures, so that the ratio of the lines corre- 
 sppnding to AP and PF shall equal their ratio. 
 
 Theorem I. 
 
 A line perpendicular to the directrix will meet a parabola 
 in one point, it may meet an ellipse in two points, and it 
 will meet a hyperbola in two points. 
 
 Take BD as the directrix, F the focus, and P a point of 
 a conic section on the line DQ perpendicular to BD. Make 
 DFE = DFP, to find the second point of the conic on DG 
 (Prob. I.). 
 
THEOEEMS I., 11. 5 
 
 (1) If P is a point of a parabola DP= PF and 
 
 PDF=PFD = DFE; 
 
 hence EF will be parallel to DGr, and there is no second 
 point. As F is without the directrix, a point P may be 
 found on any perpendicular to BD such that PD = PF by 
 erecting a perpendicular to the middle of DF; hence every 
 perpendicular to the directrix intersects a parabola. 
 
 (2) If P is a point of an eUipse DP > PF; hence 
 
 PDF< PFD = DFE, 
 
 and EF is not parallel to DG, but will intersect it in the 
 direction DO. If the line DQ be moved farther frorn F, 
 the angle PDF will increase; and as PFD > PDF, their 
 sum will finally exceed two riglit angles, when PF would 
 meet DG on the opposite side from DF, in which case PD 
 could not be greater than PF; hence a perpendicular to the 
 directrix may pass above or below the ellipse. 
 
 (3) If P is a point of a hyperbola DP < PF ; hence 
 
 PDF > PFD = DFE, 
 
 and FE must meet DG on the opposite side of the directrix 
 from the focus. 
 
 Since PFD < PDF, the triangle DPF is always possible, 
 and there will always be two points in which a line perpen- 
 dicular to the directrix intersects a hyperbola; but these 
 points are on opposite sides of the directrix. 
 
 Theorem II. 
 
 All the conic sections are symmetrical with regard to a 
 line drawn from the focus perpendicular to the directrix. 
 
 From the focus' F draw FA perpendicular to the directrix 
 DB ; then FA is an axis of symmetry. For draw BP per- 
 
b ELEMENTS OF THE CONIC SECTIONS. 
 
 pendiculai' to BD, and intersecting the curve in P. Take 
 AD = AB and draw DE perpendicular to BD. Revolve 
 
 the upper portion of the figure about AF as an axis, and BP 
 will coincide with DE, P coming at E, such a point that 
 
 DE:EF=BP:PF, 
 
 or at'a point of the curve. If BP intersects the curve in 
 two points, the second point would come on such a point of 
 DE as answers the definition of a conic section. 
 
 Dbfinitions. The perpendicular from the focus upon the 
 directrix is called tho priiicipal axis of the conic section, and 
 the point in which it intersects the curve is the vertex. 
 
 A distance measured from the vertex, or any specified 
 point, along the principal axis is called an abscissa; e.g., 
 CG, Fig. Prob. II. A line perpendicular to the principal 
 axis and meeting the conic is an ordinate; e.g.,.PG. 
 
 A tangent to a conic section is a straight line that meets 
 the curve in one point without intersecting it. If a line which 
 intersects a conic in two points be revolved about one of 
 these points till the other coincides with it, tiie line will then 
 be a tangent. 
 
 A subtangent is that portion of the principal axis included 
 between the point in which a tangent meets the axis and the 
 foot of an ordinate drawn from the point of tangency ; e.g., 
 EG, Fig. Prob. II. A normal is a line drawn perpendicular 
 to the tangent from the point of tangency ; e.g., P.K. 
 
THEOKEM II. — PROBLEM II. 7 
 
 A subnormal is that portion of the principal axis included 
 between the point in wliicb, the normal intersects the axis 
 and the foot of the ordinate from the point of tangency ; 
 e.g., GK. 
 
 A parameter is the double ordinate through the focus. 
 The parameter of a parabola equals twice the distance from 
 the focus to the directrix. 
 
 A line drawn from the focus to a point of a conic section 
 is called a focal radius. 
 
 Peoblbm II. 
 
 To draw a tangent to a parabola from a given point on 
 the parabola. 
 
 Let P be the given point. Join P and the focus F. From 
 F draw FH perpendicular to FP and join P and JI. The 
 line PH will have its two points of intersection with the para.- 
 bola coincident (Prob. I., Cor. II.) ; hence it -is a tangent. 
 
 CoEOLLARY I. The right triangles DPH and HPF have 
 DP = PF and HP common ; hence thej- are equal, and the 
 angle DPH equals HPF; or the tangent at any point of a 
 parabola bisects the angle between a line perpendicular to 
 the directrix and a line to the focus, both being drawn from 
 the point of tangency. 
 
8 
 
 ELEMENTS OF THE CONIC SBCXIONS. 
 
 Corollary II. Siuce PEF=EPD = EPF, PF=FE; 
 
 or, the distance from the intersection of a tangent to a 
 parabola with the principal axis, to the focus, equals the 
 distance from the focus to the point of tangency. 
 
 Corollary III. Since AG = DP = FP= FE, if we take 
 AC= CF from the first and last of these equals, CG = EG; 
 or, the subtangent is bisected at the vertex. 
 
 Corollary IV. Since nH is a tangent, tangents at the 
 extremities of a chord through the focus meet on the 
 directrix. 
 
 Theorem III. 
 
 The subnormal of a parabola is constant and equal to the. 
 distance from the focus to the directrix. 
 
 D 
 
 
 P 
 
 << 
 
 <0 
 
 f 
 
 \ 
 
 ^E A 
 
 
 3 C 
 
 t K ■ 
 
 Let P^be the normal at the point P; then GK=AF. 
 From the. right angles EPK and mPK take EPF and its 
 equal mPL respectively, 
 
 and FPK= KPL = PEF; 
 
 hence KF=PF=AG. 
 
 Taking out the common part FG, GK= AF. 
 
THEOEEM III. 
 
 ■PROBLEM III. 
 
 9 
 
 Corollary I. In the right triangle EPK^ PG =EG. 
 GK= 2 CO ■ AF, or the oi'dinate at any point of a parabola 
 is a mean proportional between the abscissa of the point 
 and the parameter. 
 
 Corollary II. Similarly, 
 
 BB'=2CS-AF, 
 
 and P&:BS'::GC:SC; 
 
 or the squares of any two ordinates are to each other as 
 their abscissas. 
 
 Problem III. 
 
 From a point without a parabola, to draw a tangent to the 
 parabola. 
 
 Let G be the given point. Draw GH perpendicular to 
 AS^ and with (7 as a centre, and CH as a radius, describe a 
 circle. From F draw FK a tangent to this circle, and pro- 
 duce it, if necessary, to D on the directrix. Connect D and 
 C; then DGF is a tangent. For, draw EF perpendicular to 
 DF, and EG perpendicular to AQ, from the similar tri- 
 angles we get 
 
 DG-.BE:: GK: EF::HG: GE. 
 
 But GK= GH; hence EF= GE, and E must be on the 
 parabola ; and DE is a tangent, since DEE is a right angle. 
 
10 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 CoEOLLARY I. Draw Fn tangent to the circle KHn, and 
 the tangent mB to the parabola. 
 
 Since CFn = CFK, if we add these angles to the right 
 angles BFm and DFE, GFB = GFE ; or, a line drawn from 
 the intersection of two tangents to a parabola to the focus 
 makes equal angles with the focal radii drawn to the points 
 of tangency. 
 
 Corollary II. Since DE is the perpendicular bisector 
 oiFG, GF=OG; likewise CF=CL. This will give a 
 method of drawing a tangent to a parabola. Draw a circle 
 from the point C as a centre, with a radius CF; at the points 
 G and L, where it intersects the directrix, erect perpendic- 
 ulars to meet the parabola in E and B : these points will be 
 the points of tangency. 
 
 Definition. A line from a point on a parabola parallel 
 to the principal axis is called a diameter of the parabola. 
 
 Theoeem IV. 
 
 A diameter drawn through the intersection of two tan- 
 gents to a parabola bisects the chord joining the points of 
 contact. 
 
 The line GO parallel to AF drawn from C, the inter- 
 section of the tangents GB and GF, bisects the chord EB. 
 
THEOREMS IV., V. 
 
 11 
 
 Let D be the point in which CE intersects the directrix, 
 the right triangles DFE and DOE being equal, DF= GD. 
 As 'FK\B a tangent to the circle CH (Prob. III.), DK= DH; 
 hence GH — FK. In lilfe manner, HL — Fn. Since 
 Fn = FK, HL = HG, and HO, parallel to GE and LB, 
 bisecting GL, also bisects BE. 
 
