2)^ U> IN MEMORIAM FLORIAN CAJORl 1 ELEMENTS OF ANALYTIC GEOMETRY A SERIES OF MATHEMATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By Ellery Williams Davis and William Charles Brenke. ANALYTIC GEOMETRY AND ALGEBRA By Alexander Ziwet and Louis Allen Hopkins. ELEMENTS OF ANALYTIC GEOMETRY By Alexander Ziwet and Louis Allen Hopkins. PLANE AND SPHERICAL TRIGONOMETRY WITH COMPLETE TABLES By Alfred Monroe Kenyon and Louis Ingold. PLANE AND SPHERICAL TRIGONOMETRY WITH BRIEF TABLES By Alfred Monroe Kenyon and Louis Ingold. ELEMENTARY MATHEMATICAL ANALYSIS By John Wesley Young and Frank Merritt Morgan. PLANE TRIGONOMETRY FOR SCHOOLS AND COL- LEGES By Alfred Monroe Kenyon and Louis Ingold. THE MACMILLAN TABLES Prepared under the direction of Earle Raymond Hedrick. PLANE GEOMETRY By Walter Burton Ford and Charles Ammerman. PLANE AND SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. CONSTRUCTIVE GEOMETRY Prepared under the direction of Earle Raymond Hedrick. ELEMENTS OF ANALYTIC GEOMETRY BY ALEXANDER ZIWET PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN AND LOUIS ALLEN HOPKINS INSTRUCTOR IN MATHEMATICS, THE UNIVERSITY OF MICHIGAN THE MACMILLAN COMPANY 1916 All rights reserved COPTBIGHT, 1916, bt the macmillan company. Set up and electrotyped. Published October, 1916. Norinooti ^ress J. 8. Gushing Co. — Berwick & Smith C!o. Norwood, Mass., U.S.A. PREFACE As in most colleges the course in analytic geometry is pre- ceded by a course in advanced algebra, it appeared desirable to publish separately those parts of our " Analytic Geometry and Principles of Algebra " which deal with analytic geometry, omitting the sections on algebra. This is done in the present work. In plane analytic geometry, the idea of function is intro- duced as early as possible ; and curves of the tovm y =f{x), where f{x) is a simple polynomial, are discussed even before the conic sections are treated systematically. This makes it possible to introduce the idea of the derivative ; but the sec- tions dealing with the derivative may be omitted. In the chapters on the conic sections only the most essential properties of these curves are given in the text ; thus, poles and polars are discussed only in connection with the circle. The treatment of solid analytic geometry follows more the usual lines. But, in view of the application to mechanics, the idea of the vector is given some prominence ; and the repre- sentation of a function of two variables by contour lines as well as by a surface in space is explained and illustrated by practical examples. The exercises have been selected with great care in order not only to furnish sufficient material for practice in algebraic work, but also to stimulate independent thinking and to point Hrpf\g^gr*g^M vi PREFACE out the applications of the theory to concrete problems. The number of exercises is sufficient to allow the instructor to make a choice. ALEXANDER ZIWET. LOUIS A. HOPKINS. CONTENTS PLANE ANALYTIC GEOMETRY PAGES Chapter I. Coordinates 1-21 Chapter II. The Straight Line 22-36 Chapter III. Relations between Two or More Lines . . 37-50 Chapter IV. The Circle . 51-70 Chapter V. Polynomials 71-92 Parti. Quadratic Function — Parabola . . . 71-81 PartIL Polynomials 82-92 Chapter VI. The Parabola 93-115 Chapter VII. Ellipse and Hyperbola 116-139 Chapter VIII. Conic Sections — Equation of Second Degree 140-162 Part I. Definition and Classification .... 140-147 Part II. Reduction of General Equation . . . 148-162 Chapter IX. Higher Plane Curves 163-188 Part L Algebraic Curves 163-168 PartIL Special Curves 169-177 Part III. Empirical Equations 178-188 SOLID ANALYTIC GEOMETRY Chapter X. Coordinates 189-203 Chapter XL The Plane and the Straight Line • . . 204-228 PartL The Plane 204-218 Part II. The Straight Line 219-228 Chapter XII. The Sphere 229-241 Chapter XIII. Quadric Surfaces — Other Surfaces . . 242-265 Tables 266-272 Answers 273-276 Index 277-280 vii PLANE AN^ALYTIC GEOMETRY CHAPTER I COORDINATES 1. Location of a Point on a Line. The position of a point P (Fig. 1) on a line is fully determined by its distance OP from a fixed point O on the line, if we know on which side of the point P is situated (to the right or to the left of O in Fig. 1). Let us agree, for instance, to count distances to the 4-H^ t Fig. 1 right of as positive, and distances to the left of as negative ; this is indicated in Fig. 1 by the arrowhead which marks the positive sense of the line. The fixed point is called the origin. The distance OP, taken with the sign + if P lies, let us say, on .the right, and with the sign — when P lies on the opposite side, is called the abscissa of P. It is assumed that the unit in which the distances are measured (inches, feet, miles, etc.) is known. On a geographi- cal map, or on a plan of a lot or building, this unit is indicated by the scale. In Fig. 1, the unit of measure is one inch, the abscissa of P is + 2, that of Q is ^ 1, that of i? is — 1/3. B 1 2 PLANE ANALYTIC GEOMETRY [I, §2 2. Determination of a Point by its Abscissa. Let us select, on a given line, an arbitrary origin 0, a unit of measure, and a definite sense as positive. Then any real number, such as 5, — 3, 7.35, — V2, regarded as the abscissa of a point P, fully determines the position of P on the line. Conversely, every point on the line has one and only one abscissa. The abscissa of a point is usually denoted by the letter a?, which, in analytic geometry as in algebra, may represent any real or complex number. To represent a real point the abscissa must be a real number. If in any problem the abscissa » of a point is not a real num- ber, there exists no real point satisfying the conditions of the problem. EXERCISES 1. What is the abscissa of the origin ? 2. With the inch as unit of length, mark on a line the points whose abscissas are : 3, —2, VS, — 1.26, — \/5, |, — ^. 3. On a railroad line running east and west, if the station B is 20 miles east of the station A and the station C is 33 miles east of A, what are the abscissas of A and C for B as origin, the sense eastward being taken as positive ? 4. On a Fahrenheit thermometer, what is the positive sense ? What is the unit of measure? What is the meaning of the reading 66°? What is meant by — 7° ? 6. A water gauge is a vertical post carrying a scale ; the mean water level is generally taken as origin. If the water stands at + 7 on one day and at —11 the next day, the unit being the inch, how much has the water fallen ? 6. If 051, X2 (read : x one, x two) are the abscissas of any two points Pi, P2 on a given line, show that the abscissa of the midpoint between Pi and P2 is ^ (xi + X2). Consider separately the cases when Pi, P^ lie on the same side of the origin and when they lie on opposite sides. I, § 3] COORDINATES 3 3. Ratio of Division. A segment AB (Fig. 2) of a straight line being given, it is shown in elementary geometry how to find the point C that divides ^ AB in a given ratio 7c. Thus, if A; = I, the point G such that AG^2 AB 5 is found as follows. On any line through A lay off AD = 2 and AE=5', join B and E. Then the parallel to BE through D meets AB at the required point C. Analytically, the problem of dividing a line in a given ratio is solved as follows. On the line AB (Fig. 3) we choose a point as origin and assign a positive sense. Then the abscissas Xi of A and X2 of B are known. To find a point FiG. 3 which divides AB in the ratio of division k = AC/AB, let us denote the unknown abscissa of C by x. Then we have AC^x — x-i^, AB = X2 — x^; hence the abscissa a; of O must satisfy the condition ^ — Xi __^ •*'2 — "^l whence X = Xi-\- k {X2 — Xi) ', or, if we write Ax (read : delta x) for the " difference of the ic's," i.e. Ax z=x.2 — Xi, x = Xi-\-k ' Ax. Thus, if the abscissas of A and B are 2 and 7, the abscissas 4 PLANE ANALYTIC GEOMETRY [I, § 3 of the points that divide AB in the ratios J, ^, f , f are 3, 4^, 8, 9^, respectively. Check these results by geometric con- struction. If the segments AC and AB have the same sense, the divi- sion ratio k is positive. For example, in Fig. 3, the point O lies between A and B ; hence the division ratio A; is a positive proper fraction. If the division ratio k is negative, the seg- ments AC and AB must have opposite sense, so that B and C lie on the opposite sides of A. If the abscissas of A and B are again 2 and 7, the abscissa xoi C when A; = 2, - 1, - 1, - .2 will be 12, - 3, 0, 1, respec- tively. Illustrate this by a figure, and check by the geometric construction. 4. Location of a Point in a Plane. To locate a point in a plane, that is, to determine its position in a plane, we may proceed as follows. Draw two lines at right angles in the plane ; on each of these take the point of intersection O as origin, and assign a definite positive sense to each line, e.g. by marking each line with an arrowhead. It is usual to mark the positive sense of one line by affixing the letter x to it, and the positive sense of the other line by affixing the letter y to it, as in Fig. 4. ^ These two lines are then called the axes of coordinates^ or simply the axes. We —o ~ q ~ distinguish them by calling the line Ox the yiq. 4 ic-axis, or axis of abscissas, and the line Oy the ?/-axis, or axis of ordinates. Now project the point P on each axis, i.e. let fall the perpendiculars PQ^ PR from P on the axes. The point Q has the abscissa OQ = x on the axis Ox. The point R has the abscissa OR = y on the axis Oy. The distance OQ = RP=x is called the abscissa of P, and I I, § 6] COORDINATES 5 OR = QP = y is called the ordinate of P. The position of the point P in the plane is fully determined if its abscissa x and its ordinate y are both given. The two numbers x, y are also called the coordinates of the point P. 5. Signs of the Coordinates. Quadrants. It is clear from Fig. 4 that x and y are the perpendicular distances of the point P from the two axes. It should be observed that each of these numbers may be positive or negative, as in § 1. , ^ n ^r" — The two axes divide the plane into i four compartments distinguished as in i trigonometry as the first, second, third, i ^ and fourth quadrants (Fig. 5). It is ^zr | jn readily seen that any point in the first p"' quadrant has both its coordinates posi- tive. What are the sisrns of the coordi- 1 ■^t)' nates in the other quadrants ? What are the coordinates of the origin ? What are the coordinates of a point on one of the axes ? It is customary to name the abscissa first and then the ordinate; thus the point (—3, 5) means the point whose abscissa is — 3 and whose ordinate is 5. Every point in the vlanp. hna turn dpfnifp rpnl nura ^^fn ao-rrh. o rdinates; conversel y, to every pair of real numbers correspo iids one and only one point of the plane. Locate the points: (6, -2), (0, 7), (2-V3, |), (-4, 2V2), (-5,0). 6. Units. It may sometimes be convenient to choose the unit of measure for the abscissa of a point different from the unit of measure for the ordinate. Thus, if the same unit, say one inch, were taken for abscissa and ordinate, the point (3, 48) might fall beyond the limits of the paper. To avoid this we 6 PLANE ANALYTIC GEOMETRY II, §6 may lay off the ordinate on a scale of i inch. When different units are used, the unit used on each axis should always be indicated in the drawing. When nothing is said to the con- trary, the units for abscissas and ordinates are always under- stood to be the same. 7. Oblique Axes. The position of a point in a plane can also be determined with reference to two axes that are not at right angles ; but the angle w between these axes must be given (Fig. 6). The abscissa and the ordinate of the point P are then the segments OQ = x, OR = y cut off on the axes by the parallels through P to the axes. If 0) = ^ TT, i.e. if the axes are at right angles, we have the case of rectangular coordinates discussed in §§4, 5. In what follows, the axes are always taken at right angles unless the contrary is definitely stated. 8. Distance of a Point from the Origin. For the distance r=OP (Fig. 7) of the point P from the origin we have from the right- angled triangle OQP: Fig. 7 Fig. 6 ^ r= y/x^ -f y^y where x, y are the coordinates of P. If the axes are oblique (Fig. 8), with the angle xOy = CO, we have, from the triangle OQP, in which the angle at Q is equal to ir — «,* by the cosine law of trigonometry, Fig. 8 r — y/x^ + y* — 2 xy cos (ir — «) = Vx- + y^* -f 2 xy cos w. * In advanced mathematics, angles are generally measured in radians, the symbol ir denoting an angle of 180^ I, § 9] COORDINATES 7 Notice that these formulas hold not only when the point P lies in the first quadrant, but quite generally wherever the point P may be situated. Draw the figures for several cases. 9. Distance between Two Points. By Fig. 9, the distance d = PiP2 between two points Pi{xi, y^ and P2(a72> 2/2) can be found if the coordinates of the two points are given. For in the triangle P1QP2 we have Pi Q = 3^2 - a^i , QP2 = 2/2 - yi ; hence F^^ 9 (1) d = v/(a^2-^i)2 + (2/2-2/1)2. If we write Ax (§ 3) for the " difference of the ic's " and Ay for the " difference of the y's ", i.e. Aa; = a;2 — a^i and A?/ = 2/2 — 2/1? the formula for the distance has the simple form (2) d = V(Aa?)2 + (A2/)2; or, in words, The distance between any two points is equal to the square root of the sum of the squares of the differences between their corre- sponding coordinates. Draw the figure showing the distance between two points (like Fig. 9) for various positions of these points and show that the expression for d holds in all cases. Show that the distance between two points Pi (aji, yi), P2 (X2^ ^2) when the axes are oblique, with angle w, is d = \/{X2 - Xl)2+ (^2 - yiY^ + 2(X2 - Xi){y2 - yi) cos (O = V( Aa;)2 + (A2/)2 + 2 Ax- Ay • cos w. 8 PLANE ANALYTIC GEOMETRY [I, § 10 10. Ratio of Division. If two points P^ {x^ , y^ and P^ (xo, y^ are given by their coordinates^ the coordinates x, y of any point Pon the line P1P2 can be found if the division ratio P^P/P^P^, = k is known in which the point P divides the segment P1P2' Let Qi , Q2, Q (Fig. 10), be the projections of P^, P2, P on the axis Ox-, then the point Q divides Q1Q2 in the same ratio k in which P divides P1P2' Now as OQi = Xi, 0^2= ^2? OQ = a;, it follows from § 3 that x=Xi -\-k(x2 — Xi). y R - ^ -A R, --X 1 1 X X «r Q q. In the same way we find by projecting Pi, P2, P on. the axis Oy that FiQ. 10 y = yi + ^(y2-yi)' Thus, the coordinates x, y oi P are found expressed in terms of the coordinates of Pj , Po and the division ratio k. Putting again X2 — Xi = Ax,y2 — yi=Ay, we may also write x = Xi-\-k' /iix, y = yi-\-k'Ay. Here again the student should convince himself that the formulas hold generally for any position of the two points, by selecting numerous examples. He should also prove, from a figure, that the same expressions for the coordinates of the point P hold for oblique coordinates. As in § 3, if the division ratio k is negative, the two segments P1P2 and PiP must have opposite sense, so that the points P and P2 must lie on opposite sides of the point Pj. Find, e.g., the coordinates of the points that divide the seg- ment joining (— 4, 3) to (6, — 5) in the division ratios k = ^, k=2, A:=— i, k = —l, and indicate the four points in a fiGTure. I, § 11] COORDINATES 9 11. Midpoint of a Segment. The midpoint P of a segment P1P2 has for its coordinates the arithmetic means of the corre- sponding coordinates of Pi and P^ ; that is, if x^ , y^ are the co- ordinates of Pi, ^2, 2/2 those of Pg) the division ratio being A; = i the coordinates of the midpoint P are (§ 10) a; = a?i + 1(0^2 - ajj) = Ka^i + ^2), 2/ = 2/i + i(2/2 - 2/1) = }(2/i + 2/2). EXERCISES 1. With reference to the same set of axes, locate the points (6, 4), (2, - i), (- 6.4, - 3.2), (-4, 0), (- 1, 5), (.001, - 4.01). 2. Locate the points (-3,4), (0,-1), (6, - V2), (f^ _ 10|), (0,a), (a, 6), (3, -2), (-2, V2). 3. If a and h are positive numbers, in what quadrants do the follow- ing points lie : (a, — &), (6, a), (a, a), (- 6, 6), (—6, — a)? 