UNIVEF?SITY OF CALIFORNIA DEPARTMENT OF CIVIL ENGINEER.Ma BERKELEY. CALIFORNIA CIYIL ENGINEERING U. of C. ASSOCIATION LIBRARY Digitized by tine Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofplaneaOObrigrich UMivTTKai-rv or cAt-inroi'sNiA. OF Plane Analytic Geometry. A TEXT-BOOK INCLUDING NUMEROUS EXAMPLES AND APPLICATIONS, AND ESPECIALLY DESIGNED FOR BEGINNERS. BY GEORGE R. BRIGGS, FORMERLY TUTOR IN MATHEMATICS IN HARVARD UNIVERSITY. SEVENTH EDITION, REVISED AND ENLARGED BY MAXIME BOCHER, ASSISTANT PROFESSOR IN HARVARD UNIVERSITY. UNIVERSITY OF CAUFORNIA DEPARTMENT OF CIVIL ENGINEERING BERKELEY. CALIFORNIA NEW YORK: JOHN WILEY & SONS. London : CHAPMAN & HALL, Limited, 1909 Library Library Copyright, 1881, 1903, BY G. R. BRIGGS. Sdbprt Brumuinnb ani> (Sompattg CIYiL ENGINEERING U. of C. ASSOCIATION LIBRARY PREFACE. The design of this book is to present the elements of Analytic Geometry in a simple form, for the use of beginners. With this object in view, I have tried to make the demon- strations thorough, sometimes at the expense of brevity, in order that the student may never feel that the fundamental truths have been in part assumed. Therefore it may often seem best to the teacher to present special cases of the theorems as prepa- ration for the general demonstrations of the book. The text has been interspersed with a large number of exam- ples and applications to familiar theorems of Geometry, often accompanied with the answers and hints for the solution, to stimulate the interest of the student and to teach him early how to work for himself. For the same purpose the chapter on Loci has been introduced. Although this book has been prepared for the use of the Freshman Class in Harvard College, it is hoped that it may prove useful in Schools and Academies, and to any student 800299 IV PREFACE. having a fair knowledge of Algebra and Trigonometry who wishes to prepare for a more extended course in Analytic Geometry. I gladly acknowledge my indebtedness to Prof. Byerly for his kindness in reading the manuscript and for valuable sugges- tions. I have also freely used the works of Salmon, Todhunter, Puckle, and others. G. R. BRIGGS. Cambridge, April, 1881. The changes in the present edition consist mainly in the addition of a moderate amount of new matter. Almost all of this will be found after page 107, and in particular the last half of Chapter VIII and the whole of Chapter IX are new. It is hoped that in its present form the book will be found to cover the ground of the ordinary first course in analytic geometry as given in American colleges and technical schools, and that it will prove serviceable as a text-book in such courses. Maxime Bocher. Cambridge, January, 1903. CONTENTS. ♦ Chap. Page I. Introduction. — The Point i II. Locus OF AN Equation, Equation of a Curve . 12 III. The Straight Line 26 IV. Transformation of Coordinates 61 V. The Circle 71 VI. Loci 96 VII. The Conic Sections 116 VIII. Diameters. Poles and Polars 157 IX. The General Equation of the Second Degree 175 Miscellaneous Examples 184 ELEMENTS OF ANALYTIC GEOMETRY, CHAPTER I. INTRODUCTION. — THE POINT. 1. The reader is probably familiar with algebraic statements of geometrical theorems which concern the magnitude of lines ; for example, where a, b, and c represent the legs and hypotenuse of a right triangle, ^"^ + i>^ = c^ is a statement in algebraic lan- guage of the Pythagorean theorem. There is a large class of geometrical theorems which concern the position of lines ; for example, a tangent to a circle is per- pendicular to the radius drawn to the point of tangency. Ana- lytic Geometry enables us to apply algebra to this class of problems ; and as we advance with our subject we shall see that this use of algebra shortens our work and generalizes our reasoning. 2. We shall assume that the reader is acquainted with the application of the algebraic signs plus and minus to express the upward and downward directions of vertical lines, the directions from left to right 3.ndfrom right to left of horizontal lines, and, in general, to distinguish between the opposite directions of lines which have the same general direction. The following theorem, a direct result of the distinction between positive and nega- Uve lines, will be useful for reference : — ANALYTIC GEOMETRY. If three points, Ay B^ and C, lie on the same straight line, we may write, when we consider the directions as well as the lengths of lines, AB^ BC--AC, whatever the arrangement of these points on that line, 3. Analogous to the theorem of § 2, there is a principle for the addition of angles which depends upon the distinction be- tween opposite rotations by the signs////j- and minus. We shall place it here for reference. If three lines, O A^ QBy and O C, pass through the same point O, we may write AOB-{- BOC=AOC, whatever the arrangemetit of the three lines which pass through O. 4. In considering problems which concern the position of points and lines in a plane, we must have some fixed objects in the plane to which we may refer the position of each point and line. There are several combinations of fixed objects which arc used for this purpose, the most common being two straight lines at right angles to each other and indefinite in extent. In the figure, X^ X and F' Fare the fixed straight Hnes intersecting at O. The position of P is determined if we know N P^ its distance from V v. and MP, its dis- tance from X' Xy together with the directiotts of these lines; for we know from geometry that P is the only point in the plane at a dis- tance NP to the right of y Fand at a distance MP p Y N P X' M' II P M X III P N' THE POINT. 3 above X^ X. Let P, P\ F", and F^'^ be the corners of a rec- tangle whose middle point is O, and whose sides are parallel to X' X and K' Y. Then recollecting the conventions of § 2, if we represent the length IV F by a and the length M F hy b, the distances of F from Y' F and X' ^ respectively are a and b; of F\ - a and b ; ot /^", - .^ and - b ; of 7^'", ^ and - b. 5. We see that two elements determine the position of each point. These are the straight lines which indicate by their lengths the distances of the point from X' X and Y' Y, and by their signs its directions y^-^w these lines. These elements are called the coordinates of the point which they determine : the one which indicates the distance and direction from K' Y is called the abscissa ; the other, which gives the distance and direction from X^ X, is called the ordinate. In the figure, N P, or its equal both in length and direction, O M, is the abscissa of F, MP is the ordinate of P ; or we may say that the abscissa of P is a, and the ordinate b ; or, more briefly, that the coordinates of P are a and b, the abscissa always being named first. The fixed lines are called axes of coordinates, and we speak of X' X as the axis of x, or of abscissas, while Y' Y is called the axis of j', or of ordinates. O is called the origin of coordi- nates, or simply the origin. In order to abbreviate the terms abscissa and ordinate, it is customary to denote them by the letters x and y respectively ; i. e. for the point F, x — a, y — b ; for P\ x z=z — a, y z= b, etc. 6. If we call the portions of the plane lying in the angles XOY, YOX', X'OY', Y'OX, the Jirst, second, third, and fourth quadrants respectively, we readily see that any point in the first quadrant has x positive and y positive ; any point in the second quadrant has x negative and y positive ; any point in the third quadrant has x negative and y negative ; any point 4 ANALYTIC GEOMETRY. in the fourth quadrah\f has x positive and y negative. The signs of the coordinates of a point, therefore, show in which quadrant the point lies, while the magnitudes of the coordinates fix the position of the point in that quadrant. 7. In the development of our subject we shall have to con- sider points which move in the plane and others which are sta- tionary. The coordinates of a moving point are variables^ for each assumes in turn an indefinite number of values. We shall designate a moving point by P^ its coordinates by x and j, and the foot of its ordinate by M. The fixed points may be either in known positions or un- known. We shall represent known points by P^, P^, P^, etc., their coordinates by (^1, J\), {^^.y-^, (-^sj/s)? ttc, and the feet of their ordinates by M^, M^, J/o, etc., respectively. The points whose positions are unknown, yet determined by the conditions of the problem, we shall call P\ P", etc., their coordinates (x\ y'), (x", y"), etc., and the feet of their ordinates M', M", etc., respectively. The student should be careful to accustom himself to this notation, as it is very important not to confuse the moving points with those which have determined positions. Instead of designating a point P^ as having coordinates x^ = a, ji = b, we often write the point (a, b), always understanding that the abscissa is written first; e. g. the points (i, 2), (3, — 2), are the points P^, with x^ — 1 and y^ = 2, and P^, with x^ = 3 and }'2 = - 2. 8. The system of coordinates described in §§ 4-6 is called rectangular, to distinguish it from another system often employed, in which the axes are not perpendicular to each other, but may make any angle. In the latter system the abscissa of a point is measured parallel to the axis of x, and the ordinate, parallel to the axis of ^. This system is called oblique; and, together with THE POINT. the rectangular system, forms the general class of rectilinear systems in distinction from the po/ar system, which will be men- tioned later. We shall always suppose coordinates rectangular unless other- wise stated. g. It is easy to represent the position of any given point, in a figure drawn to any scale. This is called //^///;/^ or cotistriidiiig the point. Let us plot the points (2, 3), (- 2, 4), (- 3, - 2). The first point, which we will call P^^ is at a dis- tance of two units to the right of O V, and t/iree units adove O X, as indi- cated by its coordinates. We may measure off the first distance from O along OX to M^, which must be the foot of the ordinate of P^. Now the ordinate must be drawn upward parallel to O V, and made equal to three , , ,^^^^^ ^ I 1 1 1 1 1 units of length. Its up- 012345 per extremity is P^, the required position. To plot P^, we lay off O M, = - 2 and M.P^ = 4- To plot P^, we make O Af^ = - ^ and M^P^ = - 2. In a similar way any point may be plotted or constructed. M3 p. Y f' ■ I VI 2 Ml X EXAMPLES. (i.) Plot the points (4, - 3), (6, o), (- 3, o;, (o, 4), (o, - 2), (0,0). (2.) Draw the straight line whose extremities are (— 2, (^, - 3)- and ANALYTIC GEOMETRY. (3.) Draw the triangle, the vertices of which are the points (- 4,-3). (6, i), (4, ii)- 10. We will now seek an expression for the distance between two given points, in terms of the coordinates of these points. Let Py. with coordinates x^ = O J/j, }\ = M^ P^, and P^, with coordinates x,-, ~ O M^, y^ = M^Po, be the given points. Let 8 represent the distance between P^ and P^. Through P^ draw a parallel to O X, and let it meet M^P^ (or this line extended if necessary) in the point R. Now we know that X' o"x[" M, M2 ■^ N"^ j^J-^Ji/^P^PP,^ must be a right tri- angle, and therefore that ^ 2 B'' = P^'+ PP, (i) But P^ P is equal to jW^ M,_, the opposite side of a rectangle, both in length and direction. The points M^, O, and M^ are by construction on the same straight line X' X, and therefore, by § 2, we may write P^A = Af,M, = Jf^O+ 0M,= OMo^- OM^ = X, - x^ .... (2) Again, P, M^^, and P^ are by construction on one straight line, and therefore, by § 2, PP^ = PM, + M,P, = M,P, - M,P = M,P,- M^P^ = y2-yv (3) because M^P and M^P^ 2ire equal, being opposite sides of a rectangle. Substituting in equation (i) the values of P^ R and R P^ found in equations (2) and (3), we have THE POINT. 7 or 8 = V(^2-^i)' + (j2->'iy' ; an equation which answers the requirements of our problem. We may interchange J^^ and J^^ in the value of 8 just found, as by so doing, though we change the signs of the quantities in the parentheses, we do not change their squares. Note. — The student should study the solution of this problem with great care, and satisfy himself that every step in the reason- ing is applicable to all the modified forms of the figure which result from taking F^ and P^ in different quadrants. If he con- siders the theorem stated in § 2 until its full meaning becomes clear, he will be convinced of the generality of the solution of this problem. It is important that he be so convinced, as we have here illustrated in a simple form a method of reasoning which will be often employed hereafter. EXAMPLES.* (i.) What is the length of the straight line which joins (7, - 2) with (4, - 3) .? Here we may write •^1 = 7. 7i = - 2, :^2 = 4, ^2 = - 3- Substituting these values in the expression for 8, we have 8 = V[4-7p + [-3-(-2)r = '^T^WTJpry = V' 10 Ans. (2.) Find the distance between (2, 3) and (- 2, 8). Ans. 'v/zfT- (3.) Find the length of the line given in § 9, Ex. (2). Atis. 2 V^. (4.) Show that the lengths of the sides of the triangle men- tioned in § 9, Ex. (3), are in the same ratio as V26, V29, V65' (5.) Show that the distance of a given point, P^, from the origin is \/ x^ + y^. * In solving examples, the student should construct figures, plotting each point. 8 ANALYTIC GEOMETRY. (6.) Calculate the lengths of the sides of the rectangle whose vertices are the points (a, b), (- a, b), (- a, - d), {a, - b) ; show that the vertices are equidistant from the origin ; prove that the diagonals are equal. II. Let us find the coordinates of a point F' so situated on ^-\e straight line which passes through two given points, F^ and F.,, that its distances from these points respectively are in a 7 ^1 giVen ratio, k = — -. B}^ the conditions of - , , F.F^ m, the problem are F^F' ?n„ where 77t^ and ajty two quafitities and have no connection with the points M^ and M^. Through F^ draw a line parallel to O A^and cutting M^ F^ and M^F^ (or their extensions) in R and S respectively. Since AI^ F^ and M^ F,^ are necessarily parallel, the triangles /^^ i? 7^^ and F.^SF' are always similar, and therefore FF' F,F' S F'~ F,Ff~ n, Now, by § 2, FF' = M^M' = M,0+ OAF OMf - OM. (I) and similarly SF^ = M.JF ^x' -x^. Substituting these values of F F' and S F' in equation (i), A^e have X' - x„ THE POINT. 9 Solving for x', m^ — m^ ' which gives the value of the unknown abscissa of P^ in terms of known quantities. Similarly we may find y ; for P^^S ~ P^P'- m^ ^ ^ Now, by § 2, P^P = P^M^ + M^P= M,R ~ M^P^ = MiP'-M^P^ Again, P^S = M^S-M^P^ = y-yr Substituting in (2), we have y - yx y - y-i m^ m^ m,y, - ^hy^ or v' which expresses y in terms of known quantities. We have now found both x^ and y\ and therefore solved our problem. N. B. By the conditions of the problem the distances of P^ fro77i P^ and P.^ have the ratio k. As both lines are to be meas- ured toivard P\ they will have the sa^ne direction when P' is not between P^ and P^ ; and opposite directions zvhen P' lies betiveen P^ and P^. In the first case the ratio k must ho. positive ; in the second, negative. The student should carefully notice that in the first case the R P^ P /? ratios ^ and ^ of equations (i) and (2) are necessarily positive ; in the second, negative. lO ANALYTIC GEOMETRY. Corollary. A common case of this problem is that in which P^ bisects the line /\ P^. In this case P^P^---P^P\ and the ratio we mav therefore write ^ = ^ = -,j m^ and, with this simplification, the coordinates of P^ become the formulas for bisection. EXAMPLES. (i.) On the straight line through (2, 3) and (8, - 12), find a point such that its distances from the given points respectively shall be in the ratio — \. Here .Ti=.2, J^i = 3, :x:2 = 8, JJ'2 = -I2. m^— — I*, ;;?2 = 2. Substituting these values in the expressions for x^ and y\ we have ,_2x2-(-i)8 4 + 8 (-0 3 = 4 v' - ^ >^ 3 - (- i) (- 12) _ 6-12 _ _ ■^ ~ 2 _ (- .) ~ 3 ~ The required point is, therefore, (4, — 2). Note. This problem may also be stated as follows. Find the point of trisection, nearest (2, 3), of the line which joins (2, 3) with (8, - 12). * When k is negative, it is immaterial which of the quantities nix and 7n„ is made negative. This will be readily seen by noticing that when the signs of both w, and Wo are changed in the formulas for x' and j', the signs of the numerators and denomi- nators are reversed, and therefore the signs of the fractions remained unchanged. THE POINT. II (2.) Construct the points and line given in the previous example, and verify the results by calculating the distances between the given points and the point (4, — 2). (3.) Find the middle point of the line which joins (2, 3) and (10, 7). Ans.\6, 5). (4.) Find that point of the Hne joining (- 4, 5) with (i, 2), whose distances from these points respectively are in the ratio 3 : 2. Ans. (11,- 4)- (5.) A line is dr3.wu/ropi (- 2, — 5) fo (3, 2) ; to what point must it be extended, in the same direction, that its length may be doubled ? to what point that it may be trebled ? Ans. (8, 9); (13, 16). (6 ) Find the middle points of the sides of the triangle men- tioned in § 9, Ex. (3). Ans. (i, - i), (5, 6), (o, 4). (7.) In the same triangle, find the coordinates of the points of trisection (remote from the vertices) of the lines which join the vertices to the middle points of the opposite sides. Ans. (2, 3). (8.) Solve the same problem when the vertices of the triangle are (x^, j\), {x^, 7,) , (^3 y^) ; show that your results prove a well- known theorem of Geometry. (9.) In any right triangle, the straight line drawn from the vertex of the right angle to the middle of the hypotenuse is equal to one half the hypotenuse. Prove this theorem by means of §§ 10, 11. Suggestion. — Let the vertex of the right angle be the origin and the legs of the triangle the axes of coordinates. The ver- tices of the triangle may then be expressed as the points (o, o), {x„ o), {o,y^). (10.) Show that the straight lines joining the middle points of the adjacent sides of any quadrilateral form a second quad- rilateral whose perimeter is equal to the sum of the diagonals of the first. CHAPTER 11. LOCUS OF AN EQUATION. 12. We have seen that the position of a point in a plane is completely determined when we know its abscissa and ordinate ; i. e. when we know its distances fron^ Q j\ind OX. Of course, if no information is given concerning the coordinates, x and y, of a point 7^, the point may have any position in the plane. Now we shall often have to consider points concerning whose coordinates we know so?7zet/iing, yet not enough to determine them. In this case we shall find that the point can no longer occupy any position in the plane, nor is it limited to a single position, but that // is coJifined to a series of positions, any one of ivhich it may take and, at the sa7ne time., have its coordinates satisfy the conditions imposed upon them. To illustrate : let us suppose that we know the abscissa of a point to be zero and have no in- formation concerning its ordinate ; i. e. we have given only x ^ o. This equation states that the distance of F from the axis of y is Y a B a o X c D A nothing or, in other words, that /^ lies on LOCUS OF AN EQUATION. 13 O V; but as we do not know the value of y, the ordinate of F, which expresses its distance from OX, the point may lie afiy- where on O Y. Thus we see that F may move anywhere along O V, and yet always have the equation x = o true for it. Likewise the single equation y = o states, in algebraic lan- guage, that the point F, of which y is the ordinate, must lie on O X, but allows it to take any position on OX. Again : if we have given x = a, we know that the distance of the point F from O Fmust be a units of length ; F must, there- fore, lie somewhere on the line AB parallel to O Fand at the distance a from it ; but it may occupy any position on A B. In the same way the single equation y — — a indicates that P may lie anywhere on CD, parallel to (9^ and a units below it. Let us next examine the equation x — y. We see at once that this equation does not determine x and j, the coordinates of F, but that it does show us that the coordinates of F are always equal to each other both in numerical value and algebraic sign ; or, in other words, that the point must be just as far from one axis as from the other, and that its directions from the axes are either both positive or both negative. The point will satisfy both of these conditions if it is anywhere on the line A B which bisects the angle X O Y, and it cannot fulfil them unless it is a point of A B. If the point lies on the part of the line in the angle X O Y, its coor- dinates are equal and positive ; if on the part in the angle X' O Y', they are equal and negative. 14 ANALYTIC GEOMETRY. If the coordinates of P are connected by the equation x = — y^ we readily see that P must lie somewhere on the line CD which bisects the angle Y O X\ 13. The simple examples of § 12 illustrate the fact that any single equation connecting x and 7, the coordinates of P^ though not determining these quantities, states some law which governs their changes, and, so, confines P to some series of positions in the plane. This series of positions which a point can occupy while its coordinates satisfy a given equation is called the locus of that eqiiatiofi. We may also define the locus of an equation as follows : — The locus of an equation is the path of a point which moves in the p la fie so that at each ifistant its coordinates satisfy that equatio?t when substituted therein for x and y^ the variables in that equation. 14. Looking at this relation between an equation and a line from a somewhat different point of view, we notice that the line A B m the last figure has the following geometric property : — Each point in A B \s equally distant fro?n O Y and O X; a property which (remembering the distinction between positive and negative directions) no other point in the plane possesses. Now the distances of a point from O Kand O Xare the coordi- nates x and y of that point, and therefore the equation x =y expresses algebraically the distinctive property of points on A B, or, as we may say, of the line A B. This equation, therefore, is called the equation of AB. We may define the equation of a curve * as follows : — The equation of a curve is the expression^ in an equation., of the relation which exists between the coordinates of every point of that curve, and of no other points. 15. This relation between a curve and its equation, a relation * The word curve is used in Analytic Geometry to indicate any geometric locus, whether a curve, a straight line, or an isolated point. LOCUS OF AN EQUATION. I5 SO intimate that either may be taken as a representative of the other, is the foundation of our subject, and cannot be too care- fully studied. The development of the subject will be natural and easy to the student who has so thoroughly mastered this idea that he never for an instant loses sight of it ; while to one who over- looks it, or whose conception of it is vague, the whole subject must be unintelligible and difficult. 16. In § 12 we have determined the loci of several equations by observing the geometric properties which are stated in alge- braic language in those equations. This method is useful for simple equations, where the geometric property is a simple one. For instance, we may obtain the locus of x'^ + y^ = a^ in this way ; for, by § 10, Ex. 5, we know that x'-^ + y^ represents the square of the distance of the point P, which has x and y for its coordinates, from the origin. Our equation therefore states that /'moves in the plane so that the square of its distance from O is constant and equal to a'. The distance of J^ from O is therefore constant and equal to a, which can only be the case when /^ moves on the circumference of a circle with a for a radius and O for a centre. Consequently this circumference is the locus of the equation x' + y'^ — d^. i^j. There is, however, another method of determining ap- proximately the locus of an equation, founded on the fact that the locus is the assemblage of all points whose coordinates sat- isfy that equation. We can often find enough of these points to show the general form of the locus ; for we must bear in mind that a curve drawn at random (as in the figure) is not the locus of any equation. The equation expresses some /aw which g07>erjis the changes of the coordinates of P^ and consequently controls the patn of P in the plane ; therefore the locus of the equation must be a curve governed by some law, and cannot i6 ANALYTIC GEOMETRY. be a line drawn at ran- dom. Hence the general form of the curve will be represented by the posi- tions of points upon it, chosen at reasonable in- tervals. This method of deter- mining a curve by points is called //(?///;;^ the curve. Let us plot the curve whose equation \s x — y -\- 2 = o. If we choose any value for x, we can find a value for 7 which, together with the chosen value of x, will satisfy the equation. These values are therefore the coordinates of one point of the locus of the equation. If we choose :r = i,y must equal 3 in order that x — y + 2 may equal o : therefore (i, 3) is a point of the required locus. Similarly (2, 4), (4, 6), (8, 10), (o, 2), (- i, i), (- 2, o), (- 3. - 0. (- 5, -3)y (-8. -6), are points of the lociis. Callingthese points jP^, P,, etc., to /",o, and constructing them as indicated by the figure, we observe that they all lie on a straight line ; and from this we infer that the locus of the equa- tion X — y -\- 2 — o is the straight line /Pio LOCUS OF AX EQUATION. 1 7 which cuts O i'two units above (9, and OX two units to the left of O. We shall arrive at the same result if we use the method of ^ i6. For the equation may be written _v = .v — 2. which inter- preted in geometric language means that ever}- point of the locus has an ordinate two units grctitcr than its at'sa'ssa : or, in other words, that t/ie distance of each point from O X is greater, by two I/nits, than its distance from O V. Xow in § 12 we showed that the straight line bisecting X O i'had all its points equally dis- tant from the two axes ; the required locus is therefore parallel to that bisector, and at a distance of two units from it when measured parallel to O )'. It will be seen that this result agrees with the line which we have just plotted. Let us plot the locus of y — 4 x. In this equation : if .V = o, y = o] if .V r= I, r = i 2 : if .r = 2,y = ± 2 \ 2 =r ± 2.S2 ; if x = 2,.y = -^2 \ 3 =^13.46; if .V = 4. V = = 4 : if a- = 5.j=:= 2 \ 5 =^4.48; if a- = 6, V = i 2 \ 6 = db 4.90 ; it x=.'j,y = ±2\ 7 = ^5.30; if A- = S,j—i4 \ 2 =±5.64; if X =r 9, _v = = 6 ; and so forth. We must also notice that any negative value of x \-ields imas:- inar}- values for v.- hence we conclude that there are no points of this locus with negative abscissas, i. e.. no points to the left of O y. Again, each abscissa has two ordinates. equal in length but opposite in direction, either of which may accompany it and satisfy the equation : from this we infer that for ever}- point iS ANALYTIC GEOMETRY. above (7 X there is a point symmetrically situated below OX, or we may say that the curve is symmetrical with respect to O X. It is also evident from the equation, that the larger x is, the larger _y must be ; there can be no limit, therefore, to the extent of the curve to the right of O Y; and it will go on forever in this direction both above and below (9X, gradually re- ceding from both axes. Plotting the points which we - have found, we see that the form of the curve is well represented except between the points (o, o) and (i, ± 2). it will be well, there- fore, to find more points in this interval. If ^ = 1, J = ± I ; if X — \, y — ± \/2 = ± 1. 