 Theorem V. 
 If from a point on the diameter of a parabola extended 
 two tangents be drawn, the chord of contact is parallel to 
 the tangent at the extremitj' of the diameter. 
 
 From A, on the diameter BC, draw the tangents AH and 
 AI, also the tangent LBE and the chord HI; then HI is 
 parallel to LE. 
 
 Draw the chords BH and BI, also EG and LJ parallel to 
 AG. G is the middle of HI, F of BH, and K of BI (Th. 
 IV.) ; hence, On = nP, OD = Dn, mn = mP, and Dn, the 
 half of On, equals mm, the half of 7iP ; then, in the trapezoid 
 DELm, BE = BL. Since ^C in the triangle AHI bisects the 
 base HI, and also the line LE, LE must be parallel to HI.* 
 
 * Let GB be the medial line of the triangle AGC, and let 
 
 DF= FE. If DE is not parallel to A C, take KH through 
 
 ^ i?" parallel to AC; then KF= FH, and the triangles DFK 
 
 f> and HFE will be equal. These being equal, angle DKF 
 
 equals angle FHE, and DA and IIC are parallel, which 
 
 c is impossible since they are the sides of a triangle. 
 
12 ELEMENTS OF THE CONIC SECTIONS. 
 
 Corollary I. All chords joining the points of contact 
 of pairs of tangents intei'secting on the same diameter are 
 parallel, since they are all parallel to tlie tangent at the 
 exti'emity of the diameter. 
 
 Corollary II. A diameter of a parabola bisects all 
 chords parallel to the tangent through its extremity. 
 
 Corollary III. Since 
 
 BE=CG=Gn, 
 and LB = JC=JI, LE=IHI. 
 
 Corollary IV. Since 
 
 BF= HF, AE = EH, and AB=BG; 
 
 or, the vertex of any diameter bisects that part of the diam- 
 eter included between a chord parallel to the tangent at the 
 extremity of the diameter and the intersection with a tan- 
 gent through the extremitj- of the chord. 
 
 Theorem VI. 
 
 The area included between a parabola and a chord is two- 
 thirds of the triangle formed by the chord and the tangents 
 at its extremities. 
 
 The area DIH^ lAIH. The triangle ALE = \AHI, as 
 LE bisects the lines AH and AI; liliewise, any tangent at 
 the extremity of a diameter will cut off one-fourth of the 
 triangle formed by a parallel chord and tangents at its 
 extremities. 
 
 E being the middle of AH, ADE = iADH= EDH, and 
 EnO = iEDH=iADE; likewise, LKm = ^ADL, and the 
 two exterior triangles EiiO and LKm are together 
 
 ^ALE^^AHI. 
 
THEOREM VI. 13 
 
 Continuing this process indefinitely we should find the 
 area included between AH, AT, and the curve equal 
 
 iAHI+j\AHI+ ^^AHI-\- etc. = iAHI; 
 
 hence the area between the chord and the curve is 
 
 AEI- AHDI=%AHI. 
 
 CoROLLAEY. If the tangents intersect on the principal 
 axis produced, the area between the curve and chord of 
 contact is two-thirds the rectangle of the abscissa and 
 double ordinate of the point of tangency. 
 
14 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 THE ELLIPSE. 
 
 Take F as the focus, and AD the directrix of an ellipse ; 
 there will be two points in which the principal axis intersects 
 the ellipse C and E (Th. I.). 
 
 Definitiok. The line CE is called the major axis of the 
 ellipse ; 0, the centre of the axis, the centre of the ellipse. 
 The perpendicular On is the minor semi-axis of the ellipse. 
 
 Theorem VII. 
 
 All points of an ellipse are within a circle drawn on its 
 major axis as a diameter. 
 
 ?n 
 
 » >> 
 
 m/fp-^y 
 
 --v^\ 
 
 /fK 
 
 ^ 
 
 „(i , 
 
 3 J 
 
 Describe a circle on tlie diameter CE. Take P any point 
 on the ellipse ; then P is within the circle. For, by defini- 
 tion, 
 
 AC: CF=AE:EF, 
 
 and AF is divided harmonically, and F is within the circle. 
 We know, by Geometry, that the circle on the diameter CE 
 is the locus of all points whose distances from A and F are 
 in the ratio AC: CF; hence 
 
 AC: CF= AG : GF= Am : PF=KP: PF; 
 
THEOREMS VII., VIII. 
 
 15 
 
 hence Am = KP. But Am> Km, Km being a perpendic- 
 ular ; therefore KP > Km ; or P on the line GF is farther 
 from the directrix than m on the line AO, or P is between 
 F and G ; that is, within the circle. 
 
 Definition. The circle drawn with the major axis of an 
 ellipse as a diameter is called the circumscribed circle of the 
 ellipse. 
 
 Theoebm VIII. 
 
 Half the major axis is a mean proportional between the 
 distances from the centre to the directrix and to the focus. 
 
 "/ 
 
 ^^ 
 
 n N. 
 
 vJf^ 
 
 -^y 
 
 ~~-v^\ 
 
 /(y 
 
 ^^ 
 
 %. 
 
 ( 
 
 3 
 
 CO is a mean proportional between AO and FO. For, 
 by definition, 
 
 AC:OF=AE:EF; 
 by the principles of proportion, 
 
 ' AC:CF::AC+AJE:CF+EF 
 ■.■.AE-AG:EF-GF. 
 But AG + AK = 'iAC+GE=2AC-\-'iC0=^'i,A0\ 
 likewise, CF+EF=2C0, AE-AC=2C0, 
 and EF-CF=2F0; 
 
 hence AC: OF:: AO: CO:: CO: FO. 
 
16 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 CoKOLLABT I. Let n be a point of the ellipse on the per- 
 pendicular through the centre. It is seen by tne above 
 demonstration that 
 
 Dn:nF=AO: CO. 
 
 But Dn = A0\ hence Fn = CO ; or, the distance from the 
 focus to the extremity of the minor axis of an ellipse equals 
 half the major axis. 
 
 CoKOLLAKY II. Join D and F. Since 
 
 Dn : Fn -CO :FO = Fn: FO, 
 
 and DnF=nFO, 
 
 the triangles nDF and nFO are similar, and DFn is a right 
 imgle and Dn is a tangent ; hence the minor axis is the 
 longest ordinate of an ellipse. 
 
 Theorem IX. 
 
 Ordinates of an ellipse equally distant from the centre 
 are equal. 
 
 J 
 p^^ — i 
 
 5 
 
 ■--4 
 
 I 
 
 
 
 "■^^ 
 
 ^ 
 
 cU VB--^^^ 
 
 
 
 >y 
 
 E 
 
 
 
 ^1 
 
 J 
 
 Take 
 then 
 
 Let 
 then 
 
 OB=OB'; 
 PR = HB'. 
 DP:PF=n; 
 
 PF = 
 
 DP_AB_ AO-BO ^AO BQ 
 
THEOREMS IX., X. 17 
 
 By Theorem VIII. 
 
 DP:PF=AO: 00= CO:FO = n; 
 
 hence C0 = ~, and F0= — . 
 
 n n 
 
 .-. PF=00-—, BF=FO-RO = ~-RO. 
 
 n ■ n 
 
 From the right triangle RFF 
 
 PW = PF^ - BF' = foO - — Y - f— - BO 
 
 ^VL^rcO'-BO^). 
 
 Since H is on the ellipse, 
 
 DH:HF=n\ 
 
 „_ BH AR' AO , OR' „^ , OR' 
 
 or, HF= = = 1 = C0-\ ; 
 
 n n n n n 
 
 FR' =:F0+0R' = ~ + OB'. 
 n 
 
 From the right triangle FHB' we get, as before, 
 
 IJB'^ = HF^ - FR'' = ^^^ (CO'-OR'') . 
 
 w 
 
 Comparing this value of HR' with that already found for P2?, 
 we see they are equal. 
 
 Theobem X. 
 
 An ellipse is symmetrical with respect to its minor axis. 
 
 Revolving an ellipse about its minor axis, the student will 
 see that from Theorem IX. it answers the definition of a 
 symmetrical figure. 
 
18 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 Theoeem XI. 
 
 An ordinate of the ellipse is to the corresponding ordinate 
 
 of the circumscribed circle as the minor axis is to the major 
 
 axis. 
 
 PR : ER : : OD : 00. 
 
 From Theorem IX. 
 