4. Show that the points (a, &) and («, — 6) are symmetric with respect to the axis Ox ; that (a, &) and (—a, 6) are symmetric with re- spect to the axis Oy \ that (a, &) and (— a, — 6) are symmetric with respect to the origin. 6. In the city of Washington the lettered streets (A street, B street, etc.) run east and west, the numbered streets (1st street, 2d street, etc.) north and south, the Capitol being the origin of coordinates. The axes of coordinates are called avenues, thus, e.gr., 1st street runs one block east of the Capitol. If the length of a block were 1/10 mile, what would be the distance from the corner of South C street and East 5th street to the corner of North Q street and West 14th street ? 6. Prove that the points (6, 2), (0, — 6), (7, 1) lie on a circle whose center is (3, — 2). 7. A square of side s has its center at the origin and diagonals coin- cident with the axes ; what are the coordinates of the vertices ? of the midpoints of the sides ? 8. If a point moves parallel to the axis 0^, which of its coordinates remains constant ? 10 PLANE ANALYTIC GEOMETRY [I, § 11 9. In what quadrants can a point lie if its abscissa is negative ? its ordinate positive ? 10. Find the coordinates of the points which trisect the distance be- tween the points (1, — 2) and (— 3, 4). 11. To what point must the line segment drawn from (2, —3) to (—3, 6) be extended so that its length is doubled ? trebled ? 12. The abscissa of a iwint is — 3, its distance from the origin is 5 ; what is its ordinate ? 13. A rectangular house is to be built on a comer lot, the front, 30 ft. wide, cutting off equal segments on the adjoining streets. If the house is 20 ft. deep, find the coordinates (with respect to the adjoining streets) of the back comers of the house. 14. A baseball diamond is 90 ft. square and pitcher's plate is 60 ft. from home plate. Using the foul lines as axes, find the coordinates of the following positions : (a) pitcher's plate ; (6) catcher 8 ft. back of home plate and in line with second base ; (c) base runner playing 12 ft. from first base ; (d) third baseman playing midway between pitcher's plate and third base (before a bunt) ; (c) right fielder playing 90 ft. from first and second base each. 16. How far does the ball go in Ex. 14 if thrown by third baseman in position (d) to second base ? 16. If right fielder (Ex. 14) catches a ball in position (e) and throws it to third base for a double play, how far does the ball go ? 17. A park 600 ft. long and 400 ft. wide has six lights arranged in a circle about a central light cluster. All the lights are 200 ft. apart, and the central cluster and two others are in a line parallel to the length of the park. What are the coordinates of all the Ughts with respect to two boundary hedges ? 18. With respect to adjoining walks, three trees have coordinates (30 ft., 8 ft.), (20 ft., 45 ft.), (- 27 ft., 14 ft,), respectively. A tree is to be planted to form the fourth vertex of a parallelogram; where should it be placed ? (Three possible positions ; best found by division ratio.) I, § 13] COORDINATES 11 12. Area of a Triangle with One Vertex at the Origin. Let one vertex of a triangle be the origin, and let the other vertices be Pi(a;„ 2/1) and P2(a^2> 2/2)- Draw through P^ and P2 lipes parallel to the axes (I'ig. 11). The area A of the triangle is then obtained by- subtracting from the area of the circum- scribed rectangle the areas of the three non- shaded triangles ; i.e. A^x^y^ - l^i?/! - |-i»22/2 - K^i -«^2)(2/2 -2/1) = i(^iy2 - a^22/i) ; or, in determinant notation, X, Pi Fig. 11 X2 2/2 This formula gives the area with the sign -f or — according as the sense of the motion around the perimeter OP^P^O is counterclockwise (opposite to the rotation of the hands of a clock) or clockwise. 13. Translation of Axes. Instead of the origin and the axes Ox, Oy (Fig. 12), let us select a new origin 0' (read : prime) and new ax:es O'x', O'y', parallel to the old axes. Then any point P whose coordinates with reference to the old axes are OQ = Xj QP = y will have with reference to the new axes the coordi- nates 0'Q' = x', Q'P=y'; and the figure shows that if h, k are the co- ordinates of the new origin, then x = x' -]-h, y = y' + k. 0^ y1 Fig. 12 The change from one set of axes to a new set is called a transformation of coordinates. In the present case, where the 12 PLANE ANALYTIC GEOMETRY [I, § 13 new axes are parallel to the old, this transformation can be said to consist in a translation of the axes. 14. Area of Any Triangle. Let Py{x^, y^), P2(^2)2/2), P^ix^, 2/3) he the vertices of the triangle (Fig. 13). If we take one of these vertices, say Pg, as new origin, with the new axes parallel to the old, the new coordinates of Pi , Pg will be : 3/ 2 Xi fl/g, X 2 — ~ X2 fl/g, y\=yi-y3, ^2= 2/2 -2/3. Hence, by § 12, the area of the triangle = K^i{y2 - 2/3) + 3:2(2/3 - 2/1) + 2:3(2/1 - 2/2)] ; or, in determinant notation, ^=i ^l yi 1 X2 2/2 1 X3 2/3 1 Here as in § 12 the sign of the area is + or — according as the sense of the motion along the perimeter P1P2P3P1 is coun- terclockwise. EXERCISES 1. Find the areas of the triangles having the following vertices : (a) (1, 3), (5, 2), (4, 6) ; (6) (- 2, 1), (2, - 3), (0, - 6) ; (c) (a, &), (a, 0), (0, 6) ; (d) (4, 3), (6, - 2), (- 1, 6). 2. Show that the area of the triangle whose vertices are (7, —8), (—3, 2), (— 5, —4) is four times the area of the triangle formed by- joining the midpoints of the sides. 3. Find the area of the quadrilateral whose vertices are (2, 3), (— 1, -1), (-4,2), (-3,6). I, § 15] COORDINATES 13 4. Find the area of the triangle whose vertices are (a, 0), (0, &), (-C, -c). 5. Find the area of~^he triangle (1, 4), (3, -2), (-3, 16). What does your result show about these points ? 6. Find the area of the triangle (a, 6 + c), (b^ c-\- a), (c, a + &)• What does the result show whatever the values of a, &, c ? 7. Show that the points (3, 7), (7, 3), (8, 8) are the vertices of an isosceles triangle. What is its area ? Show that the same is true for the points (a, 6), (&, a), (c, c), whatever a, 6, c, and find the area. 8. Find the perimeter of the triangle whose vertices are (3, 7), (2, — 1), (5, 3). Is the triangle scalene ? What is its area ? 9. Show that the area of a quadrilateral whose vertices are (oji, yi), (a^2, 2/2), (xs, 2/3), (Xi, y^) may be written in the form A = l x\ — xz yi — 2/3 X2 — X4 yi — yi 15. Statistics. Related Quantities. If pairs of values of two related quantities are given, each of these pairs of values is represented by a point in the plane if the value of one quantity is represented by the abscissa and that of the other by the ordinate of the point. A curved line joining these points gives a vivid idea of the way in which the two quantities change. Statistics and the results of scientific ex- periments are often represented in this manner. EXERCISES 1. The population of the United States, as shown by the census reports, is approximately as given in the following table : Teak 1800 '10 '20 '30 '40 '50 '60 '70 '80 '90 1900 '10 Million 4 5 7 10 13 17 23 31 ■ 39 50 63 76 92 14 PLANE ANALYTIC GEOMETRY [I, § 15 Mark the points corresponding to the pairs of numbers (1790, 4), (1800, 5), etc., on squared paper, representing the time on the horizontal axis and the population vertically. Connect these points by a curved line. 2. From the figure of Ex. 1, estimate approximately the population of the United States in 1875 ; in 1905 ; in 1915. 3. From the figure of Ex. 1, estimate approximately when the popula- tion was 25 millions ; 60 millions ; when it will be 100 millions. 4. Draw a figure to represent the growth of the population of your own State, from the figures given by the Census Reports. [Other data suitable for statistical graphs can be found in large quan- tity in the Census Reports ; in the Crop Reports of the government ; in the quotations of the market prices of food and of stocks and bonds ; in the World Almanac ; and in many other books. ] 6. The temperatures on a certain day varied hour by hour as follows : A.M. N. P.M. Time . . Temp. . . 6 50 7 62 8 55 9 60 10 64 11 67 12 70 1 72 2 74 3 75 4 74 5 72 6 69 7 65 8 60 9 57 Draw a figure to represent these pairs of values. 6. In experiments on stretching an iron bar, the tension t (in tons) and the elongation E (in thousandths of an inch) were found to be as follows : t (in tons) E (in thousandths of an inch) 1 10 2 19 4 38 6 60 10 103 Draw a figure to represent these pairs of values. [Other data can be found in books on Physics and Engineering.] 7. By Hooke's law, the elongation ^ of a stretched rod is supposed to be connected with the tension t by the formula E = c - t^ where c is a constant. Show that if c = 10, with the units of Ex. 6, the values of E and t would be nearly the same as those of Ex. 6. Plot the values given by the formula and compare with the figure of Ex. 6. I, § 16] COORDINATES 15 8. The distances through which a body will fall from rest in a vacuum in a time t are given by the formula s = 16 1'^, approximately, if t is in seconds and s is in feet. Show that corresponding values of s and t are 2 64 3 144 4 256 6 400 6 576 Draw a figure to represent these pairs of values. 16. Polar Coordinates. The position of a point P in a plane (Fig. 14) can also be assigned by its distance OP = r from a fixed point, or pole, 0, and the angle xOP = <^, made by the line OP with a fixed line Ox, the polar axis. The dis- tance r is called the radius vector, the angle <^ the polar angle (or also the vectorial angle, azimuth, ampli- j, tude, or anomaly) of the point P. The /r^'' radius vector r and the polar angle eft are q ^^^^ je^ called the polar coordinates of P. Fig. 14 Locate the points : (5, | tt), (6, | tt), (2, 140°), (7, 307°), (V5, tt), (4, 0°). To obtain for every point in the plane a single definite pair of polar coordinates it is sufficient to take the radius vector r always positive and to regard as polar angle the positive angle between and 27r(O^0<27r) through which the polar axis (regarded as a half-line or ray issuing from the pole 0) must be turned about the pole in the counterclockwise sense to pass through P. The only exception is the pole for which r = 0, while the polar angle is indeterminate. But it is not necessary to confine the radius vector to positive values and the polar angle to values between and 2 tt. A single definite point P will correspond to every pair of real values of r and 0, if we agree that a negative value of the radius vector means that the distance r is to be laid off in the negative sense on the polar axis, after being turned through the angle 0, and that a negative value of

, , . and y = r8in, 18. Distance between Two Points in Polar Coordinates. If two points Pi, P2 are given by their polar coordinates, r,, ^ and ra, <^2> the distance d = P^P^ between them is found from the triangle OP^P^ (Fig. 16), by the cosine law of trigonometry, if we ob- serve that the angle at is equal to ± {2—i) '- y- p tan o - <^i). ^ fiq. 16 EXERCISES 1. Find the distances between the points: (2, lir) and (4, fTr); (a, iir) and (3 a, ^tt). 2. Find the cartesian coordinates of the points (5, ^ir), ("6, — i tt), (4, i n), (2, I ,r), (7, tt), (6, - x), (4, 0), (- 3, 60°), (- 6, - 90°) . I, § 19] COORDINATES 17 3. Find the polar coordinates of the points (VS, 1), (-VS, 1), (1,-1), (-1,-1), (-a, a). 4. Find an expression for the area of a triangle whose vertices are (0, 0), (n, 00, and (ra, 02). 5. Find the area of the triangle whose vertices are (n, 0i), {1% 02), (»*3, 03). 19. Projection of Vectors. A straight line segment AB of definite length, direction, and sense (indicated by an arrow- head, pointing from A to B) is called a vector. The projection A'B' (Figs. 17, 18) of a vector AB on an axis, i.e. on a line I B 1 1 '^___ ^ 1 1 .__±_ 1 1 -> A' I B' Fig. 17 on which a definite sense has been selected as positive, is the product of the length of the vector AB into the cosine of the angle between the positive senses of the axis and the vector : A'B' = AB cos a. The positive sense of the axis makes with the vector two angles whose sum is 2 tt = 360°. As their cosines are the same, it makes no difference which of the two angles is used. With these conventions it is readily seen that the sum of the projections of the sides of an open polygon on any axis is equal to the projection of the closing side on the same axis, the sides of the open polygon being taken in the same sense around the perim- eter. c Thus, in Fig. 19, the vectors P^P^, P^P^, ••• P^Pq are in- 18 PLANE ANALYTIC GEOMETRY [I, § 19 clined at the angles cti, a2, ••• a^ to the axis I ; the closing line PiP^ makes the angle a with I ; its projection is P'iP'q ; and we have P1P2 cos tti+PgA cos 02+^3 A cos as-\-P^Ps cos u^-^-P^Pq cos a^ = P'lP'e = PiPg cos a. For, if the abscissas of Pi, P2, ••• Pe measured along I, from any origin on Z, are Xi, X2, ••• Xg, the projections of the vectors are X2 — Xi, x^ — X2, etc., so that our equation becomes the identity : X^-Xi + Xs — X2-\-X^-Xs+Xs-X^ + Xe-Xs = Xfi-Xi. 20. Components and Resultants of Vectors. In physics, forces, velocities, accelerations, etc., are represented by vectors because such magnitudes have not only a numerical value but also a definite direction and sense. According to the parallelogram law of physics, two forces OPi, OP2, acting on the same particle, are together equivalent to the single force OP (Eig. 20), whose vector is the diagonal of the parallelogram formed with OPi, OP2 as adjacent sides. The same q^.^^^ ^^ law holds for simultaneous velocities and Fiq- 20 accelerations, and for simultaneous or consecutive rectilinear translations. The vector OP is called the resultant of OPi and OP2, and the vectors OPi, OP2 are called the components of OP. To construct the result- ant it suffices to lay off from the extremity of the vector OPi the vector P^P = OP2 ; the closing line OP is the ^ resultant. This leads at once to finding the result- ant OP of any number of Fig. 21 I, § 20] COORDINATES 19 vectors, by adding the component vectors geometrically, i.e. putting them together endwise successively, as in Fig. 21, where the dotted lines need not be drawn. By § 19, the projection of the resultant on any axis is equal to the sum of the projections of all the components on the same axis. EXERCISES 1. The cartesian coordinates x, y of any point P are the projections of its radius vector OP on the axes Ox, Oy. (See § 16.) 2. The projection of any vector AB on the axis Ox is the difference of the abscissas of A and B ; similarly for Oy. 3. A force of 10 lb. is inclined to the horizon at 60° ; find its hori- zontal and vertical components. 4. A ship sails 40 miles N. 60° E. then 24 miles N. 45° E. How far is the ship then from its starting point ? How far east ? How far north ? 6. A point moves 5 ft. along one side of an equilateral triangle, then 6 ft. parallel to the second, and finally 8 ft. parallel to the third side. What is the distance from the starting point ? 6. The sum of the projections of the sides of any closed polygon on any axis is zero. 7. If three forces acting on a particle are parallel and proportional to the sides of a triangle, the forces are in equilibrium, i.e. their resultant is zero. Similarly for any closed polygon. 8. Find the resultant of the forces OPi, OP2, OP3, OP4, OP5, if the coordinates of Pi, P2, P3, P4, P5, with as origin, are (3, 1), (1, 2), (-1, 3), (-2, -2), (2, -2). (Resolve each force into its components along the axes.) 9. If any number of vectors (in the same plane), applied at the ori- gin, are given by the coordinates x, y of their extremities, the length of the resultant is = \/(Sx)2 +(2^)2 (where Sa; means the sum of the ab- scissas, Sy the sum of the ordinates), and its direction makes with Ox an angle a such that tan a = Sy/Sx. 10. Find the horizontal and vertical components of the velocity of a ball when moving 200 ft. /sec. at an angle of 30° to the horizon. 