41. Plotting these points in addition to those already found, the curve may be represented with tolerable accuracy by a line drawn through them, and having the form indicated by these points. As we have shown, it will extend indefinitely to the right. EXAMPLES. Plot the curves whose equations are : — y\i.) 2.T + 3j-6 = o; (2.) 5^+j' + 2=:o; (3.) jc2+/ = 25; ^(4.) 4Jt2 + 9/== 144; LOCUS OF AN EQUATION. I9 (5-) 4^^-9^ = 36; {6-) y^x'; (7.)/-^^; (8) x^-f = o; (9.) y = sin x; (10.) y = tan x; (11.) y = \og^^x; {i2.)y=2\ (13.) Show that equation 4-^ + SX + 2 = o has no locus. (14.) Show that the locus of the equation x' + 2y' = o consists of a single point. 18. The distance fropi the origip along the axis of .v to a point where any curve cuts the axis is called an intercept of that curve on OX. As a curve may cut the axis of x in more than one point, it may have more than one intercept on OX; each, however, is measured from O. In the same way, a curve has one or more intercepts on 6^ Y. The intercepts of a curve may be easily found from its equa- tion ; for it is evident, from the definitions just given, that the intercept on O X\?> the abscissa of the point where the curve cuts O X, and the intercept on O V, the ordinate of the point where the curve cuts O Y. Let A represent the first of these points, and B the second. The coordinates of A may be represented by X;^ and _)'a ; those of B by x^ and y^. Now since ^ is a point of the curve, its coordinates must satisfy the equation of the curve, when substituted therein for x and y ; moreover, y,, = o, because ^ is a point of OX. Making the substitutions, Xj^ is the only unknown quantity in the equation, and can be found by solving the equation. If there are two or more values for x^^, the curve has two or more intercepts on OX, — one for each value of x^^. Similarly the intercepts on O Fmay be found by substituting x^ and Jb for x and v in the equation, and then making x^ = o. The resulting values of y^ are the required intercepts. 20 ANALYTIC GEOMETRY. EXAMPLES, (i.) Calculate the intercepts of the curve whose equation is 2 X — ^y — 6=0 (i) Writing x^ for x and Va forjy, in equation (i), we have 2 Xa — 3}'a — 6 = (2) Substituting in (2) the value Ja = o, we have •^A = 3- Again ; to find the intercept on O Y, we have 2 .^B — 3 ^B — 6 = o, and Xb = o 'y therefore j;b = — 2« The curve therefore cuts OX three units to the right of the origin and O Y two units below the origin. (2.) Find the intercepts of the curve X + 3/rr 27. The intercept on O X may be found as in the last example, and is 27. To find the intercepts on O Y, we substitute x^ and y^ in the equation of the curve, and obtain, after making x^ equal to zero, ,J^'^ 'il' 3Jb^ = 27, ^ and JJ^B = ± 3. This curve, therefore cuts O Y twice ; once, three units above O, and again, three units below O, Note. — In practice we need not substitute Xj, and y^,, or ATg and j^B in the equation. We shall obtain the same result if we first put jf = o in the equation, and find the corresponding values of x ; then make :v = o in the original equation, and find the corresponding values of jj'. The results are the intercepts on 6>Xand O K respectively. We will now seek the intercepts of the locus of y" ^\x. LOCUS OF AN EQUATION, 21 Makings = o, we have and o = /^x, X = o. This shows that the intercept on OX has no length, and that the curve passes through the origin. In like manner, when a; = o in the equation, we have y — o. This, too, shows that the curve passes through the origin ; as it must cut O V a.t d. point whose distance from O is zero. ig. We may observe that a curve will always have its inter- cepts equal to zero, and therefore will pass through the origin, when every term in its equation {reduced to the simplest form) cotitains either x or y or both. For, if this is the case, when we substitute for X and y in the equation the coordinates of the origin (o, o), each term vanishes and the equation is satisfied. This shows that the origin is a point of the curve. On the other hand, if, when the equation is in its simplest form, it contains a term in which neither x nor y occurs, the curve cannot pass through the origin. For, in this case, when we substitute o for x and o for j, the equation becomes C = o, where C represents a constant which is not o. Therefore the equation is not satisfied by the coordinates of the origin, and the curve cannot contain that point. EXAMPLE. Calculate the intercepts of each of the curves whose equations are given in the examples of § 17. h^ 10* If we wish to find the intersection of two curves, given by their equations, we have only to consider that the point of inter- section of two curves is a point of ecuh curve, and, therefore, 22 ANALYTIC GEOMETRY. that its coordinates must satisfy each equation, when substituted for X and j^. If P^ represents the required point, we shall have, after substituting its coordinates x^ and y^ in the equations of the two curves, two equations between the two unknown quan- tities x^ and jv', and can readily determine them by algebraic methods. If we find only one pair of values for x^ and y, there is but one point of intersection. If there are several pairs of values for x^ and y^ , there are several positions of P\ i. e. several points of intersection. We will illustrate this method by finding the point of inter- section of the curves whose equations are 2x— y— \—o (i) and z^-y -z^^ » • (2) Let P^ be the required point. Since P^ is a point of the first curve, its coordinates satisfy equation (i). Therefore, 2 a:' - 7' - I = o (3) Similarly, we have 3 ^' - J^' - 3 = o» • (4) because P^ is a point of the locus of (2). Eliminating between equations (3) and (4), we find x^ = 2, y' =$. The curves, therefore, intersect in the point (2, 3). Let the student verify this result by plotting the curves. We will next find the intersections of the loci of and x-7y + 25 = 25- Proceeding as befo re, we have two equations to determi ^ne x^ and y : — x' - jy' + x'^+y"-- 25- = 25 - (t) , .(2) LOCUS OF AN EQUATION. 2$ Eliminating between equations (i) and (2), we find that either x' = ^ and y' = 4, °^ :»:'=-4andy :=3. The curves, therefore, intersect in the two points (3, 4) and ( - 4, 3) ; as will be seen by plotting the curves. It will often happen that some of the values of x' and v' are imaginary. The geometrical explanation of this fact may be best illustrated by an example. The points of intersection of the loci of X + y z= 4 and x'^ -\- y'^ = 4 will be found to have for their coordinates (2 + a/— 2, 2 — V— 2) and (2 — V— 2, 2 + V— 2) Since there are no real values of x' and 7', we infer that there are no real points of intersection of the curves. This is readily verified by constructing the curves, which will be found to be a straight line and a circle with no points in common. Now a straight line may cut a circle in two points, and hence we always find two sets of values for x' and y' in solving such an exam- ple. These values may give two real and different points, as in the second example of this section ; two real but coincident points, as is the case when the line is tangent to the circle ; two imaginary points, when the line falls wholly without the circumference.* Note. In practice it is not customary to substitute x' andy for X and y in the equations, but to eliminate directly between the equations of the curves, and thus find the values of x and y which will satisfy both equations, ^=^ i. e. the coordinalts of the points where the curves meet. It is safer for the begin- * It is important for the student to understand that an "imaginary point " is not a point at all. By an imaginary point we mean nothing more nor less than a pair of imaginary values of ^ and j'] and such a pair of values is not represented by any p^int whatever. 24 ANALYTIC CxEOMETRY. ner, however, to make the substitution; for he must always be careful to remember that x and y in every equation are variables^ and represent^ in turn, the coordinates of each and every poifit of the locus of that equation. EXAMPLES. Find the points of intersection of the curves represented by the following equations: — (i) 2X — 2y— \\— o, 2X— y — ']=o. ^"{2) x^+y'' = 2S, 3^ + 47 = 25. (3) •^■^+/ = 25, xy = 2. Ans. (1,-4). Ans. (3, 4). Ans. < /A/29 + -\/2I A/29 — a/2I\ (A/29 — A/2I A/29 + a/2I\ 2 ' 2 / / A/29 + V2I A/29 — a/2i\ \~ ~2 ' r / (— a/29 — A/21 A/29 4- a/2i\ ~2 ' ~ 2 / (4) y^ = 2 X, X — 2. Ans. (2, 2) and (2, — 2). (5) •^'- 5 ^+^ + 3 = O. ^^+/-5 A--37 + 6 rr o. -^^^•f- 1 (3. 3), (2,3), (4, 0, (i, i)| W :; + -^ = i. 1 + ^ = ^' ^ ^ ^ ^ a b a b /at? a \ ' \a -^ b' a -\- b) LOCUS OF AN EQUATION. 25 -^7) The equations of the sides of a triangle are jv 4- 2 J' - 5 z= o, 2x +y - y = o, y-x- 1 = o; find the vertices of the triangle, and the lengths of its sides. ( V5' V2' V5 -> (8) The equations of a circle and a chord of that circle are x^ ^y"^ = 100 and x -}- y = 14, respectively ; the centre of the circle is the origin : find the dis- tance of the chord from the centre of the circle. Ans. 7 ^"o mi ^^^ U. ot C. CHAPTER III. THE STRAIGHT LINE. 21. We have learned in the preceding chapter that any equa- tion connecting x and y has some geometric locus. Conversely, it is true that any curve which is generated by a point moving according to some law has for its equation the algebraic expres- sion of that law. We have seen that the loci of many equations are straight lines. Let us examine the distinctive properties of straight Hues and express them in an equation, — i. e. find, in a general form, an equation which will always represent a straight line, but which may in turn represent all straight lines. 22. We will first find the equation of the straight line which cuts O Xiit A and O Y at B. By changing the positions of A and B, this line may be made to take any desired po- sition in the plane. The lines O A and OB are the intercepts ~x of AB, and will be called a and b, respec- tively. Let P represent any point on A B, and draw its coordinates O M — x, M P = y. THE STRAIGHT LINE. 27 Whatever the position of P, MP is parallel to O V, and the triangles OP A and MP A are similar. We have, therefore, MP OB _ b_ Wa ^ 0A~ a (0 But and, by § 2, MP =y, MA = M0+ OA = OA- O M — a — X Substituting these values in (i), we have y J a — X a* or a y — a b — b X •••• (2) Equation (2) may be written as follows : b X ay a b a b a b ~ a b or ^-Hi = r (.) a b ^ ' The last form of the equation is the one commonly used ; it is called the equation of the straight line, in terms of its intercepts. Note. The student should study equation (i) carefully, and convince himself that it is true, wherever P may lie on the line A B, — extended indefinitely in both directions, — but cannot be true if /^ moves off of thai: line. If, in the first case, P and B lie on the same side of O A", M P 2ir\d O B have like directions and algebraic signs ; the Same is true of MA and O A ; and, therefore, the ratios in (i) have like signs, as well as *he same numerical value. If P and B lie on opposite sides of OX, MP and O B have opposite directions and algebraic 28 ANALYTIC GEOMETRY. signs ; the same is true oi MA and O A ; and, again, the ratios of (i) have like signs, together with the same numerical value. When -Pdoes not lie on A B, the triangles O B A and M FA are no longer similar, and the ratios of (i) are unequal numeri- cally. We see, then, that motion of F along the straight line A B is consistent with equation (i), but that ^P is confined to posi- tions on that line. Equation (i), therefore, expresses the dis- tinctive properties of the straight line ; and its modified forms, (2) and (3), express the invariable relation which exists between the coordinates of ajiy point of A B. It is, then, the equation of A B, according to the definition of the equation of a curve, given in § 14. EXAMPLE. Find the equation of the straight line which makes intercepts 2 and 3 on 6^ Jf and O F respectively. Here a — 2 and 6 = 3; therefore, the required equation is X y - + - = I, 2 3 or ^x-{-2y — 6 = 0. ^ 23. In this connection it is important to notice the distinc- tions between absolute constants, arbitrary constants, and variables. An absolute coiistarit has a fixed value, the same wherever that constant appears. 3, — 5, \, V2, it (the ratio of the circumfer- ence to diameter), etc., are absolute constants. An arbitrary cojistant is a quantity which, though fixed in any particular problem, has different values in different problems of the same class; r, the circumference of a circle, and r, its radius, are examples of arbitrary constants. THE STRAIGHT LINE. 29 A variable is a quantity which may have an indefinite number of values in any one connection. The length of a chord of a circle is a variable, because in the same circle it may have an indefinite number of values. In the equation of the straight line found in § 22, we have examples of each of these three kinds of quantities : i is an absolute constant ; a and b are arbitrary constants, fixed when the equation represents any particular line, but capable of other values as the equation represents other lines ; x and y are variables, for when the equation represents any particular line (i. e. with each pair of values of a and b), these quantities are susceptible of innumerable values, as they represent, in turn, the coordinates of all points on the line. 24. While the equation of the straight line found in § 22 may represent any straight line* by giving appropriate values to a and b^ we must often consider straight lines which are not known to us by their intercepts, but by means of other properties of the lines. In such cases, it is convenient to have the equations of the lines in terms of the determining quantities which are given. We shall, therefore, find several other forms of the equation of a straight line. 25. A straight line is determined if we know where it cuts O V, and also its di7'ection. The first of these ele- ments is given by the intercept on O V, b; the second is conven- iently expressed by the angle which the line makes with (9^, which we will call y. The * Provided, merely, that, tlie line does not pass through the origin. y /4 P R /y B T /' P A X 30 ANALYTIC GEOMETRY. tangent of y, which is called the slope of the line, we will indi- cate by X. In the figure, A B represents any line ; O B — b; and the angle X A Q — y. The angle y is always to be measured from O X to A Q^ and we usually measure it in the positive direc- tion. P represents any point on the line, and its coordinates, O M and M F, are x and y, respectively. Through B draw B T parallel to O X, and in the positive direction, and let this line and MP (or their extensions) meet in P. From the figure and § 2, we have y = M P= MP + P P = O B + P P (i) But O B = b, (2) and TB Q^ XA (2 = y The triangle B A' 7^ is a triangle of reference for y. From this triangle we have P P or Since we have ^^ = tany = A, P P= B P x\. BP ^ OM= X. PP=\x (3) Substituting in equation (i) the values found in (2) and (3), we have y = b -\- Xx, or y = Xx + b, the usual form of the equation of the straight line in terms of its intercept and slope. THE STRAIGHT LINE. 31 This is evidently the equation we seek ; for it expresses, in terms of the given arbitrary constants, a relation [equation (i)] which exists between the coordinates of any point of A jB, but of no other points in the plane. Remarks. — If y is in the ist or 3d quadrant, A ( = tan y) is positive ; in this case the figure shows that ^ P and B i? have like signs. If y is in the 2d or 4th quadrant, A is negative ; in this cas-j it IS easily seen that R P and B R have unHke signs. If y = o (i. e. if the line is parallel \.o O X),\ — o; the equa- tion, therefore, reduces to the form which we have already seen must represent a line parallel to OX{v.% 12). \i b — o, the equation becomes y ■=z\x, which is, therefore, the equation of a straight line through the origin ; this result agrees with the principle stated in § 19. EXAMPLES, (i.) Find the equation of the line which cuts (9 K at a dis- tance of two units above the origin, and makes an angle of 45° with OX. Here A = tan 45° = i and b = 2 ; the required equation is, therefore, 7 = jc + 2. This may be verified by plotting the locus. (2.) Find the equations of the lines which cut O Y three units below the origin, and make angles with OX of 135°, 30°, 120°, 210°, 330°, respectively. (3.) Find the equations of a pair of parallel lines making an angle of 150° with OX, one passing through the origin, the other cutting (9 Fat a distance of five units below O. (4.) x^" +y-^ = 25 is the equation of a circle with the origin as 32 ANALYTIC GEOMETRY. centre. Find the coordinates of the extremities of the diameter which has a slope |. Ans. (4, 3) and (-4,-3). 26. If a straight line passes through a given point and has a given direction (indicated by its slope), it is determined in posi- tion. The fixed point we will call F^ or {x^, y^, and the slope of the line, A. We will deduce the equation of the line in terms of these arbitrary constants, using a method which will be of great service hereafter. The equation of § 25, y ^\x + b, (i) will represent the required line, if we can find such a value for b that the line with that intercept and the slope A shall pass through Fy This we can easily do; for, that the locus of (i) may pass through F^, it is necessary and sufficient that the coordinates of F^ shall satisfy (i), when substituted therein for x and 7. Mak- ing this substitution, we have y^ = Xx^+b, (2) which may be called the equation of condition, that the locus of (i) shall pass through F^, or that F^ may lie on the locus of (1). Now, in equation (2), x^, y^, and A are the arbitrary constants which determine the position of the line whose equation we seek. We can, therefore, find b in terms of these quantities by solving equation (2). This gives b ^y^-\x^', substituting this value of <^ in (i) we have y = \x ^{y^-\ x^), which may be written y-y, = \ (x-x,). This must be the equation sought, since it expresses the inva- THE STRAIGHT LINE. 33 riable relation between the coordinates of every point of the Hne (and of no other points) in terms of the required arbitrary constants. The student should verify this equation by deducing it from a figure, without using the equations of the straight line already found. The method to be used is analogous to that of § 25. EXAMPLES, (i.) What is the equation of the straight hne which passes through the point (2, 3) and has an inclination of 45° .? ( or X — y + I =0. (2.) Find the equations of the straight lines w^hich pass through (2, — 4) and make angles of 60° with O X and 30° with O K, respectively. j + 4 = V3 (a^_ 2), J + 4 = - V3 (^ - 2). (3.) Find the equation of the straight line which passes through the point common to the loci of 2 X — y — \ — o and t^ x — y — t^ =0, and makes an angle 135° with OX. Ans. x -{- y -\- 1 = o. (4.) Find the equations of a pair of parallel lines making an angle 120° with OX, and passing through the points of inter- section of the loci of y X -\- y = ^o and x^ + y^ = 100. y+V3x-S-6V^ = o, Ans. Ans. J + V3 -^ + 6 - 8 V3 = o. 27. A straight line is also determined when it passes through two given points. Here the arbitrary constants are the coordi- nates of the given points, and we can find the equation of the 34 ANALYTIC GEOMETRY. line in terms of these quantities from the equation of the straight line found in the last section. The equation y -}'! = ^(^ -^i) (0 represents a straight line which passes through one given point /'j, and has a slope A. By giving to this line the proper direc- tion or slope, we can make it pass through the second given point P.,. Let us tind the appropriate value of A. The equation of condition which must be fulfilled in order that the locus of (i) may pass through jPg ^s y2 - yi = ^ (^2 - ^i) ', o .... (2) i.e. the coordinates of P.^ must satisfy equation (i). In equa- tion (2), A is the only unknown quantity ; and solving for A., we find _ ^ = ^^-f (3) 2 1 Therefore if A in equation (i) has this value, the locus of (i) will be a straight line through /^^ and P^. Substituting this value of A in (i), we have >'-^i = ^^(^-^.). (4) which must be the required equation. This equation is commonly written in the more symmetrical form ^2 - yx ^2 ■X (5) but the first form is often useful. Another convenient form of this equation, obtained by clear- ing of fractions and reducing, is O'l - 72) "^ - (-^1 -Xo)y-\-x,y,-x^y,^o.., (6) The student should now deduce equation (5) from a figure, without assuming any other form of the equation of the straight line. THE STRAIGHT LINE. 35 EXAMPLES, (i.) Find the equation of tlie straight line which connects the points (4, 2) and (—3, 9). We may choose either point as P^, and the other will be P^. Let x^ = 4, }\ = 2, ^2 = - 3, j^2 = 9- Substituting these values in the equation just found, we have J - 2 ^ X - a , 9-2 -3-4' or ^ - 2 _ ^ - 4 7 " -7 ' which may be written X ^ y — d = o, the required equation in its simplest form. (2.) Find the equations of the sides of the triangle which has for its vertices the points (2, i), (3, — 2), (— 4, — i). ^3^+7 -7 = 0, Ans. < X -^ J y-\- II =0, C X - sy + T- =0. (3.) In the same triangle, find the equations of the straight lines which connect the vertices with the middle points of the opposite sides. ^ X — y — 1 = o. Ans. \ X + 2y -\- I = o. C X - i2,y - 9 = o. (4.) Find the equations of the diagonals of the rectangle formed by the lines whose equations are X = a, X = a', y — b, y = b'. C (^ _ l,f) X - {a - a') y + ab' - a' b = o, ' \ (^b - b') X + {a - a') y — a b + a' b' = o. (5.) Find the equation of the straight Hne which joins the 36 ANALYTIC GEOMETRY. points of intersection of the two pairs of straight lines which have the equations 2'^ + 3J' — 4<^ = o) ^ X -{- 6y — y a = o ^ ano 2 X + y — <^=o) (^^x — 2y-\-2a = o. A?is. 4Ar + 4j_5^ = o. (6.) Find the equation of the straight Hne which joins the origin to the intersection of the loci of Ax + By -f C = o and A'x + B'y + O = o. Ans. {AC - A'C) x + (BC - B'C)y = o. (7.) Find the equations of the diagonals of the quadrilateral which has for the equations of its sides 3 ^ + 4 J ._ 13 ^ o, X - 6y + 3 = 0, 2^ + 3/ +6 = 0, 3:^+117 + 22=0. x — y — 2 = Oy A^jzs ■ 5;^+ 147+ 15 = o, (8.) Find the equations of the medial lines of the triangle whose vertices are the points B^, Br,, B^. Ans. One of the lines has the equation the other equations may be formed from this by advancing the suffixes. (9) Find the equations of the same lines when B^ is the ori- gin, and B^ B^ is the axis of x. (2y^x+ {x^- 2 x^) y-Xr,y, = o, Afis. < y^ X + {2 x^ — x^) y — x^ y^^ = o, i y^x- { x,+ ^2) y =0. (10.) Confirm the results of Ex. (9) by means of the equa- tions found in Ex. (8). (11.) Prove that the medial lines of a triangle meet in a common point. THE STRAIGHT LINE. 37 (i2.) Prove that the diagonals of a parallelogram bisect each other. Suggestion. Take one side of the parallelogram for the axis of X, and one extremity of that side as the origin. (13.) Prove that the lines which join the middle points of the opposite sides of a trapezoid bisect each other. (14.) If jE and jFare the middle points of the opposite sides, A Z>, B C, oi a parallelogram A B C Z>, the straight lines B £, D F, trisect the diagonal A C, Prove this theorem (v. §§ 10, II, 20, 27). (15.) Prove that (2, 3), ( —4, — i), (8, 7) lie on one straight line. 28. A straight line is determined in position if we know the length of a perpendicular drawn to it from the origin, and the direction of this perpendicular (represented by the angle which it makes with OX). We will next find the equation of the straight line in terms of these arbitrary constants. Let A B repre- sent any line in the plane, O Q the perpendicular upon it from O, and XO Q the angle which de- termines the di- rection of O Q. O Q IS measured from O ; and the angle is measured from O X \c^ O Q. We will call the angle a, and the perpendic- ular/; it snW be seen that/ will always be positive, for the positive diniction of the terminal line of an angle is from the vertex alon^ i^ boundary of the angle. 