 %- ^ 
 From Geometry 
 
 ER' = OR-RB = {CO-RO){00 + RO)=CO'- R0\ 
 and 
 
 Since 
 and 
 
 FD=CO 
 
 FD^n 
 F0~ l' 
 
 or, FD^:W = n^:l, 
 
 we get by division, 
 
 Fif _ FO' : W -.-.n'-l-.n'. 
 
 But W-F&^DO'. 
 
 Substituting equals for equals, 
 
 P^:ER'::D0':W; 
 or, PR : ER : : DO : CO. 
 
THEOKEMS XI., XII. 
 
 19 
 
 Corollary. The squares of ordinates to an ellipse are 
 to each other as the rectangles of the segments into which 
 they divide the major axis. > 
 
 Problem IV. 
 
 I 
 To find a second line and point which may be used as the 
 
 directrix and focus of an ellipse. 
 
 D 
 
 P^^ 
 
 
 
 ~^ff 
 
 
 B 
 
 ^/ 
 
 V 
 
 --' 
 
 \ 
 
 \ \n 
 
 
 A 
 
 1^ 
 
 
 \ 
 
 
 \\ 
 
 \ 
 
 a 
 
 C\F 
 
 L 
 
 y 
 
 K 
 
 F' 
 
 r 
 
 
 n 
 
 y 
 
 
 
 >^ 
 
 
 
 Since the ellipse is symmetrical with respect to its minor 
 axis, if the figure be revolved about KO, AD falling on BG, 
 F on -F", and P on H, these points will have the same rela- 
 tive position as before, and hence BG might be used as a 
 directrix and F' as a focus with which to describe the ellipse. 
 
 Definition. Any line drawn through the centre of an 
 ellipse, and terminated by the ellipse, is called a diameter. 
 
 Theorem XII. 
 Any diameter of an ellipse is bisected at the centre. 
 
 Draw any diameter POm ; then PO = Om. From P draw 
 the perpendicular PL, and extend it to meet the ellipse in n ; 
 
20 
 
 ELEMENTS OP THE CONIC SECTIONS. 
 
 then PL = nL. Revolve the portion at the right about OK, 
 and m will fall at some point of the ellipse on the left of 
 OK, and Om, on On. Since 
 
 m,OE = POL = LOn, 
 
 and since m falls on both the ellipse and the line On, it will 
 fall at n, and 
 
 Om=On= OP- 
 
 Thboeem XIII. 
 
 The sum of the distances from any point on the ellipse to 
 the two foci is constant and equal to the major axis. 
 
 Take P any point on the ellipse ; we are to prove 
 
 PF+ PF' = GE. 
 Let GB be the. second directrix ; then (Prob. IV.) 
 
 DP:PF=BP:PF'; 
 which, by composition, is 
 
 DP+BP: PF+PF' = DP: PF. 
 
 DP + BP=AO at any point, and BP-.PF is constant; 
 hence PF + PF' must also be constant; therefore 
 
 PF+ PF = KF+KF' = 2 GO, 
 
 which was to be proved. 
 
THBOKEM XIII. 
 
 21 
 
 Scholium. The property of the ellipse just proved is the 
 one commonly given as the definition of the curve. It fur- 
 nishes an easy method of constructing the curve. Thus, 
 
 fasten the two ends of a string, longer than FF' at F and 
 F' ; move a pencil within the string, so as to keep the string 
 taut ; the point will describe an ellipse, since at any point, as 
 P, PF+ PF' equals the length of the string. 
 
 Lemma. Two right triangles are similar if the hypothe- 
 nuse and a side of one are proportional to the hj'pothenuse 
 and a side of the other. 
 
 OK B D E 
 
 The two right triangles ABO and DEF, in which 
 AB:AO::FE:DF, 
 
 are similar. For, take BH=EF\ and draw ^/iT parallel to 
 
 AO; then 
 
 AB:AO: : HE : HK. 
 
 Comparing this with the proportion of the hypothesis, we 
 see HK= DF, and the triangles HKB and DFE are equal ; 
 and as HKB is similar to ABO, DFE is also similar to ABO. 
 
22 
 
 ELEMENTS OP THE CONIC SECTIONS. 
 
 Peoblem V. 
 To draw a taugent to an ellipse from a point on the ellipse. 
 
 Let P be the given point, connect it to the focus by the 
 line PF, draw EF perpendicular to PF, meeting the direc- 
 trix in E, join E and P ; then EP is the required tangent : 
 for, as EFP is a right angle, the line EP can meet the 
 ellipse in but one point. 
 
 CoEOLLAEY I. If Fm is perpendicular to CF\ the tan- 
 gent meets the directrix at A on the major axis extended. 
 
 Corollary II. Tangents at the extremities of a focal 
 chord meet on the directrix. 
 
 Scholium. The tangent at P might have been drawn by 
 using the focus F' and the directrix KD ; hence PF'K is a 
 right angle. 
 
 Theoeem XIV. 
 
 A tangent at any point on an ellipse makes equal angles 
 with lines drawn to the two foci. 
 
 Let ^Pbe a tangent to the ellipse; then EPF=EPF. 
 From Problem IV. we learned that 
 
 OP:PF::HP:PF\ 
 
 or OP:HP::PF:PF'. 
 
THEOREM XIV. 
 
 -PKOBLEM VI. 
 
 23 
 
 The similar triangles OPE and HPK give 
 
 GP : HP : : EP : KP ; 
 
 hence PF -. PF' :: EP: KP- 
 
 The right triangles EPF and KPF' are therefore similar 
 by the preceding lemma, and the angles EPF and KPF' are 
 equal. 
 
 Peoblem VI. 
 
 To draw a tangent to an ellipse from a point without the 
 ellipse. 
 
 p 
 
 Let E be the point ; draw ED perpendicular to the direc- 
 trix. Find a line EH, so that 
 
 AG:CF::DE:EH, 
 
 and with'-E/ff as a radius, and ^ as a centre, draw a circle. 
 From F draw FH tangent to this circle, meeting the direc- 
 trix in B. BE extended will be the tangent required. 
 
 For draw PF perpendicular to BF, meeting BE in P; 
 also P^ perpendicular to AD ; then, by similar triangles. 
 
24 ELEMENTS OF THE CONIC SECTIONS. 
 
 BE : BP : : DE : KP : : EH : PF, 
 from which KP: PF: : DE : EH:: AG: GF; 
 hence P must be on the ellipse and a point of tangency. 
 
 CoKOLLAEY. Draw the tangent EG. Since EFm = EFH, 
 adding these to the right angles GFm and HFP, we have 
 the equal angles EFG and EFP ; hence a line drawn from 
 the intersection of two tangents to an ellipse to the focus 
 makes equal angles with the focal radii drawn to the point 
 of tangency. 
 
 Theokbm XV. 
 
 The perpendicular from the focus of an ellipse upon any 
 tangent meets it on the circumscribed circle. 
 
 Draw GF perpendicular to the tangent GP, and produce 
 it to meet F'P in D. Since the angle 
 
 GPF= GPF' = DPG, DP = PF, 
 and DC =CF; 
 
 hence, as GO bisects the sides of the triangle DFF, 
 
 CO =iDF' = i(PF + PF') = iAB, 
 and C is a point of the circle on the diameter AB. 
 
THEOEEMS XV., XVI. 
 
 25 
 
 Corollary I. A line from the centre of an ellipse to tlie 
 point in which a focal perpendicular meets a tangent is par- 
 allel to the focal radius from the other focus to the point of 
 tangency. 
 
 Corollary II. The same line bisects the focal radius of 
 contact: for FC = CD ; hence FE = EP\ and CE, being 
 drawn to the middle of the hypothenuse, equals FE or EP. 
 
 Corollary III. Since CP is the perpendicular bisector 
 of DF and Zff, of FJ, their intersection is equally distant 
 from J), F^ and J. This property of the tangent furnishes 
 a means of drawing a tangent to an ellipse from a point 
 without the ellipse. From the point as a centre, with a 
 radius equal the distance to a focus, draw a circle ;" from the 
 other focus, with a radius equal the major axis, draw a circle. 
 The intersections of these circles joined to the second focus 
 will pass through the points of tangency required. 
 
 Theorem XVI. 
 