20 PLANE ANALYTIC GEOMETRY [I, § 20 11. Six forces of 1, 2, 3, 4, 5, 6 lb., making angles of 60° each with the next, are applied at the same point, in a plane ; find their resultant. 12. A particle at one vertex of a square is acted upon by three forces represented by the vectors from the particle to the other three vertices ; find the resultant. 21. Geometric Propositions. In using analytic geometry to prove general geometric propositions, it is generally conven- ient to select as origin a prominent point in the geometric figure, and as axes of coordinates prominent lines of the figure. Example. In any right triangle, the distance from the vertex of the right angle to the midpoint of the hypotenuse is equal to half the hypotenuse. Since this theorem is true, if at all, when the triangle is in any position, we may place the vertex of the right angle at the origin and the adjacent sides along the positive axes. Then the coordinates of the vertices are (0, 0), (a, 0), (0, &), where a and b are the lengths of the two sides about the right angle. The length of the hypotenuse is Va^ -j- b^. The midpoint of the hypotenuse has the coordinates (a/2, 6/2), by § 11. Hence the distance from this point to the vertex of the right angle (0, 0) is V(a/2)2+ (6/2)2 ^ i. VoH^- Since this is half the length of the hypotenuse, the theorem is proved. Sometimes greater symmetry and elegance is gained by tak- ing the coordinate system in a general position. MISCELLANEOUS EXERCISES 1. A regular hexagon of side 1 has its center at the origin and one diagonal coincident with the axis Ox ; find the coordinates of the vertices. 2. If a square, with each side 6 units in length, is placed with one vertex at the origin and a diagonal coincident with the axis Ox, what are the coordinates of the vertices ? 3. If a rectangle, with two sides 3 units in length and two sides 3V3 units in length, is placed with one vertex at the origin and a diagonal I, §21] COORDINATES 21 along the axis Ox, what are the coordinates of the vertices ? There are two possible positions of the rectangle ; give the answers in both cases. 4. Show that the points (0, - 1), (- 2, 3), (6, 7), (8, 3) are the vertices of a parallelogram. Prove that this parallelogram is a rectangle. 5. Show that the points (1, 1), (- 1, - 1), ( + V3, - V3) are the vertices of an equilateral triangle. 6. Show that the points (6, 6), (3/2, - 3), (-3, 12), (- \S 3) are the vertices of a parallelogram. 7. Find the radius and the coordinates of the center of the circle pass- ing through the three points (2, 3), (- 2, 7), (0, 0). 8. The vertices of a triangle are (0, 6), (4, - 3), (— 5, 6). Find the lengths of the medians and the coordinates of the centroid of the triangle, i.e. of the intersection of the medians. Prove the following propositions : 9. The diagonals of any rectangle are equal. 10. The distance between the midpoints of two sides of any triangle is equal to half the third side. 11. The distance between the midpoints of the non-parallel sides of a trapezoid is equal to half the sum of the parallel sides. 12. The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram. 13. If two medians of a triangle are equal, the triangle is isosceles. 14. In any triangle the sum of the squares of any two sides is equal to twice the square of the median drawn to the midpoint of the third side plus half the square of the third side. 15. The line segments joining the midpoints of the opposite sides of any quadrilateral bisect each other. 16. The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals plus four times the square of the line segment joining the midpoints of the diagonals. 17. The difference of the squares of any two sides of a triangle is equal to the difference of the squares of their projections on the third side. 18. The vertices (a:i, yi), (x^^ 1/2), (xs^ y^) of a triangle being given, find the centroid (intersection of medians). CHAPTER II THE STRAIGHT LINE M- 22. Line Parallel to an Axis. When the coordinates a;, y of a point P with reference to given axes Ox, Oy are known, the position of P in the plane of the axes is determined com- pletely and uniquely. Suppose now that only one of the coordinates is given, say, a; = 3 ; what can be said about the position of the point P? It evidently lies somewhere on the line AB (Fig. 22) that is parallel to the axis Oy and has the distance 3 from Oy. Every point of the line AB has an abscissa a; = 3, and every point whose abscissa is 3 lies on the line AB. For this reason we say that the equation a; = 3 represents the line AB; we also say that a; = 3 is the equation of the line AB. More generally, the equation x = a, where a is any real number, represents that parallel to the axis Oy whose distance from Oy is a. Similarly, the equation y = b represents a parallel to the axis Ox. Fig. 22 EXERCISES Draw the lines represented by the equations : 1. x=-2. 4. 5 a: = 7. 2. x = 0. 6. y = 0. 3. x = 12.6. 6. 2y=-7. 22 Sx+l = 10-Sy y=±2. II, §24] THE STRAIGHT LINE 23 23. Line through the Origin. Let us next consider any line * through the origin 0, such as the line OP in Fig. 23. The points of this line have the prop- « erty that the ratio y/x of their coordi- nates is the same, wherever on this line the point P be taken. This ratio is equal to the tangent of the angle a made by the line with the axis Ox, Fig. 23 i.e. to what we shall call the slope of the line. Let us put tan a = m ; then we have, for any point P on this line : y/x = m, i.e. : (1) y = mx. Moreover, for any point Q, not on this line, the ratio y/x must evidently be different from tan a, i.e. from m. The equa- tion y = mx is therefore said to represent the line through whose slope is m; and y = mx is called the equation of this line. We mean by this statement that the relation y — mx is satis- fied by the coordinates of every point on the line OP, and only by the coordinates of the points on this line. Notice in partic- ular that the coordinates of the origin 0, i.e. a; = 0, y = 0, satisfy the equation y = mx. 24. Proportional Quantities. Any two values of x are proportional to the corresponding values ofyiiy= mx. For, if (a?!, 2/i) and {xz, 2/2) are two pairs of values of x and y that satisfy (1), we have yi = mxi, y2 = mx2; hence, dividing, 2/1/^2 = a;i/iC2- * For the sake of brevity, a straight line will here in general be spoken of simply as a line ; a line that is not straight will be called a curve. 24 PLANE ANALYTIC GEOMETRY [II, § 24 The constant quantity m is called the factor of propoHionality. Many instances occur in mathematics and in the applied sciences of two quantities related to each other in this man- ner. It is often said that one quantity y varies as the other quantity x. Thus Hooke's Law states that the elongation ^ of a stretched wire or spring varies as the tension t ; that is, E = kt^ where k is a constant. Again, the circumference c of a circle varies as the radius r ; that is, c = 2 Trr. EXERCISES 1. Draw each of the lines : (a)y = 2x. (c)y=-^^x. (e)5x4-3y = 0. (g)y=-x. (b)y=-3x. (d) 6y = Sx. {f)y = x. (h) x - y = 0. 2. Show that the equation ax -\- by = can be reduced to the form y = mx, if & T^ 0, and therefore represents a line through the origin. 3. Find the slope of the lines : (a) x + y = 0. (c) 3x_- j\y = 0. (b) x-y = 0. (d) V2 X + y = 0, 4. Draw a line to represent Hooke's Law E = kt, if A; = 10 (see Ex. 7, p. 14), Let (be represented as horizontal lengths (as is x in § 23) and let E be represented by vertical lengths (as is y in § 23). 6. Draw a line to represent the relation c = 2 Trr, where c means the circumference and r the radius of a circle. 6. The number of yards y in a given length varies as the number of feet / in the same length ; in particular, f=3y. Draw a figure to represent this relation. 7. If 1 in. = 2.64 cm., show that c = 2.64 i, where c is the number of centimeters and i is the number of inches in the same length. Draw a figure. II, § 26] ,THE STRAIGHT LINE 25 25. Slope Form. Finally, consider a line that does not pass through the origin and is not parallel to either of the axes of coordinates (Fig. 24) ; let it intersect the axes Ox, Oy at A, B, respectively, and let P (x, y) be any other point on it. The figure shows that the s lope m of the line, i.e. the tangent of the . angle a at which the line is in- clined to the axis Ox, is m tana = RP BR ' or, since RP=:QP-QR= QP- ' _2/ Fig. 24 m OB=y. -b b3ind BR=OQ=x y = mx-\- b„ that is, (2) where h — OB is called the intercept m ade by the line on the axis Oy, or briefly th ejji^intercept. The slope angle a at which the line is inclined to the axis Ox is always understood as the smallest angle through which the positive half of the axis Ox must be turned counterclockwise about the origin to become parallel to the line. 26. Equation of a Line. On the line AB of Fig. 24 take any other point P' -, let its coordinates be x', y', and show that y' = nnx' + b. Take the point P* (x/ y') outside the line AB and show that the equation y = mx -f 6 is not satisfied by the coordinates x', y' of such a point. For these reasons the equation y = mx + & is said to represent the line whose y4ntercept is b and whose slope is m; it is also called the equation of this line. The ^/-intercept OB = b and the slope m = tan a together fully determine the line. 26 PLANE ANALYTIC GEOMETRY [II, § 26 Every line of the plane can he represented by an equation of the form y = mx + b, excepting the lines parallel to the axis Oy. When the line be- comes parallel to the axis Oy, both its slope m and its y-inter- cept b become infinite. We have seen in § 22 that the equa- tion of a line parallel to the axis Oy is of the form x — a. Reduce the equation 3x—2y=5 to the form y = mx + b and sketch the line. EXERCISES 1. Sketch the lines whose y-intercept is 6 = 2 and whose slopes are in = i, 3, 0, — f ; write down their equations. v 2. Sketch the lines whose slope is m = 4/3 and whose y-intercepts are 0, 1, 2, 5, — 1, — 2, — 6, — 12.2, and write down their equations. 3. Sketch the lines whose equations are : (a) y=2x+S. (c) y=x-^. (e) x-y=l. (gr) 7a;-y+12=0. (6) y=-ia;+L (d) x-^y = l. (/) x-2y-\-2=0. (h) 4a;+3y+5=0. 4. Do the points (1, 5), (-2, -1), (3, 7) lie on the hne y=2x-\-S ? 5. A cistern that already contained 300 gallons of water is filled at the rate of 100 gallons per hour. Show that the amount A of water in the cistern n hours after filling begins is ^ = 100 n +300. Draw a figure to represent this relation, plotting the values of A vertically, with 1 vertical space = 100 gallons. 6. In experiments with a pulley block, the pull p in lbs., required to lift a load I in lbs., was found to be expressed by the equation |>=. 16 1-\-2. Draw this line. How much pull is required to operate the pulley with no load (i.e. when Z = 0) ? 7. The readings of a gas meter being tested, T, were found in compari- son with those of a standard gas meter S, and the two readings satisfied the equation T = SOO + 1.2 S. Draw a figure. What was the reading T when the reading S was zero ? What is the meaning of the slope of the line in the figure ? II, §27] THE STRAIGHT LINE 27 27. Parallel and Perpendicular Lines. Two lines y = m-ipc 4- 6i , y = m<^ + &2 are obviously parallel if they have the same slope, i.e. if (3) mi = m^. Two lines y = m-^x -\-bi, y = m^ + h^ are perpendicular if the slope of one is equal to minus the reciprocal of the slope of the other, i.e. if (4) mYm2 = — 1. For if mj = tan «! , mg = tan «£ , the condition that m^m2 = — 1 gives tan a2 = — 1/tan Oi = — cot «! , whence otg = «i + i t. EXERCISES 1. Write down the equation of any line : (a) parallel to y = 3 ic — 2, (6) perpendicular to y = 3 x — 2. 2. Show that the parallel to y = Zx — 2 through the origin is y = 3 x. 3. Show that the perpendicular to y = 3 x — 2 through the origin is y=-\x. 4. For what value of h does the line y = 3 a; + & pass through the point (4, 1) ? Find the parallel to ?/ = 3 x — 2 through the point (4, 1). 6. Find the parallel to y = 5 a; + 1 through the point (2, 3). 6. Find the perpendicular to y = 2x — \ through the point (1, 4). 7. What is the geometrical meaning of 6i = 62 in the equations y = m\x + 61 , y = mix + &2 ? 8. Two water meters are attached to the same water pipe and the water is allowed to flow steadily through the pipe. The readings B\ and Bi of the two meters are found to be connected with the time t by means of the equations i?i = 2.5f, i?2 = 2.5« + 150, where i?i and ^2 are measured in cubic feet and t is measured in seconds. Show that the Unes that represent these equations are parallel. What is the meaning of this fact ? 9. The equations connecting the pull p required to lift a load w is found for two pulley blocks to be Pi = .05w> 4- 2, Pi = .05w + 1.5 Show that the hnes representing these equations are parallel. Explain. 28 PLANE ANALYTIC GEOMETRY [II, § 27 10. The equations connecting the pull p required to lift a load w is found for two pulley blocks to be Pi = .15w + 1.5, p2 = .05w> + 1.5. Show that the lines representing these equations are not parallel, but that the values of pi and p2 are equal when lo = 0. Explain. 28. Linear Function. The equation y = mx-\-b, when m and b are given, assigns to every value of x one and only one definite value of y. This is often expressed by saying that mx -|- 6 is a function of x ; and as the expression mx + 5 is of the first degree in Xj it is called Si function of the first degree or, owing to its geometrical meaning, a linear function of x. Examples of functions of x that are not linear are 3 a* — 5, aa^ + bx-^Cf x{x — l), 1/a;, sin a;, 10*, etc. The equations y = 30^ — 5, y = ax^ -\- bx -{• c, etc., represent, as we shall see later, not straight lines but curves. The linear function y = nix -h b, being the most simple kind of function, occurs very often in the applications. Notice that the constant b is the value of the function for x = 0. The con- stant m is the rate of change of y with respect to x. 29. Illustrations. Example 1. A man, on a certain date, has $ 10 in bank ; he deposits $ 3 at the end of every week ; how much has he in bank x weeks after date ? Denoting by y the number of dollars in bank, we have 2^ = 3a; + 10. His deposit at any time a; is a linear function of x. Notice that the coefficient of x gives the rate of increase of this de- posit ; in the graph this is the slope of the line. Example 2. Water freezes at 0° C. and 32° F. ; it boils at 100° C. and at 212° F. ; assuming that mercury expands uni- formly, i.e. proportionally to the temperature, and denoting II, § 29] THE STRAIGHT LINE 29 by X any temperature in Centigrade degrees, by y the same temperature in Fahrenheit degrees, we have y-32 _ 212-32 _9 . „_8j,,32 If the line represented by this equation be drawn accurately, on a sufficiently large scale, it could be used to convert centi- grade temperature into Fahrenheit temperature, and vice versa. Example 3. A rubber band, 1 ft. long, is found to stretch 1 in. by a suspended mass of 1 lb. Let the suspended mass be increased by 1 oz., 2 oz., etc., and let the corresponding lengths of the band be measured. Plotting the masses as ab- scissas and the lengths of the band as ordinates, it will be found that the points (x, y) lie very nearly on a straight line whose equation is y = ^^x-\-l. The experimental fact that the points lie on a straight line, i.e. that the function is linear, means that the extension, 2/ — 1, is proportional to the tension, i.e. to the weight of the suspended mass x (Hooke's Law). Notice that only the part of the line in the first quadrant, and indeed only a portion of this, has a physical meaning. Can this range be extended by using a spiral steel spring ? Example 4. When a point P moves along a line so as to describe always equal spaces in equal times, its motion is called uniform. The spaces passed over are then proportional to the times in which they are described, and the coefficient of pro- portionality, i.e. the ratio of the distance to the time, is called the velocity v of the uniform motion. If at the time ^ = the moving point is at the distance Sq, and at the time t at the dis- tance s, from the origin, then S = So + vt. Thus, in uniform motion, the distance s is a linear function of the time t, and the coefficient of t is the speed : v= (s— s^/t. 30 PLANE ANALYTIC GEOMETRY [II, § 29 EXERCISES 1. If the constants m and b (§28) are given numerically, any number of points of the line can be located by arbitrarily assigning to the ab- scissa X any series of values and computing from the function the corre- sponding values of the ordinates. This process is known as plotting a line by points. Two points are sufficient to determine the line. Plot by points the following functions : (a) y = ^x, {b)y = 2x-5, (c)2/=-3x + 6, (d)y=-^x-i, (e)y = x(ix-l), {f)y = x\ {g)y = x\ ih)y = 2'. 