38 ANALYTIC GEOMETRY. Let P be any point of A B, with coordinates x = O M, y — 3f P. From M draw perpendiculars to 6^ ^ and A B meeting tliem in R and S respectively. From the figure and § 3 we have QO V= QOX^ XO Y = XO V- XOQ o = 90 — a. If we always take 7" and ^on J/^ and MP respectively, so that J/ 7" shall have the same direction as its parallel O Q, and M [/ 3.S O V, we may write TMU = QO V=go° - a. Now, O R M is always a triangle of reference for the angle QO X, or (—a); and MSP is a triangle of reference for the angle T M U, or (90° — a). From the figure we have, by § 2, p = OQ= 0R + RQ = OR + MS (i) OR , . Q-^= cos (-a) = cos a; Now therefore, Again, therefore, OR = OMx cos a — X cos a (2) j^ = cos (90 - a) = sm a ; MS = MP X sin a = J sin a (3) Substituting in equation (i) the values found in (2) and (3), we have p = X cos a 4- 7 sin a, THE STRAIGHT LINE. 39 or X COS a + y sin a = /, the usual form of the equation. Since this equation expresses the necessary relation between the coordinates of every point of the straight line AB (and of no other point), it is the equation of that line. Let the student obtain this form of the equation of the straight line from the equation in terms of the intercepts, X y - + 7 = I. This may be done by expressing a and b in terms of/ and a. Note. — It is important to notice that the equation which we have deduced in this section will remain the equation of A B, if we consider Q O the positive direction of the perpendicular, instead of O Q. P'or the angle a is measured from O X \.o the positive direction of the perpendicular, and (90° — a) is measured from the positii'e diredioJi of the perpendicular to O Y. Therefore, by reversing the direction of the perpendicular, we change each angle by 180°, and only change the algebraic signs of their cosines. Each term in the equation, therefore, has its sign changed, and the equation is still true for all points on A B, and for no other points. EXAMPLES. (i.) Find the equation of the straight line such that the per- pendicular upon it from the origin bisects the angle X O V, and equals V^* Here / = vT ^^^ ^ — 45° > therefore cos a = Vg and sin a = -v/J. The required equation is, therefore, X Vi +y Vh = Vh or X -\-y = I. 40 ANALYTIC GEOMETRY. (2.) A circle, with a radius 4, has the origin for its centre. Find the equations of the tangents to the circle at the extremi- ties of the diameter which makes an angle 120° with O X. 29. We have found, in §§ 22-28, the equation of a straight line in different forms ; and we have in several cases shown how one of these forms is merely a modification of others. It is noticeable that in each form the equation is of the Jirst degree. We will now examine the general form of the equation of the first degree between x and j, and show that it always expresses a law governing these quantities such that the point {x, y) will be confined to motion on a straight line. The general form of the equation of the first degree may be written : — Ax-\.By+C=o', (i) where A represents the sum of the coefficients of all the terms which contain x ; B, of all which contain y ; and C, the sum of all terms which contain neither x nor y. Xf we solve equation (i) for_r, we have AC , ^ which is in the form y=\x\b, (3) ' A where \ — —-— and b — -^ - Bl whic h is known to be the equation of a straight line. Now, whatever the values of A^ B, and C, — — is the slope of B 8om* line. For the slope of a line is the tangent of the angle whi( h the line makes with O X ; and as this angle changes from 0° t) 180°, its tangent passes through all possible values from 00 THE STRAIGHT LINE. 4I to — 00. There is some angle, therefore, between o° and i8o°, which has its tangent equal to — — , and a straight line having this inclination has - -^ for its slope. C Again, — — is the intercept of some line on O V; for a line may cut O F anywhere, and so have any intercept. If, then, we form the equation of the straight line which has A C the slope — -^, and — ^- for its intercept on O V, the equation must be =(-i)-(-f) or A X -ir ^ y -\- C = o. Therefore^ whatever the values of A, B, and C, equation (i) has for its locus some straight line. But (i) represents in turn all equations of the first degree between x and y. Therefore, all such equations are equations of straight lines. From the preceding discussion, we see that the values of the slope and intercept on O V can be easily obtained from the equation of the line. For, if this equation is put in the form A X -{- By + C = o, we have A , C Note. The student should notice that the principle of this section will be proved if we can put the general equation in any of the forms which we know to represent straight lines. EXAMPLES, (i) Put the general equation of the first degree in the form X y a o 42 ANALYTIC GEOMETRY. and obtain values for a and b in terms of the coefficients A, B, and C. A ^ J ^ Ans.a = --, b = -- (2) Calculate the intercepts and slope of the straight line whose equation is 2^ + 37 — 6 = 0. Here ^ = 2, ^ = 3, C = -6, and we easily find « = 3, b^2, A = --. o (3) Find the intercepts and slope for each of the lines which have the following equations : — X cos o. ■\- y sin a z= p. 30. It is obvious that two equations, of which the second can be obtained from the first by muliiplication by a constant factor, will have precisely the same locus, since any pair of coordinates which satisfies one of these equations also satisfies the other. Thus the two equations, 2^ — 3^— 6 = 0, 6x — gy — 18 = 0, both represent the line whose intercepts are 3 and — 2. No less important, though perhaps less obvious, is the fol- lowing fact // iivo equattofis of the first degree have the same locuSy the sec- ond can be obtained frofn the first by 7nultiplication by a constant factor. For, let Ax-^By-\-C = o, (i) A^x ^ Bj + Cj = o, (2) THE STRAIGHT LINE. 43 be the two equations of the first degree which are supposed to have the same locus. This locus, as we know, is a straight line whose slope A and intercept b may be computed from the first equation by the formulae from the second equation by the formulae When we remember that the two equations are supposed to rep- resent one and the same line, we see that the two values of A just obtained must be equal; and the same is true of the two values of b. Therefore B B^' B B^* ABC (3) A Let us now multiply (i) by the constant -— ^. This gives A B or, when we replace — ^ in the second term by its value ^, and ^ A B * C in the third term by its value ~ [v. (3) ], 44 ANALYTIC GEOMETRY. and this reduces at once to equation (2). Thus we see that A (2) may be obtained by multiplying (i) by the constant -j- , 31. Let us reduce the general equation of the first degree to the form found in § 28. Whatever the locus of A X +B y +C=o, (i) its equation may be expressed in the form X cos a+jsina— /z=o; (2) for every straight line can have its equation written in any of the forms found in §§ 22-28. Regarding (i) and (2) as equations of the same straight line their first members must either be identical, or one must be some multiple of the other (v. § 30). Let the unknown multiplier, which will make (i) identical with (2), be represented by Ji. '^^^^"' RAx-^RBy-\.RC=o, (3) and . , ^ X cos a + J' sm a — / = o (4) are identical. Therefore A' ^ :^ cos a, B B = sin a, R C = -/. We have from Trigonometry cos'^ a + sin^ a r= I ; substituting the values of cos a and sin a, and solving for A^, ^2 I -A' + B^' and R ± Va' 4- B^ THE STRAIGHT LINE. 45 Using this value of R, we hav^e cos a — Sin a = / = - c We may use either sign with the radical, but must, of course, use the same one in the values of cos a, sin a, and /. We have decided to take / positive ; therefore, we will in each case C choose that sign with the radical which will make — . ^m ^/a^ + B positive. This must be the opposite sign to that of C. Note. — The reason why we may choose either sign with "s/ A^ + B' (i. e. have two multipliers, R)^ is found in the note to § 28. It is there shown that the same line will have two equa- tions in the form X cos a -\- y sin a — p, if we allow either algebraic sign with // and that the values Ol cos a, sin a, and/, will be the same numerically, but have oppo- site signs in the two equations. Now, if we choose that sign for \/A^ + B'^ which makes p positive, we get certain values for cos a and sin a ; these values will only have their signs reversed when we change the sign of VA'"^ + B'^, so that the second value of a differs from the first by 180°, as was shown in the note to § 28. EXAMPLES. (i.) Reduce the equation ^x-sy-y =0 to the form X cos a + J sin a =/. 46 ANALYTIC GEOMETRY. Here therefore ^ = 4, ^ = -3, C= -7; /?- ' ' ^ + V^2 + ^2 y j6 ^. ^ 5 We choose the positive sign with the radical, b^ of C is negative. ^.-^7 = 7 5 5 5 is, therefore, the required equation, and 4 . 3 7 cos Ci = -, SUl a = -",/= i-. (2.) Find the distance from the origin to the line which has the equation 2 ^3 .r — 2 _y + 7 = o, and the angle which this perpendicular makes with O X. Afis. - : ii;o°. 4 ^ (3.) Find the length of the perpendicular from the origin on the line whose equation is a {x — a) ■\- b {y — b) =0, 32. In the preceding sections we have found those forms of the equation of the straight line which will be most useful here- after. We have also shown that any equation of the first degree between x and^ must represent a straight line. We will now make use of the equations which we have found to solve several problems relating to straight lines and points, the results of which are convenient for reference. 33. Let us find the angle between two straight lines given by their equations : — THE STRAIGHT LIX 47 P^rst, suppose the equations of the Hues to be in terms of inter- cept and slope. We may write the equa- tions : y = Xx + b, . . (i) y = A, X + /?, . . (2) Here, A =tan ■; , and Aj = tan y,, 7 and y^ being the angles which the lines (i) and (2) re- spectively make with O X, as represented in the figure. Let O be the angle which the line (2) makes with the line (i) ; i. e. t> is measured y)'6'w (i) to (2). This is the angle we seek. By § 3» 7 + r'^ = y„ (3) is is evident when we draw through T, the point of intersection of the lines, a line parallel to O X. From (3), we have ^ = vi - y- Therefore, - tan 71 - tan 7 tan t9- = '— . I + tan 7^ tan 7 Substituting the values of tan y^ and tan 7, we have tan 19^ A, -A I + A, A (4) This gives tan (y in terms of known quantities, and ^ can be easily found from its tangent by Trigonometry. 48 ANALYTIC GEOMETRY. If the equations of the lines are Ax+By-rC=o, (i') A,x + B,y + C^ = o, (2') we have only to find the values of X and Aj in terms of A, B, A^^ B^, and substitute the values thus found in (4). By § 29, Therefore, 1 or X = - ^andA,=_-. tan ^ - A A -i-i) -(- U-i)' tan i -ff \-A,B ■ + -5 ^i (5) Note. — It should be carefully remembered that ^ is meas- ured from (i) to (2), and so will be positive or negative accord- ing as -/i is greater or less than y. By interchanging the lines (i) and (2), therefore, we shall change the sign of i^, and so change the sign of its tangent. The latter fact is also evident from the values of tan i>, in either of which the proposed inter- change will reverse the sign of the fraction. Corollary 1. If the lines (i) and (2) are parallel, the angle i9- is 0°, or 180°, and tan Or =.0, That this may be true, we must have the numerators of the fractions in (4) and (5) equal to o. The conditions for parallel lines are, therefore, Aj — A = o, or Aj = A ; or, A B^-A,B = 0, or AB,= A^B. THE STRAIGHT LINE. 49 Corollary 2. If the lines are mutually perpendicular, & is 90°, or 270°, and tan 19- = 00 . Making the denominators of the fractions in (4) and (5) equal to o, we have for the conditions of perpendicular lines I + X^kz=o, ovX^ = --; or, A A^ + B B^ = o,or A A^ = - B B^. We may state the results of these corollaries, thus : Parallel lines have the same slope. Lines at right angles to each other must have the slope of one equal to the negative of the reciprocal of the slope of the other. EXAMPLES. (i.) Find the angle which the line makes with the line Here, by the statement of the question, the angle is to be measured /r^;;? the last line. We have, then, X = 4, Ai = - 3, and I +(-3) 4 11' therefore ^ = tan - ' — If we use the other formula we have ^ = 4, B = - 1, A,=s, A = i> and 4 X 3 + (- I) X I II which agrees with the previous result. 50 ANALYTIC GEOMETRY. (2.) Find the angle which the line 3-^+7 = 7 makes with the line 2x^y = ^. Ans. — . 4 (3.) Find the angle which the last line mentioned in Ex. (2) makes with 3^+^ = 5. (4.) Find the angles of the quadrilateral mentioned in § 27, Ex. 7. C 22 15 I 21 > Ans. < tan~' — — , tan"' -^, tan"' , tan~' — V C 21 16 3 535 (5.) Find the equation of the straight line which cuts O V four units below the origin, and is parallel to the line 2x- sy + 4=0; find the equation of a line which has the same intercept on O Y, and is perpendicular to the line S 2 '2Jt: — 5^— 20 = 0, '-\ Ans. - t^x + 2y + 5 = 0. (6.) Recollecting that the angle between two lines is equal to the angle between the perpendiculars upon those lines from the origin, find an expression for the tangent of the angle which the line X cos a.^+ y sin 04 =/j makes with the line X cos o' + y sin or = ^, (7.) By the same method, deduce the formula already found for the tangent of the angle which the line A^x + B^y + C, ^o makes with the line Ax^By^C=o. THE STRAIGHT LINE. 5l (8.) Find the angle between the lines 4^-3J-f5=o» and 6x + Sy + ly = o. (9.) Show that the diagonals of a square are perpendicular to each other. (10.) Show that the second quadrilateral mentioned in § 11, Ex. 10, is a parallelogram. 34. It often happens that we have a line Ax + By+Cr=o, (i) and wish to find the equation of another line parallel or per- pendicular to it. The following considerations are very useful in such cases. The slope of any line parallel to (i) is — (§ t,^, Cor. i). Accordingly the equation of such a line may Ije written (§ 25) y = -^x -h ^ or Ax + By - dB =z o. 7 he equation of any line parallel to ^i) may be so written that its equation differs from ( i ) only in the value of the constant term. Similarly the equation of any line perpendicular to (i) may be written (§ 33, Cor. 2) B or B X— Ay -V b A — o. The equation of any line perpendicular /^ ( i ) may he obtained hv interchanging the coefficients of x and y in ( i ) and changing the sign of one of them^ and (hen suitably alte*-ing the constant term. These two rules enable us to write down the equation of the desired line completely, except for the constant term. The 52 ANALYTIC GEOMETRY. fact that the constant term is not determined by these rules should not surprise us, for the mere statement that the desired line is to be parallel or perpendicular to (i) does not deter- mine the line completely. Some further statement with re- gard to the position of the line must be made before the con- stant term can be determined. This may be done, for instance, as follows: Let us consider the problem: To find the equation of the line which passes through the point (x^, }\) and is parallel to (i). Since this line is to be parallel to (i), its equation can, by the first of the above rules, be written in the form Ax + By — k, (2) where, for convenience, we have written the constant term k on the right-hand side of the equation. The question which remains to be settled is: What is the value of the constant k ? Since the line (2) is to pass through (a:,, y^, the following equation of condition must be satisfied: A x^ -\- B y^ — k^ and it will be seen that this equation gives the value of the unknown constant k in terms of the known constants A^ By a:,, y^. Substituting this value of k in (2), we find as the equa- tion of the desired line Ax ■{■ By = A x^ + By^. By precisely the same method we find as the equation of the line which passes through the point (^j, y^) and is perpen- dicular to (i) Bx — Ay = Bx^ — Ay^, EXAMPLES, (i.) Find the equations of two straight lines through (2, — 1\ parallel and perpendicular, respectively to the line 5x + 3j+7 = o. Ans. \2 X -y - 1 = o 2y + 3 = o. THE STRAIGHT LINE. 53 (2.) Find the equations of two lines through the origin, par- allel and perpendicular, respectively, to the line 3-^+5>'-4 = o- (3.) Find the equation of the line which passes through the point common to the lines X — 2y — a = o, and X -{- ^y — 2 a ^= o, and is parallel to the line 30: + 47 = o. Ans. 3^ + 4_y — 5^ = 0. (4.) Find the equations of the perpendiculars from the ver- tices to the opposite sides of the triangle given in § 20, Ex. 7. (x-\-y-4 = o V X (5.) Through the vertices of this triangle lines are drawn parallel to the opposite sides ; find the equations of the sides of the triangle thus formed. ^ X — y — 2 = o Ans. \ X + 2 y - d> =0 \2X-\-y — ^=0. (6.) Find the condition that the lines X + [a + b) y + c = o, and a [x + ay) + b (x — by) -f ^/ = o KKiy be parallel; that tliey may be mutually perpendicular. Ans. b = o, b' — a'' =^ u (7.) Choosing the axes as in § 27, Ex. 9, prove that the per- pendiculars erected at the middle points of the sides of a tri- angle meet in the point (-, — — ^- — -). \ 2 2 y^ ' (8.) In a similar manner, show that the perpendiculars from the vertices to the opposite sides of a triangle meet in the point (^,, fL^^). 54 ANALYTIC GEOMETRY (9.) Prove that the common intersection of the medial lines of a triangle, the intersection of the perpendiculars at the middle points of its sides, and the intersection of the perpendiculars from the vertices to the opposite sides lie on one straight line. (10.) Show that the first of these points is one of the points of trisection of the line which joins the other two. To which of these points is it nearer? (v. § ji.) 35. Let us find the distance of a given point P^from a given straight line. Let ^^ be the given line, and let its equation be X cos a Ary sin a =/>(l) Draw a line, (2), through /'p parallel to (i) ; and from O draw a perpendicular to the lines (i) and (2), meeting them in Q and Q', respec- tively, ^/^i, perpen- dicular to A B, measures the required distance of P^from A B. In the figure, O Q =p and X O Q = a. We have chosen p, measured from O to Q, positive ; this is, therefore, the positive direction of lines parallel to O Q. Since (i) and (2) are parallel lines, the perpendiculars upon them, from (9, make the same angle, a, with OX; the equation of (2) is, therefore, X cos a + jj^ sin a = /', (2) where/' = O Q'. From the figure and § 2, we have J^Pi= QQ'= O Q' - OQ =/'-/ (3) THE STRAIGHT LINE. 55 Now, P^ is a point of the line (2) ; therefore its coordinates must satisfy equation (2), and we have x^ cos a^- y^ sin a =/'. Substituting this value of/' in equation (3), we have R F^ — x^ cos a + j'^ sin a — /, (4) which expresses the required distance in terms of known quan- tities. Note. We see from the figure, that R P^ is positive or nega- tive according as P^ and O are on opposite sides oi A P or on the same side. The same result is obtained from equation (4). For, in the first case,/' is greater than/, and both are positive, in the second, when P^^ is nearer A B than O is,/' is positive and less than/; when P^ is farther from A B than O is,/' will itself be negative, according to the principle explained in § 28, Note. 36. We may solve the problem of the last section when the given line has the equation Ax^By^C^Q (i) by reducing equation (i) to the form X cos a -\- y sin a =/, as explained in § 31. The value of the required distance is, therefore, A B _ ( C_ ^^ VA-' + B' "^ '^' VA" + B' ~ ^~ \/A^+~B' or Ax^ + By.^C VA'' + B' where the radical has the opposite sign to that of C in order to make ..(2) ^[=-v^Jp°"'"" 56 ANALYTIC GEOMETRY. EXAMPLES, (i.) How far is the point (3, - 2) from the line 3^-4J + 2 = o? The formula of § 36 shows that the required distance is 3X3-4(-2)+2 __ - Vf+i-^r The negative sign indicates that the point lies on the same side of the line with the origin. (2.) Calculate the distance of the point (7, 4) from the line 8 X + 6 7 = 45. Ans. 3i ; and the line lies between the point and the origin. (3.) Find the distances of the origin from the lines X cos a -\- y sin a =p, A x + B y + C = o. C (4.) In the triangle given in § 20, Ex. 7, find the lengths of the perpendiculars from the vertices to the opposite sides. • (5.) Find the lengths of similar lines in the triangle given in § 9, Ex 3, and show by your results that the origin lies within the triangle. (6.) We may obtain the distance between a given point and a given line by finding, first, the equation of the perpendicular from the point to the line ; then, the point where this perpen- dicular meets the given line ; and, lastly, the distance between this point and the given point. By this method, calculate the distance of the point (— S, 3) from the line 3^7-27 + 4 = 0. Confirm your result by applying the formula of § 36. THE STRAIGHT LINE. 57 (7.) Using the same method, show that the distance between /\ and the Hne A X -{- By + C •=. o is A x^ ■\- B y^^ C (8.) If the given point lies on the given line, its distance from the line is nothing. Show that the formulas of §§ 35, 36, give the same result. (9.) Two parallel lines are drawn at an inclination & to the axis of X : one through the point {a^ b), the other through (a', b'). Show that the distance between these lines is {a' - a) sin {> - {b' - b) cos ^. 37. The area of a triangle can be found when the coordi- nates of its vertices are given. For we have only to calculate the length of one side, by § 10, and the perpen- dicular upon this side from the opposite vertex, by § 36. The half-prod- uct of these quantities is the measure of the area. Let F,, P,, P., be the vertices of the triangle. The equation of the line P,P.^{\.^. of the indefi- nite line determined by P.^ and P3) is, by § 27, The distance of P^ from this line is, by § 2)^^ 0'2 - J'3) -^1 - (^2 - ^3) }\ + ^2 ->^3 - ^3 y-1 (0 p p - V(/2 - i's)' + (^2 - ^3)' (2) 58 ANALYTIC GEOMETRY. By § 10, we have p, p, = VO; -y^' + {X, - x,y (3) If A represents the area of the triangle, = (J2 - y^^ ^1 - (-^2 - ^3) yi + ^2^3 - ^3 J2 • • • • (4) Equation (4) may be written 2 ^ = (^1 - X,) 7, + (at, - ^3) y^ + {x^ - x^ y^. This form is evidently symmetrical with respect to the coor- dinates of i",, P^, and P.^, as should be the case ; for it is, of course, immaterial which side is used as the base of the triangle, when calculating the area. Note. As only the Icngi/i of P^ P^ is needed in calculating the area of the triangle, wc need only notice the numerical value of our result. The algebraic sign will generally differ when we take different sides for the base. EXAMPLES. (i.) Find the area of the triangle which has for its vertices the points (2, 3), (- I, 4) (6, - 5). Calling the points P^, P.^, and iPj, respectively, we have 2 ^ = [2 - (- I)] (- 5) + (- I - 6) 3 + (6 - 2) 4 rr — 15 — 21 + 16 = — 20. Therefore the area is 10. The unit of area is, of course, the square which has each side equal to the unit of length in terms of which the coordinates of the given points are expressed. (2.) What is the area of the triangle given in § 20, Ex. 7. (3.) F'ind the area of the quadrilateral given in § 27, Ex. 7. Ans. y. (4.) Obtain the formula for the area of the triangle P, P,, P^ [§ 37, Figure] by subtracting the trapezoid J/g P.^ P.^ J/, from the sum of the trapezoids J/^ iR, ^\M^ and M^ P^ P. M.^ THE STRAIGHT LINE. 59 (5.) Prove that the three straight lines joining tiie middle points of the sides of a triangle divide the triangle into four equal triangles. 38. In all our work hitherto, we have used the rectangular system of coordinates. It is often more convenient to denote the position of a point by means of its di)'ection and distance from a fixed point in the plane, than by means of its distances from two fixed lines. If O X\x\ the figure is a fixed line, and O a fixed point of that line, the position of P is completely given by the angle X O F = (\>, and the dis- tance O P = r. The two elements, or coordinates^ which determine the position of the point P are, therefore, r, called the radius vector, and (/>, called the vectorial angle ; ^ is measured from the fixed straight line, called the initial line, and r from O, called the pole. This system of coordinates is called /^'/^r. It will be readily seen that the simple equation r — a has for its locus the circumference of a circle with (9, the pole, for its centre, and a for its radius. This example is sufficient to show the student that certain problems may be conveniently treated by this system of coordinates. 39. We will find the equation of the straight line, using the polar system. In the figure on the next page, O is the pole, O X the initial line, and AB any straight line in the plane. The position of the line is determined by /, equal to O Q, the perpendicular from the pole to the line, and a, equal to X O Q, the angle which / makes with the initial line. 6o ANALYTIC GEOMETRY. Let P be any point on A B, with coordinates r = OF, (f> = XOP. Now from the figure and § 3 we may write QOP= QOX+ XOP = XOP- XOQ = ct>-a (i) Again ; the triangle Q O P must be a triangle of reference for the angle Q O P, and there- fore we may write OQ OP = cos QOP = cos {4, - a) (2) Substituting in (2) the values oi O Q and OP, and clearing of fractions, we have r cos ((}) — a) = p, which must be the required equation, since it expressejs a neces- sary relation between r and <^, the coordinates of any point on A B (and of no other points), in terms of the chosen arbitrary constants,/ and a. \ CHAPTER IV. TRANSFORMATION OF COORDINATES. 40. When the eruation of a curve referred to one system of coordinates is known, it is often desirable to obtain the equation of the same curve referred to a different system. We shall be able to find formulas which connect the coordinates of any point in the plane in Xhejirst system with those of the same point in the s£co?id, and thus render easy the desired change. 41. We will first compare the coordinates of a point referred to one set of axes with the coordinates of the same point referred to a parallel set. Let6>Xand OY be the original set of axes ; O' X' and O' y the new set, parallel respective- ly to the old. O', the new ori- gin, has coordi- nates in the old system which we will call x^ and y^. Then, Y Y P M X M, 1 VI X Let P represent any point referred to XO Fare in the plane ; its coordinates ::} (3) 62 ANALYTIC GEOMETRY. X = OM, y r=.MF, Referred to X^ O' V, the coordinates of I' are O' M' and M^ P; these we will call x^ and y^ respectively, to distinguish them from x and_>^; but we must remember that x^ and y are here variables in the sa?ne sense that x and y are variables. From the figure and § 2 we have C>J/= OM^ + M^M=. OM, + O^ M^ • • • • (i) MP = MM' + M'P= 3f,0' + M'P. .... (2) Substituting in these equations the values of the lines as given above, we have X = Xq -{- X' y = yo + y' which are evidently the equations sought, since they enable us to find values for either pair of coordinates for P, in terms of the other pair and the constants x, and y, which determine the relative positions of the two sets of axes. We can best illustrate the use of these formulas by an example. The equation of a line referred to rectangular axes is 2x + sy^5', find the equation of the same line, referred to a parallel system which has the point (2, 4) for its origin. Here Xq = 2 and y, = 4. The equations for transformation are, therefore, X = 2 + x', y = 4 -\- y. These equations mean that a/iy point in the plane has its abscissa in the Jirsf system greater than its abscissa in the second by two units of length, and that the old ordinate exceeds the new hy four units. TRANSFORMATION OF COORDINATES. 63 Now the locus of the given equation comprises all the points which have twice their old abscissas plus thrice their old ordinatcs equal to five ; consequently it consists of those points which have twice the new abscissa increased by two plus thrice the new ordinate increased by four equal to five. Its equation in the new system is, therefore, 2 (^' + 2) + 3 {y^ + 4) = 5, or 2;v' + 4 + 3/ + 12 - 5, or 2 ^' + 3 y + II = o. Since in this equation the accents are only used to distinguish the new coordinates from the old, now that we have changed to the new system and there is no longer opportunity for confusion, we will use the usual symbols for the coordinates of the point which generates the locus, and write the equation 2^ + 37+ 11=0. The student should construct the two sets of axes, and then plot the loci of the two equations, and show that these loci are one and the same straight line. Note. — Since the results of this section have been obtained by the principle of § 2, and without reference to the angle X O V, the same formulas may be used for transforming from one oblique system to 2, parallel oblique system, [v. § 8.] 42. We will next find the formulas by which we may change from one rectangular system to another, keeping the same oj-igin. Let OX and 6> F be the original axes; O X^ and O V\ the new axes. Let the angle XO X, which the new axis of x makes with the old, be called i>. The coordinates of P (any point in the plane) in the old system are x - O M and y - MP; in the new system they are x^ ^ O M' and / = M' P. From M' draw M' P perpen- 64 ANALYTIC GEOMETRY. dicular to O X, and J/' T parallel to O X. Let the latter line (or its extension) meet MP (or its extension) in S. Take U on the line M^ P so that the angle X^ M' U shall be 90° of positive ro- tation. T and X are, of course, to be chosen sothatJ/Tandil/'^' shall have the direc- tions of O X2.x\6. OXK From the figure and § 3, we have TM'U= TMX> + X'M'V = {f + 90°. The triangles O R M' and M' SP are therefore triangles of reference for if and (90° + i>), respectively. Now, from the figure and § 2, we have x=OM=OR + RM^OR+M'S....{i) y = AfP= AfS+ S P =RM' + SP (2) But O /? = O Af cos = x' cos {y M' S = M' Pcos (90° + d) = - M' Ps\n if = -/ sin & RAP = OAf' sin 0^ = x' sin (f .S^ P = Af'P sin (90° + {y) = M' Pcos r^ = / cos & . Substituting in equations (i) and (2) the values found in (3)-(6), we have X = x' cos i> — y sin & y = a;' sin i*)- + y' cos which are the required equations of transformation, since they '.[ (3) (4) (5) (6) (7) TRANSFORMATION OF COORDINATES. 