 If a tangent and a diameter be drawn from a point on an 
 ellipse, a line from the focus perpendicular to the tangent 
 will meet the diameter on the directrix. 
 
 c K 
 
 Let EP and IP be the tangent and diameter drawn from 
 the point P- Draw EF perpendicular to EP ; then EF will 
 
26 ELEMENTS OP THE CONIC SECTIONS. 
 
 meet IP on the directrix. Eepresent by D' the point in 
 which EF intersects OP. Since E is on the circle BG 
 (Th. XV.), OE is parallel to F'PG, and 
 
 D'0:D'P::E0:OP=PF. (1) 
 
 Let D be the point in whlcli OP intersects the directrix ; 
 
 then AO -.HP •.-.DO: DP. 
 
 
 (2) 
 
 But AO -.BO -.-.HP-.PF; 
 
 
 (Th. VIII.) 
 
 or, AO : HP : : BO : PF. 
 
 
 (3) 
 
 From (2) and (3) 
 
 
 
 DO:DP::BO:PF; 
 
 
 (4) 
 
 and from (1) and (4) 
 
 
 
 DO:DP::D'0:D'P; 
 
 
 
 bence D and D' must be the same 
 
 point. 
 
 
 Corollary. By extending the diameter PI to meet the 
 second directrix in L, it is easily seen that the triangles 
 DFO and LF'O are equal and the lines DF and LF' parallel, 
 and hence that the tangents at P and I, the extremities of a 
 diameter, are parallel. 
 
 Definition. A diameter drawn parallel to the tangent at 
 the extremity of another diameter is called a conjugate 
 diameter. 
 
 Theorem XVII. 
 
 If one diameter is conjugate to a second, the second is 
 conjugate to the first. 
 
 Draw Jn parallel to the tangent EP and produce it to 
 meet the directrix in K; join D, the point in which the diame- 
 ter PL intersects the directrix with the focus F; then DF is 
 perpendicular to EP and Jn. Join S and the second focus 
 
THEOEEMS XVII., XVIII. 
 
 27 
 
 O, producing the line to meet Jn in I. DF is parallel to 
 GS (Cor., Til. XVI.) ; hence 81 is, perpendicular to Jn. 
 
 In the triangle OKS, Oin being perpendicular to KS and 
 SI to OK, KG must be perpendicular to OS. But KG is per- 
 pendicular to JB (Th. XVI.) ; hence the diameter PL is 
 parallel to the tangent at the extremity of Jn. 
 
 CoEOLLARY. A line from the point iu which a diameter 
 intersects the directrix to the focus is perpendicular to the 
 conjugate diameter. 
 
 Theorem XVIII. 
 
 A secant of an ellipse and a secant of its circumscribed 
 circle crossing the curves on the same ordinates meet on the 
 major axis, or the major axis extended. 
 
 EB and FC meet at G on the major axis extended. For, 
 . major axis : minor axis : : CA : BA : : FD : ED ; hence, FC, 
 EB, and DA meet in a point. 
 
28 
 
 ELEMENTS OP THE CONIC SECTIONS. 
 
 CoROLLABY. If the secants be revolved about jEJ and F 
 respectively till B comes to E and C to F, we have the tan- 
 gents HE and HF\ hence a tangent to an ellipse and a 
 tangent to its circumscribed circle and tangent on the same 
 ordinate meet on the major axis extended. 
 
 This principle affords a convenient method for drawing a 
 tangent to an ellipse at a given point on the ellipse by first 
 drawing an ordinate arid a tangent to the circumscribed 
 circle. 
 
 Theorem XIX. 
 
 A diameter of an ellipse drawn to the intersection of two 
 tangents bisects the chord joining the points, of tangency. 
 
 Draw tangents to the ellipse at E and K, and extend the 
 ordinates EI and JK to meet the circumscribed circle in D 
 and L. Draw tangents at D and L, intersecting in A, and 
 from A drop a perpendicular to the major axis, meeting the 
 tangent from E in JS, and the tangent from ^in B'. Since 
 the tangents from D and E meet on the major axis (Th. 
 XVIIl., Cor.), 
 
 DI:EI ■.:AG:BC; 
 
 likewise, the tangents at ^and L meeting on the major axis, 
 
 LJ:KJ::AC:B'0; 
 
 but 
 
 DI : EI : : LJ : EJ. 
 
THEOREMS XIX., XX. 
 
 29 
 
 From these proportions it appears that the tangents from K 
 and E meet on AQ. 
 
 Join AO and BO, and from F, the point in which AO 
 intersects DL, drop a perpendicular to the major axis inter- 
 secting BO in Q ; then 
 
 AC : BG : : FH : OH. 
 
 Let (?' represent the point in which' the perpendicular FH 
 intersects the ciiord KE. As DL and KE intersect on the 
 major axis (Th. XVIII.), 
 
 FH: G'H::DI:EI::AC:BG; 
 
 hence Q and (?'_coincide. As DI, FH, and LJ axe parallel, 
 and F is the middle of DL, G is the middle of KE. 
 
 Theorem XX. 
 Parallel chords are bisected by the same diameter. 
 
 D 
 
 
 A — 
 
 T 
 
 2 
 
 R-,/ 
 
 F 
 
 B 
 
 J iT^^ 
 
 f/n 
 
 1 
 
 ( 
 
 
 
 1 
 
 t 
 
 OfK 1 
 
 I ] 
 
 Draw BE and G/S, two parallel chords of the ellipse ; erect 
 ordinates from their extremities, producing them to meet the 
 circumscribed circle. Join D and L, the extremities of two 
 corresponding ordinates, and through ^ where GS intersects 
 the major axis draw FT parallel to DL, meeting Gn in F, 
 and mS in T. From the similar triangles thus formed 
 
 DI iFn ■.-.DH-.FE, 
 
 EI -.Gn:: EH: GK, 
 
30 ELEMENTS OF THE CONIC SECTIONS. 
 
 and DH: FK: : EH: GK; 
 
 hence DI : Fn :: EI : On, 
 
 or DI -.EI -.-.Fn -.On; 
 
 and i?' must be a point of the circumscribed circle (Th. XI.). 
 Let A be the point in which tangents from D and L inter- 
 sect, and B the point in which tangents from E and B 
 intersect. As FT is parallel to DL, tangents at F and T 
 will intersect at some point Xon ^0, and tangents from O 
 and 8 will intersect on a perpendicular to the major axis 
 from X, and the two intersections will divide this perpen- 
 dicular in the ratio DI: EI or AC: BG (Th. XIX.) ; hence 
 the two tangents from G and S intersect on BO, which will 
 bisect G8. 
 
 CoEOLLART I. If the parallel chord be moved toward V, 
 the vertex of the diameter BO, the points of intersection 
 with the ellipse approach each other till at V they coincide, 
 and the chord becomes a tangent ; hence chords parallel to 
 a tangent at the extremity of a diameter are bisected by that 
 diameter. 
 
 Corollary II. A diameter parallel to the tangent is the 
 conjugate to VO ; hence chords parallel to a diameter are 
 bisected by its conjugate diameter. 
 
 Theoeem XXI. 
 
 If two pairs of tangents be drawn from two points equally 
 distant from the centre of an ellipse and on the same diam- 
 eter, they will intersect on the diameter conjugate to that on 
 which the first points are taken. 
 
 Let the points A and H be taken on the diameter KL 
 equally distant from ; the intersection E of the tangents 
 AB and HF, and D of AC and HG, are on the conjugate 
 diameter to KL. 
 
THEOREMS XXI., XXII. 
 
 31 
 
 Revolving the figure half way around on the point as a 
 pivot, it is easily shown that DO = EO. But BC is bisected 
 by AO ; hence BC is parallel to DE. BC is also parallel to 
 the tangent at ^(Th. XX., Cor. I.), and DE answers the 
 definition of a conjugate diameter to EL. 
 
 Theorem XXII. 
 
 The distance from the extremity of any diameter to its 
 conjugate measured along a focal radius is constant, and 
 equal to half the major axis. 
 
 Let OE and AB be two conjugate diameters. Since the 
 focal radius at (7 is parallel to OD, 0C= OD. 
 
 Definition. Supplemental chords in an ellipse are two 
 chords passing through the opposite extremities of a diameter 
 and intersecting on the ellipse. 
 
32 
 
 ELEMENTS OP IHE CONIC SECTIONS. 
 
 Theorem XXIII. 
 
 If a diameter bisects one of two supplemental chords, its 
 conjugate will bisect the other. 
 
 Let AP and PB be two supplemental chords ; draw tan- 
 gents to the ellipse at A, P, and B. From C, the intersection 
 of two of these tangents, draw CO ; it will bisect AP (Th. 
 XIX.). From Z>, the other intersection of the tangents, 
 draw DO; it will bisect PB. The tangents AC and BD 
 at the extremities of the diameter AB being parallel, and 
 AO = OB, EO = OD. CO bisecting AP and ED, these lines 
 are parallel. But AP is parallel to the tangent at Q ; hence 
 HO is the conjugate to CO. 
 