2. Draw the line represented by the equation y = fx + 32 of Ex- ample 2, § 29. What is its slope ? What is the y-intercept ? What is the meaning of each of these quantities if y and x represent the tempera- tures in Fahrenheit and in Centigrade measure, respectively ? 3. Represent the equation y = ^^ x + 1 of Example 3, § 29, by a figure. What is the meaning of the y-intercept ? 4. Draw the line 8 = SQ + vt ot Example 4, § 29, for the values Sq = 10, V = 3. What is the meaning of t) ? Show that the speed v may be thought of as the rate of increase of s per second. 5. If, in the preceding exercise, v be given a value greater than 3, how does the new line compare with the one just drawn ? 6. If, in Ex. 4, v is given the value 3, and Sq several different values, show that the lines represented by the equation are parallel. Explain. 7. In experiments on the temperatures at various depths in a mine, the temperature (Centigrade) T was found to be connected with the depth d by the equation r=G0 + .01^, where d is measured in feet. Draw a figure to represent this equation. Show that the rate of increase of the temperature was 1° per hundred feet. 8. In experiments on a pulley block, the pull p (in lb.) required to lift a weight w (in lb.) was found to be j9 = .03 w> + 0.5. Show that the rate of increase of j3 is 8 lb. per hundredweight increase in w. 9. The velocity tj of a body falling from rest is proportional to the time: V = gt, where.gr is a constant (about 32 in English units). If the body is thrown down with an initial velocity t?o, the velocity at any time tis V = Vq-{- II, §30] THE STRAIGHT LINE 31 Draw a figure to represent this equation for g = 32, v^ = 10. Show that g is the rate of increase of the velocity (called the acceleration), 30. General Linear Equation. The equation in which. Ay B, C are any real numbers, is called the general equation of the first degree in x and y. The coefficients A, B, G are called the constants of the equation ; x, y are called the variables. It is assumed that A and B are not both zero. The terms Ax and By are of the first degree ; the term C is said to be of degree zero because it might be written in the form Csfi ; this term C is also called the constant term. Every equation of the first degree, (5) Ax-{^By ^ C=0, in which A and B are not both zero, represents a straight line; and conversely, every straight line can be represented by such an equation. For this reason, every equation of the first degree is called a linear equation. The first part of this fundamental proposition follows from the fact that, when B is not equal to zero, the equation can be reduced to the form y = mx + 6 by dividing both sides by B ; and we know that y = mx + b represents a line (§ 25). When B is equal to zero, the equation reduces to the form x = a^ which also represents a line (§ 22). The second part of the theorem follows from the fact that the equations which we have found in the preceding articles for any line are all particular cases of the equation Ax -\- By +0=0. This equation still expresses the same relation between x and y when multiplied by any constant factor, not zero. Thus, any one of the constants A, B, C, if not zero, can be reduced to 1 by dividing both sides of the equation by this constant. 32 PLANE ANALYTIC GEOMETRY [II, § 30 The equation is therefore said to contain only two (not three) essential constants. 31. Conditions for Parallelism and for Perpendicularity. It is easy to recognize whether two lines whose equations are Ax + By -\-C=0 and A'x -f- B'y + C" = are parallel or per- pendicular. The lines are parallel if they have the same slope7\ and they are perpendicular (§ 27) if the product of their slopes is equal to —1. The slopes of our lines are — A/B and — A/B' ; hence these lines are parallel if — A/B = — A!/Bf, i'e. if A.B = A''.B'', and they are perpendicular if A A b' B , = -1^ i.e. if AA' -hBB' = 0. 32. Intercept Form. If the constant term C in a linear equation is zero, the equation represents a line through the origin. For, the coordinates (0, 0) of the origin satisfy the equation Ax + By = 0. If the constant terra C is not equal to zero, the equation Ax -{- By -\- C=0 can be divided by (7; it then reduces to the form A , B , ^ n —X H — v + 1 =0. C If A and B are both different from zero, this can be written : -C/A' - C/B or putting — C/A = a, — C/B = b : (6) a o The conditions A=^0, B =^ mean fig. 25 evidently that the line is not parallel to either of the axes. Therefore the equation of any line, not passing through the II, §32] THE STRAIGHT LINE 33 origin, and not parallel to either axis, can be written in tlie form (6). With y = this equation gives x=: a; with x = it gives y = b. Thus are the intercepts (Fig. 25) made by the line on the axes Ox, Oy, respectively (see § 25). EXERCISES 1. Write down the equations of the line whose . intercepts on the axes Ox, Oy are 5 and — 3, respectively ; the line whose intercepts are — I and 7 ; the line whose intercepts are — 1 and — |. Sketch each of the lines and reduce each of the equations to the form Ax+By-\- (7=0, so that A, B, C are integers. 2. Find the intercepts of the lines : Sx — 2y = l,x-{-Ty-\-l=0, — Sx+ly — 5 = 0. Try to read off the values of the intercepts directly from these equations as they stand. 3. In Ex. 2, find the slopes of the lines. 4. Prove (6), § 32 by equality of areas, after clearing of fractions. 6. What is the equation of the axis Oy ? of the axis Ox ? 6. What is the value of B such that the line represented by the equa- tion 4:X + By — li = passes through the point (—6, 17). 7. What is the value of A such that the line Ax i-l y=10 has its oj-intercept equal to — 8 ? 8. Reduce each of the following equations to the intercept form (6) , and draw the lines : (a) 3 X - 5 ?/ - 16 = 0. (b) X -\- ^ y -\- 1 = 0. 4x-_3j^^-6^2 (^d) 5x=:Sx + y-10. 9. Reduce the equations of Ex. 8 to the slope form (2), § 25. 10. Find the equation of the line of slope 6 passing through the point (6, - 5). D 34 PLANE ANALYTIC GEOMETRY [II, § 32 11. Show that the points (- 1, - 7), (i, - 3), (2, 2), (-2, - 10) lie on the same line, 12. Find the area of the triangle formed by the lines x -\- y = 0, x — y = OfX — a = 0. 13. Show that the line 4(aj — a) -f- 5(y — 6) = is perpendicular to the line 5x — 4y — 10 = and passes through the point (a, b). 14. A line has equal positive intercepts and passes through (—5, 14). What is its equation ? its slope ? 16. If a line through the point (6, 7) has the slope 4, what is its y-intercept ? its x-intercept ? 16. The Reaumur thermometer is graduated so that water freezes at O'^ and boils at SO*^- Draw the line that represents the reading i? of the Reaumur thermometer as a function of the corresponding reading C of the Centigrade thermometer. 17. Express the value of a note of $ 1000 at the end of the first year as a function of the rate of interest. At 6 % simple interest its value is what function of the time in years ? 33. Line through One Point. To find the line of given slope 7% through a given point Pi{xi, yi), observe that the equation must be of the form (2), viz. since this line has the slope m^. If this line is to pass through the given point, the coordinates ajj, yi must satisfy this equa- tion, i.e. we must have Vi = iriiXi + h. This equation determines b, and the value of b so found might be substituted in the preceding equation. But we can eliminate b more readily between the two equations by subtracting the latter from the former. This gives as the equation of the line of slope mj through Pi(x^, y^. II, § 34] THE STRAIGHT LINE 35 The problem of finding a line through a given point parallel^ or perpendicular, to a given line is merely a particular case of the problem just solved, since the slope of the required line can be found from the equation of the given line (§ 27). If the slope of the given line is m^ = tan a^, the slope of any parallel line is also m^, and the slope of any line perpendicular to it is 1 mg = tan (a^ + ^tt) = — cot Wj = — m, 34. Line through Two Points, To find the line through two given points, Pi{x^, Pi), P^ix^, y^, observe (Fig. 26) that the slope of the required line is evi- dently m, = _ y2-yi _^y Xa Xi Ax if, as in § 9, we denote by Ax, Ay the projections of P^Pz on Ox, Oy; and as the line is to pass through (x^, y{), we find its equation by § 33 as y-yi = y2-yi Xt> — 3/1 or y-yi = ^{x Ax (x - X,), 0. The equation of the line through two given points (xi, yi), (^2j 2/2) can also be written in the determinant form X y 1 xi yi 1 =0, ^2 2/2 1 which (§ 14) means that the point (x, y) is such as to form with the given points a triangle of zero area. By expanding the determinant it can be shown that this equation agrees with the preceding equation. 36 PLANE ANALYTIC GEOMETRY [II, § 34 EXERCISES 1. Find the equation of the line through the point (—7, 2) parallel to the line y = Sx. 2. Show that the points (4, — 3), (— 5, 2), (5, 20) are the vertices of a right triangle. 3. Find the equation of the line through the point (—6, —3) which makes an angle of 30° with the axis Ox ; 30° with the axis Oy. 4. Does the line of slope | through the point (4, 3) pass through the point (-5, -4) ? 6. Find the equation of the line through the point (— 2, 1) parallel to the line through the points (4, 2) and (—3, — 2). 6. Find the equations of the lines through the origin which trisect that portion of the line 6 x — 6 y = 60 which lies in the fourth quadrant. 7. What are the intercepts of the line through the points (2, — 3), C-5,4)? 8. Show that the equation of the line through the point (a, b) per- pendicular to the line Ax + By -}- C = ia (x — a) /A = (y — bys. 9. Find the equations of the diagonals of the rectangle formed by the lines x-{-a = 0, x — b = 0, y + c = 0, y — d = 0. 10. Find the equation of the perpendicular bisector of the line joining the points (4, — 5) and (—3, 2). Show that any point on it is equally distant from each of the two given points. 11. Find the equation of the line perpendicular to the line Ax — Sy-\-G=0 that passes through the midpoint of (— 4, 7) and (2, 2). 12. What are the coordinates of a point equidistant from the points (2, — 3) and (— 5, 0) and such that the line joining the point to the origin has a slope 1 ? 13. In an experiment with a pulley-block it is assumed that the rela- tion between the load I and the pull p required to lift it is linear. Find the relation if p = 8 when I = 100, and p = 12 when I = 200. 14. In an experiment in stretching a brass wire it is assumed that the elongation E is connected with the tension t by means of a linear relation. Find this relation if i = 18 lb. when E = A in., and < = 58 lb. when ^ = .3 in. CHAPTER III RELATIONS BETWEEN TWO OR MORE LINES 35. Intersection of Two Lines. The point of intersection of any two lines is found by solving the equations of the lines as simultaneous equations. For, the coordinates of the point of intersection must satisfy each of the two equations, since this point lies on each of the two lines ; and it is the only point having this property. Thus, by solving the equations 3 a; + 5^-34 = 0, we find X = S, y = 5] hence (3, 5) is the point of intersection of the two lines represented by these equations. 36. Particular Cases. The equations of any two lines being given, say (1) aix + hy = fci, a^x -f- 622/ = ^2) we find by the usual method, that is, first multiplying by 62? &i and subtracting, then multiplying by 02, ai and subtracting : (2) , (ai&2 — a2bi)x = fci&2 — ^2^) {a^2 — <^^i)y = GbJ^2 — Ct2^1« The expression aih^ — a^bi is called the determinant of the equations. Two cases must be distinguished according as this determinant is :?i: or = 0. (a) If a^2 — <^2^i =9^ ^> which means by § 31 that the lines are not parallel, we can divide the equations (2) by this determi- nant and thus find x and y. If, in particular, ki and k2 are both 37 38 PLANE ANALYTIC GEOMETRY [III, § 36 zero, tiiat is, if the equations (1) are homogeneous and hence represent two lines through the origin, we find from (2) a; = and 2/ = 0, as was to be expected. (6) If ai&2 — ct2^i = 0, that is, if the lines (1) are parallel, we cannot in (2) divide by 0162 — (h^i ; the equations (2) then become 0'X = ^162 — k2bif . 2/ = aik2 — 02^1, and cannot be satisfied by any values of x and y unless the right-hand members are both zero. In the latter case we have rt'2 ^ ^2 that is, the second equation is merely a multiple of the first. In this case the two equations (1) represent the same line and have therefore all points in common. EXERCISES 1. Find the coordinates of the points of intersection of the following lines ; and check by a sketch : (a') J5x-7 2/ + ll=0, ,^. f 3x-}-2y=0, ,. f 2.4a;+3.1 y= 4.6, ^^ [3ic+2y-12=0. ^^ [6x-4y+4=0. ^^ [ .8x + 2y = 6.2. 2. Do the following pairs of lines intersect, or are they parallel or coincident ? . . f3a;-6y-8=0, ... f2x-6y-4=0, , . f x + iy = 0, ^^ I x-2y+l = 0. ^^ [ x-3y-2=0. ^^ |2x + 3y = 0. 3. Show that the condition that the three lines Ax -{- By + (7=0, A'x 4- B'y + C" = 0, A"x + B"y + C" = meet at a point is ABC A' B' C =0. A^- B" C" 4. Show that the straight lines 3x + y — 1=0, jc-3y + 13 = 0, 2x — y -^6 = have a common point. Ill, § 37] TWO OR MORE LINES 39 6. Show that the lines joining the midpoints of the sides of any tri- angle divide the triangle into four equal triangles. 6. Show that the altitudes of any triangle meet in a point. 7. Show that the medians of any triangle meet in a point. 8. Show that the line through the origin perpendicular to the line through the points (a, 0) and (0, b) meets the lines through the points (a, 0\, {—b, b) and (0, &), (a, -va) in a common point. Xejt. &XX M^, y-o -^ 37. Angle between Two Lines. We shall understand by the angle (/, l')= 6 between two lines I and V the least angle through which I must be turned coun- terclockwise about the point of inter- section to come to coincidence with V. This angle is equal to the differ- ence of the slope angles a, a' (Fig. 27) of the two lines. Thus, if a' > a, we have 6 = a' — a, since a' is the exterior Fig. 27 angle of a triangle, two of whose interior angles are a and 6, It follows that /o\ i. /I i. / f \ tan a' — tana. (3) tan 6 = tan (a' — a) = • ^ ^ ^ ^ 1 H- tan a tan a' If the equations of I and V are y = mx -^ b, y = m'x -|- 6', respectively, we have tan a = m, tan «' = m' ; hence (4) t^ne^^^^^^^r ^ ^ 1 -f mm' If the equations of I and Z' are Ax -\- By + 0=0, A'x + B'y + C" = 0, respectively, we have tan a = — A/B, tan a' = — A'/B' ; hence AB' - A'B (5) tan 6 = AA! + BB' 40 PLANE ANALYTIC GEOMETRY [III, § 38 38. It follows, in particular, that the two lines I and V, § 37, are parallel if and only if m' = m, or AB' - A'B = ; and they are perpendicular to each other if and only if "^ m' = --,ovAA' + BB' =0. m (Compare §§27, 31.) Hence, to write down the equation of a line parallel to a given line, replace the constant term by an arbitrary constant ; to write down the equation of a line per- pendicular to a given line, interchange the coefficients of x and y, changing the sign of one of them, and replace the constant term by an arbitrary constant. EXERCISES 1. Determine whether the following pairs of lines are parallel or per- pendicular : Sx + 2y-Q = 0, 2a;-3y + 4 = 0; 6a; + 3y-6 = 0, 10x + 6y + 2 = 0;2ic-f6y-14=0, 8a;-3y + 6=0. 