65 connect the old and new coordinates by means of the arbitrary constant & which determines the relative positions of the two sets of axes. EXAMPLE. The equation of a curve is x'+f = a'; '. (I) find the equation of the same curve when referred to axes which make an angle of 30° with the original axes, but retain the same origin. Here ^ = 30° ; sin ^ = -; cos ^ = ^. 2 2 The formulas for transformation are, therefore, \/'X I I V "? X = x' — - y and y z= - x' + y'. 2 2 2 2 Since the coordinates of any point of the curve, when referred to the old system, must satisfy equation (i), their equivalents in terms of the new coordinates of those points must satisfy equa- tion (i) ; therefore we have or 3^'2_ 2\/^x'y' +y^ , x'"^ -f 2 \^3x'y -f 3 /^ _ 2 4 which reduces to the form x'^ +y' = a' (3) If we drop the accents — because we have now left the old system and need no longer fear a confusion of the old and new coordinates of any point — we may write equation (3) x"^ + y^ = a"" (4) 66 ANALYTIC GEOMETRY. Since (4) is identical in form with (1), we see that the curve has the same law governing the coordinates of its points when referred to the new axes as when referred to the old. This is evidently true from the geometrical properties of the curve ; for the locus of (i) is a circle with its centre at the origin, [v. § 16.] Corollary. — The formulas of §42 become much simplified when {^ = 90°. In this case sin i> = i and cos {> = o : the equations for transformation therefore become X — — y and y = x'. We may obtain the same result by considering that in this case we merely interchange the axes of x and_>^/ at the same time changing the positive direction of the horizontal axis. 43. To change from a rectangular to a polar system, with the origin for pole^ and the axis of x for initial line. Let P be any point in the plane. Its coordinates in the rectangular system are X ^ OAf,y = MP; in the polar system, they are x" r=OP,cj> = XOP; O M P is a triangle of reference for y (i) M P = O P X sm (f), or y = r sm ff>^ . . . . (2) which are the required equations for transformation, because they connect the rectangular and polar coordinates of any point in the plane. TRANSFORMATION OF COORDINATES. 67 EXAMPLE. Find the polar equation of the circle referred to the centre as pole. We have seen in the example of § 42 that the position of the axis of X does not change the equation of this curve, provided the centre is origin. Therefore, if the centre is the pole, what- ever the direction of the initial line, the formulas for transforma- tion are X = r cos (f), y = r sin <^. Substituting for x and y, in the equation of the circle, their values in terms r and <^, we have r^ cos^ (fi + r^ sin^ = a^, or r^ (cos^ ) = a% or which may be written r = ± a, a Jesuit which agrees with the polar equation of the circle found in § 38. The double sign is explained by the fact that, with any value of '^ = 31. (2.) The equation of a curve is what is its equation when referred to lines which bisect the angles between the old axes ? Ans. X y = 2)' (3.) When we change from one rectangular system to another without changing the origin, x^ + f- must equal ^^ +y^ be- 70 ANALYTIC GEOMETRY. cause each denotes the square of the distance of the same point from the origin. Verify this by squaring and adding the expres- sions for X and_y, in terms of x' and j', given in § 42. (4.) The equation of a curve is ^xy - ^ x'^ z=i a^ ; turn the axes through an angle whose tangent is 2, and find the new equation of the curve. Ans. xP" — 4^^ = a^. (5.) Find the polar equation of the line Ax-\-By-\-C-=o^ when the origin is the pole of the new system, and the axis of x the initial line. Ans, {A cos (/) + -^ sin <^) r 4- C = o. (6.) Under similar circumstances, find the polar equation of the curve which has for its rectangular equation X" V2 Ans. r^ = a^ sin^ ffi -\- d^ cos^ <^* (7.) The equation of a curve is 2 ^' + 3 / -16^+18^ + 53 = 0; find its polar equation, when (4, — 3) is the pole, and the initial line is parallel to the axis of x. Ans. r^ - — r^. ^— ■ . 3 sin'' 9 + 2 cos'' Pf with coor- dinates O M — X and M P ^ y, is the point which generates the curve. Letting a represent X the constant distance of P from C, we may ex- press the definition of the circle by the equation CP=a , (i) By § 10, the distance between the two points C and P may be expressed in terms of their coordinates, and we have 72 ANALYTIC GEOMETRY. Substituting this value in (i), we have and clearing of radicals, , (x - x,y -{. (j - je)^ = ^^ (2) which must be the equation of the circle, for it expresses in an equation, connecting x andj^, the essential property of the circle generated by the point {x, y). Corollary 1 . When the centre is at the origin Xc = o and y^ = ^ i the equation becomes in this case x^ +y = a^, a result already obtained in § 16. Corollary 2. When a diameter is taken as the axis of x^ and its left-hand extremity as the origin, Xc = a and y^ = o ; in this case the equation of the circle is {x - ay + / = a% or. x^ -\- y^ = 2 a X. EXAMPLES. (i.) Write the equation of the circle whose radius is 5, and which has for its centre the point (3, 4) ; reduce the equation, and show from its form that the circle passes through the origin. (2.) A circle, with radius 4, has its centre on O X and is tangent to O Y : what is its equation if it lies on the right of O Y? What if on the left? y^ -^x r^ o, y'^ -{- d> X = o. (3.) Obtain the equation of Cor. i, from the first form of the Ans. ^ o 2 THE CIRCLE. ^ 73 equation of the circle, by transforming the axes to a parallel set through the point (^o^c)* (4.) Obtain the equation of Cor. 2 from each of the preced- ing forms, by transformation of coordinates. (5.) Prove analytically that any angle inscribed in a semi- circle is a right angle. Suggestion. — Take the diameter bounding the semicircle as axis of x. (6.) Find the equations of the three circles whose diameters are the sides of the triangle formed by the lines 2x -\- y -7=0, X- y +1=0. x' +/ - 5 ^ - 4 i' + 9 = 0, x^ +/ -3-^ -5J + 8 — 0, x" +/ - ^x - 3 J + 5 = 0. (7.) Find the equation of the circle whose centre is the ori- gin, and which is tangent to the line 2-^-7 + 3=0' Ans. 5 (^2^/) = 9. (8.) The equation of a chord of the circle x'^ -\- y'^ = 100 is 1 X -\-y = 50; find the equation of the circle which has this chord for a diameter. (9.) Find the points of meeting of the circles x'^ + y- — 2 X — /^ y — 20 =0, :v^ +y — 14 a" - i6_y + 100 = o. Ans. (4, 6), (5, 5). 74 ANALYTIC GEOMETRY. (lo.) Find the equation of the chord common to the circles given in the last example, and show that this chord is perpen- dicular to the line joining their centres. (ii.) Show that the circles (^ - 2)' + (y- 3)' = 36. (x - -,2)' + (y - ^y = i6, are tangent at the point (8, 3). (i 2.) Prove that the perpendicular upon a diameter of a circle from any point of the circumference is a mean proportional be- tween the segments of the diameter. (13.) Prove that if two circumferences are tangent internally, and the radius of the larger is the diameter of the smaller, any chord of the larger drawn from the point of contact is bisected by the circumference of the smaller. Suggestion. — Take the point of contact for the origin, and a diameter for the axis of x ; the equation of any chord may then be written y = \ X. (14.) If two straight lines are drawn through the point of contact of two circles, the chords of the intercepted arcs are parallel. Prove when the circles are tangent externally, taking the origin at the point of contact. 46, Equation (2) when expanded becomes This equation has the form x' +/ + ax + by + c=o (i) where a, b, c are constants. The circle is therefore a curve of the second degree. Conversely any equation of the form (i) can be shown to represent a circle, provided its locus is a curve of any sort.* * The reason for this restriction will be apparent presently. THE CIRCLE. 75 Before proving this fact we will illustrate it by a numerical example. Consider the equation ^2 ^ y _ ^j^ _|_ ^ y — I =o. This equation is of the form (i), and we will prove that it repre- sents a circle., and at the same time find the centre and radius of this circle by the following device. We arrange the terms of the equation in three groups. First, the two terms involving .T, then two terms involving y, and then, transposed to the right-hand side, the constant term; thus: x^ — 4x + y + 4>' = !• Now apply to the first group of terms the process, familiar from algebra, of completing the square by adding the square of half the coefficient of x. In the same way complete the square for the second group of terms by adding the square of half the coefficient of ;'. The two quantities which we have just added to the left-hand side of the equation must, of course, also be added on the right, and we thus get (x - 2y -{- (y + 2y = 9. This, however, is [§ 45, (2)] the equation of the circle with centre at (2, — 2) and radius 3. Precisely this method of completing the square, which we have here applied in a numerical case, can be applied without change to the general equation (i), and gives Provided the right-hand side of the equation is positive, we /a ^\ have here the equation of the circle with centre at ( , ) and radius \l— c -\ -\ . If, however, the right-hand \ 4 4 side of (2) is negative, this reasoning is no longer possible, 76 ANALYTIC GEOMETRY. since the radius we should thus obtain is an imaginary quan- tity. In fact in this case equation (2) has no locus, i.e. there is no real point whose coordinates satisfy this equation. For the left-hand side of (2), being the sum of two squares, is essen- tially positive, while the right-hand side is now supposed negative. There remains to be considered the case in which the right- hand side of (2) is zero. In this case the circle consists of a single point ( , — — j, since the left-hand side of (2), being the sum of two squares, can vanish only when each of these squares vanishes. This point is commonly regarded as a circle of radius zero. Summing up, we may say that equation (i) represents a circle, a point, or has no real locus according as the right- hand side of (2) is positive, zero, or negative. It is usual to designate these three cases by saying we have respectively a real, null, or imaginary circle. The general equation of the second degree in x and_)' is much more general than equation (i), for not only may the terms in x^ and^ have coefficients different from unity, but there may be a term involving the product xy. Thus the general equation of the second degree may be written Ax" + Bxy + Cf + £>x + Ey + F = o. . ... (3) This equation usually represents a curve which is much more complicated than a circle. It will represent a circle (real, null, or imaginary) ii B = o and ^ = C ^^ o, for then (3) has the form Ax"" + Ay"" + Dx + Ey + F = o, and this equation reduces to (i) if we divide it through by A THE CIRCLE. 77 EXAMPLES. Find the centres and radii of the following circles : (i) A"" + jj'' + 6 ^ — 4 jv 4- 9 = o. Ans. (— 3, 2), 2. (2) X'' -f j' — 6^ — 10;' + 34 = o. Ans. (3, 5), o. (3) ^' +/-2,x+y + 2 = o. Ans. (|, - -^) , | V~2' I 2 1 (4) \2 x"^ ■\- 12 y^ -\- \(i X — \2 y -\- 7^ ^^ o. Ans. 1 , — 2 1 \ 2 3" 47. We know from Geometry that a circle is determined by any three points on the circumference. We may, therefore, obtain the equation of a circle which passes through any three given poifits. Let ^p P^, and P^ be the three given points. Since equation (2), § 45, represents in turn all circles, it will represent the required circle when we give appropriate values to the arbitrary constants, x^, y^, and a, in that equation. That the locus of may pass through P^^ P^ and Py, it is necessary and sufficient that the coordinates of these points should satisfy equation (i)j these conditions are expressed by the three equations : — (pc^-x^y ^ {y^ -y^y = a% {x^^x-.y-h {y, -y^y = a\ In these equations the arbitrary constants, x^, jc, and a, are the quantities whose values we seek, while the other quantities are the known coordinates of the given points. We have, then, 78 ANALYTIC GEOMETRY. three equations between three unknown quantities, and can deter- mine them all by elimination. Substituting in (i) the values thus found for x^, yc) ^nd a, we have the equation of the required circle. EXAMPLES. (i.) Find the equation of the circle which passes through the points (9, 6), (10, 5), (3, - 2). Calling these points J^^, P.^, and Z^^, the three equations from which we may obtain the proper values of x^, j'c, and a are (9-^„)^+ (6_j,)= = «^ (i) (io-x,)^+ (5_jg^=«^ (2) (3-^c)'+(-2-a)^ = «' (3) We can easily eliminate a from these equations, and obtain two new equations connecting x^ and y^.^ from which we shall find x^ = 6, 7c = 2. Substituting these values in either of the equations (i), (2), (3), we obtain a" = 25. We have now determined all the constants in the required equation, which must be (^-6)«+(^-2)«=2S, or 0^ ^y^ - 12 :v - 4>' + 15 = o. (2.) Find the circle through (2, - 3), (3, - 4), ( - 2, - i). Ans, x^ +y^ + S X + 207 + 31 = 0. (3.) Find the equation of the circle circumscribed about the triangle given in § 20, Ex. 7. Ans, ^x^ + ^y — 13 jp — II 7 + 20 s= o. THE CIRCLE. 79 (4.) Solve the last three problems by starting from equation (i) of § 46. 48. The term secant is used in Analytic Geometry to denote a straight line which cuts a curve in two or more points. If one of the points in which a secant cuts a curve is moved along the curve towards a second of these points, the secant approaches the position of a tangent to the curve at the second point, and it is by thus regarding the tangent as the limit of a secant that we are able in Analytic Geometry to obtain its equation. A normal to a curve at any point is the straight line perpen- dicular to the tangent to the curve at that point. As it is often necessary to consider tangents and normals to curves, we shall seek the equations of these lines for each curve that we study. 49. Let us find general forms for the equations of a tangent and a normal to a given circle at a given point of the curve. We will use the simplest form of the equation of the circle : — ^^ +/ = <'" (i) and let P^ be the given point. By § 26, the equation of any line through P^ may be written y -yx^^ (^-Xi) (2) Equation (2) will represent a secant^ a tangent, or a normac\ according to the value given to A. Let Ag represent the slope of a secant which cuts the curve in P^ and a second point P^\ A^, the slope of the tangent at P^\ Ajj, the slope of the normal at Py If we obtain values for Ag, A^, and A^., in terms of known quantities, and substitute them, in turn, in equation (2), we 8q ANALYTIC GEOMETRY. shall have the equations of the secant, tangent^ and normal^ respectively. Instead of denoting tlie coordinates of P,_ by {_x^^y^ it will be convenient to introduce the following notation: Let M ^ and M^ be the feet of the ordinates of P^ and P^ V ] / 1 ^\ ^2 respectively, and draw through P^ a line parallel to the axis of A' cutting J/, P^ in N, and let P^N=h, NP^ = k, According to this notation the coordinates of P^ may be writ- ten (jCj -j- //, y\ -\- k). The student should satisfy himself by drawing other figures that this is true for all possible positions of P^ and P^. The slope of the secant is K = k (3) Now the difficulty of our problem consists in this, that as jP, moves down the circle towards P^ both numerator and THE CIRCLE. 8i denominator of this fraction approach zero as a limit, and it is therefore not possible to see, without further examination, what limit the fraction itself is api)roaching. This is not sur- prising, as we have not yet made any use of the fact that F ^ and P^ are both on the circle (i). This fact gives us the two equations of condition: {x, + hf + (j- + Iz)^ = a- {.) By subtracting the first of these equations from the second we get the relation 2 /^ ^1 + /^^ -f 2 ky^ + k' = o. k In order to get another expression for the fraction — we have merely to divide this equation by /i : 2 Jt', + /^ + 2 or 2 x^ + /i + (2 ;', + ^) Accordingly, We thus have A.= - 2 X, (5) 2 ;', + ^ UNIVERSfTY OF CALIFORNIA DEPARTMENT OF CIVIL ENGINEERING BERKELEY, CALIFORNIA 82 ANALYTIC GEOMETRY. This expression for Ag is less simple than that given by equa- tion (3), but it has the advantage of being in a form in which we can easily determine its limit. As P^ approaclies 7^,, both h and k approacli zero, so that the numerator of the fraction in (5) approaches 2 :r,, the denominator 2 }\. The 2 X slope Ag itself therefore approaclies the limit — — ^ . There- fore ^.= -5 (6) Since the normal is perpendicular to the tangent, vi^e have, by § 2>T,, Cor. 2, A. = -f = f (7) At X^ The equations of the required tangent and normal are, there^ fore, .>'-.>'i= -y|(^-^i), (8) and y -yx = JC-^ -^1) (9) Clearing (8) of fractions, we have y^y - y^ ^ - x^x ^r x'' 1 y transposing and reducing by means of equation (4), we have x^x +y^y = a^, (10) the simplest form of the equation of the tangent. Reducing equation (9), we have y^x - x,y = o, (11) the equation of the normal. The form of equation (11) shows us that the normal must THE CIRCLE. 83 pass through the origin (v. § 19). Now the origin is, in this case, the centre of the circle, so that we have proved the well- known theorem of geometry, that a perpendicular to a tangent to a circle, at the point of tangency, passes through the centre. Note. This method of finding the equations of tangent and normal should be carefully studied, not only on account of the importance of the tangent and normal to the circle, but because it will be used in finding the equations of these lines for each of the curves which we shall study. Equation (i) will change as we change our curve, and so equations (4) and (5) will have different forms, and the algebraic work by which we obtain the value of Ag will vary; but, in the essential steps, the same method may be used to find the equations of the tangent and normal to any curve. EXAMPLES. (i.) Find the equations of the tangent and normal to the circle x^ ■\- f" ^ 25 at the point (3, — 4). The given point is a point of the circle, because its coordi- nates satisfy the equation ; these coordinates therefore corre- spond to .Tj and j/j in the general equations of tangent and normal, while d^ — 25. The equations sought are, therefore, 3^ - 4j' = 25, and _4^_3j = o, respectively. (2.) Find the equations of tangents and normals to the circle x^ + yr- — 1 69 at all points of the curve which have their abscissas numerically equal to five. (3.) Find the equation of the normal to the circle, at a point 84 ANALYTIC GEOMETRY. jP,, by means of the fact that it passes through the centre ; find the equation of the tangent from that of the normal by means of the mutual relation existing between these lines. (4.) Find the equations of the tangent and normal to the circle x^ -\- y = 2 a X at a point J^^ of the curve. ij\x- (x^ -a)y-ay, = o. (5.) Confirm these results from equations (10) and (11), by transformation of coordinates. (6.) Obtain, by the general method, the equations of tangent and normal at jP^ to the circle {x-x,y + (y-y^)^ = a^ Ans. < [ x^ - x^^ (7.) Show that the same equations, obtained by transforma- tion of coordinates from (10) and (11), are {x, - x^) (x - x^) + (j'l - Jc) (y - Jc) = a\ and (j'l - j'c) X - (-^1 - -^c) y - -^c J'l + ^1 yc = o. (8.) Reduce the results of the last two examples to identical forms. (9.) Find the equations of the tangent and normal to the circle (x - 2)2+ (y _ sY = 10. at a point (5, 4). Ans. ;^ X + y - ig = o, ^ - sy + 7 = Q- THE CIRCLE. 85 (10.) Obtain, by the general method, the equation of the tangent at F^ to the circle x^-Yy^^-ax-^by-Vc^o. Ans. x^x + y,y + ~{x + x) + -(y + y,) + c = o. 2 2 (11.) Find the equations of the tangent and normal at the origin to the circle x"^ + y^ — 7, x — 2 y = o. 2 X — 2>y — o. (12.) The projection on the axis of x of that part of the tan- gent which is contained between (9Xand the point of tangency is called the sub-tangent ; in like manner the projection of that part of the normal which lies between its points of meeting with the curve and O X'\s called the sub-nor7nal. Find the lengths of the sub-tangent and sub-normal for the circle x^ ^ ^2 _ ^2^ Suggestion. It follows from the definition that the sub-tan- gent equals the intercept of the tangent on O X, diminished by the abscissa of the point of tangency. Ans. — , X,. (13.) Prove that the two circles given in § 45, Ex. 11, have a common tangent at the point (8, 3) ; show that the line which joins their centres is perpendicular to this tangent and passes through the point of tangency of the circles. (14.) What must be the relation between a^, b^, and a in order that the line may touch the circle X y x" +y^ = a"} Afis. a = V^i' 4- V 86 ANALYTIC GEOMETRY. (15.) When will the line y z= X X -\- b touch the circle Ans. When ^2 = ^2 (j _j. ;^2)^ (16.) When will the line X cos a + J' sin a =/ be tangent to the circle ^2 _^ ^2 _ ^2^ . Ans. When/ = ^. 50. The equation of the tangent to a circle from a given poirtS, without the curve may be found as follows : — Let P^ be the given point, and J^2 _J. j;2 _ ^2 ^jj the equation of the circle ; let P' be the unknown point ol tangency. By § 49, we may write the equation of the required tangent x' X + yy = a% (2} where x' and y' are at present unknown. Since the locus of (2) must, by the conditions of the problem, contain the point P^, the coordinates of P^ must satisfy (2) , therefore x' X, + y y,^a^ = . (3} Again, P' is by supposition a point of the circle, and, there fore, its coordinates satisfy (i), and x'-^+y'^ = a^ (4} We have now two equations, (3) and (4), connecting b} means of the known quantities, x^, y^, and a, the unknown quan ties x' and y' ; consequently we can find both x' and j'' by elimi- nating between (3) and (4). It is evident that there will be two, and only two, pairs ol 1 THE CIRCl E. 87 values for x' and y, because (3) is of the first, and (4) of the second, degree. From this we infer that two tangents can be drawn to the circle from a point without, a fact w'ell known in Geometry. If we call the two pairs of values mentioned above {x', y') and (x", y"), we shall have, by substituting them in turn in (2), the equations of the required tangents x' x + y' y = a'^, x" X -^ y" y = a% touching the circle at jP' and J^", respectively. Note, In this discussion we have supposed J^^ a point with- out the circle, because we know from Geometry that no tangent can be drawn to a circle from a point within. If we apply the same method when /^j lies within the circle, the coordinates of /" and P" will have imaginary values, showing that there are no real tangents in this case. If J^^ lies on the circle, and the same method is used, the values of the coordinates of J^' and J^" will be alike and equal to {x\, y^) ; the two tangents, therefore, will coincide in this case, a result which Geometry also teaches. EXAMPLES, (i.) Find the equations of the tangents from the point (7, i) to the circle Here, ^1 = 7, yi = I, ^' = 25. Making these substitutions in equations (3) and (4), we have 7^' +y = 25, and x'^ +y^ = 25. From these equations we find that either x' = 3 and y' = 4, or x^ = 4. and y = — 3, 55 ANALYTIC GEOMETRY. The required equations are, therefore, 3 ^ + 47 = 25, and 4^-37 = 25. (2.) Find the equations of the tangents from (—7, — i) to the circle x^ -l-y = 25. C 4-^ -37 + 25 = o. (3.) What are the equations of the tangents to the same circle from the point ( — , — ) ? Prove that the lengths of these tan- gents are equal. (4.) Find the equation of the chord which joins the points where tangents from P^ touch the circle This may be done by substituting the coordinates of /" and P^^ in the equation of a line which passes through two points, or by the following instructive method : — The equations of the tangents to the circle at P^ and P^^ are x^ X -\- y' y = a"^ and x" X + y" y =z a^. Since by supposition P^ lies on each of these lines, we have and x^ Xi + y^ _)', r= a^ x'^ Xj + y" J, = a^. These last equations may also be regarded as the equations of condition which make the line x^x +y^y^a^ THli ClRCLt;. i>9 pass through the points {x',y') and {x",y"). But we are seek- ing the line joining F' and P" ; therefore is the required equation. Remark. If F^ lies on the circle, this chord becomes the tangent to the circle at F^, as is evident from its equation. 51. A diameter of any curve is the locus of the middle points of a set of parallel chords. We will find the equation of a diameter of the circle ^2 + y2 ^ ^2 ^^^ Let the inclination of each of the parallel chords be y^, and let their slope be A^. The equations of these chords may then be written : — y ^\x ^b^ . . (i) y = \x + d, . . (2) y = X^x-^b, . . (3) etc. In the figure the line (i) represents the locus of the first of these equations ; the extremities of this chord are F^' and F^", and its middle point is F'. In like manner, F.J FJ', etc., are other chords having the equations (2), etc. By § II, corollary, the coordinates of F' are ,/ _ ,y_y'+y''. Now/'/ and 7^/' are the points where the chord (i) cuts the circle ; therefore, by § 20, their coordinates may be found by eliminating between equations (o) and (i). 9© ANALYTIC GEOMETRY. Squaring (i) and substituting in (o), we have x^ + Xj^ x^ + 2 l?i X^ X + l^j^ = a% or (I + A/-) X" + 2 /^i Xj jx: + (<^j^ — / = o the roots are - ;3 ± | -)3- ^ -4^7 2 a a Accordingly the sum of the roots is — — , and (a fact which we add for future refer- TIIF. CIRf'T^E. 91 in the (luantities /^i, A., etc., whicli enter as factors into the vahies of each coordinate ; dividing y' by x we obtain y I S = -V <5) a rehition which is independent of b^, and consequently must be true where P' represents the middle point of the chord (2), or (3), or any other of the system of parallel chords. liquation (5) expresses an invariable relation between the coorcHnates of each point of the required locus, and dropping the accents we have i=-i °'" y=-\''^ (^) tile equation of the diameter which bisects all chords which have a slope A,. Since equation (6) is of the first degree^ its locus must be a strai<;iit line ; as there is 710 constant tcrtn^ the origin is a point of tile locus. Equation (6) may be written y^^x (7) where ^ = -f (8) By §25, RemaT.v, equation (7) represents a line through the origin with a slope A; and, by §33, Cor. 2, and equation (8), this line must be perpendicular to the chords which it bisects. NoiK. 'I'liese properties are well-known pro|)erties of the (liatneter of the circle, and the equation of the diameter might have been easily found by means of them ; but we have used the general method of finding the diameter of any curve, in order to illustrate it by a simple example. 92 ANALYTIC GEOMETRY. 52. We may readily show that every chord through the centre of a circle is a diameter; for by § 25, Remark, any such chord may have its equation written in the form y = ^x (i) Now, whatever the value of A, we can find a quantity X^, such that K = -{, or A=-^. Since any quantity, from 00 to —00, may be the slope of a line, it is always possible to draw a set of chords which shall have the value of Aj, thus found, for their common slope ; then, by §51, these chords will be bisected by the locus of equation (i), and that locus must be a dia?neter of the circle. EXAMPLES. (i.) A system of straight lines parallel to 2x -^Ty = S intersects the circle x^ + / = 40- Find the equation of the line which bisects the chords thus formed. 