 Corollary. If a chord is parallel to a diameter, its sup- 
 plemental chord is parallel to the conjugat^e diameter. 
 
 Theorem XXIV. 
 
 The abscissa of the extremity' of any diameter of an ellipse 
 is equal to the ordinate of the circumscribed circle passing 
 through the extremity of the conjugate diameter. 
 
 arc 
 
THEOREMS XXIII.-XXV. 
 
 33 
 
 'Let OP and OG be conjugate diameters ; then OB equals 
 EH. Apply DBG to HEO, so that BC comes on EO, 
 placing at ; since PG is parallel to GO, P will fall on 
 GO ; and, as 
 
 DB : PB : : HE : GE, 
 
 D will fall on OH, making the angle DGO equal HOE, 
 and hence DOB equal OHE. The triangle DOB will equal 
 HOE, and BO = EH. 
 
 Theorem XXV. 
 
 If a normal be drawn intersecting the major axis of an 
 ellipse, and the diameter conjugate to that through the point 
 of tangency, the product of its two segments is constant and 
 equals the square of the semi-minor axis. 
 
 To prove PL ■ PK = square of semi-minor axis. It is 
 easily proven that EG equals the semi-minor axis ; and, by 
 Geometry, 
 
 F&^FG- FI= FG ■ F'D. 
 
 Since PH is the bisector of the exterior angle, and PL 
 of the interior angle of the triangle F'PF, ^ 
 
 HF:HF'::LF:LF'. 
 By a double composition, 
 HF: HF+LF: : HF+ HF' : HF+ LF+ HF' + LF', 
 
34 ELEMENTS OF THE CONIC SECTIONS. 
 
 or HF :HL::HO: HF>. 
 
 Because of the parallels F'D, OF, PL, and FG, 
 HF : HL : : FG : PL, 
 and HO : HF' : : OF : F'D ; 
 
 hence FG : PL ■.-.OF: F'D, 
 
 or PL ■ OF = PL- PK= FG ■ F'D = FC\ 
 
 Theorem XXVI. 
 
 The area of the parallelogram formed by the tangents at 
 the extremities of two conjugate diameters of ah ellipse is 
 constant, and equals the rectangle on the axes. 
 
 Let PO be the normal at P ; then the triangles PCH and 
 EOD, having their sides respectively perpendicular, are 
 similar, and 
 
 PC: OF:: PH: OD : : PH: KH: : On : Om, 
 
 multiplying the first and last ratios by PB and On, respec- 
 tively, 
 
 PG-PB:OE-PB:: On : On ■ Om. 
 
 But PC-PB=On; 
 
 hence PB-OE = On- Om. 
 
 (Th. XXV.) 
 
 The first member represents one-fourth the area of the 
 parallelogram FG, and the second one-fourth the rectangle 
 on the axes. 
 
THBOKBMS XXVI.-XXVIII. 
 
 35 
 
 Theorem XXVII. 
 
 A circle is an ellipse in which the foci are at the centre, 
 the directrix is infinitely distant, and the minor axis equals 
 the major axis. 
 
 E 
 
 Draw the ellipse BEG. If B, F, and are given, A may 
 be found by dividing BO in accordance with the proportion 
 of the definition. 
 
 BF:BA::GF:CA. 
 
 Take on the indefinite line OK, 011= OF, and HG-- 
 draw BG, and HA parallel to BG ; then 
 
 HG:BA::HC:AO, 
 
 ■■BF; 
 
 and A is thus fixed. Suppose F to move toward ; the 
 points H and G preserving the relations above assumed will 
 move toward 0, and A must thus move from B. When F 
 reaches 0, G will be at 0, and AH parallel to BO, and the 
 directrix infinitely distant. The foci being at the centre, 
 the distance from any point to either is the same, or half the 
 major axis ; consequently, the ellipse has become the circle on 
 the major axis, and the minor axis is evidently equal the 
 major axis. 
 
 Theoeem XXVIII. 
 
 The area of an ellipse is to the area of the, circle formed 
 on its major axis as the minor axis is to the major axis. 
 
 Inscribe a regular -polygou ^KGHT, etc., in the circle, and 
 from its vertices drop ordinates. Connect the points DEF, 
 
36 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 etc., in which these intersect the ellipse. The trapezoids 
 ABHG and ABED having the same altitude, 
 
 But 
 and 
 or, 
 
 ABED : ABHG :: AD + EB: AG + BH. 
 AD:AG::BE:BH::LO:KO, (Th. XI.) 
 
 AD + BE:AG + BH::LO:KO; 
 ABED : ABGH: -.LO : KO. 
 
 Likewise it may be shown that each trapezoid in the semi- 
 ellipse is to the corresponding trapezoid in the semicircle 
 as LO : KO. Taking the sum of the antecedents and the 
 sum of the consequences in the continued proportion thus 
 formed, we find the polygon formed in the ellipse is to the 
 polygon formed in the circle as the minor axis is to the 
 major axis. If the number of sides of the polygon inscribed 
 in the circle be increased indefinitely, the polygon approaches 
 the circle as its limit ; likewise the polygon inscribed in the 
 ellipse will approach the ellipse as its limit : hence the area 
 of the ellipse is to the area of the circle formed on its major 
 axis as the minor axis is to the major axis. 
 
 CoEOLLART. Writing the theorem as a proportion, 
 area of the ellipse : ttKO :: LO: KO ; 
 hence area of the ellipse = irKO-LO. 
 
THEOEEM XXVIII. 
 
 37 
 
 Scholium. If from A two tangents AK and AL be drawn 
 and circles for the construction of these tangents be made, 
 AEH with reference to F and BC, and AE'H' with reference 
 to F' and B'C, the difference of their radii is equal the major 
 axis. 
 
 For 
 and 
 
 As 
 
 AE:An = AE' : Am' = BB' : mn\ 
 AF' - AE : Am' -An:: BB' : mn', 
 AE' -AE:nm':: BB' : mn'. 
 nm'= mn', AE' — AE = BB'. 
 
 From F and F' as centres with radii equal to the major 
 axis draw circles F'DO and FD'G'. Tangents at D and G, 
 D' and O' pass through C, B, C", and B', the points in which 
 tangents to the ellipse intersect the directrices. Each of 
 these tangents is a tangent to one of the circles used in con- 
 
38 ELEMENTS OP THE CONIC SECTIONS. 
 
 structing the tangents to the ellipse. Angle DEC = CKF; 
 hence 
 
 KCD = KCF = EC A = ACL 
 
 Draw -<d J perpendicular to DC; the triangles ACI &n&ACE 
 being equal, AE = AI, and / is a point of tangency on the 
 circle AEH. In the same way it might be shown that GBJ, 
 C'D'Y, and B'O'J' are common tangents. 
 
 Since AE and D'F are both perpendicular to EF, and D' 
 is a point of tangency, AE comes to Y, the point of tangency. 
 Likewise, AJ, AI, and AH come to points of tangency on 
 the circle AE'H'. 
 
THEOREMS XXIX., XXX. 
 
 39 
 
 THE HYPERBOLA. 
 
 Take F as the focus, and AD the directrix, of a hyperbola ; 
 there will be two points on the principal axis, such that 
 
 AG: CF::AE:EF::DP:PF. 
 
 CE is called the transverse axis; 0, its centre, the centre, 
 and any line drawn through and terminated by the hyper- 
 bola, a diameter, of the hyperbola. 
 
 Theoeem XXIX. 
 
 Half the transverse axis of a h3'perbola is a mean propor- 
 tional between the distances from the centre to the focus and 
 to the directrix. 
 
 By definition, 
 
 AC: CF::AE:EF. 
 Following the same line of reasoning as in Theorem VIII., 
 the student can prove 
 
 AO:CO::CO: OF:: AC: OF. 
 
 Theorem XXX. 
 
 Ordinates to the hyperbola equally distant from the centre 
 are equal. 
 
 Take OB = OR' ; 
 
 then PR = HR'. 
 
 Let DP:PF = n. 
 
40 ELEMENTS OF THE CONIC SECTIONS. 
 
 ^ 
 
 
 K 
 
 D P/ 
 
 V 
 
 
 
 
 B' IE 
 
 A 
 
 cIfV B 
 
 As in Theorem IX., the sides PF and BF may be found 
 equal respectively to 
 
 ^0 ^^ , „^ CO 
 CO and BO 1 
 
 n n 
 
 and from these values the perpendicular PB of the right 
 triangle FPB, or 
 
 PB? 
 