2. Find the point of intersection of the line 6a; + 8y-f-17 = with its X)erpendicular through the origin. 3. Find the point of intersection of the lines through the points (6, —2) and (0, 2), and (4, 6) and (- 1, -4). 4. Find the perpendicular bisector of the line-seginent joining the point (3, 4) to the point of intersection of the lines 2x — y-\-l = and 3 x + y - 16 = 0. 6. Find the lines through the point of intersection of the lines 5 x— y =0, x-t-7y — 9 = and perpendicular to them. 6. Find the area of the triangle formed by the lines 3 a; + 4 y = 8, 6 a; — 5 2/ = 30, and x = 0. 7. Find the area of the triangle formed by the lines a; + y — 1 = 0, 2 X + y + 6 = 0, and X - 2 y - 10 = 0. 8. Find the point of intersection of the lines a ha Ill, §39] TWO OR MORE LINES 41 9. Find the area of the triangle formed by the lines y = m\X + 6i, y = mix, + 62 and the axis Ox. 10. The vertices of a triangle are (5, - 4), (— 3, 2), (7,6). Find the equations of the medians and their point of intersection. 11. Find the angle between the lines 4 x— 3 y— 6=0 and x—1 1/+6=0. 12. Find the tangent of the angle between the lines (a) 4ic— 3y+6=0 and9a; + 2y-8 = 0; (6) 3a;4-6y- 11 = and a; + 2y-3 = 0. 13. Find the two lines through the point (6, 10) inclined at 45° to the line 3 a; - 2 ?/ - 12 = 0. 14. Find the lines through the point (— 3, 7) such that the tangent of the angle between each of these lines and the line 6a; — 2y + ll=0isJ. 15. Show that the angle between the lines Ax + By + = and {A + B)x-(A-B)y + D = is 45°. 16. Find the lines which make an angle of 45° with the line 4x — 7 y + 6 =0 and bisect the portion of it intercepted by the axes. 17. The hypotenuse of an isosceles right-angled triangle lies on the line 3a; — 6y— 17=0. The origin is one vertex ; what are the others ? 39. Polar Equation of Line. The position of a line in the plane is fully determined by the length p = ON (Fig. 28) of the perpendicular let fall from the origin on the line and the angle ft = xOJSf made by this perpendicular with the axis Ox. Then p and ft are evidently the polar coordinates of the point -A'' (§ 16). Let P be any point of the line and OP = r, xOP= its polar coordinates. As the projection of OP on the perpendicular ON is equal to ON, and the angle NOP — — ft, we have (6) rcos(-ft)=p. This is the equation of the line NP in polar coordinates. 42 PLANE ANALYTIC GEOMETRY [III, § 40 40. Normal Form. The last equation can be transformed to Cartesian coordinates by expanding the cosine : r cos cos )8 + r sin <^ sin )3 =p and observing (§ 17) that r cos = «, r sin = ?/ ; the equation then becomes (7) xco8p+ y8inp=:p. This equation, which is called the normal form of the equation of the line, can be read off directly from the figure ; it means that the sum of the projections of x and y on the perpendicular to the line is equal to the projection of r (§ 20). Observe that in the normal form (7) the number p is always positive, being the distance of the line from the origin, or the radius vector of the point N. Hence x cos fi-\-ysin ft is always positive ; this also appears by considering that xcos (S-^y sin /? is the projection of the radius vector OP on ON, and that this radius vector makes with ON an angle that cannot be greater than a right angle. , The angle p = xON is, as a polar angle (§ 16), always under- stood to be the angle through which the axis Ox must be turned counterclockwise about the origin to make it coincide with ON; it can therefore have any value from to 2 tt. By drawing the parallel to the line NP through the origin it is readily seen that, if a is the slope angle of the line NP, we have fi = a-{-^Tr or ^ = a + |7r according as the line lies on one side of the origin or the other, angles differing by 2 tt being regarded as equivalent. Thus, in Fig. 28, a = 120°, /5 = a + | tt = 120° + 270° = 390°, which is equivalent to 30°. For a parallel on the opposite side of the origin we should have y8 = «+ i tt = 120° + 90° = 210°. Ill, § 41] TWO OR MORE LINES 43 41. Reduction to Normal Form. The equation Ax-{-By+C=0 is in general not of the form (7), since in the latter equation the coefficients of x and y, being the cosine and sine of an angle, have the property that the sum of their squares is equal to 1, while in the former equation the sum of the squares of A and B is in general not equal to 1. But the general equation Ax + By-^C=0 can be reduced to the normal form (7) by multiplying it by a factor k properly chosen ; we know (§ 30) that the equation kAx + kBy + kO=0 represents the same line as does the equation Ax-{-By+C=0. Now if we select k so that kA = cos p, kB = sin /3, kC=—p, the equation Ax -{- By -{- G = reduces to the normal form X cos p -{-y sin y3 — p = 0. The first two conditions give k^A^ + fc2jB2 ^ cos2 p + sin2 yS = 1, whence A; = ± ■yjA^ + B^ ^ Since the right-hand member p in the normal form (7) is posi- tive, the sign of the square root must be selected so that kC becomes negative. We have therefore the rule : To reduce the general equation Ax -\- By -\- C =0 to the normal form xcos p -\-y sin /3 —p= 0, divide by — VA^ + B^ when C is positive and by -{-■\/AF+^ when C is negative. Then the coefficients of x and y will be cos ft sin /3, respec- tively, and the constant term will be the distance p of the line from the origin. 44 PLANE ANALYTIC GEOMETRY [III, § 41 Thus, to reduce Sx-\-2y-\-5 = 0to the normal form, divide by -V3M^ = -Vl3; this gives cosy8 = =, smy8 = z=j —p = Vl3' ' Vl3' ^ V13' i.e. the normal form is 3 :2/ = V13 Vl3 V13 The perpendicular to the line from the origin has the length 6/Vl3 ; and as both cos ft and sin ft are negative, this perpen- dicular lies in the third quadrant. Draw the line. Reduce the equation 3x-{-2y — 5 — 0to the normal form. 42. Distance of a Point from a Line. If, in Fig. 28, we take instead of a point P on the line any point P^ {xi, y^) not on the line (Fig. 29), the expression Xi cos /3 + yi sin p is still the projection on ON (produced if necessary) of the radius vector OPi. But this projection OS differs from the normal ON = p to the line. The figure shows that the difference Xi cos p + yisin p ^ p = OS — 0N= NS fio. 29 is equal to the distance NiPi of the point P^ from the line. Thus, to find the distance of any point P^ (xi, yi) from a line whose equation is given in the normal form xcos p + ysinp — p= 0, it suffices to substitute in the left-hand member of this equa- tion for Xf y the coordinates x^, y^ of the point Pj. The expression Xicos p + yisin p —p then represents the distance of P^ from the line. If this expression is negative, the point Pi lies on the same side of the line as does the origin ; if it is positive, the point in, § 43] TWO OR MORE LINES 45 Pi lies on the opposite side of the line. Any line thus divides the plane into two regions which we may call the positive and nega- tive regions ; that in which the origin lies is the negative region. To find the distance of a point P^ (aji, ^i) from a line given in the general form Ax + By+C=0, we have only to reduce the equation to the normal form (§ 41) and then apply the rule given above. Thus the distance is Ax^ 4- By^ + O ^^ Ax^±ByjJ-C_ according as C is positive or negative. 43. Bisector of an Angle. To find the bisectors of the angles between two lines given in the normal form xGos ft-\-y sin ^8 — p = 0, a; cos /?' + ?/ sin ^' — p' = 0, observe that for any point on either bisector its distances from the two lines must be equal in absolute value. Hence the equations of the bisectors are xGos ^-{-ysm (i—p=±(x cos /3' -\-y sin /8' —p'). To distinguish the two bisectors, ob- serve that for the bisector of that pair of vertical angles which contains the origin (Fig. 30) the perpendicular dis- tances are, in one angle both positive, in the other both negative ; hence the plus sign gives this bisector. If the equations of the lines are given in the general form Ax-^By-^C = 0, A'x+B'y-[-C' = 0, first reduce the equations to the normal form, and then apply the previous rule. N^ / / N fv Xn— /''>-' \ "n ~\\ 1 \/ 1 ^ ^ 1 ^x • / / / 1 ' Fig. 30 46 PLANE ANALYTIC GEOMETRY [HI, § 43 EXERCISES 1. Draw the lines represented by the following equations : (a) r cos (0 - ^ ir) = 6. (e) r cos (0 + f tt) = 3. (6) r cos (0 — t) = 4. (/) r sin (0 - ^ tt) = 8. (c) r cos = 10. ($r) r sin (0 + | ir) = 7. (d) r sin = 5. (Ji) r cos (0 — | tt) = 0. 2. In polar coordinates, find the equations of the lines : (a) parallel to and at the distance 6 from the polar axis (above and below) ; (6) per- pendicular to the polar axis and at the distance 4 from the pole (to the right and left) ; (c) inclined at an angle of ^ tt to the polar axis and at the distance 12 from the pole. 3. Express in polar coordinates the sides of the rectangle ABC it OA = 6 and AB = 9, OA being taken as polar axis. 4. What lines are represented by (7) when p is constant, while /3 varies from zero to 2 tt ? What lines when p varies while /3 remains con- stant? 6. The perpendicular from the origin to a line is 5. units in length and makes an angle tan-i ^ with the axis Oac. Find the equation of the line. 6. Reduce the equations of Ex. 8, p. 33, to the normal form (7). 7. Find the equations of the lines whose slope angle is 160° and which are at the distance 4 from the origin. 8. What is the equation of the line through the point ( — 3, 5) whose perpendicular from the origin makes an angle of 120^ with the axis Ox ? 9. For the line 7x — 24y — 20=0 find the intercepts, slope, length of perpendicular from the origin and the sine and cosine of the angle which this perpendicular makes with the axis Ox. 10. Find by means of sin /3 and cos/3 the quadrants crossed by the line 4ic — 5y = 8. 11. Put the following equations in the form (7) and thus find p, sin /3, cos /9: (a)y=:mx + b. (6)?+| = l. (c)3a; = 4y. a IS. Is the point (3, — 4) on the positive or negative side of the line through the points (— 5, 2) and (4, 7) ? Ill, § 43] TWO OR MORE LINES 47 13. Is the point (—1, — f ) on the positive or negative side of the line 4:X-9y-8 = 0? 14. Find by means of an altitude and a side the area of the triangle formed by the lines Zx + 2y + 10 = 0, ^x-3y+lQ = 0, 2x + y-4: = 0. Check the result with another altitude and side. 15. Find the distance between the parallel lines (a) Sx— 5y— 4 = and 6 a; — 10 y + 7 = ; (6) 5 x + 7 2/ + 9 = and 16 re + 21 y — 3 = 0. 16. What is the length of the perpendicular from the origin to the line through the point (—5, — 4) whose slope angle is 60° ? 17. What are the equations of the lines whose distances from the origin are 6 units each and whose slopes are f ? 18. Find the points on the axis Ox whose perpendicular distances from the line 24 a: — 7 t^ — 16 = are ±5. 19. Find the point equidistant from the points (4, — 3) and (—2, 1), and at the distance 4 from the line Sx — Ay — 6 = 0. 20. Find the line parallel tol2a; — 5y — 6 = and at the same distance from the origin ; farther from the origin by a distance 3. 21. Find the two lines through the point (1, ^^) such that the perpen- diculars let fall from the point (6, 5) are of length 5. 22. Find the line perpendicular to 4 a; — 7 y — 10 = which crosses the axis Ox at a distance 6 from the point (— 2, 0). 23. Find the bisectors of the angles between the lines: (a) x—y —4= and 3 x + 3 2^ + 7 = 0; (6) 6a:-12y-16 = and 24x + 7y + 60 = 0. 24. Find the bisectors of the angles of the triangle formed by the lines 5x + 12 2/ + 20 = 0, 4x-32/-6 = 0, 3x-4y+5 = and the center of the circle inscribed in the triangle. 25. Find the bisector of that angle between the lines 3 x — VSy+ 10 =0, V2 x + y — 6 = 0in which the origin lies. 26. If two lines are given in the normal form, what is represented by their sum and what by their difference ? 27. Show that the angle between the lines x + y = and x — y = is 90° whether the axes are rectangular or oblique. 48 PLANE ANALYTIC GEOMETRY [IIL § 44 44. Pencils of Lines. All lines through one and the same point are said to form a pencil; the point is called the center of the pencil. If ^^ \A'x-j-B'y-j-C = are any two different lines of a pencil, the equation (9) Ax-{-By+ C-^k{A'x + B'y + C) = 0, where k is any constant, represents a line of the pencil. For, the equation (9) is of the first degree in x and y, and the coeffi- cients of X and y cannot both be zero, since this would mean that the lines (8) are parallel. Moreover, the line (9) passes through the center of the pencil (8) because the coordinates of the point that satisfies each of the equations (8) also satisfy the equation (9). All lines parallel to the same direction are said to form a pencil of parallels. It is readily seen that if the lines (8) are parallel, the equation (9) represents a line parallel to them. EXERCISES 1. Find the line : (a) through the point of intersection of the lines 4x— 7y+6 = 0, 6x-f-lly — 7 = and the origin ; (&) through the point of intersection of the lines 4x — 2y — 3=0, x + y — 5 = and the point (—2, 3); (c) through the point of intersection of the lines 4x— 6y4-6 = 0, y— x — 3 = 0, of slope 3 ; (d) through the intersection of5x— 6y-f-10 = 0, 2x + 3y — 12 = 0, perpendicular to 4 y + x = 0. 5. Find the line of the pencil x — 5 = 0, y + 2 = that is inclined to the axis Ox at 30°. 3. Determine the constant b of the line y = Sx + b so that this line shall belong to the pencil 3x — 4y + 6 =0, x = 5. 4. Find the line joining the centers of the pencils x — 3 y = 12, 5 X - 2 y = 1 and x + y = 6, 4x — 5y = 3. 6. Find the line of the pencil 4x-5y-12 = 0, 3x + 2y-16 = that makes equal intercepts on the axes. Ill, § 45] TWO OR MORE LINES 49 45. Non-linear Equations representing Lines. When two lines are given, say A'x + B'y + O' = 0, then the equation {Ac + By-^Cf){A'x + B'y-{-C) = 0, obtained by multiplying the left-hand members (the right-hand members being reduced to zero) is satisfied by all the points of the first given line as well as all the points of the second given line, and by no other points. The product equation which is of the second degree is there- fore said to represent the two given lines. Similarly, by equat- ing to zero the product of the left-hand members of the equations of three or more straight lines (whose right-hand members are zero) we find a single equation representing all these lines. An equation of the nth degree may therefore represent n straight lines, viz. when its left-hand member (the right-hand member being zero) can be resolved into n linear factors, with real coefiicients. EXERCISES 1. Find the. common equation of the two axes of coordinates. 2. Show that n lines through the origin are represented by a homo- geneous equation (i.e. one in which all terms are of the same degree in X and y) of the nth degree. 3. Draw the lines represented by the following equations : (or) (X -a)(y-b)= 0. (/) xy - ax = 0. (&) 3 a;2 - xy - 4 2/2 = 0. (g) y^ - ^y^ + Qy = 0. (c) a;2_9y2-o. (h) x^y-xy = 0. (d) ax2 4- &?/2 = 0. (0 2/3 _ 6 a:2/2 + 11 a;2y - 6 x3 = 0. (e) x2 - X - 12 = 0. 4. What relation must hold between a, h, &, if the lines represented by ax2 + 2 hxy + by^ = are to be real and distinct, coincident, imag- inary ? 50 PLANE ANALYTIC GEOMETRY [III, § 45 MISCELLANEOUS EXERCISES 1. Find the angle between the Hnes represented by the equation ax^ + 2 hxy + hy'^ = 0. What is the condition for these Unes to be per- pendicular ? coincident ? 2. Reduce the general equation ^x + J?«/ 4- C = to the normal form x cos /3 + 2/ sin /3 = p by considering that, if both equations represent the same line, the intercepts must be the same. 3. Find the line through (xi , y{) making equal intercepts on the axes. 4. Find the area of the triangle formed by the Unes y — m\x + 6i , y = miX -\-b2,y = b. 6. What does the equation