2 Here Aj, the slope of the system of chords, is ; conse- 7 ^ quently A, the slope of the diameter, must be -, and the equation of the diameter is 7 y = -x, or 'J X — 2 y z^ o. (2.) Show, by transforming coordinates, that the equation of the diameter of the circle {x-x,Y ^ {y-y,Y = a\ THE CIRCLE. 93 which bisects all chords with a slope A,, is y — jc = ~ Y ^^ ~ ■^''•^* (3.) Find the equations of the diameters of the circles x' +/ — 4.r + 4>' — I — o, .r' -f-/4-6.T - 3J^- I =0, which bisect all chords with an inclination of 135°. (4.) Show that the extremities of a diameter of a circle whose centre is the origin are the points (a a X \ / a a\ \ (5.) Prove that the lines joining the extremities of two diam- eters of a circle are parallel. (6.) Prove that the tangents to a circle at opposite extremi- ties of any diameter are parallel. (7.) Show that the extremities of a diameter of the circle {x -x^)^ -^r {y -y,f=r'' are the points l^c ± ■ , . y. ± -j=T.X ^ V 1 + /^ V I + ^^ (8.) If through one of the points of intersection of two circles a diameter of each circle is drawn, the straight line which joins the extremities of these diameters passes through the other point of intersection, and is parallel to the line joining their centres. Prove this for the circles given in § 45, Ex. 9. 53. We will close this chapter by finding the polar equation of the circle. Let Ox be the initial line, and O the pole. Let C be the 94 ANALYTIC GEOMETRY. centre of the circle with coordinates (r^, i'>e), and let a be the radius. The coordi- nates of P, any point of the curve, are In the triangle O C P, we have by Trigonometry Now C P'-^ O C'-V OP'- 2 OCX OPx cos COP. . (i) OP^r. CP=a, OC=r, From the figure and § 3, we have C O P= XOP- XOC = {y - x%. Substituting these values in (i), we obtain a^ = 7-^ _|_ r^ — 2 ^c r cos {& — &q\ which must be the required equation. This may be written in the form r' — 2 r^ r cos (i> — t>c) + ^'^ — a^ = o. It is evident from the form of this equation that with each value of ft there will be two values for r ; a fact which agrees with the geometric properties of the circle. Corollary 1. If C is on OX, 0^ = o, and the equation becomes r^ — 2 7\ r cos i> + 7\' — a'^ = o. Corollary 2. When the pole is on the circumference, and the initial line passes through the centre, THE CIRCLE. 95 r^ — a and ^^ — o\ in this case the equation reduces to the form r — 2 a cos 0-. This equation gives but one value of r for each value of ^ ; as we have divided each term by r, however, r may equal o with each value of ^, a fact which is apparent from the hypothesis. Corollary 3. If the centre is the pole, r^. = o, and the equa- tion becomes r = ± a^ as shown in § 38. EXAMPLES, (i.) Show that the polar equation of the circle, the origin being on the circumference and the initial line a tangent, is r =: 2 ^ sin i9-. (2.) Obtain the equation of Cor. 3 from the rectangular equa- tion referred to the centre, by transformation of coordinates. (3.) Obtain the equation of Cor. 2 from that of § 45, Cor. 2. (4.) Obtain the general form of the polar equation from the rectangular equation (5.) Prove that the two values of r, given by the general polar equation, are real and different, real and equal, or imaginary, according as sin (& — d-^) is less than — , equal to — , or greater than — , in numerical value. Show that these results mav also be obtained from the figure by means of well-known properties of the circle. (6.) Prove the theorem of § 45, Ex. 13, using polar coordi- nates. CHAPTER VI. LOCI. 54. In finding the equation of the circle, we expressed in an equation its most familiar property. But the circle may also be defined by any other essential property, and we can find its equation by expressing that essential property in an equation connecting the coordinates of any point of the curve. In § 51, a diameter of a circle is defined as the locus of the middle points of a set of parallel chords, — a definition different from the familiar one of Geometry, but expressing a well-known property of diameters. From this new definition, however, we found the invariable relation which exists between the coordi- nates of every point of the locus, and we obtained the equation of a diameter by expressing this invariable relation in an equa- tion. The result was recognized as representing a straight line through the centre, and in this way we obtained from the equation the properties by which a diameter is commonly defined. If these properties were unknown to us, we should be no less certain of their truth. In like manner, if a point moves according to a given geo- metric law, we can find the equation of its path, or locus, though ignorant of the form of that path. This equation may be recog- nized as the equation of some known curve, in which case we can determine at once the locus of the moving point ; if, how- ever, the equation is not familiar, we can deduce from it the form and properties of the curve. 97 In this chapter we shall apply this method to several loci defined by properties more complicated than those mentioned above, and we shall add a number of other examples leaving the work wholly or in part to the student. 55. Let us find the locus of the vertex of a triangle when the base and the difference of the squares of the sides are given. Take the base for the axis of X, and its left-hand ex- tremity for the origin. Call the vertex /^, with coordi- nates, X = O M, y = M P ; let the length of the base, O £>, be r, and let m- repre- sent the constant difference between the squares of the sides. The definition of the locus may be expressed by the equation O^' - nT^ =pi\ (i) which will be the equation of the required locus when we express O P and D P\x\ terms of .r, r, and c. Since the coordinates of D are (<^, o), we may write, by § 10, 'OP^ = x' + y\ Wp'' = {c^xY + y\ Substituting these values in equation (i), we have .v-^+/_[(r-.r)2+/] ^m\ which may be reduced to the form 2 c X — r — m' or X — c^ + tn^ UMfVrrFcSlTY OF G,^LJFCV^NIA -ARTMENT O- CIVIL Ei-^G.:,-;.. 98 ANALYTIC GEOMETRY. From the form of this equation, we know that the locus is a straight line parallel to O F, and therefore perpendicular to the base of the triangle, at a distance from O equal to <:^ + ni^ 2 c It should be noticed that the other segment of the base is , a result which may also be obtained by taking the origin 2 c at D. (2.) Show that if the sum of the squares of the sides is given instead of their difference, the locus is a circle with the centre at the middle of the base. Find the length of the radius of this circle. (3.) Solve problems (i) and (2), taking the origin at the middle of the base (v. Ex. 5, Fig.), and show that the results agree with those already found. (4.) Given the base of an isosceles triangle, find the locus of the vertex. Ans. A straight line perpendicular to the base and bisect- (5.) Given the base and the ratio of the sides of a triangle, find the locus of the vertex. Let the base be the axis ^ of X, and its middle point the origin. Let the length of the base equal 2 c, and the ratio of C P Xo DP he m : n. With this notation, the coordinates C and Z> are ( — r, o) and (c, o) ; the vertex iPhas coordinates {x,y). We have, by § 10, CP Z>P= ^{x-cf ^y\ LOCI. 99 The equation of the required locus may, therefore, be written squaring and clearing of fractions, this becomes nr \{x - c-y- +/] = 71" \{x + cY +/], or (w- - if) x^ - 2 {7n^ + it^) c X -{■ {m^ — ti^) y + {nv^ — rF) c^ = o, which may be written x^ - 2 —^ 2 ^ ^ + y^ + r z= o. This may be readily shown to be the equation of a circle, with ^ -, c, o I and its radius equal to 2 m n m^ — n^ (6.) Given the base of a triangle, and m times the square of one side plus n times the square of the other ; find the locus of the vertex. Afis. Using the same axes and notation as in the last example, . , . /m — ft \ the locus IS a circle with its centre at i c, o i. \m + 11 J (7.) Given the base and sum of the sides of a triangle; find the equation of the locus of the vertex. Ans. If 2 n^ — ^^), where each side of the square is 2 a and the given constant is represented by 8 nt^. (10.) Find the locus of a point whose distances from two straight lines, given by their equations, have a given ratio, m : n. Let the equations of the lines be Ax-^By^C— o, (i) A^ x-\. B^ y + C'^o; (2) and let P^ with coordinates {x\y^) be the moving point. The distances of P^ from lines (i) and (2) are, by § 36, Ax^ ^ By^ ^ C ^ A' x' + B' y' + C - — and — V^' + £' Va'-' + ^" Putting m times one of these equal to n times the other, we have the equation of the locus, where x' andy are the variable coordinates. Clearing of fractions and writing x for x' and y for y', we evidently have an equation of the Jirsf degree; there- fore the required locus is a straight line. (11.) Show that the equations of the bisectors of the angles between the lines A X -\- By + C= o and A^ x ^ B^ y ^ C = o are Ax + By+C ^ A'x + B'y+a VA^ + B' V'A'^'^B"' ' (12.) A point moves so that the sum of the squares of its distances from the four sides of a given square is constant ; LOCI. I C T. show that the locus of the point is a circle ; find its , ':eritre, ,and> radius. ^ ^' ' > J . . > ^ ' , >' ; *-.', ,^ '. \ (13.) Find the locus of a point, the square of whose distance from a given point is proportional to its distance from a given line. A^is. A circle. Suggestion. Take the fixed line for the axis of y, and the perpendicular upon it from the fixed point for the axis of x. (14.) Find the equation of the locus of a point whose dis- tance from a fixed point always equals its distance from a fixed line. Ans. y = 2 Xj^x — x^ [axes as in 13]. 56. There are many problems in whfch it is difficult to express at once the given conditions in terms of the coordinates of the moving point and known quantities. In such cases we may introduce other variable quantities, and then find enough equa- tions expressing necessary relations of the parts of the figure to enable us to eliminate the variable quantities thus introduced. We shall then have an equation which fulfils the requirements of the definition of the equation of a curve. (i) To illustrate, let us find the locus of the middle points of rectangles inscribed in a given triangle. We will take the base and perpendicular from the vertex as axes, and will suppose the tri- angle given by means of the in- "x tercepts which the sides cut from these axes. / Y F / \ F / -.p/'"' "'["^^-. \ \ A I D M ( a B X Let OA=a, OB OC = b. 102 ANALYTIC GEOMETRY. Let I) E, F G represent any rectangle inscribed in the tri- angle,, and iec its middle point be P. We can easily express the coordinates of P in terms of those of E and F; for, from the figure and § ii, it is evident that x must equal the half sum of x^^ and x^ (the abscissas of E and F), and y^ one-half the common ordinate of these points. Call this common ordinate k; then the equation of ^ i^ may be written y-^k (i) By § 22, the equations of ^C and B C are and - + i=I, (2) a P ^-f=' (3) Since E and F are the points where the line (i) meets the lines (2) and (3), we have by § 20 therefore x^ + Xp (a -f. a') (b -k) •^ = ~~2~~ = Vb ^ ^4^ also for the point P y = i (5) Equations (4) and (5) contain the coordinates of P, the vari- able k, and the known constants a, a^ and b. We may, there- fore, by combining these equations eliminate k and obtain an equation connecting x and y by means of the given quantities ^, a\ and b. This equation is _ (^ + a') {b - 2 y^ LOCI. 103 or 2 X 2y the equation of a straight line bisecting the base and perpendic- ular of the triangle, for its intercepts are a + a^ , b — ^ and Remark. The student will remember that, in finding the equation of a diameter of a circle, we first expressed the coor- dinates of any point of the locus in terms of another variable, the intercept of any chord on O Y. That problem, therefore, is one of the class we are now considering. (2.) A line is drawn parallel to the base of a given triangle and its extremities joined transversely to those of the base ; find the locus of the point of intersection of the joining lines. With the same axes and notation as in the last example, the equations of the joining lines may be formed by § 27. Finding the coordinates of the point where these lines meet, we have, after eliminating k^ the equation of the locus, 2 X y which represents the medial line drawn from the vertex of the triangle. (3.) A line of constant length moves with its extremities on two strai2:ht lines at right angles to each other ; find the locus of its middle point. This may be done conveniently by introducing as a third vari- able the angle which the moving line makes with one of the fixed lines. Ans. A circle. Compare §11, Ex. 9. (4.) A square is moved so as always to have the two extrem- ities of one of its diagonals upon two fixed straight lines at right I04 ANALYTIC GEOMETRY. angles to each other ; show that the extremities of the other diagonal will at the same time move upon two other fixed straight lines at right angles to each other. (5.) From one extremity A' oi a fixed diameter ^'^ of a given circle, a secant is drawn through any point P' of the cir- cumference ; B.t I" 3. tangent is drawn to the circle, and a perpen- dicular to the tangent is drawn from A, and extended to meet the secant in J^. Find the locus of jP. Take the fixed diameter as the axis of x, and the centre as origin. We shall use as auxiliary variables the coordinates of J^', The points A' and A have coordinates (- a, o) and (a, o). We can now write the equation of A' J^' by § 27, and oi A F by § 2)2>^ Cor. 2 ; for we know the slope of the tangent at F' by § 49. Since F is the point of intersection of A' F and A F, its coordinates may be found by § 20, and will be X — 2 x' -If- a, y = 2 y, Expressing x^ and y in terms of x, y, and a, by means of these equations, and substituting the values thus found in the equation of condition which puts F' upon the circle, x^"" +y^ = a\ we have for the equation of the locus {x - ay +y'^ = 4a% which represents a circle with its centre at A, and A A^ for its radius. (6.) Given the base of a triangle and the length of the medial line drawn from one of its extremities ; find the locus of the vertex. Take the base as the axis of x, and the end from which the medial line is drawn as the origin. Introduce as auxiliary vari- ables the coordinates of the other extremity F^ of the given medial line. LOCI. 105 Ans. A circle with centre at the point (— c, o) and radius 2 w, where c and m represent the lengths of the base and medial line. (7.) To find the locus of the centre of a circle which has a given radius and passes through a given point. Let F' be the unknown centre. The equation of the circle may then be written {x-x'Y + {y-yy = a\ Making this pass through the given point F^, we have changing the signs of the quantities in the parentheses (but not of their squares) and writing x and y for the variables x' and y', this equation becomes (x-x,y + (y -y,y = a\ the equation of a circle with P^ for the centre, and a for its radius. Remark. This equation may be easily obtained from the fact that the distance of the moving centre from the fixed point must be equal to the given radius ; but the method used above is instructive. (8.) Find the locus of the middle points of chords of a given circle, drawn from a fixed point on the circumference. Ans. A circle with the radius of the given circle for its diameter. Compare with § 45, Ex. 13. 57. Instead of expressing in an equation the geometrical definition of the locus, and then substituting for each quantity in this equation its value in terms of the coordinates of the moving point and known quantities, it is often more convenient to use as the primary equation one which expresses some necessary relation of the parts of the figure other than the one explicitly io6 ANALYTIC GEOMETRY. given. From our hypothesis we may be able to express all quantities in this equation in terms of the variable coordinates and known quantities, and so find the equation of the locus. We will illustrate by an example in which this method is used in connection with that of § 56. (i.) Given the base and sum of the sides of a triangle, if the perpendicular be produced beyond the vertex until its whole length is equal to one of the sides, find the locus of the extrem- ity of the perpendicular. Take the axes as in § 55, Ex. I ; let the base be r, and the sum of the varying sides be m J let M F ht made equal to O C, whatever the position of C. We may, therefore, write OC = MF=y, CD^m Now, from the figure, it is evident that y- and MC" = CZ>^ (9 J/2 ^ ^2 MD'' = (m - yf (c - x)\ Putting these values oi M C^ equal to each other, we have the expression of a necessary relation between x and j', in an equa- tion which contains only those quantities and the given con- stants m and c. Expanding and reducing, this equation becomes 2 c X — 2 7ny + m^ €" = which is of the first degree, and therefore represents a straight line. (2.) A line of constant length moves with its ends on two straight lines at right angles to each other, and perpendiculars LOCI. 107 to the lines are raised at its extremities; find the locus of their intersection. Ans. A circle. 58. Polar coordinates may often be employed to advantage, (i.) To tind the locus of the middle points of chords of a circle, which pass through a given point. Let the given point p ^„^ "N^ be the pole, and the ^,^^ X line joining it with the centre be the initial 19 [ \ line. Let F' F" be O \ C j X any chord which passes through (9, and let^Pbe \^ yr its middle point, with coordinates /' and &. It is evident from the figure that i> will be the same ior F, F\ and F^\ and that ^ = —J- (0 The equation of the circle may be written, by § 53, Cor. i, r- — 2 ;-c r cos i> + Tc^ — equals X O F", the roots of this equation are r' and r"; therefore, by equation (i),* r = r^cos d-, (3) and similarly with every value of i> ; for though for certain val- ues of l^, r' and r" are imaginary, yet r, their half-sum, is always real and equal to r^ cos />. Equation (3) is, therefore, the equation of the required locus, which, by § 53, Cor. 2, must be a circle with r^ for a diameter. * V. foot note on p. 90. lo8 ANALYTIC GEOMETRY. (2.) Find the locus of the middle points of chords drawn from the extremity of any diameter of a circle. Compare with § 56, Ex. 8. (3.) Find the locus of a point which moves so as to divide in a given ratio all lines joining a fixed point with a fixed straight line. 59. A great many problems in loci lead us to equations of an unfamiliar form. We have had an illustration of this in Ex. 7, on page 99. In such a case the locus will usually, as in the example just referred to, not come under any of the forms of curves we have so far studied. There is, however, one method by which we can sometimes determine what the locus of an unfamiliar equation is. Consider, for example, the equation xy = (i) which is in an unfamiliar form since it is of the second degree and yet is not of the form of the equation of a circle (§ 46). It is clear that any point whose abscissa or whose ordinate is zero lies on the curve (i), but that no other points lie on it. The locus of (i) therefore consists of two parts : first the axis of X, second the axis of Y. As a second example let us consider the equation x'^ — X y ■= Q ' . • • . (2) This equation may be written x{x - y) = o ^ • • . (3) It will therefore be satisfied by the coordinates of any point whose abscissa is zero, i.e., of any point on the axis of Y. There is, however, another way in which (3) can be satisfied, namely by the equality o{ x and>'. The locus of (2) or (3) LOCI. lor therefore consists of two parts: first the axis of K, a: = o second the straight line x — y = o. Moreover it is clear thai a point not lying on either of these two lines does not belon*; to the locus, since for such a point neither x nor x — jv is zero, and therefore (3) is not satisfied. The general principle of which the two foregoing examples are illustrations may be stated as follows: Suppose that all the terms of the equation we wish to investigate have been trans- posed to the left-hand side, and suppose that this left-hand side can then be factored into two factors which we will de- note by the letters u and v. The equation we wish to investi- gate may therefore be written uv = o (4) where we must not forget that the letters ?/ and v stand for certain more or less complicated expressions involving the variables x and r. Then the curve (4) consists of two and 071/ y two parts, first the curve ^ = o (5) and second the curve (6) Let (^ijJi) be any point on the curve (5) ; {x^,y^ any point on the curve (6) ; and (^v,, j,) any point which does not lie on either of the curves (5) or (6). The statement we have made will be proved if we can show that (a",, J,) and {,x^^y^ both lie on (4), but (jCg, y^ does not. In order to find out whether ix^^y^ lies on (4) or not, we must replace the variables {x,)') in (4) by the constant values (;c,,>',) and then see whether the equation is fulfilled. In do- ing this we must remember that u is merely an abbreviation for a certain expression in the variables x and y. This expression will therefore take on a certain constant value, which we will 110 ANALYTIC GEOMETRY. call u^, when these variables are replaced by the constants x^, y\. In the same way the expression v takes on a constant value i\ when x, y are replaced by ^, , )\. Thus equation (4) takes on, after this substitution, the form z/j z^j = o (7) and it remains to be proved that this is a true equation. By hypothesis [x^,y^) lies on the curve (5). Accordingly its coordinates satisfy (5) and we have and from this the truth of (7) follows. We have thus proved that (x^, y^) lies on (4), and by pre- cisely the same method we show that ('^2jJ^'a) does so. In order finally to show that (^^3, ^3) does not lie on (4) let us indicate by ti^ the value of the expression tt when the vari- ables x,y are replaced by the constants x^,y^, and by z^, the value of V after the same substitution. The result of substi- tuting ^3,>', in (4) is then ^^3 ^s = o (8) This equation, however, is not true, since, the point (^3,^3) lying on neither (5) nor (6), neither u^ nor v^ is zero. The result of this section may be stated by saying that if we have two curves whose equations are so written that the right- hand -members are zero,* the two curves may be represented by a single equation obtained by multiplying these equations together. * This restriction is very important, for if the right-hand members are not both zero the fact here stated will not be true. Thus if we have the two lines x = \, x = 2 and multiply these equations together as they stand, the resulting equation x"^ = 2 repre- sents the two wholly different lines x = ± V2. In order to get an equation represent- ing the two lines we started with we must write their equations x — 1=0, ^ — 2 =0, and then multiply; getting,^^ — 3^ 4- 2 = o as the equation of the pair of lines. LOCI. Ill EXAMPLES. What are the loci of the following equations ? — (i.) x"^ —y = o. (2.) X' + 2 xy —3^ = 0. (3.) x'^y +y' — ^y — o. (4.) Jt^* — / — 9 ^' + 9/ = o. (5.) Prove that if «, v, w are three expressions in x^ y, the equation 7^vu' = o has as its locus the three curves : u = o, V = O, IV = o. (6.) Show that the first members of the following equations can be factored : ax'^ + bx + c=^o^ ax"^ + b xy + cy"^ = 0. What do these equations represent ? 60. In the last section we have seen what the result is of multiplying together the two equations u =^ o • • (i) ^' = o (2) Let us now consider what the result of adding these equa- tions is. That is, we wish to find out as much as we can about the locus of the equation u ■\-v^o (3) Let (^i,jV,) be any point common to the two curves (i) and (2), i.e. any point of intersection or of contact of these curves, and let us, as in the last section, denote by u^ and i\ the values of the expressions u and v when the variables x^y are replaced by the constant values x^^y^. Now since (:v,, y^ lies on (i) we have 7/, = o. 112 ANALYTIC GEOMETRY. Since (jc,, jv,) lies on (2) we have v^ = o. Accordingly u, -{-v^ = o, i.e. {x^,y^) lies on (3). Thus we see that (3) passes through all the points common to (i) and (2). Now let (^,, y,) be a point of (i) which does not lie on (2). Then, letting «,, v^ denote the values of u and Z' when the vari- ables X, y are given the values -x^^y^, we have u, = o, ^, 7^ o- Therefore u^ + v^ ^ o. That is (.r„ y^ does not lie on (3). In the same way we see that a point (x^, j',) which lies on (2) but not on (i) will not lie on (3). Thus we have established the fact that if we add together the equations of two curves, the resulting equation will have a locus which passes through all the points common to the two curves, but meets neither curve in any other point. We can easily throw this result into a slightly more general form. The equations of the curves (i) and (2) may, if we wish, be written in the form a u =^ o (4) bv-o • . (5) where «, b are any constants other than zero. By applying the principle just established to the curves (4) and (5) we get the theorem : LOCI. 113 Ifu=^o and v = o are the equations of two gii en curves^ then a u -^ bv= o represents a curve which passes through all the points common to the two given curves, and ?neets neither of them in any other point. Just what curve of this sort we get depends, of course, on the values of the constants a and b. 61. As an illustration of the way in which this principle can be applied let us consider tlie problem of finding the equation of the common chord of two intersecting circles. Consider, for instance, the two circles. x^ + y^ — 2 X — 4y + 4 = o (i) x'^+y^ — 4x — 2y + i=o. . . . '. . . . (2) These circles intersect, as the reader will see if he draws a rough figure. In order to find the equation of their common chord let us denote the first member of (i) by «, the first member of (2) by v. Then the equation a u -h b V = o (3) will represent some curve through the two points of intersec- don of (i) and (2). But it is clear that, in general, the equa- tion (3) is of the second degree and therefore does not repre- sent a straight line. If, however, we give to a the value i, to b the value — i, the terms of the second degree cancel and the equation (3) becomes u — V = o or 2X — 2y -^ S = o (4) 114 ANALYTIC GEOMETRY. This, being of the first degree, represents a straight line, and, being merely a special case of (3), this line must pass througii the points of intersection of (i) and (2). Thus (4) is the equa- tion we were seeking. The equation (4) was obtained by subtracting the equation of one of the given circles from the other ; and it is clear that the same reasoning applies to any similar case. That is, to find the equation of the common chord of two given intersect- ing circles, write the equations of these circles so that the coefficients of x"^ and j''^ are t, and then subtract one equation from the other. EXAMPLES. (i.) A circle has its centre at the origin and radius i, and a second circle has its centre at the point ( — 1,2) and radius 2. Find the equation of the common chord of these circles. (2.) Show that the three lines 2 X + y ~ ^ =■- o, X - 2y + 5 = o, 4x — 2,y + T = meet in a point. [Suggestion: Show that if the first two equations are de- noted by « — o, 7' = o, tlie third can be written in the form ^ // + <^ 7' = o ] (3.) Find the equation of the line which connects tlie origin with the point of intersection of the lines X — 2y — 3 = 0, 2 X — J' + 5 =r O. [Suggestion : Determine the constants a and b in tlie equa- tion a u -\- b Z! ^= o, so as to eliminate the constant term.] LOCI. 115 (4.) Prove that the three common chords of three intersect- ing circles meet in a point. [Suggestion : Let the circles be x^ + y"" + a^ X + if^y + <^i — o> ^' + y + a^ X + t?