 {GO -BO'). 
 
 In the right triangle HB'F 
 
 HF=-^ + CO, B'F=B'0 + — , 
 
 n n 
 
 and 
 
 EB'^ = ^?^ {Cd - BO^) ; 
 
 hence PB and HB', being equal to the same thing, are equal 
 to each other. 
 
 Theoeem XXXI. 
 
 A hyperbola is symmetrical with respect to a perpendicular 
 drawn through the middle of Its transverse axis. Proved by 
 revolving half the figure about such a perpendicular. 
 
THEOEEMS XXXI.-XXXIII. 
 
 41 
 
 Problem VII. 
 To find a second directrix and focus to a hyperbola. 
 
 Since the hyperbola is symmetrical with respect to OK-, if 
 the figure be revolved on OK, DA will fall on some line 
 OB, F at-F', and P on H; these parts having the same rela- 
 tive position as before the revolution, BQ and F' might be 
 used as the directrix and focus of the hyperbola. 
 
 Theokem XXXII. 
 
 Any diameter of a hyperbola is bisected at the centre. 
 Proof similar to Theorem XII. 
 
 Theorem XXXIII. 
 
 The difference of the distances from any point of a hyper- 
 bola to the two foci is constant, and equal to tlie transverse 
 axis. 
 
 Let P be any point of a hyperbola whose foci are F and F', 
 and whose directrices are GB and AD (see Fig. above) ; 
 
 then - PF'-PF=EG. 
 
 For BP:PF'::DP:PF; 
 
 and, by division, 
 
 BP-DP: PF' - PF: : DP: PF. 
 
42 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 DP : PF being the same for all points of the hyperbola, 
 and, as BP — DP = BD, PF' - PF must be of the same 
 value wherever P is situated ; hence 
 
 PF< -PF= CF' -CF= EC. 
 
 Scholium. This property is commonly given as the defi- 
 nition of a h3-perbola. It gives a simple method of con- 
 structing a hyperbola. 
 
 Fasten a ruler so that it may revolve about a point F', to 
 the other end attach a string shorter than the ruler, and 
 fasten its end at F. Revolve the ruler, holding the string 
 tight by a pencil-point P against the ruler ; the pencil will 
 describe one branch of a hj-perbola. By fastening the ruler 
 at F, and the string at JF", the other branch can be made. 
 
 Problem VIII. 
 
 To draw a tangent to a hyperbola from a point on the 
 hyperbola. Let P be the given point. Join P and F; draw 
 ^i?" perpendicular to PF; then the line ^Pis the tangent. 
 
PKOBLEM VIII. — THEOREM XXXV. 
 
 43 
 
 Theoeem XXXIV. 
 
 A tangent at any point of a hj-perbola bisects the angle 
 between two lines drawn from ttie point to the two foci. 
 
 Prove the triangles EPF and KPF' of the last figure 
 similar, as in Theorem XIV. 
 
 Peoblbm IX. 
 
 To draw a tangent to the hyperbola from a point without 
 the hyperbola. Let E be, the point, draw Em perpendicular 
 to the directrix, and with ^ as a centre and a radius EB such 
 that EB -.Em:: HE: AH, 
 
 construct a circle. . 
 
 The entire construction and demonstration are the same 
 as for the corresponding problem in the ellipse. 
 
 Theoeem XXXV. 
 
 The perpendicular from the focus upon any tangent meets 
 it on the circle drawn on the transverse axis as a diameter. 
 
 Draw FH perpendicular to the tangont PD ; 
 
44 ELEMENTS OP THE CONIC SECTIONS. 
 
 theii PE = PF, 
 
 and EH=HF; 
 
 hence 0H= ^F'E = ^ CA. 
 
 Corollary. The radius drawn to the point in which a 
 tangent intersects the circle on the transverse axis is parallel 
 to the focal radius of contact. 
 
 Theorem XXXVI. 
 
 If a tangent to a hyperbola be drawn from a point on the 
 transverse axis, half the transverse axis will be a mean pro- 
 portional between the distance from the centre to the point, 
 and the abscissa of the point of tangency.' 
 
 From B draw the tangent BP ; then OG is a mean propor- 
 tional between OB and OA. Since OQ is parallel to PF 
 (Th. XXXV., Cor.), 
 
 OB iBF : lOG : PF: : 00: PF. 
 By Th. XXIX. 
 
 OH: OG ::HA: PF, 
 or OH:HA::O.C :PF; 
 
 hence OB : BF :: OH : HA, 
 
 Taking this by composition, 
 
 OF: OB::OA: OH; 
 and hence OB x 0A= OFx '0H= 0C\ (Th. XXIX.) 
 which was to be proved. 
 
THEOREMS XXXVl-XXXVm. 
 
 45 
 
 Theoeem XXXVII. 
 
 If from a point on the transverse axis, a tangent be drawn 
 to a hyperbola and an ordinate to the circle on the trans- 
 verse axis, the tangent to the circle at the extremity of this 
 ordinate will pass through the foot of the ordinate to the 
 point of tangency on the hyperbola. 
 
 Draw the tangent BP to the hyperbola and the ordinates 
 BE and PA. Join E and A ; then, by the previous theorem, 
 
 OB : OE : : OE : OA, 
 
 and the triangles OEB and OEA are similar, and OEA is 
 a right angle making EA a tangent to the circle. 
 
 Theorem XXXVIII. 
 If a circle be drawn on the transverse axis of a hyperbola 
 as a diameter, a line connecting its point of intersection with 
 the directrix to the focus is a tangent to the circle. 
 
 Let E be the point in which the circle B'EB intersects the 
 
46 
 
 ELEMENTS OF THE CONIC SECTIONS. 
 
 directrix AU ; then EF is a tangent to the circle. For OB 
 being a mean proportional between AO and OF, 
 
 OF: OB:: OB : OA, 
 
 or OF:OE::OE:OA, 
 
 and the triangles OFF and OEA must be similar, and, as 
 OAE is a right angle, OEF is also a right angle, or EF is 
 a tangent at E. 
 
 Theorem XXXIX. 
 
 If a circle be drawn on the transverse axis of a hyperbola 
 as a diameter, and another be drawn with the transverse axis 
 as a radius, and one focus at its centre, the common tangent 
 to these two circles will pass through the other focus. 
 
 Draw F^ parallel to OF. Since is the middle of FF, 
 
 F'K= 2 0E = BB', 
 
 and JT would be on a circle having F as its centre and BB' 
 as its radius ; and, as FE is tangent to the circle B'EB, it is 
 also tangent to th§ circle F'l^- 
 
 Theorem XL. 
 
 The line joining the middle of the transverse axis to the 
 point in which the circle on the transverse axis intersects the 
 
THEOREMS XXXIX., XL. 
 
 47 
 
 directrix is the limit to which tangents to the hj'perbola 
 approach as the point of tangency is carried from the vertex. 
 Tangents to the hyperbola approach the position 0^ as 
 the point of tangency is carried from the vertex B. 
 
 X- 
 
 Let PG be a tangent. Draw PF and PF' ; also the circle 
 FKH with F'K= BB'. Since 
 
 PF'-PF=BB' = Fm, 
 Pm = PF\ and PO, bisecting the angle mPF, is the perpen- 
 dicular bisector of niF. The line mF making with the 
 radius F'm an angle PmF less than a right angle, is within 
 the tangents FK and FH, and will intersect the directrix 
 between F and m, as at D. Suppose the line mF to move 
 toward the tangent KF; the angle P?Bii^ approaches a right 
 angle, or the angle TKF and its equal 7nFP must approach 
 a right angle also ; as these angles increase, the point of 
 intersection of their sides recedes until mF becomes KF, 
 when they become parallel, and at this position the tangent, 
 or perpendicular bisector of mF, becomes the perpendicular 
 bisector of KF or the line OF. Hence OE is the limit of 
 tangents as the point of tangency recedes from B. 
 
 CoROLLAHY. The distance from the centre to the directrix 
 measured along the line OE is equal to the semi-transverse 
 axis. 
 
 Definition. A limit of tangents to a curve is called an 
 asymptote. 
 
48 
 
 ELEMENTS OF THE CONIC SECTIONS, 
 
 CONJUGATE HYPERBOLA. 
 
 At the vertex A of the transverse axis draw a tangent 
 meeting the asymptotes in C and H. Make DO = CA and 
 E0= F0\ and, with E as a. focus, and OD as the semi- 
 transverse axis, draw a hyperbola. The hyperbola so con- 
 structed is called a conjugate hyperbola to the first. 
 