is a function of d. Represent this function graphically ; interpret the part of the hne in the third quadrant. 9. A train, after leaving the station J., attains in the first 6 minutes, IJ miles from A, the speed of 30 miles per hour with which it goes on. How far from A will it be 50 minutes after starting? (Compare Ex- ample 4, § 29.) Illustrate graphically, taking s in miles, t in minutes. 10. A train leaves Detroit at 8 hr. 25 m. a.m. and reaches Chicago at 4 hr. 5 m. p.m. ; another train leaves Chicago at 10 hr. 30 m. a.m. and arrives in Detroit at 5 hr. 30 m. p.m. The distance is 284 miles. Regard- ing the motion as uniform and neglecting the stops, find graphically and analytically where and when the trains meet. If the scale of distances (in miles) be taken 1/20 of the scale of times (in hours), how can the velocities be found from the slopes ? 11. A stone is dropped from a balloon ascending vertically at the rate of 24 ft. /sec; express the velocity as a function of the time (Example 6, § 29) . What is the velocity after 4 sec. ? 12. How long will a ball rise if thrown vertically upward with an initial velocity of 100 ft. /sec. ? CHAPTER IV THE CIRCLE 46. Circles. A circle, in a given plane, is defined as the locus of all those points of the plane which are y at the same distance from a fixed point. Let G (h, Tc) be the center, r the radius (Fig. 31) ; the necessary and sufficient condition that any point P (x, y) is at the distance r from C (h, k) is that Fig. 31 1-* (1) (aJ - ^)2 + (2/ - A;)2 = *'2. This equation, which is satisfied by the coordinates x, y of every point on the circle, and by the coordinates of no other point, is called the equation of the circle of center C (h, k) and radius r. If the center of the circle is at the origin (0, 0), the equation of the circle is evidently (2) a?-\-y'^ = i\ EXERCISES Write down the equations of the following circles : (a) center (3, 2), radius 7 ; (6) center at origin, radius 3 ; (c) center at (— a, 0), radius a ; {d) circle of any radius touching the axis Ox at the origin ; (e) circle of any radius touching the axis Oy at the origin. Illustrate each case by a sketch. 51 52 PLANE ANALYTIC GEOMETRY [IV, § 47 47. Equation of Second Degree. Expanding the equation (1) of § 46, we obtain the equation of the circle in the new form x^^y'i-2hx-2 ky + li" + Ic" - r^ = 0. This is an equation of the second degree in x and y. But it is of a particular form. The general equation of the second degree in X and y is of the form (3) Ax'-{-2Hxy-\-By^-\-2Gx + 2Fy-\-C=0'y i.e. it contains a constant term, (7; two terms of the first de- gree, one in x and one in y ; and three terms of the second de- gree, one in x^, one in osy, and one in y\ If in this general equation we have it reduces, upon division by Aj to the form ii^ + y^ + --^x + —y-\-- = 0, which agrees with the form (1) of the equation of a circle, ex- cept for the notation for the coefficients. We can therefore say that any equation of the second degree which contains no xy-term and in which tlie coefficients of a^ and y^ are equal, may represent a circle. 48. Detennination of Center and Radius. To draw the circle represented by the general equation (4) Ax^ + Ay^ + 2 «a; + 2 Fy + C = O, where A, G, F, Q are any real numbers while ^ ^ 0, we first divide by ^ and complete the squares in x and y ; i.e. we first write the equation in the form The left-hand member represents the square of the distance of the point (x, y) from the point (— 0/A, — F/A) ; the right- IV, §49] THE CIRCLE 53 hand member is constant. The given equation therefore repre- sents the circle whose center has the coordinates 7. ^ T. ^ and whose radius is This radius is, however, imaginary if G^ -^ F"^ < AC; in this case the equation is not satisfied by any points with real co- ordinates. If G"^ + F^^ AG) the radius is zero, and the equation is satis- fied only by the coordinates of the point ( — G/A, — F/A). If G^+F"^ > AG, the radius is real, and the equation repre- sents a real circle. Thus, the general equation of the second degree (3), § 47, repre- sents d circle if, and only if, A = B=^0,H=0, G' + F'>AG. 49. Circle determined by Three Conditions. The equation (1) of the circle contains three constants h, 7c, r. The general equation (4) contains four constants of which, however, only three are essential since we can always divide through by one of these constants. Thus, dividing by A and putting 2 G/A = a, 2 F/A — b, C/A = c, the general equation (4) assumes the form (5) a^-f2/' + a^-h&2/ + c = 0, with the three constants a, b, c. The existence of three constants in the equation corresponds to the possibility of determining a circle geometrically, in a variety of ways, by three conditions. It should be remembered in this connection that the equation of a straight line contains two essential constants, the line being determined by two geometrical conditions (§ 30). 54 PLANE ANALYTIC GEOMETRY [IV, § 49 EXERCISES 1. Draw the circles represented by the following equations: (a) 2 a;2 + 2 2/2 _ 8 X + 5 y + 1 = 0. (b) Sx^ + Sy^ + 17 x - 15y-6 = 0. (c) 4a;2 + 4y2_6x-10y +4 = 0. (d) x^ + y^ + x - iy = 0. (e) 2 x2 + 2 y2 _ 7 x = 0. (f) x^ + y'^-Sx-e=0. 2. What is the equation of the circle of center (A, k) that touches the axis Ox ? that touches the axis Oy ? that passes through the origin ? 3. What is the equation of any circle whose center lies on the axis Ox ? on the axis Oy ? on the line y= x? on the line y = 2 x ? on the line y = mx? 4. Find the equation of the circle whose center is at the point (— 4, 6) and which passes through the point (2, 0). 5. Find the circle that has the points (4, — 3) and (— 2, — 1) as ends of a diameter. 6. A swing moving in the vertical plane of the observer is 48 ft. away and is suspended from a pole 27 ft. high. If the seat when at rest is 2 ft. above the ground, what is the equation of the path (for the observer as origin)? What is the distance of the seat from the observer when the rope is inclined at 45° to the vertical ? 7. Find the locus of a point whose distance from the point (a, 6) is k times its distance from the origin. Let P (x, y) be any point of the locus ; then the condition is V(x - a)2 + (y - 6)2 = kVx^ + y* . upon squaring and rearranging this becomes : (1 - k2)x2 + (1 - K'^)y^ - 2 ax - 2 &y + a2 + 62 = 0. Hence for any value of k except k = 1, the locus is a circle whose center is a/(l - K^), 6/(1 — /c2) and whose radius is k VaM^/(l — k^). What is the locus when k = 1 ? 8. Find the locus of a point twice as far from the origin as from the point (6, — 3). Sketch. 9. What is the locus of a point whose distances from two points Pi, P2 are in the constant ratio k ? IV, § 50] THE CIRCLE 55 10. Determine the locus of the points which are k times as far from the point (—2, 0) as from the point (2, 0). Assign to k the values V5, V3, V2, I VS, I V3, I \/2 and illustrate with sketches drawn with respect to the same axes. 11. Determine the locus of a point whose distance from the line 3x — 4y+l=0 is equal to the square of its distance from the origin. Illustrate with a sketch. 12. Determine the locus of a point if the square of its distance from the line x + y — a = is equal to the product of its distances from the axes. 50. Circle in Polar Coordinates. Let us now express the equation of a circle in polar coordinates. If C(ri, <^i) is the center of a circle of radius a (Fig. 32) and P(r, ) any point of the circle, then by the cosine law of trigo- nometry o^"'A^9i. r^ + ri^ — 2 Tir cos (<^ — <^i) = a\ Fig. 32 This is the equation of the circle since, for given values of rj, <^i, a, it is satisfied by the coordinates r, of every point of the circle, and by the coordinates of no other point. Two special cases are important : (1) If the origin be taken on the circumference and the polar axis along a diameter OA (Fig. 33), the equation becomes r"^ -\- a^ — 2 ar cos = a% i.e. r = 2a cos <^. This equation has a simple geometrical interpretation : the radius vector of any ^^' point Pon the circle is the projection of the diameter OA =2 a on the direction of the radius vector. (2) If the origin be taken at the center of the circle, the equation is r = a. 56 PLANE ANALYTIC GEOMETRY [IV, § 50 EXERCISES 1. Draw the following circles in polar coordinates : (a) r = 10 cos 0. (6) r = 2a cos (0 - iir). (c) r = sin0. id) r = 6. (e) r = 7 sin (0 _ ^ «■) . (f)r = 17 cos 0. 2. Write the equation of the circle in polar coordinates : (a) with center at (10, ^ ir) and radius 5 ; (6) with center at (6, \ir) and touching the polar axis ; (c) with center at (4, f w) and passing through the origin ; (d) with center at (3, tt) and passing through the point (4, |ir). 3. Change the equations of Ex. (1) and (2) to rectangular coordinates with the origin at the pole and the axis Ox coincident with the polar axis. 4. Determine in polar coordinates the locus of the midpoints of the chords drawn from a fixed point of a circle. 51. Intersection of Line and Circle. To solve two equa- tions in X and y of which one is of the first degree (linear) while the other is of the second degree, it is generally most convenient to solve the linear equation for either x or y and to substitute the value so found in the equation of the second degree. It then remains to solve a quadratic equation. The method for solving a quadratic equation consists in completing the square of the terms in x^ and x. The equation aa;2 + 6a; -f c = has the roots — b± V62 — 4 ac X = • 2a The quantity h^ — 4:ac is called the discriminant of the equation. According as the discriminant is positive, zero, or negative, the roots are real and different, real and equal, or imaginary. An equation of the first degree represents a straight line. If the given equation of the second degree be of the form IV, §51] THE CIRCLE 57 described in § 47, it will represent a circle. By solving two such simultaneous equations we find the coordinates of the points that lie both on the line and on the circle, i.e. the points of intersection of line and circle. Let us find the intersections of the line y = mx + h with the circle about the origin ic2 + 2/^ = r"^' Substituting the value of y from the former equation into the latter, we find the quadratic equation in x : fl;2 + {mx H- by = r2, or (1 + m2) a;2 + 2 mbx + &2 _ ,^2 ^ q. The two roots iCi, x^ of this equation are the abscissas of the points of intersection ; the corresponding ordinates are found by substituting x-^, x^ in ?/ = mx + h. It is easily seen that the abscissas x^, X2 are real and differ- ent if (1 + m2) r2 - 62 > 0, ... b ^ I.e. II — ==^ < r. Vl -1- m2 Since m = tan a, and hence 1/ Vl + m2 = cos a, the .preceding relation means that b cos a 2/2)* P3 (x^ , 2/3), observe that the coordinates of these points satisfy the equation of the circle (§ 49) * (6) x' + if + ax + hy + c^O) hence we must have (J) ^i + Vi + «^i + &2/1 + c = 0, ^2 + 2/2^ + ax^ + &?/2 4-c = 0, a^3^ + 2/3^ + «i«3 + ^2/3+ c = 0. From the last three equations we can find the values of a, &, and c ; these values must then be substituted in the first equar tion. In general this is a long and tedious operation. What we actually wish to do is to eliminate a, h, c between the four equations above. The theory of determinants furnishes a very simple means of eliminating four quantities between four homogeneous linear equations. Our equations are not homo- geneous in a, 6, c. But if we write the first two terms in each equation with the factor 1 : {x^ -f y^) • 1, (x^ -f y^) • 1, etc., we have four equations which are linear and homogeneous in 1, a, b, c; hence the result of eliminating these four quantities is the determinant of their coefficients equated to zero. Thus the equation of the circle through three points is a^ -j- 2/' X y 1 ^i + 2/1' «i 2/1 1 ^i + 2/2' ^2 2/2 1 ^z+Vz ^z 2/3 1 = 0. Compare § 34, where the equation of the straight line through two points is given in determinant form. 62 PLANE ANALYTIC GEOMETRY [IV, § 55 EXERCISES 1. Find the equations of the circles that pass through the points : (a) (2,3), (-1,2), (0,-3). (6) (0,0), (1,-4), (5,0). (c) (0, 0), (a, 0), (0, 6). 2. Find the circles through the points (3,-1), (—1,-2) which touch the axis Ox. 3. Find the circle through the points (2, 1), (- 1, 3) with center on the line 3x-y + 2=0. 4. Find the circle whose center is (3, — 2) and which touches the line 3a; + 4y-12 = 0. 6. Find the circle through the origin that touches the line 4a; -5y- 14 = Oat (0, 2). 6. Find the circle inscribed in the triangle determined by the lines 24x-72/+3=0, 3x-4y-9 = 0, 5a; + 12y-50 = 0. 7. Two circles are said to be orthogonal if their tangents at a point of intersection are perpendicular ; the square of the distance between their centers is then equal to the sum of the squares of their radii. If the equations of two intersecting circles are x^ -i-y^ + aix + biy + Ci = 0, and x^ + y^ + a^x + 62^ + c^ = 0, show that the circles are orthogonal when aiOa + &162 = 2(Ci -f Ca). 8. Find the circle that has its center at (—2, 1) and is orthogonal to the circle x^ + y «_ q a; + 3 = 0. 9. Find the circle that has its center on the line y = 3 a; + 4, passes through the point (4, — 3), and is orthogonal to the circle a;2 + y^ + 13 a; + 5 y + 2 =0. 66. Inversion. A circle of center O and radius a being given (Fig. 35), we can find to every point P of the plane (excepting the center 0) one and only one point P on OP^ produced beyond P if necessary, such that / y/^f OP . OP' = a\ The point P' is said to be inverse to P with respect to the circle (0, a) ; and as the relation is not Fig. 35 IV, §57] THE CIRCLE 63 changed by interchanging P and P', the point P is inverse to P^ The point is called the center of inversion. It is clear that (1) the inverse of a point P within the circle is a point P' without, and vice versa ; (2) the inverse of a point of the circle itself coincides with it ; (3) as P approaches the center 0, its inverse P' moves off to infinity, and vice versa. The inverse of any geometrical figure (line, curve, area, etc.) is the figure formed by the points inverse to all the points of the given figure. 57. Inverse of a Circle. Taking rectangular axes through O (Fig. 36), we find for the relations between the coordinates of two in- verse points P{x, y), P' (x", y')^ if we put OP = r, OP' = r' i x' _y' _r' _ rr' _ a^ X y r f^ r^ since rt' =:a^; hence / . a^^ yf^ a^y a;2 + y2' ^ x^ + y^ and similarly x = These equations enable us to find to any curve whose equation is given the equation of the inverse curve, by simply substituting for x, y their values. Thus it can be shown that by inversion any circle is transformed into a circle or a straight line. For, if in the general equation of the circle A(x^ -{-y'^)+2Gx-^2Fy-\- C = we substitute for x and y the above values, we find {x'^ + 2/'2)2 ^ ic'2 + ?/'2 ^ x'2 + y'^ ' that is, Aa^ + 2 Ga'^x' + 2 Fa^ + C{x'-^ + y'^) = 0, which is again the equation of a circle, provided G =^0. In the special case when C = 0, the given circle passes through the origin, and its in- verse is a straight line. Thus every circle through the origin is trans- formed by inversion into a straight line. It is readily proved conversely that every straight line is transformed into a circle passing through the origin ; and in particular that every line through the origin is transformed into itself, as is obvious otherwise. 64 PLANE ANALYTIC GEOMETRY [IV, § 57 EXERCISES 1. Find the coordinates of the points inverse to (4, 3), (2, 0), (— 5, 1) with respect to the circle x^ + i/2 = 25. 2. Show that by inversion every hue (except a line through the center) is transformed into a circle passing through the center of inversion. 3. Show that all circles with center at the center of inversion are transformed by inversion into concentric circles. 4. Find the equation of the circle about the center of inversion which is transformed into itself. 6. With respect to the circle x^-\-y^ = 16, find the equations of the curves inverse to : (a) x=5, {h) x-y=0, (c) x^-\-y^-6x=0, (d) x'^-\-y^-10y-{-l=0, (e) 3x-4y+6=0. 6. Show that the circle Ax^ -\- Ay^ + 2 Gz + 2 Fy + a^A = is trans- formed into itself by inversion with respect to the circle x^ -\- y"^ = a*. 7. Prove the statements at the end of § 57. 