^y + c^ = o, x"^ + f + a^ X -\- b^ y + c^ = o. Find their common chords by § 61, and then use the method of example 2.] CHAPTER VII. THE CONIC SECTIONS. 62. An ellipse is a curve generated by a point moving in a plane so that t^^ sum of its distances from two fixed points in that plane is constant. An hyperbola is a curve generated by a point moving in a plane so that the difference of its distances frojn two fixed points in that plane is constant. A parabola is a curve generated by a point moving in a plane so that at each instant its distances from a fixed point and a fixed line in that plane are equal. These three curves are called the conic sections, because, if a right circular cone is cut by a plane which does not pass tlirough the vertex of the cone, the section will be bounded by one of these curves. We will find the equations of the conic sections. Y 63. In finding the equation of the ellipse, we will take the straight line joining the two fixed points, which are called foci, for the axis of x, and the point midway between the foci, called the centre, for the origin. In the figure, F^ and F are the foci ; P is any point THE CONIC SECTIONS. II7 of the curve; and r' and r, called /^^^/ radii^ measure the dis- tances of P from F' and F^ respectively. Let 2 c represent the constant distance between the foci, and 2 a the sum of the focal radii, which is constant by the defini- tion of the curve. In the triangle F' FP the side F' F — 2c is less than 2 a, the sum of the other two sides. Tlierefore c < a (1) The coordinates of F' and F are {— c, o) and {c, o), and the equation which expresses the definition of the locus is r' + r ■= 2 a. Expressing / and r in terms of x, y. and c, by § 10, r' = ^{x^ cY +J-, (2) r ^ y/yx - ^^f~Vy' (3) Substituting these values, we have the equation of the ellipse, a/(^ + cf .f / + ^/{x -cy ^ y = 2a, (4) which will be in a more convenient form when cleared of radi- cals. Equation (4) may be written V{x + cy^ +f = 2 a- V{x-cy +f', squaring, . {^x + cy + y'' = 4a^ - 4a V{x - r)- + y^ + (x - cf + / ; expanding, combining, and transposing, 4 a ^/{x — c)~ -Y y'^ — \a^ — ^c X'y dividing by 4, and squaring again, d^ x^ — 2 or c X -\- a} c^ 4- (r y^ — a^ — 2 a^ c x -\- c^ x^ \ reducing, {a^ - c') x' + a'y' = a'' (^^ - c') (5) We will represent the constant quaniity {a* — c^), which, as we see from (i), is positive, by d"^; with this notation, (5) becomes ii8 ANALYTIC GEOMETRY. or, dividing by d^ b"^. b'^x' + a'^y' = a^b\ X' y* ~2 + A2 a" b the usual form of the equation (6) (7) 64. With the same clioice of axes, the equation of the hyper- bola may be found in a similar manner. In this case we represent the constant difference between the focal radii by 2 a. Since the side F' F — 2 c oi the triangle F' FP is greater than the difference 2 a oi the other two sides, we have a (i) Accordingly we use F' as an abbreviation for / — ^^ With this notation show lliat llie equation of the hyperbola is (2) Note. Since the equations of the ellipse and hyperbola differ only in the sign of z^"", the results of smiilar operations performed on each will differ in the same way. One of these results may, therefore, be obtained from the other by changing the sign of ^^, a method which will enable us to infer many properties of the hyperbola from the corresponding properties of the ellipse, or vice versa. 65. In finding the equation of Xheparabo/a, we will take the fixed straight line, called the directrix, as the axis of y, and the perpen- dicular upon it from the fixed point, or focus, as the axis of x. In the figure O V is the direc- trix, F the focus, and F any ^ point of the curve. Letting m THE CONIC SECTIONS. IIQ represent the constant distance of the iooM^from the directrix, the coordinates of F are (w, o). The equation which expresses the definition of the locus is FP ^ EP. , . (i) By § lo FP ^ ^/{x - my + /, Also FP = O M ^x. Substituting these values in (i), we have, after squaring, {x — 77iy- -i- y^ = x'^f . . . c (2) which reduces to y^ — 2 m X -{- m^ = o, (3) the equation sought. 66. A simpler form of the equation of the parabola may be found by changing the origin to the point midway between the focus and directrix. The formulas for transformation are m x = x'-\.-, y=y, and the new equation is, therefore, which becomes, after reducing and dropping the accents, f- — 2 m X, the usual form of the equation of the parabola. Note. The student will remember that we have already found the equations of the conic sections in the chapter on loci. It will be useful for him to compare the definitions of the conies and their equations with examples 7, 8, 14, of § 55. 6y. We will now investigate the forms of the conic sections by discussing their equations. y''^ — 2 m ( ^^ + — I + w^ t^O ANALYTIC GEOMETRY. Beginning with the ellipse, we find its intercepts on O X to be a and — a\ on 6> K, b and — b. Solving equation (6), § 63, forjv, we have or y= ±^7^1^?, (I) from which it appears that with each value oi ^ there are two values of _>', equal numerically but opposite in sign; this shows that the curve is symmetrical with respect to OX. Similarly solving for x^ we have X=±j V'^^ (2) From this equation we infer that the curve is symmetrical with respect to O Y. From (^i) we see that the values of y are real when x is less than a, null when x equals a, and imaginary when x is greater than a, in numerical value ; also, as x increases, y must decrease. From these considerations we infer that the curve cuts the axis of y at two points equally distant from the origin ; that it extends from these points both to the right and the left, gradu- ally approaching the axis of x in each direction, and meeting that axis at points whose distances from the origin are a and — a; and, also, that there are no points of the curve farther from O Kthan the points in which it cuts OX, Similarly, by examining equation (2), we see that the curve is limited in the direction of the axis of y by two points at dis- tances b and — b from the origin, lying wholly between these points. Since the ellipse is symmetrical with respect to each axis, it must be symmetrical with respect to their intersection, the origin or centre of the curve ; therefore every chord drawn through the centre is bisected at that point. It is easy to prove this THE CONIC SECTIONS. I2t analytically ; for the equation of any line through the origin is, by § 25, y = ^x> (3) and the equation of the ellipse is ,2 (4) a- A' Now if (^', y) is one point of intersection of the loci of (3) and (4), it is evident, from the form of these equations, that ( _ x\ — y^) will also satisfy both and be the other point of intersection. Since the distance of each of these points from the origin is V-^'" + J''^ the origin or centre must be the middle point of each chord drawn through it. B^ are at distances a^ — a, b, — b from O respec- tively. The ellipse must cut the axes at these points, and will be in- scribed in the rectangle whose sides are parallel to the axes and pass through the points A, A', B, B'. Its form, which is shown in the figure, may be deter- mined with more precision by plotting a number of points, or by the following method based on the definition of the curve. Let the extremities of a flexible string, whose length is 2a, be fixed at the foci, F' and F; let F, the point of a pencil, move in the plane, keeping the string tight ; its path must be the ellipse sought. The transverse, or major, axis of an ellipse is that part of the straight line through the foci, contained by the curve ; it is repre- sented in the figure by A^ A. 122 ANALYTIC GEOMETRY. The co?ijugate, or minor, axis is that part of the perpendicular to the major axis, through the centre, which is included by the curve ; it is shown in the figure as B^ B. The student has probably noticed that a, the half-sum of the focal radii, is equal to the semi-major axis of the ellipse, and that b is the semi-minor axis. In § 63 we introduced b as an ab- breviation for ^/ a" — ^'; hence b < a. Thus we see the reason for the terms major axis and minor axis. These results may be obtained directly from the figure ; for by the definition of the curve, F^ A-\- FA z=2a, and, therefore, since A' F' equals FA, A' A = 2 a and O A = a ; also, since F' B ^- FB = 2 a, and i^'^ equals FB, we have, from the right triangle O FB, OB = a/^2 - c' = b. The eccentricity of an ellipse is the ratio oi c to a ; this will always be less than unity [v. (i), § 63]. EXAMPLES. (i.) The distance between the foci of an ellipse is 6, and the sum of the focal radii of any point is 10 ; find its equation. x" 1'^ Ans. — + -^ = I. 25 16 (2.) Determine the semi-axes of the ellipse x" + 2/ = 8. This may be put in the form of equation (7), § 63, by dividing each term by 8, the constant term. This gives THE CONIC SECTIONS. 123 which is the equation of an ellipse for which cr — 8 and <^^ = 4. The required semi-axes are, therefore, a — 2 V^ and b = 2, (3.) What are the semi-axes of the ellipse ' 16 A-^ + 257^ = 1600? Letting e represent the eccentricity of the elllpserSfeew-WM (4.) Determine a, b, and e for the ellipse 169 25 c^ rS (5.) The equation of a chord of the ellipse ^2 _^ 7 / _ 16 = o, IS ^-7j + 4 = o; find the length of the chord. rr-3 c^ Aas^Ji,/t^sz (6.) The /cz///i- rectum, or parameter, of a conic is the breadth of the curve through a focus ; it is, therefore, equal to twice an ordinate erected at a focus. Show that the latus rectum of an ellipse is equal to — . (7.) Prove that the minor axis of an ellipse is a mean propor- tional between the major axis and latus rectum. (8.) Find the equations of the focal radii drawn to a point P^ of the ellipse. Ans. \y,x- {x, +c) y + cy^ = 0, \ y^x - {x^ - c) y-c y^z= o. (9.) Find the points of intersection of the circle and ellipse whose equations are 124 ANALYTIC GEOMETRY. x^ -^ y^ — lo, x^ + y y^ = i6c Ans. (3, i), (3,-1), (-3, i), (-3» -i)- (10.) Find the points of intersection and the length of the common chord for the curves whose equations are x^ + y'^ -{■ 20 X -\- ^o = o, x'^ -{- J y'^ — 16 = o. Ans. (-3, i), (_ 3, _ i), 2. (11.) Prove that the semi-major axis of an ellipse is a mean proportional between the intercepts on that axis of the lines which join the extremities of the minor axis with any point of the curve. 68. It is worth while for us to notice that the circle is merely the special case of the ellipse in which the two foci coincide, for in this case c — o^b — a^ and the equation of the ellipse reduces to the familiar form of the equation of a circle. It may be noticed that the eccentricity c of the circle is zero. 69. To investigate the form of the hyperbola, we first find its intercepts on the axes. Obtaining these from the equa- tion (2) of § 64, we find the intercepts on the axis of x to be a and —a; those on the axis of j to be /^ V— i and — b V" i, which are imaginary. The hyperbola, therefore, does not cut the axis of _y, but since it has points on each side of that axis, it must comprise at least two parts or bra7iches. Solving equation (2), § 64, for y and x, we have y ^±- \/x- -^a\ (i) x = ±- Vb-' +y\ (2) equations which show that the curve is symmetrical with respect to each axis. From equation (i), we see that the values of ^ are imaginary THE CONIC SECTIONS. 125 when X is less than a, null when x equals a, and real when x is greater than a, in numerical value. These facts show that there are no points of the curve, either above or below O X, between two points at distances a and — a from the origin, while beyond these points the curve extends indefinitely in each direction of that axis. We learn from (2), that all values of y give real values of x ; consequently we infer that the curve runs without interruption in the direction of the axis of 7, extending indefinitely in the positive and negative directions. In either (i) or (2), we ob- serve that X and J' must increase together ; therefore, the curve recedes from both axes at the same time. We may show, as for the ellipse, that every chord drawn through the centre of an hyper- bola is bisected by the centre. The curve represented in the figure agrees with these results. It may be constructed by finding a number of its points, or, by a method * based on the definition of the curve, as follows: — A pencil F is tied to a point near the middle of a piece of string. Two pegs are driven into the drawing-board at the points 7^ and F\ and the string is looped around these pegs in the manner indicated in the figure. The two parts of the string are grasped in one hand at H and drawn away from F'. The pencil P will then describe an arc of an hyperbola with foci at F and F\ for the two distances F F and F' F are evidently decreased by the same amounts, so that their differ- * See pnpers by Ransom and Huntington in the Annals of Mathematics, Second Series, Vol. Ill, p. 164, and Vol. IV, p. 50. 126 ANALYTIC GEOMETRY. ence remains constant. The otlier l^ranch may be generated in the same manner by placing the pencil F at some point of the other branch before drawing the string tight at H, In the equation of the hyperbola, § 64, we used b to represent ^c^ — a- ; it will not, as for the ellipse, be the length of the in- tercepts on O V, for the hyperbola does not cut O V. We shall, however, call that portion of the axis of y contained between points at distances b and — b from the origin^ or centre, the con- jugate axis of the hyperbola ; its relation to the curve will be seen hereafter. The definitions of the transverse axis and eccentricity are the same for the hyperbola as for the ellipse. The transverse axis will lie between the two branches of the curve ; the eccentricity must be greater than unity, because c is greater than a (v. § 64), and, for the same reason, the foci F^ and F lie one within each branch of the hyperbola. THE CONIC SECTIONS. I27 EXAMPLES. (i.) The distance between the foci of an hyperbola is 10, and the difference between the focal radii of any point is 8 ; find its equation. Ans. -^ - — = 1. 16 9 (2.) Determine the semi-axes and eccentricity of the hyper- bola 9 x^ — \6y'^ = 576. Using the same method as in Ex. 2, § 67, we have a = 8, d = 6, e =^. 4 (3.) The equation of a chord of the hyperbola ^ x^ — g y'^ — 64 = o is 2 X — gy — 8 = 0; find the length of the chord, and determine the semi-axes and eccentricity of the hyperbola. A//S. Length of chord is A/85 ; e - — — . 3 2 //-^ (4.) Show that the latus rectum of the hyperbola is ; and state the theorem for the hyperbola analogous to that of § 67, Ex. 7. (5.) Show that the equations of the focal radii drawn to a point J^^ of the hyperbola are the same as the equations of like lines for the ellipse. (6.) Find the points in which the following curves meet: — x^ 5 y^ x^ 7 y'^ ^"^"87 "' ^"'sT " '• Ans. (4, ±3), (- 4, ±3)- 128 ANALYTIC GEOMETRY. 70. We will now consider the equation of the hyperbola in polar coordinates, as in this way a good deal of light will be thrown on the shape of the curve. Using the formulae of §43, we have as the equation of the hyperbola r' cos* r' sin' __ a' d' ""^' accordingly r= ± = ± cos" sin' \a^ cos (p ^^ ~2~ tan^ Using the upper sign here and starting with the value = o, we have r = a. As increases, the denominator of r de- creases, and therefore r itself increases. When has the value 00 for which tan 00 = - ' the denominator of ;' vanishes. Accordingly as increases and approaches the value 0^ as a limit, r increases indefinitely. When is greater than 0^ the expression under the radical sign is negative and r is therefore imaginary. Thus we see that in the first quadrant the hyperbola lies wholly below the line through the origin with slope -• In a similar manner we could discuss the shape of the curve in each of the other quadrants, or we can even more easily deduce the result from THE CONIC SECTIONS. 129 the symmetry of the curve. By either method we find that the hyperbola lies wholly between the two lines through the origin with slopes ± — , while any line through the origin cuts the curve if it makes with the axis of X a smaller angle than these lines do. These two lines y = ± —x a are called the asymptotes of the hyperbola. 71. The most remarkable property of the asymptotes is that the hyperbola approaches them indefinitely as it is extended from the centre, but never reaches them. We may prove this prop- erty as follows : — Draw corresponding ordinates for the hyperbola and the asymptote O L. These are represented in the figure by M^ P^ = y^ and M^ P; — y^. The equation of O Z, the asymptote in the first and third quad- rants, is b y=-x, a From this equation and that of the hyperbola, we obtaiff 2 *' / 2 therefore j,^=^«^ = *'; 130 ANALYTIC GEOMETRY. from which we may write or p.p:=y. ' ' y; + )■, Now as P^ moves away from the centre, jc, increases indefi- nitely, and the equations of the hyperbola and asymptote show that the accompanying values of y^ and jj^/ must also increase indefinitely. The value of P^P^' must therefore approach zero as its limit, for the numerator, ^^ is constant, while the denominator increases indefinitely. The distance of P^ from the asymptote, measured on a perpendicular, is equal to P^ P ^ multiplied by a constant quantity, the sine oi P^P^' 0\ therefore this distance varies with P^P^', and approaches zero together with it. This proves the theorem. It may be shewn in like manner that a point below the axis of X, or one on the other branch of the hyperbola, approaches one of the asymptotes as it recedes from the centre, but can never really reach it. EXAMPLES. (i.) Find the equations of the asymptotes of the hyperbola x^ -4/ = 36. Ans. X — 2 y — o, x -\- 2 y — o. (2.) Show that the distances of either focus from the asymp- totes equal half the conjugate axis. (3.) Discuss the shape of the ellipse by means of polar co- ordinates. (4.) Prove that if from any point on an hyperbola a line is drawn parallel to either axis, the product of the parts cut off between the point and the asymptotes equals the square of half that axis. THE CONIC SECTIONS. 131 72. Since in the hyperbola c is greater than a^ and ^' = c^—a^, b may be less than a^ equal to «, or greater than a^ according as c a is greater than, equal to, or less than /=, V2 When we make b equal to a^ the equation of the hyperbola becomes oc" - / = a\ and the curve is called an equilateral hyperbola. The asymptotes of the equilateral hyperbola are y - ± X, and since these lines are perpendicular to each other, this hyperbola is sometimes called the rectangular hyperbola. 73. To investigate tlie form of the parabola, we will discuss the equation found in § 66. The intercepts of this curve are all o. If we solve the equa- tion for y and x^ we have (•) and y — -^ Vs m x^ X = 2 711 (2) Y P E —p^ D "Oi F X We learn from equation (i) that the curve is symmetrical with respect to OX. Also, if in is positive, all positive values of x give real values of j-, but negative values of x make y imaginary ; in this case the curve must lie wholly on the right of O Y. In like manner we may show that when in is negative, the curve is wholly to the left of O Y. From (2) we see that all values of j^ 132 ANALYTIC GEOMETRY. give real values of x ; consequently the curve is continuous and indefinite in extent, in the direction of the axis of j. Either (i) or (2) shows that x and j must increase together; therefore the curve recedes from both axes at the same time. The parabola represented in the figure agrees with these results. It may be constructed by points, or by the following method based upon the definition of the curve : — Let the leg E F^ of the right-angled triangular ruler E F G slide along the directrix D' D ; fix one end of a string, equal to E G, at the focus F, and attach the other end to the ruler at G. A pencil point F, which presses the string tightly against E G, must move on a parabola. The straight line through the focus and perpendicular to the directrix is called the axis of the parabola, and the point in which it cuts the curve is called the ve7-tex of the parabola. The equation of the parabola found in § 66, is called the equation referred to the vertex and axis, or simply, to the vertex. EXAMPLES. (i.) What is the equation, referred to the vertex, of the para- bola whose focus is at a distance 7 from its directrix ? What is the equation of the same curve referred to its axis and directrix ? Ans. y^ — \\x; y"- — 14:^ + 49 = o. (2.) Show that the second locus constructed in § 17 is a parabola ; how far is its focus from the directrix } how far from the vertex ? (3.) Find the latus rectum of the parabola. Ans. 2 m. THE CONIC SECTIONS. ^33 (4.) Find the equations of the chords which join the vertex to the extremities of the latus rectum, the vertex being the origin. Ans. y = 2 X, y -\- 2 x = o. (5.) What is the equation of the circle which passes through the three points mentioned in the last example ? Ans. 2 x^ -j- 2 y^ = c^m x. (6.) The equations of two curves are 2,x'^ + 4^ = 48 and ()x — 2y^ — o. What are the curves, and where do they intersect ? 74* We will next find the equations of a tangent and a normal ^o an ellipse at a point /\ of the curve. y In the figure, P^ T is the tangent, and P^ JV the normal. The coordinates of P^ are x^ = O M^ and y,^M,P,. Using the same me- thod as in § 49, we have, corresponding to equation (4) of that section, the following equations, obtained from equation (6), § 63: — b'x;" + a'y^ ^a-'b\ (i) b\x^ + hY + a\y, + ky=aH' (2) Subtracting (i) from (2), we may write 21?' hx^ + b' h'' + 2a^ky^ + k'a' = o, from which we obtain 134 ANALYTIC GEOMETRY. Taking the limit of this as ^ and /i both approach zero, we have P X and I a'^y^ ^^-y, =^ (5) The equation of the tangent is, therefore, /y'x, , y-yx ^-^/-^-^i)' • • (6) and of the normal ^ - ^1 = 1^ (^ - ^i) (7) Clearing (6) of fractions, and transposing, we have b^ x^x ■\- a'^j\ y — b^ x^ + ^^ J'l^ which, by equation (i), may be written b'^ x^x + a^y^y — d^b'^, (8) or -^+7^='' (9) the usual form of the equation of the tangent. Clearing equation (7) of fractions, we may write it as fol- lows : — d^ y^ X - Ir x^y - {a^ - b'^) x^y^ = o, (10) a form which may readily be obtained from (8) by the method of § 34- EXAMPLES. (i.) Find the equations of tangent and normal at (i, 2) to the ellipse 3 x'' -f 4/ = 19. THE CONIC SECTIONS. 135 Since the coordinates of the point satisfy the equation of the curve, the point lies on the ellipse, and, finding the semi-axes, we may write the equations of tangent and normal as follows : — X 2 y 1^ + -yT = '• or 3^ + 87 = 19, and 8^ - 3 J = 2. x^ y^ (2.) Find the equations of tangents and normals to the ellipse 12+ 6 - ' at all points of the curve whose abscissas are numerically equal to 2. (3.) Calculate the intercepts on the axis of x of the tangent and normal drawn at a point J^^ of the ellipse. Ajis. — , — — 5— X, or e^ x.. x^ a^ ^ ^ (4.) Show that the lengths of the sub-tangent and sub-normal (i~ — X '^ b^ for a point F^ of the ellipse are ^ and -^^j, respectively. (v. § 49, Ex. 12, and the figure of this section, where J/, T and N M^ are the required lines). (5.) By means of these results, prove that if a series of el- lipses have the same major axis, tangents drawn to them at cor- responding points (i. e., points which have the same abscissa) meet on the axis of x. (6.) Find those points on an ellipse where the tangent is equally inclined to the axes. / a"- . b^ \ 136 ANALYTIC GEOMETRY. (7.) Find the equations of the tangents to the ellipse, which are incHned at an angle of 45^ to the major axis. Ans, x—y + 2=0, x—y — 2 = 0. (8.) Solve the last example when 45° is replaced by f tt. Ans. X -\- y — 2 =0, x-\-y + 2=o, (9.) Find the equation of a tangent to the ellipse -^ + i:^ = I. at the extremity of the latus rectum. Ans. ± Vs X ± ^ y = g a represents the four tangents. (10.) From the points where the circle on the major axis is intersected by the minor axis produced, tangents are drawn to the ellipse ; find the points of contact. Ans. The extremities of the latera recta. (11.) Find the condition which must be fulfilled that the line f +Z = i + m n may be tangent to the ellipse oc^ y ~ + I2 = '• Ans. __+__ = I. m- n^ (12.) Prove that if with the coordinates of any point of an ellipse as semi-axes, a concentric ellipse be described with its axes in the same direction, the lines which join the extremities of the axes of the first ellipse are tangent to the second. 75- The method of finding the equations of a tangent and normal to an hyperbola at a point of the curve \? entirely analo- THE CONIC SECTIONS. 137 or gous to that shown in the last section for the ellipse. Show by this method that the equation of the y^ tangent at J^^ is •^1 ^ _j',y_ J . a' "If- '' ' and of the normal. (j-7i)=-^j(^-^i), (0 (2) (3) Note. These equations evidently differ from the correspond- ing equations for the ellipse only in the sign of b^, according to the principle stated in the note to § 64. EXAMPLES. (i.) Find the equations of tangent and normal to the hyper- bola at the point (3, a/z). 2 x^ — 3 j^ = 12, Ans. \ ^^-A^/ = 4, ( V2 -^ + 27 = 5 V2. (2.) Find the equations of tangents and normals to the hyper- bola ^2 y^ = I. 9 3 at all points of the curve whose ordinates are numerically equal to 3. (3.)^ Show that the expressions for the intercepts on O X oi a tangent and normal at a point J^^ of the hyperbola are the same as for the analogous lines in the ellipse. \ 138 ANALYTIC GEOMETRY. (4.) What is the property of hyperbolas analogous to that stated in § 74, Ex. 5 ? (5.) Show that the lengths of the sub-tangent and sub-normal of the hyperbola have the same expressions as for the ellipse. because the change in the sign of b^ does not affect the numeri- cal value of the sub-normal. (6.) Find those points on the hyperbola at which a tangent has an inclination of 45° ; of 135°. (7.) Is the problem of the last example possible for every hyperbola ? If not, what are the exceptions ? (8.) Find the equations of normals to the hyperbola gx^ - 2sy" = 225, which have an inclination of 45°. (9.) Find the equations of tangents and normals to an hyper- bola at the extremities of the parameters. Ans. For the upper extremity of the right-hand latus rectum, ex — y = a, ax + aey — e{d^-{- b'^). (10.) Find the condition which must be fulfilled that the line X y — -f — = I m n may touch the hyperbola x^ y^ d' ~ i^ ^ ^' 76. Let us find the equations of a tangent and a normal to a parabola at a point F^ of the curve. Using the simplest form of the equation of a parabola THE CONIC SECTIONS. 139 which gives and y"^ = 2 m X . . . (i) we have for the equations from which to find the value of A., y{' = 2mx^. . . (2) and (7,+^)-^-2^K+^). (3) PVom these we easily find 2Jr -\- k X.-"^ ^-J. m The required equations are, therefore, m or and or y-y^=-^(x -x^\ y^x + my = {m -\- x,) y\, . (4) .(5) . (6) . (7) , (8) . (9) (10) EXAMPLES. (i.) Find the equations of tangent and normal to the para- bola at the point (i, 2). r = 4-^, Ans. ^ (x -y + I =0, \x + y - 2 =: o. 146 ANALYTIC GEOMETRY. (2.) Obtain the equations of tangents and normals to the parabola y =. 