 Theorem XLI. 
 
 The asymptotes to a hj-perbola are also the asymptotes 
 to the conjugate hyperbola. 
 
 \1 
 
 
 B 
 
 < 
 
 if 
 
 V 
 
 
 
 
 V 
 
 f\ 
 
 
 ^ 
 
 ^L 
 
 \ 
 
 
 a! / 
 
 D' 
 
 } 
 
 V 
 
 // 
 
 K 
 
 
 ^ 
 
 Iv 
 
 Let Km, be the directrix of the hyperbola JLd.', and QL of 
 its conjugate. 
 
 As the tangent AG is parallel to Km,, 
 
 OK: Om = OA: OC ; 
 
 but 
 
 OK: OA=OA: OF. 
 
 (Th. XXIX.) 
 
 m being taken on the asymptote OC, Om= OA (Th. XL. 
 
 Cor.) ; hence 
 
 00=OF=OE. 
 
49 
 
 OL being parallel to DC, and L being taken on OC, 
 
 OG : OL : : OD : 00. 
 
 But OG : OD : : OD : OJE ; 
 
 hence OL = OD, 
 
 and the circle with DD', the transverse axis, as its diameter, 
 will intersect the directrix in L, and OL is an asymptote to 
 the hyperbola DD' (Th. XL.). 
 
 CoROLLAET. The diagonals of the rectangle formed by 
 drawing tangents at the vertices of two conjugate hyperbolse 
 are asymptotes to the hyperbolfe and equal the distance 
 between the foci of either hyperbola. 
 
 Theorem XLII. 
 
 The distance from the centre of a hyperbola to the direc- 
 trix equals the distance from the focus to the directrix of its 
 conjugate hyperbola. 
 
 % — 4 ^>7iO 
 
 Let F be the focus, and AC the directrix of a hyperbola, 
 E the focus and BD the directrix of its conjugate. Then 
 OA equals BE. 
 
 Let G be the point in which DB intersects a circle drawn 
 with OE as a diameter. As OG is a mean proportional 
 between OB and OE, it equals the semi-transverse axis 
 (Th. XXIX.), and hence OG is the asymptote to the hyper- 
 bola with E as its focus. Let H be the point in which AC- 
 intersects a circle drawn with FO as a diameter ; then, as 
 before, OH is the asymptote to the hyperbola with F as its 
 
60 
 
 ELEMENTS OP THE CONIC SECTIONS. 
 
 focus, and OH and OQ are the same line (Th. XLI.) . The 
 angle HOE intercepting the arcs HO and OE of. the equal 
 circles FHO and EGO, these arcs are equal, and 0A= BE. 
 
 CoEOLLABY. Since FOE and KBE are isosceles right- 
 angled triangles, FE will pass through K; or, the line joining 
 the foci of two conjugate hyperbola passes through the in- 
 tersection of their directrices. 
 
 Theobem XLIII. 
 
 A line from the focus of a hyperbola perpendicular to a 
 tangent meets the diameter through the point of tangeucy on 
 the directrix. 
 
 Draw FE from the focus perpendicular to the tangent PE, 
 and let D' represent the point in which it intersects the diam- 
 eter through OP. As OE is parallel to F'P (Th. XXXV., 
 Cor.),D'0:D'P::0E:OP=PF. 
 
 Let D represent the point in which the diameter intersects 
 the directrix ; 
 
 then A0:HP::D0 -.DP. 
 
 But AO : BO : : HP : PF, 
 
 or AO:HP::BO:PF::DO:DP; 
 
 hence DO: DP:: D'O: D'P; 
 
 and D and D' are the same point. 
 
THEOREMS XLIII.-XLV. 
 
 61 
 
 Theorem XLIV. 
 
 A perpendicular from the focus to a diameter of a hyper- 
 ■bola intersects the directrix on a line through the centre and 
 parallel to the tangent at the extremity of the diameter. • 
 
 Draw OB parallel to the tangent EP, join B and F, let 
 D be the point in which the diameter OP intersects the 
 directrix ; then DF is perpendicular to EP or OB, and BA 
 being perpendicular to OF, OP, through their intersection, 
 must be perpendicular to the third side, BF, of the triangle. 
 
 As an asymptote is the limit of the tangents to a hyper- 
 bola and passes through the centre, a line through the centre 
 and parallel to a tangent will not Intersect the hyperbola, 
 but will meet its conjugate. 
 
 Definition. A diameter of a hyperbola parallel to a tan- 
 gent to its conjugate hyperbola is called a conjugate diameter 
 to- the one passing through the point of tangency. 
 
 Theorem XLV. 
 
 If one diameter is conjugate to a second, the second is 
 conjugate to the first. 
 
 Draw OB parallel to the tangent at K. OB is the diameter 
 conjugate to OK; then OK will be the conjugate diameter 
 to OB. Draw FH perpendicular to OK; irwill intersect 
 OB at G on the directrix (Th. XLIV.) . Draw GL perpendic- 
 
52 
 
 ELEMENTS OP THE CONIC SECTIONS. 
 
 ular to OB, and let it intersect OK in m. On OF and OG 
 as diameters construct semicircles and place the semicircle 
 OG with the lines it contains on the semicircle OF, on F, 
 
 and G on 0. As the arcs On and FH measure equal 
 angles, they will coincide, and the line On will fall on FH, 
 m coming somewhere on FC; likewise, GL will come on OL, 
 and hence m must come at C. The perpendicular niE will 
 coincide with CA, and OF = FA ; hence mE is the directrix 
 of the hyperbola whose focus is G (Th. XLII.). As GL 
 perpendicular to OB intersects OK on the directrix, OK is 
 the conjugate diameter to OB. (Th. XLIV.) 
 
 Theorem XL VI. 
 
 The distance from the extremity of any diameter to its 
 conjugate, measured along the focal radius, is constant, and 
 equal to the transverse semi-axis. 
 
 Draw OA parallel to the tangent PB ; then PA equals the 
 
THEOREMS XLVI., XLVI 53 
 
 transverse semi-axis. Let B be the point in which the tan- 
 gent PB intersects the circle on the transverse axis. As OB 
 is parallel to FP, OB equals AP. 
 
 Theohem XL VII. 
 
 Any plane section of a cone of revolution is a conic section. 
 
 Let the plane Om intersect the cone AnC. Inscribe in 
 the cone a sphere OD tangent to the plane Gm at D ; and 
 through the circle of contact of the sphere and cone pass 
 the plane LF, intersecting Om in GH; then will GH be 
 the directrix and D the focus of the conic section in which 
 Gm intersects the cone. 
 
 i /A ,G 
 
 Take AC any element, and let it intersect mG in P; draw 
 PH and DF perpendicular to GH, denoting by E the point 
 in which DF intersects the surface of the cone ; also draw 
 PK and EI perpendicular to LF, and intersecting it in iiT 
 and I respectively. Join IF and KH. From the similar 
 triangles lEF and KPH 
 
 EF:PH::EI:KP. (1) 
 
54 ELEMENTS OF THE CONIC SECTIONS. 
 
 Join ER and SI. Since KP is parallel to AB, 
 
 KPR = BAG\ 
 
 similarly, lES =BAE. 
 
 But all elements make equal angles with the axis ; hence 
 
 BPK=IES, 
 
 and the right triangles RKP and SIE are similar, and 
 
 SE:BP::EI:KP. (2) 
 
 Comparing (1) and (2), 
 
 SE:RP::EF:PH. 
 
 _ All tangents drawn from a point without a sphere to the 
 sphere are equal ; hence RP= DP, and ES = ED, and 
 
 ED : DP : : EF : PH, 
 
 or ED:EF::DP:PH, 
 
 and P answers the definition of a point on a conic section 
 whose focus is D and directrix GHi 
 
 CoEOLLAET. If the plane Om is parallel to an element 
 An, 
 
 angleSFE = AWS = AS W= ESF, 
 
 and EF=ES = ED, 
 
 and the curve is a parabola. 
 
 If Gm intersects all the elements below the vertex, then 
 
 SEE <AWS = ASW= FSE, 
 
 and FE>ES = ED, 
 
 •and the curve is an ellipse. 
 
 If Gm intersects some elements above the vertex, 
 
 SEE > A WS = AS W= FSE, 
 
 and FE<SE=DE, 
 
 and the curve is a hyperbola. 
 
EXERCISES. 55 
 
 Exercises. 
 