58. Pole and Polar. Let P, P' (Fig. 37) be inverse points with respect to the circle (O, a) ; then the perpen- dicular I to OP through P' is called the polar of P, and P the pole of the line Z, with respect to the circle. Notice that (1) if (as in Fig. 37) Plies within the circle, its polar I lies outside ; (2) if P lies outside the circle, its polar intersects the circle in two points ; (3) if P lies on the circle, its polar is the tangent to the circle at P. Referring the circle to rectangular axes through its center (Fig. 38) so that its equation is x2 + 2/2 = a2, we can find the equation of the polar I of any given point P (x, y). For, using as equation of the polar the normal form X cos /3 + F sin /3 = p, we have evidently, if P' is the point inverse toP: Fig. 37 r /»^>i\ ^ I ^y\ l\ JC M Fig. 38 \ IV, § 59] cos/3 = THE CIRCLE 65 sin/3 y therefore the equation becomes xX . yY _ , p= 0P' = ■N/aj2 + 2/2 or simply Va;2 + 2/2 Va;2 + 2/2 Vx^ + 2/2 xX+2/r=a2. This then is the equation of the polar I of the point P (x, y) with re- spect to the circle of radius a about the origin. If, in particular, the point P (x, y) lies on the circle, the same equation represents the tan- gent to the circle x^ + 2/^ = a^ at the point P (x, y), as shown previously in § 53. 59. Chord of Contact. The polar l of any outside point P with respect to a given circle passes through the points of contact Ci , C2 of the tangents drawn from P to the circle. To prove this we have only to show that if Ci is one of the points of intersection of the polar I of P with the circle, then the angle OCiP (Fig. 39) is a right angle. Now the triangles OCiP and OP'C\ are similar since they have the angle at in common and the including sides proportional owing to the relation OP- 0P' = a\ OP ^ a a OP'' i.e It follows that '^ OCiP = Fig. 39 where a = OCi. OP'Gi = ^w. The rectilinear segment C1O2 is sometimes called the chord of contact of the point P. We have therefore proved that the chord of contact of any outside point P lies on the polar of P. It follows that the equations of the tangents that can he drawn from any outside point P to a given circle can be found by determining the intersections Ci , d of the polar of P with the circle ; the tangents are then obtained as the lines joining C\ , C2 to P. 66 PLANE ANALYTIC GEOMETRY [IV, § 60 60. The General Case. The equation of the polar of a point P (x, y) with respect to any circle given in the general form (4), § 48, viz., (4) Ax'^ + Ay^-\-2Gx-Sr2Fy + C = ^, is found by the same method that was used in § 54 to generalize the equation of the tangent. Thus, with respect to parallel axes through the center the equation of the circle is ^^~A^^A^ A' the equation of the polar of P (x, y) with respect to these axes is by 58 xX+yY: A^ A^ Hence, transferring back to the original axes, we find as equation of the polar of P (a;, y) with respect to the circle (4) : AxX+AyT+Gix + X)+F{y+ r)+C = 0. If, in particular, the point P (x, y) lies outside the circle, this polar contains the chord of contact of P ; if P lies on the circle, the polar be- comes the tangent at P (§ 54). 61. Construction of PolarS. if a point Pi describes a line I, its polar h with respect to a given circle (0, a) turns about a fixed point, viz. , the pole P of the line I (Fig. 40) . Conversely, if a line h turns about one of its points P, its pole Pi with respect to a given circle (O, a) describes a line I, viz. the polar of the point P. For, the line I is transformed by in- version with respect to the circle (0, a) into a circle passing through O and through the pole P of I; as this circle must obviously be symmetric with respect to OP it must have OP as diameter. Any point Pi of I is transformed by inversion into that point Q of the circle of diameter OP at which this circle is in- tersected by OPi . The polar of Pi is the perpendicular through Q to OPi ; it passes therefore through P, wherever Pi be taken on I. The proof of the converse theorem is similar. Fia. 40 IV, §61] THE CIRCLE 67 The pole Pi of any line h can therefore be constructed as the intersec- tion of the polars of any two points of h ; this is of advantage when the line h does not meet the circle. And the polar li of any point Pi can be constructed as the line joining the poles of any two lines through Pi ; this is of advantage when the point Pi lies inside the circle. EXERCISES 1. Find the equation of the polar of the given point with respect to the given circle and sketch if possible ; (a) (4, 7), 0^2 + y2^ 8. (6) (0, 0), x2 + 2/2 - 3 a; - 4 = 0. (c) (2,l),x^ + y^-ix-2y+l=0. (d) (2, - 3), x2 + ?/2 4- 3 a; + 10 2/ + 2 = 0. 2. Find the pole of the given line with respect to the given circle and sketch if possible : (a) X + 2 y- 20 = 0, x^ + y"^ = 20. (b) x + y + l=0, x^ + y^ = 4. (c) ^x-y = 19, a;2 + y2 = 25. (d) Ax + By + C = 0, x^ + y^ = r^. (e) y = mx -\- b, x^ -\- y^ = r^. 3. Find the pole of the line joining the points (20, 0) and (0, 10), with respect to the circle x^ + y^ = 25. 4. Find the tangent to the circle x2+?/2_io a:+4 ?/+9=0 at (7, - 6). 6. Find the intersection of the tangents to the circle 2 a;2 + 2 ?/2— 15 cc + 2/ — 28 = at the points (3, 5) and (0, — 4) . 6. Find the tangents to the circle x^-{-y^ — 6x — 10y + 2 = that pass through the point (3, — 3) . 7. Find the tangents to the circle a;2 + ?/2_3ic + ?/— 10 = that pass through the point ( — |, — V) • 8. Show that the distances of two points from the center of a circle are proportional to the distances of each from the polar of the other. 9. Show analytically that if two points are given such that the polar of one point passes through the second point, then the polar of the second point passes through the first point. 10. Find the poles of the lines x — y — S = and x + y + 8 = with respect to the circle x'^ + y'^ — Qx + iy + S=0. 68 PLANE ANALYTIC GEOMETRY [IV, § 62 62. Power of a Point, if in the left-hand member of the equa- tion of the circle we substitute for x and y the coordinates Xi , j/i of a point Pi not on the circle (Fig. 41), the expression {xi — h)^ + {yi — ky — r^ is different from zero. Its value is called the power of the point P\ (xi , y{) with respect to the circle. As {x^ — hy + (yi - ky is the square' of the distance PiC = d be- tween the point Pi (a;i, yi) and the center C{h, k), the power of the point -Pi (a^ii yO with respect to the circle is cP — r^ ; and this is positive for points without the circle (d > r) , zero for points Fig. 41 on the circle (d = r), and negative for points within the circle (d < r). If the point lies without the circle, its power has a simple interpretation ; it is the square of the segment PiT = t oi the tangent drawn from Pi to the circle : t'i=cP-r^=(^Xi- hy -\- (yi - ky - r^. Hence the length t of the tangent that can be drawn from an outside point Pi {xi , j/i) to a circle x^ + y"^ + ax -\- by -^ c = is given by «2 = xi^ + yi* -f- aa;i -f &yi + c. Notice that the coeflBcients of x^ and y^ must be 1. Compare the similar case of the distance of a point from a line (§ 42). 63. Radical Axis. The locus of a point whose powers with respect to any two circles a;2 + r/2 + aix -\- Iny -I- Ci = 0, x2 -}- y2 -f aax + b-2y -|- C2 = 0, are equal is given by the equation a;2 + y2 ^ aix + biy -{- ci = x^ -\- y^ + a^x -H b^y + c^., which reduces to (ai - a2)x -}- (61 - b2)y + (ci - C2) = 0. This locus is therefore a straight line ; it is called the radical axis of the two circles. It always exists unless ai = a-z and bi = bo, i.e. unless the circles are concentric. IV, §64] THE CIRCLE 69 Three circles taken in pairs have three radical axes which pass through a common point, called the radical center. For, if the equation of the third circle is x^ + y^ + a,x + h,y + C3 = 0, the equations of the radical axes will be (.052 - «3)a; + (62 - 63)2/ + (C2 - C3) = 0, {az - a{)x + (63 - hi)y +(03 - Ci) = 0, («! - a2)a; + (&i — &2)y +(ci — C2) = 0. These lines intersect in a point, since the detenninant of the coefficients in these equations is equal to zero (Ex. 3, p. 38). 64. Family of Circles. The equation (8) (x^ + 2/2 + aix + hiy + ci) + k{x^ + y^ + a2X + h^y + C2) = represents a family^ or pencil^ of circles each of which passes through the points of intersection of the circles (9) x^ + y^ + aix + biy + Ci = 0, and (10) x'^-\-y^+ a^x + h^y + Ca = 0, if these circles intersect. For, the equation (8) written in the form (1 + /c)x2 +(1 + /c)y2 +(ai + Ka'i)x +(&i + Khi)y + Ci + kC2 = represents a circle for every value of k except /c= — 1, as the coefficients of x2 and y"^ are equal and there is no x?/-term (§47). Each one of the circles (8) passes through the common points of the circles (9) and (10) if they have any, since the equation (8) is satisfied by the coordinates of those points which satisfy both (9) and (10). Compare § 44. The constant k is called the parameter of the family. In the special case when k =— 1, the equation is of the first degree and hence represents a line, viz. the radical axis (§ 63) of the two circles (9), (10). If the circles intersect, the radical axis contains their com- mon chord. EXERCISES 1. Find the powers of the following points with respect to the circle x2 + y2 _ 3 3. _ 2 y = and thus determine their positions relative to the circle: (2,0), (0,0), (0, -4), (3,2). 2. What is the length of the tangent to the circle : (a) x"^ -\- y^ -\- ax + 62/ + c = from the point (0, 0), (6) (x - 2)2+ (^ - 3)2 - 1 = from the point (4, 4) ? 70 PLANE ANALYTIC GEOMETRY [IV, § 64 3. By § 62, t^ = d:^ — r'^=(d-\- r){d — r); interpret this relation geometrically. 4. Find the radical axis of the circles x^ '\- y^ -\- ax -\- by -\- c = and x^ + y^ + bx -\- ay + c = and the length of the common chord. 6. Find the radical center of the circles x^ + y2 ^ Sx-\-4y — 'J=0, x^ -{■y^ = l6, 2{x^ + y^) -{■ Q X -\- 1 = 0. Sketch the circles and their radi- cal axes. 6. Find the circle that passes through the intersections of the circles x^ -\-y'^+ 5x = and x^ +y^-{-x — 2y-6 = 0, and (a) passes through the point (— 5, 6), (6) has its center on the line 4x — 2y — 16 = 0, (c) has the radius 6. 7. Sketch the family of circles x^ + y^ - 6 y + k(x^ -\-y'^ -\- By) = 0. 8. What family of circles does the equation Az -\- By -\- C + k{x^ 4- y2 4- (jx + &?/ + c) = represent ? 9. Find the family of curves inverse to the family of lines y = mx + b\ (a) with m constant and b variable, (6) with m variable and b constant. Draw sketches for each case. 10. Show that a circle can be drawn orthogonal to three circles, pro- vided their centers are not in a straight line. 11. Find the locus of a point whose power with respect to the circle 2x2 + 2y2_5a; + lly — 6=0is equal to the square of its distance from the origin. Sketch. 12. Find the locus of a point if the sum of the squares of its distances from the sides of an equilateral triangle of side 2 a is constant. 15. Show that the circle through the points (2, 4), (- 1, 2), (3, 0) is orthogonal to the circle which is the locus of a point the ratio of whose distances from the points (2, 3) and (— 1, 2) is 3. Sketch. 14. Show that the circles through two fixed points, say (—a, 0), (a, 0), form a family like that of Ex. 8. 16. The locus of a point whose distances from the fixed points (—a, 0), (a, 0) are in the constant ratio *c ( ^ 1) is the circle x^ + y'^ + 2^-^ax + a^ = 0. 1 — k'^ Compare Ex. 9, p. 54. Show that, whatever k{=^1), this circle inter- sects every circle of the family of Ex. 15 at right angles. CHAPTER V POLYNOMIALS PART I. QUADRATIC FUKCTION — PARABOLA 65. Linear Function. As mentioned in § 28, an expression of the form mx -\- h, where m and 6 are given real numbers (m^O) while a; may take any real value, is called a linear function of x. We have seen that this function is represented graphically by the ordinates of the straight line y = mx -\- b ; b is the value of y for x = 0, and m is the slope of the line, i.e. the rate of change of the function y with respect to x. 66. Quadratic Function. Parabola. An expression of the form ax"^ + &a; + c in which a :^ is called a quadratic func- tion of Xj and the curve y — ax"^ -{- bx -\- c, whose ordinates represent the function, is called a parabola. If the coefficients a, 6, c are given numerically, any number of points of this curve can be located by arbitrarily assigning to the abscissa x any series of values and computing from the equation the corresponding values of the ordinates. This process is known as plotting the curve by points ; it is some- what laborious; but a study of the nature of the quadratic function will show that the determination of a few points is sufficient to give a good idea of the curve. 71 7^ PLANE ANALYTIC GEOMETRY [V, § 67 Fig. 42 67. The Form y = ax\ Let us first take 6 = 0, c = ; the resulting equation (1) y = cia? represents a parabola which passes through the origin, since the values 0, satisfy the equation. This parabola is symmet- ric with respect to the axis Oy ; for, the values of y correspond- ing to any two equal and opposite values of x are equal. This line of symmetry is called the axis of the parabola ; its intersection with the parab- ola is called the vertex. We may distinguish two cases accord- ing as a > or a < ; if a = 0, the equa- tion becomes y = 0, which represents the axis Ox. (1) If a > 0, the curve lies above the axis Ox. For, no matter what positive or negative value is assigned to a;, y is positive. Furthermore, as x is allowed to increase in absolute value, y also increases indefinitely. Hence the parabola lies in the first and second quadrants with its vertex at the origin and opens upward, i.e. is con- cave upward (Fig. 42). (2) If a < 0, we conclude, similarly, that the parabola lies below the axis Ox, in the third and fourth quadrants, with its vertex at the origin and opens down- wardf i.e. is concave downward (Fig. 43). Draw the following parabolas : y = x',y = Sx'',y = -\y?,y=\x'. 68. The General Equation. The curve represented by the more general equation (2) y = aa? -{■hx-\-c differs from the parabola y = aa? only in position. To see this Fig. 43 V, § 69] THE PARABOLA 73 we use the process of completing the square in a;; i.e. we write the equation in the equivalent form = K'^^J 4a + c; I.e. If we put 2a 4a Fig. 44 the equation becomes and it is clear (§ 13) that, with reference to parallel axes OiXij Oi2/i through the point Oi (h, k) the equation of the curve is 2/1 = <^^i (Fig. 44). The parabola (2) has therefore the same shape as the parabola y = ax^ ; but its vertex lies at the point (h, k), and its axis is the line x = h. The curve opens upward or downward according as a > or a < 0. 69. Nature of the Curve. To sketch the parabola (2) roughly, it is often sufficient to find the vertex (by completing the square in x^ as in § 68, and the intersections with the axes. The intercept on the axis Oy is obviously equal to c. The in- tercepts on the axis Ox are found by solving the quadratic equation aar^ + 6a; + c = 0. We have thus an interesting interpretation of the roots of any quadratic equation: the roots of aa;^ -f- 6a; -h c = are the abscissas of the points at which the parabola (2) intersects the axis Ox. The ordinate of the vertex of the parabola is evidently the least or greatest value of the function y — ax^ + hx -\- c according as a is greater or less than zero. 74 PLANE ANALYTIC GEOMETRY [V, § 69 EXERCISES 1. With respect to the same coordinate axes draw the curves y = ax'^ for a =2, f, 1, i, 0, — i, — 1, - |, — 2. What happens to the parabola y = ox2 as a changes ? 2. Determine in each of the following examples the value of a so that the parabola y = ax^ will pass through the given point : (a) (2,3). (b) (-4, 1). (c) (-2, -2). (d) (3, -4). 3. A body thrown vertically upward in a vacuum with a velocity of v feet per second will just reach a height of h feet such that h = ^v^. Draw the curve whose ordinates represent the height as a function of the initial velocity. (a) With what velocity must a ball be thrown vertically upward to rise to a height of 100 ft. ? (6) How high will a bullet rise if shot vertically upward with an ini- tial velocity of 800 ft. per sec. , the resistance of the air being neglected ? 