8 ;^ at all points of the curve whose abscissas are equal to 8. (3.) Show that the intercepts on O X oi the tangent and nor- mal at a point /\ of the parabola y'^ = 2 mx are — x^ and m + x^, respectively. (4.) Find the lengths of the sub-tangent and sub-normal for the parabola ; show that the first is bisected at the vertex, and that the second is constant. (5.) Prove that the tangents at the extremities of the latus rectum are equally inclined to the axes of x and y, and meet the directrix on the axis of ^ (v. § 67, Ex. 6). (6.) Form the equations of normals to the parabola at the extremities of the latus rectum ; show that they are mutually perpendicular and intersect on the axis of the curve. What kind of quarilateral is bounded by these normals and the tangents mentioned in the last example ? What is its area ? r 2 {x +y) =2>^, Ans. •< 2 {x — y) =z ^ m, ( Area = 2 fn^. (7.) Show that if the focus of a parabola is the origin and the axis of the curve, the axis of x, the equation of the tangent at J^j, a point of the curve, is y^^y z= m (x -}- x^ + m). (8.) With the same axes as in the last example, find the tan- gents and normals at the extremities of the latus rectum. :F>'+W = o, ±y — m = o. Ans. < Ix THE CONIC SECTIONS. I4t (o.) Write the equations of the directrix and iatus rectum, re- ferred to the vertex. (lo.) Show that the tangent at any point of the parabola meets the directrix and Iatus rectum produced at points equally distant from the focus, the common distance being (ii.) Find the condition that the line may touch the parabola a b y^ — 2 m X, Ans. 2 b^ + a 771 77. Instead of finding the tangent to the ellipse *>-■ (0 at the give7i poi7tt (:r,, j,), it is sometimes necessary to find the tangent to this ellipse which has a givoi slope A. It is obvious, when we consider the shape of the ellipse, tiiat there are two such tangents. Their equations can be found as follows. Denoting the coordinates of the point of contact of one of these tangents by (^,, J^.), its equation may be written x^ X = I, (2) P'rom (3) we have X, b' h' y = ^x -\ y^a and the equation of the tangent can be written y = Xx -\- (3) (4) " rrW CIVIL ENGINE U. ol C. ASSl!fUTl(i\' I ER1N( Mkl 142 ANALYTIC GEOMKTRV. It remains merely to replace jVj in this equation by its value in terms of A. This value we find as follows. Squaring (4) and clearing of fractions, we have I b'x^ - a'X'y,' = o (6) On the other hand, since (Xi,J\) is a point of the ellipse (i), it satisfies this equation, and we have d'x; + a\v^' = a'd' (7) By multiplying (7) by ^' and subtracting (6) from it we elimi- nate x^ and get or Substituting this value of jk, in (5), we get as the desired equa- tion of the tangent y = Xjc ± Va' X' + b' (8) The two signs here give us the two tangents with slope A. In the same way we find that the equations of the tangents to the hyperbola which have the slope A are y^Xjc ± Va'A:' - b' (9) It should be noticed that this equation becomes imaginary when A is numerically less tliAt - , so that the hyperbola has a no tangents which make a smaller angle with the transverse axis than that made by the asymptotes. In the case of the parabola it is easily seen that there is only one tangent with a given slope A. By using the method ex- plained above, its equation is found to be y = Xx+~ (10) 2/L THE CONIC SECTIONS. I43 EXAMPLES. (i.) Deduce the results of this section by starting from the equation y ^= X x -]- p and imposing the condition that the points in which this line cuts the conic should coincide. (2.) A right angle moves so that both of its sides touch an ellipse. Find the locus of its vertex. (3.) Solve the last problem if instead of an ellipse we have a parabola. (4.) Find the equations of the tangents from the point (14, i) to the ellipse x"^ + 4}''^ = 100. Suggestion: Call the coordinates of the point of contact of one of these tangents {x^,y^), and then determine the values of these coordinates. 78. Let us find the lengths of the focal radii (r and r' in the figure of § 6;^) drawn to a point /^j of the ellipse. By § 10, we have r^ = (x,-cy+y,\ (I) r'-' = {x, +cy+y,' (2) Also we know that the coordinates of I'j^ must satisfy the equa- tion of the ellipse, and therefore y,' = ^,(a^-x,% (3) Substituting this value of j'/^ '^^ equation (i), we have after expanding and collecting, r^ = ""'-/ x,'-2cx, + c^^ + P (4) c Now a- - b' - r and therefore c^ + /^'^ = a^ ; also e = - and consequently c - a e. We may therefore write equation (4) as follows : — r^ — e^ x^ — 2 a e x^-^ a^ ', and r —e X. — a, or a — e X, (5) 144 ANALYTIC GEOMETRY. As we only wish the length of r, we shall use the second of these values, because the first is necessarily negative, e being less than unity and x^ less than a. In like manner show that r^ — a ^ e Xy (6) This may also be proved from (5) and the definition of the curve. EXAMPLES. (i.) Prove that if on any ellipse three points be taken whose abscissas are in arithmetical progression, the corresponding focal radii of these points are also in arithmetical progression. (2.) Prove that if the ordinate of any point of an ellipse be extended to meet the tangent drawn at the extremity of the latus rectum, the length of the extended ordinate equals that of the corresponding focal radius of the point. (3.) The ordinate of any point of an ellipse is extended to meet the circumscribed circle, and a tangent is drawn to the cir- cle at its extremity ; prove that either focus is equally distant from the point and the tangent. 79. We may easily find the segments, F^ TVand N F^ cut by the normal from the base of the triangle F^ P^ F, in the figure of § 74, by adding and subtracting O TV^and c. They are F^ N ^ a e -^e^ x„ JV F = a e - e^ x^. Comparing these values, we have F' JV _ a e + e^ x^ _ a a- e x^ _r* JV F a e — e^ x^ a — e x^ r Since the base is divided into segments proportional to the adjacent sides, we know, from Geometry, that tJie angle between the focal radii drawn to any point is bisected by the normal at that point. THE CONIC SECTIONS. I45 The normal is the internal bisector of the angle between the focal radii, consequently the tangent perpendicular to the normal must bisect the supplementary external angle. 80. For the hyperbola we may show, as in § 78, that for the right-hand branch (when x^^ is positive) the focal radii are r' = e Xj^ + a, and r — e x^ — a ; (i) for the left-hand branch (where x^ is negative) r^ ^ — e x^— a, and r — a — e x^ (2) EXAMPLE. Show that the theorems stated in the examples i and 2 of §78 are true for the hyperbola. 81. The tangent to an hyperbola at a point P^ will be found to cut the base of the triangle F^ P^ P {% 75? Figure) into segments a" a"- pf T^ae + -. TP=ae -—- Comparing these segments, we have P' T _a e x^ + d^ _e x^ j^ a _r* T P a e x^ — (T- e x^ — a r ' The tangent therefore bisects internally the angle between the focal radii of the hyperbola drawn to the point of tangency, and consequently the normal at that point bisects the supplementary external angle. 82. The results of §§ 79 and 81 show us that when an ellipse and hyperbola have the same foci, the curves cut each other at right angles. For, at the points of intersection, the curves have the same focal radii, the tangent to the hyperbola bisects the in- ternal angle between these radii, and the tangent to the ellipse bisects the external angle. These tangents are therefore mutu- 46 ANALYTIC GEOMETRY. ally perpendicular; and, since a tangent to a curve at any point has the direction of the curve at that point, we learn that co7i- focal co?iics intersect at right a?igles. 83. It follows from the definition of the parabola, that the focal radius F F^ (v. figure of g 76) drawn to a point F^ of the 7n curve is equal to Now we may easily show, by the results of Ex. 3, § 76, that TF is equal to x^ + therefore. the triangle T F F^ is isosceles, and the tange?it to a parabola at any point must make equal angles with the axis and the focal radius drawn to that point. From this it follows that if at any point of the curve the focal radius is drawn and a line parallel to the axis, the tangent and normal at that point bisect the external and internal angles be- tween these lines. 84. If we describe a circle on the major axis of an ellipse as a diameter, corresponding or- dinates of the ellipse and cir- cle will have a constant ratio, b : a. In the figure, y^ {= ^i F^) and _>'/ (= Ml jP/) are corre- sponding ordinates, because they are drawn with the same abscissa x^ (= O M^). From the equations of the ellipse and circle, we have for the values of these ordinates y, = ^-^/^::l^*, I _ v^ . 3 . '1 S THE CONIC SECTIONS. 147 therefore, remembering that a and x^ are by hypothesis the same in these equations, which was to be proved. It may be shown in like manner that when a circle is inscribed in an ellipse, with the minor axis as a diameter, corresponding abscissas of ellipse and circle have the constant ratio, a : b. By means of these properties v/e can construct an ellipse by shortening each ordinate of the circumscribed circle in the ratio ^ : ^, or by lengthening each abscissa of the inscribed circle in the ratio a \ b. We can also draw a tangent at any point of an ellipse by drawing the tangent to the circumscribed circle at the corre- sponding point, and joining the point in which this tangent meets the major axis produced, with the point of the ellipse at which the tangent is to be drawn ; for the circumscribed circle may be regarded as one of the series of ellipses mentioned in § 74, Ex. 5. EXAMPLE. Two tangents to the circle whose equation is x^ -f y- — 100 meet in the point (14, 2) ; find the equations of tangents to the corresponding points of the ellipse or + 4 y^ = 1 00. Ans. 3 .r + 8 J' = 50, 2 ^ - 3 j = 25. 85. By means of the theorem of § 84, we may calculate the area of an ellipse. Let A' A^ in the figure of § 84, be divided into any number of equal parts, and suppose M^ and M.^ two of the dividing points. Erect corresponding ordinates at these points, and at their ex- J4S ANALYTIC GEOMETRY. tremities draw parallels to ^^'^4, forming pairs of corresponding rectan;;ks, of which J/, ^ and J/, ^' are one pair. Now from the figure we see that M^R^ ~ M^M^ X yl ~ // " a Sinnilarly we may show that the area of any corresponding rectangles of ellipse and circle are in the ratio h -. a ; therefore the sum of the rectangles of the ellipse is to the corresponding sum for the drcle as ^ : dr. Now if we increase indefinitely the number of equal divisions of A' A, the sum of the rectangles of the ellipse approaches the \\7iM'ZXt2i of the ellipse as a limit, and the rectangles of the circle approach the half-area of the drcle as a limit ; therefore, by the ITieory of Limits, the area of the ellipse is to the area of the circle as ^ : '^ ~ loo. (2.) Find the ratio of the areas of the ellipses rjx^ + 16 y= 144, 4^^+ 9/ ^ 144- ^;2j. 2:3:4. 86. J.ct us find the equation of an hyperbola whose foci lie- on the axis of r, whose transverse axis is 2 <^, and whose conju- gate axis is 2 ^/. We will first find the equation of an hyperbola, with its foci on O V, wi h 2 i . iri ly be (AA'dlned from the cqwdUon of § 64, by turning the •nil', CONIC sKC'iioNS. 149 axes through an an^lc of 90". Usin;^ tlic forniuhis for Iraiis- formation given in § 42, (Jor., we have for tiie equali(jn of this hyi)erl)ohi , ^2 -,-77 = 1 (I) The transverse axis of the recjuired hyperl)ohi has tin; same length as the conjugate axis of this one, and lucc iwrsa ; (luM-e- fore we shall obtain the required ecjuaticMi by intercliaii;;iii;; a and b in (i), Tliis gives which must be the equation sought. This hyperbola is said to l)e conjugate lo llic one wIios(; (-(jua- tion was found, and llic two are closely connected. We may therefore delinc; lonju^atc hyperbolas as lliose which have the same axes, the transverse axis of one being the conjugate axis of the other, and the conjugate axis of the one the transverse axis of the other. EXAMPLES, (i.) I'uid the ecjuation of the hyperbola conjugate to X' - 4/ =z 36, and determine its foci, axes, and eccentricity. Ans. 4/ x' rz: 36, (o, 1 3 V.S), 3, 6, Vs, are the eqiialion, foci, semi transverse axis, semi-conjugate axis, and eccentricity, res[)ectively. (2.) Find the equations of tangent and normal Xo the hyper bola conjugate to x^ _ y' _ a" ~ b' ~ '' at a point /\ of the curve {.Vxy x^x Ans. y^^ "a^-^ ' '''"^ (b'x,y + a'y, x = {a"" + b^) x, y^. ISO ANALYTIC GEOMETRY. 87. C 071 jugate hyperbolas have the same asymptotes. For, turning the axes through 90°, and interchanging a and b the equations of the asymptotes to the hyperbola a' found in § 70, we have X = for the equations of the asymptotes to the hyperbola a , b x = ± -y, or y = ±-x, (0 21 Comparing (i) with the equations of the asymptotes in § 70 we see that the equations are alike, which proves our proposi- tion. 88. In the accompanying figure,we showa pair of conjugate liy- perbolas with their common asymptotes 6^Z and O L'. A' A and B^ B are axes of both curves; the one with A^ A for its transverse axis has foci 7^ and F^ ; the one which has B B' for its transverse axis has foci F^ and F^ . The reader may easily deduce the following properties from those already proved and the definitions which have been given : The four foci lie on the circumference of a circle whose radius is c. The rectangle which has its sides parallel and equal to the axes, and the common centre of the hyperbolas for its centre, ^ %^ F, ^^ ^' \ ^"--^ i \^ / 1 A' ^^ b' A ,F \ ^ F^' X. THE CONIC SECTIONS. I51 lies wholly between the curves, having each side tangent to one of the branches of these curves, and the asymptotes of the hyperbolas for its diagonals. An ellipse with the same axes as the hyperbolas will be in- scribed in this rectangle and will touch the hyperbolas at A, A' , 89. The definitions of the conic sections which we gave at the beginning of this chapter fail to bring out the fact, which appears on closer study, that these three curves are merely different species of a single genus. It will therefore be of in- terest to give a new definition which covers equally the three cases of the ellipse, the hyperbola, and the parabola. We shall in this way gain a more comprehensive view of these curves and learn some new facts about them. Boscovich's Definition. — A Conic Section is the locus of a point which moves in a plane so that its distance from a fixed point of the plane bears a constant ratio to its distance from a fixed line in the plane. The fixed point is called \.\\q focus, the fixed line iht direc- trix, and the value of the constant ratio the eccefitrtcity of the conic section. We will now examine this definition and show that it is not in contradiction with the definitions previously laid down. Let us take the directrix in the above definition as axis of Y and the perpendicular dropped upon it from the focus F as axis of X. We will call the eccentricity c, and the distance from the directrix to the focus ;;/, so that the coordinates of F are (w, o). Denoting by {x,y) the coordinates of the moving point P which traces out the locus, we obtain at once as the equation which expresses the definition of the curve V{x- mf +7^ (i) ^WrVEF^SITY OF CALJF01c,NiA 152 ANALYTIC GEOMETRY. This equation, when simplified, takes the form (i — e^)x^ f y — 2mx -\- m^ = o (2) This is the equation of the conic section. If (? = I this equation reduces to equation (3), § 65, and in fact it will be seen that in this case our present definition of a conic section is the same as our old definition of a parabola, and our use of the terms focus and directrix coin- cide in these cases. We have merely introduced the new term eccentricity, which up to this time we had not used in the case of the parabola, in such a way that the eccentricity of every parabola is i. Considering now the cases where ^ ^ i,let us find the inter- cepts of our curve on the axis of X. This is done by letting ^ = o in (2): ( I — i we see that (5) represents a hyperbola whose semi-transverse axis is a, while its semi- conjugate axis is b = a VT^^i (9) Moreover, we have em FC==OC- OJ^= ^=ae= Va' -^ d\ . (10) I — e so that F is really a focus of the hyperbola according to our earlier definition. As in the case of the ellipse we see that our new definition of the eccentricity coincides with the old. We also see that the hyperbola has two directrices whose equations are (8). The ellipse and hyperbola are sometimes called the central conies in distinction to the parabola, which has no centre. THE CONIC SECTIONS. 155 90. From the point of view of the last section the three kinds of conic sections are distinguished from one another by the value of the eccentricity, and we have An Ellipse when i. Thus we see that the parabola may be regarded as a form of curve intermediate between the ellipse and the hyperbola. If in equation (2) we keep m, the distance from the directrix to the focus, unchanged and allow e, starting with a value less than I, to increase, we have an ellipse whose major axis (v. (4)) and whose minor axis (v. (6)) are both increasing. If we allow which is in the form where ^=-4 (7) From the form of the equation, we know that the diameter passes through the origin or centre , but it is not, as for the cir- cle, perpendicular to the chords which it bisects, for the slopes of chord and diameter do not satisfy the test of perpendicularity. It may be shown in the same manner as for the circle (v. § 52) that every chord through the centre is a diameter. 93- Using \ to represent the slope of that diameter which bisects chords with a slope \, let us find the slope of the diame- ter which bisects all chords whose slopes are A^. Calling the required slope A, we have by equation (7) of the last section '=-^; ^'^ where Substituting this value of Ag in (i), we have DIAMETERS. POLES AND POLARS. 159 ^ = K (2) the slope of the original set of chords. Therefore, if one of two diameters bisects chords parallel to tht other ^ the second will also bisect all chords parallel to the first. Such diameters are called co7iju- gate. In the figure, P^ P^ and P^ P^ represent a pair of conjugate diameters of the ellipse, and each of these lines will bisect all chords parallel to the other. The product of the slopes of a pair of conjugate dia- meters is constant; for we readily obtain from equations (i) and (2) \\ = -r. (3) Note. The results of this section may be obtained at once from equation (7) of the last section by writing it in the form b"" \ = where Xj bears the same relation to A which, in the original form, X bore to \. From this we may draw the inferences stated in this section. Equation (3) shows that the product of the slopes of conjugate diameters is constantly negative ; therefore one of these slopes must be positive and the other negative. It follows that the in- clination of one of the diameters must be between 0° and 90°, and of the other between 90° and 180°, except in the special case where the diameters are the axes of the ellipse, (v, § 25, Rem.) 6o ANALYTIC GEOMETRY. Therefore conjugate diameters of an ellipse lie on opposite sides oj the minor axis. 94. Changing the sign of IP- in the equations of the last two sections, we have for the equation of a diameter of the hyperbola bisecting chords with a slope \. a^\ X : and for the product of the slopes of conjugate diameters ^1 ^2 = -2* (i) (2) Let the student verify these equations by finding the equation of a diameter of an hyperbola, and show that if one diameter of an hyperbola bisects all chords parallel to the other, the second bisects all chords parallel to the first. In the figure P^ P^ and P^ P^ represent a pair of conjugate diameters ; P^ P^ extended bisects all chords of either branch of the hyperbola, which are parallel to P^ P^; and P.l P^ extended bisects all chords lying between the two branches of the curve and parallel to P^ P^. We limit Pc^ P^ by the conjugate hyperbola, as will be explained later. It follows from equation (2) that conjugate diameters of an hyperbola lie on the same side of the conjugate axis, as shown in the figure. Also, the numerical value of the slope of one diameter must be less than -, and of the other greater than -, except in the case DIAMETERS. POLES AND POLARS. l6l where each equals — and the diameters coincide with one of the ^ a asymptotes. We learn from this that one diameter has a less inclination than the asymptote, and the other a greater; therefore one lies in the angle Z' O Z, and the other in the supplementary angle. We may also state this fact as follows : — Of two conjugate diameters of an hyperbola, oJily one meets the hyperbola, the other meeting the conjugate hyperbola. EXAMPLES. (i.) Find the equation of the diameter of the ellipse X- y^ which bisects the chord whose equation is 3 a: + _>' + 2 = o. Ans. 3 ^ — i6 j^ = o. (2.) What is the equation of the diameter conjugate to the one found in the last example ? Ans. 3 X -\-y — o. (3.) Find the equations of a pair of conjugate diameters of the hyperbola x^ — 12 y^ = 48, one of which bisects the chord whose equation is 2^-9 J = 3. Ans. ^x T^S y and 2 x = g y. (4.) Find the equation of the diameter of the ellipse 4x^ + gy^ = 144, which has for one extremity the point (3 Vs, 2) ; what is the other extremity of this diameter ? Ans. 2x = ^Vsy, ( - 3 V3» - 2). 1 62 ANALYTIC GEOMETRY. (5.) Find the equation and extremities of the diameter con- jugate to the one found in the last example. Ans. 2x + V3y = o, (3.-2 V3) ( -3) 2 V^). (6.) One extremity of the diameter of the ellipse ,2 +-72- = ! is the point {x^, y^ ; what is the other extremity ? what the equation of the diameter ? Ans. (-x^, -y,),y^x = x,y. (7.) What is the equation of the diameter conjugate to the one found in the last example ? . x,x y^y (8.) Remembering that (x^, y^) is a point of the curve, prove that the extremities of the diameter of Ex. 7 are the points Oi.-^^.) and (-f/p^^i). (9.) Show that the tangent to an ellipse at any point of the curve is parallel to the diameter conjugate to the one which passes through that point ; and that tangents at opposite extrem- ities of any diameter are parallel. Prove the same property for the hyperbola. (10.) Find the equation of the diameter of the hyperbola x^ y^ which meets the curve in the point -Pj ; also, the equation of the conjugate diameter. X, X y^y Ans. y,x-x,y=o, -^^ ^ = °- (11.) Show that the coordinates of the points where the sec- DIAMETERS. POLES AND POLARS. 163 ond diameter meets the hyperbola are imaginary, their values being (12.) Show that the second diameter meets the hyperbola conjugate to the one given in example (lo) in the points (a b \ ( a b \ (13.) Find the equation of the diameter which bisects all chords with a slope A^ in the hyperbola conjugate to and show that it coincides with the corresponding diameter of this hyperbola. (14.) Show that in the figure of § 94, /*/ P^ and P/ ^2 ^^^ ^ pair of conjugate diameters for each of the two conjugate hyper- bolas. (15.) Find the equation of the diameter of a parabola, which bisects all chords whose slope is \ ; show that all diameters of a parabola are parallel to the axis. m Ans. y = —. 1 (16.) Prove that when a parabola lies on the right oi O Y (i. e. when in is positive), the diameter bisecting parallel chords whose inclination is less than 90° lies above the axis of the curve ; the diameter bisecting chords whose inclination is greater than 90° lies below the axis. State the corresponding theorem when ;;/ is negative. What is the diameter which bisects chords perpendicular to the axis ? 95- By means of the results of example (8) of the last section we can find the lengths of a pair of conjugate diameters of an l64 ANALYTIC GEOMETRY. ellipse in terms of the coordinates of an extremity P^ of one of them. For, since the centre is the middle point of each diam- eter, if we let a^ and b^ represent the halves of the conjugate diameters, we have a^^ = x:^^y,\. (I) and a V" '"--wy'-^^'^' (^) Now F^ is a point on the ellipse, and therefore yi' = ^, (a' - X,') (3) Substituting this value in equations (i) and (2), we have a''^ = X,' + d^' - ^^x,' = b' + e' x,\ (4> and ^d'-e'x^ (5J Equations (4) and (5) give the squares of the half-diameteis in terms of .Tj, y^^ and quantities which are know^n when the ellipse is given, and from them the lengths of the diameters may be found. 96. For the hyperbola, one of the semi-diameters is imaginary and the expression for its length contains the factor V — i, for we have shown in § 94> Ex. 11, that only one of a pair of coniu- gate diameters meets the hyperbola in real points. Using , having re- gard to sign, if CA_DA , . ~CB~ DB ^'^ This equation may also be written AC BC AD B D (2) The left-hand side of (2) is the ratio in which the point A divides the line CD, while the •4 1 i ! ' right-hand side is the negative of the ratio in which B divides this line. Accordingly (2) expresses the fact that A and B divide the line CD harmonically. Hence the theorem : If the points C, D divide the line A B harmonically^ then con- versely the poi?its A, B divide the line CD harmonically, 98. If the three points A, B, C are given, we can in general construct a fourth point D so that C, D divide A B harmoni- cally. This is evident from Plane Geometry, since if C divides A B internally all we have to do is to construct the point di- viding this line externally in the same ratio^ and vice versa. Analytically we can see the same fact as follows. Let the coordinates of A be {x, , )\), of B^ {x^,y^, and let m^ : m^ be the ratio in which C divides A B. Then (v. p. 9) the coordi- nates of C are hn^x^ — m,x^ m,y, — fn,y \ ^ ^ ^ , \ 7n, — ?n. ' m^ — m, I DIAMETERS. POLES AND POLARS. 1 67 and the point D which divides A B \n the latio — m^ : m^ has the coordinates \ m^ + in^ ' m^ + 7n^ j ^ ^ We thus see that there is no difficulty in finding the point D except in the one case in which w, = — tn^', i.e.^ when C bisects the line A B. In this case it is also obvious geometrically that the point D cannot be constructed, since every point on the line A B produced is nearer to one end than to the other. It is often more convenient to use a single letter to denote the ratio in which C divides A B. If we write the formulae (i) and (2) for the coordinates of C and £> take the form V I - /^ -' I-/. ] (2) \"TT7^' 1. ^ fx] ' " ' ^4) 99. By the />o/ar of a fixed point B with regard to a given conic is meant the locus determined as follows. Through B a straight line is passed cutting the conic in C and B>. On this line a fourth point Q is constructed so that B, Q divide C B> harmonically. The line B C B> is then allowed to revolve about B, the point Q being constructed on every position of the line in the manner just indicated. The locus of Q, which we shall prove to be in all cases a straight line iS called the polar of B. Let us consider first the case in which the coriic is an ellipse. i68 ANALYTIC GEOMETRY. As usual we will use the major and minor axes of this ellipse as coordinatvf axes, writing its equation in the form -H a'y' = a'b\ (i) The coordinates of P shall be [x^.y'). Call the coordinates of the moving point Q, whose locus we want to find {x\y'). Since by hypothesis F^ Q divide C Z> harmonically, conversely C, D must divide F Q harmonically (v. § 97). We will intro- duce as an auxiliary variable (v. § 56) the ratio in which the point C divides the line F Q. Calling this ratio //, we have by formulae (3) and (4) of § 98 as the coordinates of C and D respectively (x,-^ix' y, - ^y \ , /x^ /Ax^ y, + ^iy'\ ,. V I + /( ' I + yw / ^^^ DIAMETERS. POLES AND POLARS. 