 1. In a parabola, the perpendicular dropped from the 
 focus on anj' tangent to the curve meets it on the tangent 
 through the vertex. 
 
 2. Given the directrix and two points of a parabola, to 
 find the focus. 
 
 3. Given the focus and axis of a parabola, to draw the 
 curve so that it shall touch a given straight line. 
 
 4. "The perpendicular distance from the focus of a para- 
 bola to any tangent to the curve is a mean proportional 
 between the distance from the focus to the vertex, and the 
 distance from the focus to the point of contact. 
 
 5. To find the locus of the centre of a circle that passes 
 through a given point and touches a given straight line. 
 
 6. The ordinate of a point on a parabola is 8, its abscissa 
 6 ; find the area included between these and the curve. 
 
 7. To draw a tangent to a parabola parallel to a given 
 straight line. 
 
 8. Tangents at the extremities of a focal chord intersect 
 at right angles on the directrix. 
 
 9. The tangent at any point of a parabola meets the 
 directrix and latus rectum produced in points equally distant 
 from the focus. 
 
 10. If the diameter PE meet the directrix in jB, and the 
 chord drawn through the focus parallel to the tangent at its 
 vertex in E, prove PB = PE. 
 
 11. The distance from the focus to the directrix of a 
 parabola is 4. Find the ordinate whose abscissa is 9. 
 
56 ELEMENTS OF THE CONIC SECTIONS. 
 
 12. The double ordinate through the focus of a parabola 
 is 12 ; a tangent is drawn to the parabola at a point wliose 
 double ordinate is 24. Find how far from the focus the 
 tangent intersects the principal axis. 
 
 13. Find the area of the triangle formed bj' the tangent 
 in the above example, the normal from the same point, and 
 the principal axis. 
 
 14. The area of a parabolic segment cut off by a double 
 ordinate is 96, and the corresponding abscissa is 6. Find 
 the distance from the focus to the vertex. 
 
 15. Find the locus of the centre of a circle that touches a 
 given straight line and a given circle. 
 
 16. The distance from the vertex of the major axis of an 
 ellipse to the directrix is 4, and to the focus is 3. Find the 
 major axis. 
 
 17. The major axis of an ellipse is 40, the distance from 
 the focus to the directrix 9. Find the distance from the 
 focus to the centre. 
 
 18. The major axis of an ellipse is 10, and the distance 
 between the foci 8. What is the distance from the centre 
 to the directrix ? What is the length of the minor axis ? 
 
 19. In the same ellipse the distance of a point from the 
 focus is 3. What is its distance from the directrix? 
 
 20. The ordinate of the circumscribed circle passing 
 through the focus is equal to the minor axis of the ellipse. 
 
 21. Lines drawn from the extremities of a diameter of an 
 ellipse to its two foci form a parallelogram. 
 
 22. The focal perpendiculars upon any tangent of an 
 ellipse are proportional to the adjacent focal radii of contact. 
 
 23. The normal of an ellipse cuts the distance between 
 the foci into segments proportional to the adjacent focal 
 radii. 
 
"EXEECISES. :'57 
 
 24. The major axis of an ellipse is 12, the distance 
 between the foci 8. What is the length of a diameter 
 drawn so that one focal radius to its extremity is 5 ? 
 
 25. The minor axis of an ellipse is 6, the sum of two 
 focal radii to a point on the curve is 16. What is the major 
 axis? the distance between the foci? and the area? 
 
 26. The two segments of the tangents to an ellipse 
 formed by the circumscribed circle are 15 and 12, the focal 
 perpendicular upon the shorter segment is 5. Find the 
 major axis. 
 
 27. In an ellipse whose major axis is 20, a tangent is 
 drawn at a point whose distance from one focus is 12; Find 
 the ratio of the two segments of the tangent formed by the 
 directrices and the point of tangency. 
 
 28. If the distance from the centre to the focus in the 
 above example is 8, find the length of the segments of the 
 tangent. 
 
 29. If an ordinate be drawn to an ellipse and its circum- 
 scribed circle, a parallel to the major axis through the ex- 
 tremity of the ordinate to the ellipse will intersect the radius 
 of the circumscribed circle drawn to the extremity of the 
 ordinate, on a circle drawn with the minor axis as its diam- 
 eter and the centre of the ellipse as its centre. 
 
 30. Lines drawn from the foci bisecting the angles be- 
 tween focal radii drawn to the extremities of a chord meet 
 at the point of intersection of the tangents drawn at the 
 extremities of the chord. 
 
 31. Having given the major axis of an ellipse, the foci, 
 and the direction of a diameter, to find its extremities. 
 
 32. Having given the focus of an ellipse, its centre, and 
 a tangent, to construct the ellipse. 
 
58 ELEMENTS OF THE CONIC SECTIONS. 
 
 33. To find the foci of an ellipse, having given the major 
 axis and a tangent to the curve. 
 
 34. To find the foci of an ellipse, having given its major 
 axis and a point on the curve. ^ 
 
 35. To draw a tangent to an ellipse without using the 
 foci. 
 
 36. If a perpendicular be drawn from a focus of an ellipse 
 to any tangent, it is required to find the locus of its inter- 
 section with a line drawn from the other focus through the 
 point of tangency. 
 
 37. All ellipses described upon a common major axis will 
 have a common subtangent for any given abscissa of contact. 
 
 38. The minor axis of an ellipse is 10, and its area 251 .328. 
 What is its major axis, and the ordinate through the focus? 
 
 39. Lines drawn from the intersection of two tangents to 
 the foci of an ellipse, are respectively perpendicular to the 
 chords joining the intersections of the tangents with the 
 circuniscribed circle. 
 
 40. Focal perpendiculars upon any tangent meet the 
 directrices on the same diameter. 
 
 41. If a tangent be drawn to meet the major axis ex- 
 tended, the semi-major axis will be a mean proportional 
 between the distance from the intersection to the centre, and 
 the abscissa of the point of contact. 
 
 42. If a tangent to an ellipse cuts the two directrices, one 
 point of intersection is on the common tangent to two circles 
 drawn from the other intersection and the opposite focus as 
 centres, and with the major axis as the radius of each. 
 
 43. The semi-axis minor is a mean proportional between 
 the segment of the major axis formed by either focus. 
 
 44. The diagonals of the rectangle formed on the axe's of 
 an ellipse are conjugate diameters and equal. 
 
EXERCISES. 69 
 
 45. In an ellipse whose axes are 8 and 6, what is the 
 length of a diameter that makes an angle of 45° with the 
 major axis ? what is the length of its conjugate ? 
 
 46. In an ellipse whose axes are 60 and 80 a tangent is 
 drawn at a point whose ordinate is 15. Find the length of 
 the tangent between the point of tangency and its intersec- 
 tion with the major axis produced. 
 
 47. Diameters drawn through tlie intersection of any tan- 
 gent of an ellipse with two parallel tangents are conjugate. 
 
 48. A tangent drawn from the point in which a diameter 
 of an ellipse meets the directrix intersects the conjugate 
 diameter on the circumscribed circle. 
 
 49. A tangent is drawn intersecting the two tangents at 
 the extremities of the major axis of an ellipse. Show that 
 the circle described on this as a diameter will pass through 
 the foci. 
 
 50. The transverse axis of a hyperbola is 3, the distance 
 between the foci is 5. Find the distance from the focus to 
 the directrix, and the length of the ordinate through the 
 focus. 
 
 61. Tangents at the extremities of a diameter of a hyper- 
 bola are parallel. 
 
 52. Diameters which make supplemental angles with the 
 transverse axis of a hyperbola are equal. 
 
 53. To draw a tangent to a hyperbola parallel to a given 
 straight line. 
 
 54. The perpendicular from either focus to an asymptote 
 is equal to the semi-axis of the conjugate hyperbola. 
 
 55. If on the portion of any tangent intercepted between 
 the tangents at the vertices, a circle be described, it will pass 
 through the foci of the hyperbola. 
 
60 ELEMENTS OE THE CONIC SECTIONS. 
 
 56. The transverse axis of a hyperbola is 8, the distance 
 from the centre to the focus 6. What is the length of the 
 transverse axis of the conjugate hyperbola? 
 
 57. If an asymptote meets the directrix in B and the tan- 
 gent at the vertex in C, then the line from the vertex to B is 
 parallel to the line from the focus to C. 
 
 58. The distance between the vertices of two conjugate 
 hyperbolae equals the distance from the centre to a focus. 
 
 59. If an ellipse and a hyperbola have the same foci, tan- 
 gents drawn to the two curves at their point of intersection 
 are perpendicular to each other.