169 Since C and D lie on the ellipse, we have the following two equations obtained by substituting the coordinates (2) and (3) in (i) and then clearing of fractions : H'ix, - )i x'Y + a^ (/. - l^y'Y = a-f(i-iAY . . (4) nx, + ^ix'Y + a^ (y, + M/r = «'^' (I + My . . (s) Between these two equations we must eliminate /a. This can be done by subtracting (4) from (5), getting 2// i^' X, x' -^ 2 }Aa ^r, y' = 2 }.ia''b'' (6) The factor 2 yw can be cancelled out from this equation, and we then have as the et^uation of the polar, after dropping the accents, b'x.x + a'yj^a'b' (7) or This equation, being of the first degree, represents a straight line. Replacing b"^ by — b'^, we find as the polar of the point {x^,y^) with regard to the hyperbola -^a"- ¥ = ^^ (9) the straight line ~a^-l^ = ^ (^°^ Show by the same method that the polar of the point (^lO'i) ^'it^^ regard to the parabola y = 2 /;/ jc (11) is the straight line y^y — ni {x + x^) (12) I70 ANALYTIC GEOMETRY. If a = l>f the ellipse (i) becomes the circle ^'+/ = a^ (13) Accordingly the polar of (:v„ y^) with regard to this circle is the straight line x,x+y,y=a'' (14) EXERCISES, (i.) Find the polar of the point (2, — i) with regard (a) to the ellipse — + <- = i ; 9 4 (^) to the hyperbola 2 x'' — 3^ = i ; (c) to the parabola j'^ = 6 x. (2.) Prove that the polar of a point with regard to a circle is perpendicular to the line joining the point to the centre of the circle ; and that the radius of the circle is a mean propor- tional between the distances of the point and its polar from the centre of the circle. (3.) Find the equation of the polar of the point {x^,y^) with regard to the hyperbola conjugate to (9). Hence prove that the polars of a point with regard to two conjugate hyperbolas are parallel. What more can you say about the relative position of these two lines ? (4.) Find the equation of the polar of the point (x^,y^) with regard to the conic (2) of § 89. Hence show that the polar of a focus of a conic is the cor- responding directrix. 100. It will be noticed that fornuike (8),(ro), (12), and (14) for the polars of (x,, y^) with regard to the ellipse, hyperbola, parabola, and circle are identical with the formulae we obtained in the last chapter (§§ 74, 75, 76) and in § 49 for the tangents DIAMETERS. POLES AND POLARS. 171 to these curves at the point (x^,yj. The difference is that in finding the tangent we assumed that (^,, y^) lies on the curve, while we have now assumed that (^i,J,) does not lie on the curve. In fact the definition of the polar which we have given obviously breaks down if jR lies on the conic. It is cus- tomary, however, in this case to speak of the tangent of the conic at I* as being the polar of I^. In this way the formulae of the last section represent the polar of (-^i ,71) even if this point lies on the conic. There is one other case in which our definition of a polar breaks down, namely when P lies at the centre of the conic ; for in this case all chords through P are bisected there, so that the point Q can never be constructed. T/i£ centre of a cofiic has no polar. lOI. If P lies on the convex side of the conic, we can draw two tangents from it touching the conic at P^ and P^ (v. the 172 ANALYTIC GEOMETRY. figure). In this case the lineP^P^ is the polar of P. For if we allow the secant P C D \o turn around P and approach the position of the tangent P P^diS a. limit, both the points C and D approach the point P^ as their limits, and since Q always lies between C and P>, it must also approach P^ as its limit. Therefore the locus of Q points directly towards P^ and, if continued, passes through P^. In the same way we see that the polar passes through P^ ; so that, being a straight line, it must coincide with the line P^ P^. 102. We have seen that every point (with the single excep- tion of the centre of the conic) has a definite polar. If the line A B is the polar of Py then conversely P is called the po/e of A B\ and now it is a fact that every line {ivith the excep- tion of Hues through the centre of the conic) has a definite pole. In order to prove this theorem let us suppose first that our conic is the ellipse 7' + !=' (') Let the given line be Ax ^ By -^ C= o (2) In order that (^lyyi) should be the pole of (2) it is necessary and sufficient that its polar ^^ + ^ - ^ =° (3) should coincide with (2), and the condition for this (v. § 30) 's that the coefficients of (2) and (3) be proportional : •^1 _ ^ Ji _ ^ Z~~"C' l?'~~'C' Accordingly Aa' Bb"" DIAMETERS. POLES AND POLARS. 173 is the pole of (2), — a perfectly definite point if C is not zero, that is, if (2) does not pass through the centre of the ellipse. A similar proof holds in the cases of the hyperbola and the parabola. 103. The most important property of poles and polars is expressed in the following : Theorem. — // two points P^ and F^ and their polars A^B^ and A^B^ are given, and if A^B^ passes through /\, then A^B^ will pass through Fi. We will prove this theorem for the case of the ellipse, lie proofs for the hyperbola or the parabola being almost identica,. Let the coordinates of F^ and F^ be (.r,, y,) and (^2, J'J re- spectively. Then the equations of A^ B ^ and A^ B^ are ■^1 ^ ji_y\y ( \ a' + 7^ = '' (') "-#■+-^=1 (a) a o Since by hypothesis (a:,, jj lies on (i) we have d' "^ b-^ ~ ' but this is precisely the condition that the point (a-,,)-,) should lie on (2). This proves the theorem. From tliis theorem it follows at once that if several points lie on a line, their polars intersect on the pole of that line, and if several lines tneet in a point, their poles lie on the polar of this point, EXAMPLES. I. Find the pole of the line x + ji' = i with regard {a) to the ellipse x"^ + ^f = i; {b) to the parabola/" = 4 .t. I74 ANALYTIC GEOMETRY. 2. Prove that the tangents at the extremity of any focal chord of a conic intersect on the directrix (v. Ex. 4, § 99). 3. If i^ is a focus of a conic, and /^ is a point on the cor- responding directrix, prove that the line jP /^ is perpendicular to the polar of P. 4. Show that the fact established in §101 is a special case of the theorem of § 103. CHAPTER IX. THE GENERAL EQUATION OF THE SECOND DEGREE. 104. We saw in Chapter III that a straight line is a curve of the first degree, and, conversely, that every curve of the first degree is a straight line. In Chapter VII we have seen that the three conic sections are curves of the second degree. These, however, are not the only curves of this kind, since two straight lines may be rep- resented by a single equation of the second degree (v. § 59). It is our purpose in this chapter to show that every equation of the second degree, if it has a real locus, represents either one of the conic sections or two straight lines. The general equation of the second degree is Ax"" ^ B xy ^- Cy' + I?x + £y + J^ = o. . . . (i) We begin by examining the special case of this equation in which B =z o: Ay+ Cy' + £>x + £y + J^=o (2) 105. Before considering equation (2) of the last section in all its generality, let us look at two special cases. Suppose first that D = £ =^ o while neither A nor C is zero. The equation becomes Ax' + 'Cy' + J^=o, (i) 175 176 ANALYTIC GEOxMETRY. or, as we may write it, F or, finally, - ^-^^ - ^/= ^' (2) t A C (3) If tliese denominators are positive we may denote their square roois by a and b^ and (3) takes the standard form for the equa- ticm of an ellipse (v. §63, (7)). If the first denominator is positive and the second negative we have in the same way the standard form for the equation of the hyperbola. If the first denominator is negative and the second positive equation (3) takes the form ^ _/ _ _ a o which, as we know (v. § 86), also represents a hyperbola, but one whose transverse axis lies on the axis of Y, Finally, if the two denominators in (3) are negative, the equation (3) has no locus since its right-hand side, being essentially negative, cannot be equal to i. In this case (3) is often spoken of as an imaginary ellipse. In this discussion we have tacitly assumed that F is not zero, since in passing from (i) to (2) we divided by F. If F -= o equation (i) becomes Ax' + Cy' = o (4) [f A and C have the same sign, this equation is obviously satisfied only by the values x ^= o, y = o. The locus of (4) is therefore a single point, the origin; and (4) is spoken of as a nu/i ellipse (cf. the use of the term null circle in § 46). \i A THE GENERAL EQUATION OF THE SECOND DEGREE. I 77 and C have different signs Ave may suppose ^ })ositive and C negative (as otherwise we could change the signs throughout in (4)), and then (4) may be written a' x' - b'f = o, an equation which represents the two straight lines ax — by = o, a X + by = o, which intersect at the origin. Thus we have seen that, it being assumed that neither ^ nor C is zero, equation (i) represents an ellipse (real, null, or imaginary), an hyperbola, or two intersecting straiglit lines. 106. As a second special case of equation (2), § 104, let us suppose that A = E = o, while Cis not zero: Cy''+Dx + F=o (i) If Z> = o this equation may be written ^=±/-f (^) This equation has no real locus if C and F have the same sign. It represents two lines parallel to the axis of x if C and i^ have opposite signs; and finally if F= o we have no longer two distinct lines, but only the single linej' = o. These three ^ases may be conveniently characterized by saying we liave two parallel lines, real and distinct, real and coincident, or imaginary. If, however, D is not zero, we write (i) in the form cf = -r>[x + ^, (3) and now we make use of a device, which we shall have to em- 178 ANALYTIC GEOMETRY. ploy again presently. We make the transformation of coordi- nates that is (v. § 41 (3)) we shift our co5rdinate axes, without turn- ing them, so that the origin of the new system lies at the point (— — , o 1. After this transformation of coordinates the V ^ / equation of the curve (3) has the form /" = -f^' (4) and this is in the standard form,y = 2 mx, for the equation of a parabola. Accordingly Equaiio7i (i) represents either a parabola or two parallel straight lines {distinct, coincident, or imaginary). Precisely the same reasoning applies to the equation Ax' + Ey + F = o (5) (where we assume that A is not zero) except that the x siwdy coordinates are interchanged. 107. We are now in a position to take up the consideration of equation (2), § T04. We will first illustrilte the method to be used by a numerical example. Suppose we have the equation 2^' - 3/ + 4^ + 5J^^ + 3 = o- Let us rewrite this equation, taking first the two terms in x, then the two terms in y, and last the constant term. We will also take out the factor 2 (the coefficient of x"^) before the two THE GENERAL EQUATION OF THE SECOND DEGREE. Ijg terms in x, and the factor — 3 (the coefficient of y) before the two terms iny: 2(x^ + 2x) - s (/ - I7) + 3 = o- Let us now complete the squares for the two expressions in parentheses by adding 2 to the first term and — f| to the second. 2 (^ + ir - 3 (^ - 1)^ + (3- 2 + fi) = o. If we now make the transformation of coordinates x' = X -{- I, i.e., if we shift the coordinate axes without turning so that the origin lies at the point (— i, f) the equation of the curve takes the form 2 x'-'- 3/^ + ^ = 0. This equation is of the form considered in § 105. Following the method there indicated we throw it into the form ^ _ /_' _ _ Accordingly our original equation represents an hyperbola whose centre is at the point (— i, f), whose transverse axis is parallel to the axis of V and is of length 2^11- = i'^^37, and whose conjugate axis is of length 2V'||- = -^1^222. Precisely the method just explained applies to any equation of the form (2), § 104, provided neither A nor C is zero. We first write the equation in the form ('■ |,Uc(/ + |>.)+i^ = o. l8o ANALYTIC GEOMETRY. and then, by completing the square in each of the parentheses, in the form (-^.r-(^-i) 4A 4C The transformation of coordinates 2 A reduces this equation to and this equation is of the form considered in § 105. 108. If either A or C is zero the method must be slightly modified. Suppose, for instance, A — o. Then we can write equation (2), § 104, in the form C^y + I^) ^Dx^ F^o (i) Completing the square, this equation becomes After the transformation of coordinates ^ =^ + 7c' E \ (3) the equation of the curve reduces to Cy"' + Dx' + {f - ^ =0, (4) and tliis equation is of the for-.ii ('), § ^06. THE GENERAL EQUATION OF THE SECOND DEGREE. l8l Similarly if C — o the equation can be reduced to the form (5), § io6. 109. Thus we have proved that if neither A nor C is zero equation (2), § 104, represents an ellipse (real, null, or imagin- ary), an hyperbola, or two intersecting straight lines; while if A or C is zero it represents either a parabola or two parallel straight lines (distinct, coincident, or imaginary). Moreover, the method we have used enables us in every case to determine not only what kind of curve is represented by the equation, but just how it is situated. EXERCISES. (i.) What curves are represented by the following equations ? Draw diagrams to show how they are situated. {a.) 2 jc^ + 3^ — 5 = o- ib.) 2 x"" — 3j;^ + 5 = o. ((T.) / — 2 .V — 3 = o. id.) y"" + 7,x + 2 — o, ie.) x" - 3 J' + 5 33 0. (/.) 2 .t' + 5 J'"' — 3 X + 2JF — 4 = O. ig.) 9 .V- + 4j'^ — 18 ^ + 2^y + 27 = 0. ih.) 2 jc^ + 3 .T — 5 _)' + 2 = o. (2.) Show that the condition that (2), § 104, represents two straight lines is ^ACF - CD'' - A E' = o. 110. We come now to the general equation of the second degree, (i), ?' 104, where B is not zero. Let us try to simplify this equation by turning the coordinate axes through the l82 ANALYTIC GEOMETRY. angle ^. The formulae for this transformation (v. (7), § 42) are X = x' cos ^ — y' sin t9, ) , ^ J = :r' sin ^ + >■' cos -9 ) ^^^ Equation (i), § 104, now takes the form A' x" -^ B' x'y' + a y'-'^D' x' + E' y' + F=o, (2) where ^' = y^ cos^ i9 + ^ sin ^ cos ^ + C sin' d, (3) B' =z - 2 A sin ^ cos ^ 4- ^ cos' ^ — B sin"-* t9 4- 2 C sin ^ cos ^, . . . (4) a = A sin' ^- B sin ^ cos -^ 4- C cos' ^, (5) Z)' = i:> cos -^ 4- -^ sin i&, (6) Z' = - Z> sin + ^ cos ^ (7) The second of these coefficients may be written more simply as follows: B' ^B cos 2 i9- (^ - C) sin 2 ^ (8) Accordingly B' will be zero if B co^2^={A — C) sin 2 ^, i.e., if cot 2 t9 =: — ^— (9) If we remember that as an angle increases from o to 180° its cotangent decreases from 4- 00 to — 00 , we see that there is just one value of 2 i9 less than 180° which satisfies (9), and therefore just one acute value of t9. If, then, we turn the coordinate axes through the acute angle ^ determined 1)y (9), equation (1), § 104, takes the form A' x" -Y a y' ' + D' x' 4- E' y' -V F ^ o. THE GENERAL EQUATION OF THE SECOND DEGREE. 1 83 This equation is of the form (2), § 104. Accordingly we have proved that every equation of the second degree which has a real locus represents either a conic section or two straight lines. EXAMPLES, (i.) Determine what curves are represented by the follow- ing equations and how they are situated. Illustrate your answers by diagrams: (a) x' + 4xy +/ +-5 = 0- (^) x"" + 2 xy — / -\- 1 = 0. {c) 2 x"" -\- 4xy + 2y'^ — ^x + y = o. (d) 2 x' — 3 xy + 2/ - 3 .T + 2^ — 5 = o. This last equation, like those which precede it, can be treated by the method of § no. Let us, however, shift the coordinate axes without turning them so that the origin lies at the point {a, b). X = x' + a, y ~ y 4- b. Equation id) then takes the form 2 ^'* — 3 x' y' + 2 y + (4 ^ — 3 /^ — 3) .t' — (3^ — 4^ — 2)y + {2 a"^ — ^a b + 2 F — 3 ^ + 2 ^ — 5) = o. Now a and b can be so chosen that the coefficient of x^ and y are zero: a = i, b = \. Thus if we shift to the new origin (f , 4^), equation {d) takes the form 14 x^ — 2\ X y 4- 14 jf'' — 43 = o, and this equation can be much more easily treated than equa- tion {d) by the method of § no. (2.) Discuss by the method just explained the equation 3 x:^ -h 5 X y + / + 2 .T — 3_>' + I =: o. MISCELLANEOUS EXAMPLES. (i.) Prove that the foot of the perpendicular from a focus of an hyperbola to an asymptote is at distances a and b from the centre and focus respectively. (2.) A circle has its centre at the vertex (9 of a parabola whose focus is F^ and the diameter of the circle is 3 (9 F ; show that the common chord of the curves bisects OF. (3.) Prove that the semi-minor axis of an ellipse is a mean proportional between the parts of a tangent at an extremity of the major axis, cut off by conjugate diameters produced. (4.) Prove that the parts of a normal to an hyperbola cut off by the transverse and conjugate axes respectively, are in the ratio U^ : a^. (5.) Prove that if the major axis of an ellipse is equal to twice the minor axis, a straight line, equal to half the major axis, and which moves with one end on the upper half of the curve and the other on the lower half of the minor axis, is bisected by the major axis. (6. ) A line of constant length moves with its extremities on two straight lines at right angles to each other : show that the locus of a fixed point on the moving Une is an ellipse with the segments of the line for semi-axes. (7.) Adapt the last example to the case in which the fixed point is taken on the extension of the moving line. (8.) Two equal rulers, A B, B C, are connected by a pivot at B ; the extremity A is fixed, while C moves on a given straight line : find the locus of any fixed point of B C, taking the origin at A and A C for the axis of x. MISCELLANLOUS EXAMPLES. 185 (9.) Normals are drawn to an ellipse and the circumsciibed circle at corresponding points ; find the locus of their inter- section. Aus. A circle whose radius is ^ + ^. (10.) The ordinate of a point on an hyperbola is extended until its length equals a focal radius of the point; find the locus of its extremity. (11.) Any straight line through the origin meets a parallel to OX drawn through a fixed point, B^ on the axis of 7, in P' ; on OF',F is taken so that its ordinate equals B F' j find the (ocus of F. A, 'IS. A parabola. (12.) Find the locus of the centre of a circle which passes through a fixed pomt and touches a given Ime. (Axes as in § 55' Ex. 13.) A71S. A parabola. (13.) Given the base and the product of the tangents of the base angles of a triangle ; find the locus of the vertex. Suggestion. Taking the axes as in § 55, Ex. 5, the tangents y J y of the base angles are -^ — and . C + X C — X A?is. Letting m represent the given product, the locus is an ellipse or hyperbola, according as m is positive or negative. (14.) Given the base and the product of the tangents of the halves of the base angles ; find the locus of the vertex. If we express the tangents of the half angles in terms of the sides, we shall find that the sum of the sides, 2 i" is given; the locus must therefore be an ellipse with the extremities of the base for its foci. (15.) Show by means of the results of the last two examples that when the base and the sum of the sides of a triangle are given, the centre of the inscribed circle moves on an ellipse whose major axis is the base of the triangle. M ANALYTIC GEOMETRY (i6.) Show that the equation of the elHpse referred to the major axis as the axis of x, and its left-hand extremity as origin, may be written in the form y^ — ^^ {2 a X — x^'). a" (17.) By means of this equation, prove that the locus of the middle points of chords of an ellipse, which pass through the left-hand extremity of the major axis, is another ellipse. What are its axes, and how is it situated ? (18.) Prove that if the right-hand focus of an ellipse is the origin (the major axis being the axis of x)^ the radius connecting that focus with any point of the ellipse equals a {\ — e^) —ex; and when the left-hand focus is the origin, the radius joining that focus with a point of the ellipse \^ a {\ — e^) ^ e x. (19.) By means of the results of the last example, obtain the polar equations of the ellipse referred to the right-hand and left- hand focus, respectively. Ans. r — ^ (i -^') ^ _ ^ (i - g^) r z= I + e cos cfi I — e cos <^ (20.) Show that for the right-hand branch of the hyperbola the focal radii, when referred to the respective foci (v. Ex. 18), are e X — a (i — e^) and e x -\- a {\ — e^) ; changing the signs of these expressions, we have the focal radii for the left-hand branch. (21.) From the results of the last example, show that the polar equations of the right-hand branch of an hyperbola re- ferred to the foci are a i\ — e^\ a {\ — e'-') r = ^^ ^~ and r = — ^^ i-. 1 — e cos The negative values of r derived from these equations give MISCELLANEOUS EXAMPLES. I 87 points on the left-hand branch, or we may use for that branch the following equations, similarly obtained : — a (i — e-) , a (i — e"^) r— — ^^^ i- and r — ^^ '—. I + e cos (f> I + ^ cos (/) (22.) Obtain the polar equation of the parabola referred to the focus, by transformation of coordinates. A71S. r = ^ . I — cos cfi (23.) Find the polar equation mentioned in the last example from the definition of the curve, without assuming any other form of the equation of the parabola. (24.) Discuss the polar equation of the parabola referred to the focus. (25.) From the equation of the parabola referred to axis and directrix, obtain the polar equation when the intersection of these lines is the pole. (26.) Find the locus of the middle points of lines drawn to a parabola from (i) the vertex, (2) the focus, (3) the intersec- tion of axis and directrix.' • Ans. Parabolas whose parameters are one half that of the original curve. (27.) Prove that if a straight line is drawn from the intersec- tion of the directrix and axis of a parabola cutting the curve, the rectangle of the intercepts made by the curve is equal to the rectangle of the parts into which the parallel focal chord is divided by the focus. (28.) Two parabolas have the same focus and the same transverse axis. A tangent to the first curve and a tangent to the second move so as to remain at right angles to each other. Find the locus of their intersection. Suggestion. Take the common focus as origin. A?is. The straight line parallel to the two directrices and half way between them. ibS ANALYTIC GEOMETRY. (29.) Show, by the method of § s;^, that the tangent to an ellipse or hyperbola at any point makes equal angles with tlie focal radii drawn to that point, (v. §§ 79, 81.) (30.) Prove, as in the last example, that the tangent to a parabola at any point makes equal angles with the axis and the focal radius of that point. (31.) Prove that tangents to the parabola at P^ and J^^ in- tersect at the point whose coordinates are 2 2 m (32.) From the results of the last example, show that the area of a triangle formed by three tangents to a parabola is one half the area of the triangle whose vertices are the points of tangency. (^S-) Prove that the triangle formed by a tangent to an hy- perbola and the asymptotes has a constant area. (34.) If from any point F' oi 2l line M' F', perpendicular to the axis C O M' of a parabola whose vertex is O, a parallel to the axis is drawn meeting the curve in F'^ ; show that, if C (7 is made equal to O M', the locus of the intersection oi O F' and C F" is the original curve. (35.) F' P" is any chord of an ellipse at right angles to the major axis A^ A ; find the locus of the intersection of A^ F' and AFf'. Ans. An hyperbola with the same axes as the ellipse. (36.) The right-hand extremity of a diameter of a circle is joined with the middle point of a parallel chord ; find the locus of the intersection of this line with the radius drawn to the right- hand extremity of the chord. Ans. A parabola. (37.) Prove that the perpendicular from the centre of an el- lipse upon a straight line which joins the ends of perpendicular MISCELLANEOUS EXAMPLES. 189 diameters is of constant length. What is the locus of its extremity ? (38.) From a point P* of an ellipse straight lines are drawn to A^ and A^ the extremities of the major axis, and from A and A^ perpendiculars are erected to A^ F' and A F' ; show that the locus of their intersection is another ellipse, and find its axes. (39.) Find the locus of the intersection of the ordinate of any point of an ellipse, produced, with the perpendicular from the centre to the tangent at that point. (40.) A perpendicular is drawn from a focus of an ellipse to any diameter ; find the locus of its intersection with the con- jugate diameter. A?is. A straight line perpendicular to the major axis. (41.) Tw^o perpendiculars are drawn from the extremities of a pair of conjugate diameters of an ellipse to the diameter whose equation is y z= X tan a ; show that the sum of the squares of the perpendiculars is a^ sin'^ a + b'^ cos^ a. (42.) Prove that the tangents to conjugate hyperbolas at the extremities of conjugate diameters intersect on the asymptotes. (43.) Show that the locus of the intersection of tangents to an ellipse at the extremities of conjugate diameters is an ellipse ; find its axes. (44.) Show that the equation xy = c represents a rectangular hyperbola whose asymptotes are the coordinate axes. 190 ANALYTIC GEOMETRY. (45.) Obtain by the method of § 49 the equation of the tangent to the rectangular hyperbola xy = }'i)- ^^^^' J\ ^ + ^xy = 2 <;. (46.) Obtain by the method of § 99 the equation of the polar of the point (^i,J'i) with regard to the hyperbola X V — c. (47.) Prove that if two rectangular hyperbolas are so situated that the axes of one are the asymptotes of the other, the polars of any point with regard to the two hyperbolas are perpendicular to each other. Hence show that these hyper- bolas intersect at right angles. (48.) Two concentric ellipses have axes lying on the same lines. Tangents are drawn to one ellipse. Find the locus of their poles with regard to the other. (49.) Prove that the circle described on any focal radius of an ellipse as diameter is tangent to the circle described on the transverse axis of the ellipse as diameter. (50.) State and prove the proposition analogous to (49) in the case of the parabola. Of the hyperbola. (51.) A set of lines terminated by two rectangular axes pass through a fixed point. Find the equation of the locus of their middle points, and determine what this equation represents. Ans. A rectangular hyperbola. (52.) In an equilateral hyperbola show that focal chords parallel to conjugate diameters are equal. (53.) Parallel tangents are drawn to a set of confocal ellip- ses. Find the equation of the locus of the points of contact, and determine what this equation represents. Ans. A rectangular hyperbola. (54.) Find by the method of § 51 the equation of the diam- eter of the hyperbola xy — c which bisects the chords whose slope is /^. Hence show that the angles between any pair of conjugate diameters of a rectrngular hyperbola are bisected by the asymptotes. MISCELLANEOUS EXAMPLES. I91 (55.) Find by the method of § 49 the equation of the tangent to the conic A x' + Bxy + Cf + nx + Ey + F = at the point Gv,j',). /? Z) Ans. A x^ X + - {y^ x + x^y) + Cyj' + -{x + x) 2 2 + -iy+y.) + F=o. 2 (56.) Find the polar of (.r, ,j',) with regard to the conic of Ex. (55). (57.) Show that the condition that the conic of Ex. (55) be a rectangular hyperbola is ^ = — C. CIYIL ENGINEERING U. of C. ASSOCIATION LIBRARY UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. OCT 2 6 . v^.'^,3 1 ^ 1951 ^ ao'Ji 28 1952^^ AUG OCT 20 195 LD 21-100m-9,'47(A5702sl6)476 800299 Lib- UNIVERSITY OF CALIFORNIA UBRARY