4. The period of a pendulum of length I (i.e. the time of a small back and forth swing) is 7"= 2iry/l/g. Take g = S2 ft. /sec. and di-aw the curve whose ordinates represent the length I of the pendulum as a function of the period T. (a) How long is a pendulum that beats seconds (i.e. of period 2 sec.) ? (6) How long is a pendulum that makes one swing in two seconds ? (c) Find the period of a pendulum of length one yard. 6. Draw the following parabolas and find their vertices and axes : (a) y = \x^-x + 2. (b) y = - \ x^ -h x. (c) y = 5x^ + 16x + 3. (d)y = 2-x-x2, (e) y=x2-9. (f)y = -9^x^. (fir) y =3x2- 6x -1-5. (A) y = ix2-|-2x-6. (i) x^ - 2x -y = 0. 6. What is the value of b if the parabola y = x^ -\-bx — 6 passes through the point (1, 6) ? of c if the parabola y = x2 — 6x-|-c passes through the same point ? 7. Suppose the parabola y = ax^ drawn ; how would you draw y = a(x-|-2)2? y = a(x-7)2? y = ax'^ + 2? y = ax2 - 7 ? y = ax2-f 2x -|- 3 ? 8. What happens to the parabola y = ax^ ■}- bx -\- c as 'c changes ? For example, take the parabola y = x2 — x + c, where c = — 3, — 2, — 1, 0, 1, 2, 3. V, §69] THE PARABOLA 75 9. What happens to the parabola y = ax^ -\-hx -^ c as a changes ? For example, take y = ax'^ — x — Q, where a = 2, 1, ^, 0,-^,-1, — 2. 10. (a) If the parabola y = ax^ + bx is to pass through the points (1, 4), (—2, 1) what must be the values of a and b ? (&) Determine the parabola y = ax^ + bx -\- c so as to pass through the points (1, ^), (3, 2), (4, f ) ; sketch. 11. The path of a projectile in a vacuum is a parabola with vertical axis, opening downward. With the starting point of the projectile as origin and the axis Ox horizontal, the equation of the path must be of the form y = ax^ + bx. If the projectile is observed to pass through the points (30, 20) and (50, 30), what is the equation of the path? What is the highest point reached ? Where will the projectile reach the ground ? 12. Find the equations of the parabolas determined by the following conditions : (a) the axis coincides with Oy, the vertex is at the origin, and the curve passes through the point (— 2, — 3) ; (&) the axis is the line x = 3, the vertex is at (3, — 2), and the curve passes through the origin ; (c) the axis is the line x =— 4, the vertex is (— 4, 6), and the curve passes through the point (1, 2). 13. Sketch the following parabolas and lines and find the coordinates of their points of intersection : (a) y = 6x^,y = 7x + S. (6) y -2x'^ ^Sx, y = x + 6. (c) y = 2-Sx^,y = 2x + S. (^a) y = 3 -\-x- x^, x + y - 4 = 0. 14. Sketch the following curves and find their intersections : (a) x^ + y^ = 25, y = f a;2. (6) 3^^ +y^ ^ 6y = 0, y = ^x^ - 2x + 6. 15. The ordinate of every point of the line y = | ic + 4 is the sum of the corresponding ordinates of the lines y = ^x and y = 4. Draw the last two lines and from them construct the first line. 16. The ordinate of every point of the parabola y ^^x"^ + ^x—l is the sum of the corresponding ordinates of the parabola y = ^x^ and the line y = ^x — 1. From this fact draw the former parabola. 17. The ordinate of every point of the parabola y = ^x^ — x + Sis the difference of the corresponding ordinates of the parabola y = ^x^ and the line y = x — S. In this way sketch the former parabola. 76 PLANE ANALYTIC GEOMETRY [V, § 70 70. Symmetry. Two points P, , Pg are said to be situated symmetncally with respect to a line Z, if / is the perpendicular bisector of P^P^ ; this is also expressed by saying that either point is the reflection of the other in the line h Any two plane figures are said to be symmetric with respect to a line I in their plane if either figure is formed of the reflec- tions in I of all the points of the other figure. Each figure is then the reflection of the other in the line I. Two such figures are evidently brought to coincidence by turning either figure about the line I through two right angles. Thus, the lines y = 2x-\-S and y = — 2x — 3 are symmetric with respect to the axis Ox. A line I is called an axi's of symmetry (or simply an axis) of a figure if the portion of the figure on one side of / is the reflection in Z of the portion on the other side. Thus, any diameter of a circle is an axis of symmetry of the circle. What are the axes of symmetry of a square ? of a rectangle ? of a parallelogram? In analytic geometry, symmetry with respect to the axes of coordinates, and to the lines y= ±x, is of particular importance. It is readily seen that if a figure is symmetric with respect to both axes of coordinates, it is symmetric with respect to the origiuy i.e. to every point Pi of the figure there exists another point P2ot the figure such that the origin' bisects PiP2- A point of symmetry of a figure is also called center of the figure. EXERCISES 1. Give the coordinates of the reflection of the point (a, 6) in the axis Ox ; in the axis Oy; in the line y = z ; in the line y = 2x ; in the line y =—x. 2. Show that when x is replaced by — jc in the equation of a given curve, we obtain the equation of the reflection of the given curve in the y-axis. V, § 70] THE PARABOLA 77 3. Show that when x and y are replaced by y and x, respectively, in the equation of a given curve, we obtain the equation of the reflection of the given curve in the line y = z. 4. Sketch the lines y = - 2 ac + 5 and a; = - 2 y + 5 and find their point of intersection. 6. Sketch the parabolas y = a^ and x = y^ and find their points of intersection. 6. Find the equation of the reflection of the line 2 a; — 3 y + 4 = in the line y = x\ in the axis Ox ; in the axis Oy ; in the line y- — x. 7. What is the reflection of the line a; = a in the line y = a; ? in the axes? 8. Find and sketch the circle which is the reflection of the circle x2 + y2 _ 3 X - 2 = in the line y = a;, and find the points in which the two circles intersect. 9. Find the circle which is the reflection of the circle a;2-}-y2 — 4x4-3 = in the line y = x\ in the coordinate axes. Sketch all of these circles. 10. What is the general equation of a circle which is its own reflection in the line y = a: ? in the axis Ox ? in the axis Oy ? What circle is its own reflection in all three of these lines ? 11. What is the equation of the reflection of the parabola y =— x2+4 in the line y = x ? in the line y =— x ? Are these reflections parabolas ? 12. What is the reflection of the parabola y=3x2 — 5x + 6 in the axis Ox ? In the axis Oy ? Are these reflections parabolas ? 13. If the cartesian equation of a curve is not changed when x is re- placed by — X, the curve is symmetric with respect to Oy ; if it is not changed when y is replaced by — y, the curve is symmetric with respect to Ox ; if it is not changed when x and y are replaced by — x and — y, respectively, the curve is symmetric with respect to the origin ; if it is not changed when x and y are interchanged, the curve is symmetric with respect to y = x. 78 PLANE ANALYTIC GEOMETRY [V, § 71 71. Slope of Secant. Let P(x, y) be any point of the parabola (1) y = a^. If Pi {xi, yi)be any other point of this parabola so that (2) yi = axi\ the line PPj (Fig. 45) is called a secant. For the slope tan a^ of this secant we have from Fig. 45 : (3) tan„. = ^' = -^ y^^y QQi Ax' or, substituting for y and yi their values : (4) tan «! = ^ W - a?') ^ „/^ _j_ ^ X ari — X 72. Slope of Tangent. Keeping the point P (Fig. 45) fixed, let the point P^ move along the parabola toward P; the limiting position which the secant PPi assumes at the instant when Pi passes through P is called the tangent to the parabola at the point P. Let us determine the slope tan a of this tangent. As the secant turns about P approaching the tangent, the point Qi ap- proaches the point Q, and in the limit OQi = Xi becomes OQ=x. The last formula of § 71 gives therefore tan a if we make Xi = x: tan a = 2 ax. The slope of the tangent at P which indicates the "steep- ness " of the curve at P is also called the slope of the parabola at P. Thus the slope of the parabola y = ax^ at any point whose abscissa is a; is = 2 ox ; notice that it varies from point to point, being a function of a?, while the slope of a straight line is constant all along the line. V, § 73] THE PARABOLA 79 The knowledge of the slope of a curve is of great assistance in sketching the curve because it enables us, after locating a number of points, to draw the tangent at each point. Thus, for the parabola y = | ic^ ^^ find tan a=^x -, locate the points for which a; = 0, 1, 2, — 1, — 2, and draw the tangents at these points ; then sketch in the curve. 73. Derivative. If we think of the ordinate of the parab- ola y = aa^ as representing the function ax^, the slope of the parabola represents the rate at which the function varies with X and is called the derivative of the function ax\ We shall denote the derivative of y by y'. In § 72 we have proved that the derivative of the function y = ax^ is y' = 2 ax. The process of finding the derivative of a function, which is called differentiation, consists, according to §§ 71-72, in the following steps: Starting with the value y=ax^ of the func- tion for some particular value of x (say, at the point P, Fig. 45), we give to x an increment Xi^—x = Ax (compare § 9) and calculate the value of the corresponding increment yi--y = Ay of the function. Then the derivative y' of the function y is the limit that Ay / Ax approaches as Ax approaches zero. In the case of the function y = ax^ we have Ay=y^-y = a{x^ - a;^) = a\_{x -h Axf - x^] = a[2 xAx + {Axf] ; hence -^ == a(2 a; + Ax). The limit of the right-hand member as Ax approaches zero gives the derivative : ?/' = 2 ax. Thus, the area y of a circle in terms of its radius xiay = ttx^. Hence the derivative y', that is the slope of the tangent to the curve that rep- resents the equation y = irx^, is y' = 2Trx. This represents (§72) the rate of increase of the area y with respect to x. Since 2 wx is the length 80 PLANE ANALYTIC GEOMETRY [V, § 73 of the circumference, we see that the rate of increase of the area y with respect to the radius x is equal to the circumference of the circle. 74. Derivative of General Quadratic Function. By this process we can at once find the derivative of the general quad- ratic function y = aoi? -{- hx -\- c (§ 66), and hence the slope of the parabola represented by this equation. We have here Ay = a{x + Aa;)2 + h{x + Aa;) + c — {ao? -\-hx-{-c) = 2 ax^x + a(Aa;)2 -}- h^x ; hence — ^ = 2 aa; + 6 + a Aa;. Ax The limit, as Aa; becomes zero, is 2ax-\-h\ hence the deriva- tive of the quadratic function y =ax^ -\-hx -\- c is y^ =^2 ax •\- b. 76. Maximum or Minimum Value. It follows both from the definition of the derivative as the limit of Ay/ Ax and from its geometrical interpretation as the slope, tana, of the curve that if, for any value of x, the derivative is positive, the function, i.e. the ordinate of the curve, is (algebraically) increasing; if the derivative is negative, the function is decreasing. At a point where the derivative is zero the tangent to the curve is parallel to the axis Ox. The abscissas of the points at which the tangent is parallel to Ox can therefore be found by equat- ing the derivative to zero. In this way we find that the abscissa of the vertex of the parabola y=aa? -{-bx -^ cis x = — 6/2 a, which agrees with § 68. We know (§ 68) that the parabola y = ax^ -\- bx -{- c opens upward or downward according as a is > or < 0. Hence the ordinate of the vertex is a minimum ordinate, i.e. algebraically less than the immediately preceding and following ordinates, if a > ; it is a maximum ordinate, i.e. algebraically greater than the immediately preceding and following ordinates, if a < 0. This enables us to determine the maximum or minimum of V, § 75] THE PARABOLA 81 a quadratic function ax^ + &« -f- c ; the value of x for which the function becomes greatest or least is found by equating the derivative to zero ; the quadratic function is a maximum or a minimum for this value of x according as a < or > 0. Thus, to determine the greatest rectangular area that can be inclosed by a boundary {e.g. a fence) of given length 2 A;, let one side of the rectangle be called x ; then the other side m k — x. Hence the area A of the rectangle \b A = x{k — x) =kx — x'^. Consequently the derivative of A is k— 2 x. If we set this equal to zero, we have 2x = kj whence x = k/2. It follows that k—x=k/2; hence the rectangle of greatest area is a square. EXERCISES 1. Locate the points of the parabola y = x^—4:X + ^ whose abscissas are -- 1, 0, 1, 2, 3, 4, draw the tangents at these points, and then sketch in the curve. 2. Sketch the parabolas 4:y= — x^ + 4:X and y = x^ — S by locating the vertex and the intersections with Ox and drawing the tangents at these points. 3. Is the function y = 6(x^ — 4x + 3) increasing or decreasing as x increases from x = I? from x = | ? 4. Find the least or greatest value of the quadratic functions : {a)2x^-Sx + 6. (&) 8-6x-a;2. (c)x^-5x-6. (d)2-2a;-x2. (g)4 + x-^x2. (/) 5x2 - 20 a; + 1. 6. Find the derivative of the linear function y = mx + b. 6. The curve of a railroad track is represented by the equation y = I x2, the axes Ox, Oy pointing east and north, respectively ; in what direction is the train going at the points whose abscissas are 0, 1,2, — ^ ? 7. A projectile describes the parabola y = fx—Zx^, the unit being the mile. What is the angle of elevation of the gun ? What is the greatest height ? Where does the projectile strike the ground ? 8. A rectangular area is to be inclosed on three sides, the fourth side being bounded by a straight river. If the length of the fence is a con- stant k, what is the maximum area of the rectangle ? Q 82 PLANE ANALYTIC GEOMETRY [V,§76 Fio. 46 4 . 18 . PART 11. POLYNOMIALS 76. The Cubic Function. A function of the form a^a^ + OiX^+azX -|- ttg is called a cubic function of x. The curve repre- sented by the equation y =0^ + a^x^ + a^x + a^ can be sketched by plotting it by points (§ 66). For example, to draw the curve repre- sented by the equation y = a;3-2a^-6a;-f 6, we select a number of values of x and com- pute the corresponding values of y : a;=-3-2-101 23 2/=- 24 860-40 These points can then be plotted and connected by a smooth curve which will approximately represent the curve corre- sponding to the given equation (Fig. 46). 77. Derivative. The sketching of such a cubic curve is again greatly facilitated by finding the derivative of the cubic function ; the determination of a few points, with their tan- gents, will suffice to give a good general idea of the curve. To find the derivative of the function y = a^ + a^a? + a^ -f as the process of § 73 should be followed. The student may carry this out himself ; he will find the quadratic function 2/' = 3 a^"^ -I- 2 OiX + aj. 78. Maximum or Minimiun Values. The abscissas of those points of the curve at which the tangent is parallel to the axis Ox are again found by equating the derivative to zero ; they are therefore the roots of the quadratic equation V, §80] CUBIC FUNCTIONS 83 3 tto^ + 2 aiX + tta = 0. If at such a point the derivative passes from positive to nega- tive values, the curve is concave doimiward, and the ordinate is a maximum; if the derivative passes from negative to posi- tive values, the curve is concave upward^ and the ordinate is a minimum. 79. Second Derivative. The derivative of a function of X is in general again a function of x. Thus for the cubic function y = aocc^ -\- a^pi? -{• a>^ + a^ the derivative is the quad- ratic function 2/' = 3 a^^ + 2 a,a; 4- «2. The derivative of the first derivative is called the second deriva- tive of the original function ; denoting it by y'\ we find (§ 74) 2/" = 6 aoo; + 2 a^. As a positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function (§ 75), it follows that if at any point of the curve the second derivative is positive, the first derivative, i.e. the slope of the curve, increases ; geometrically this evidently means that the curve there is concave upward. Similarly, if the second de- rivative is negative, the curve is concave downward. We have thus a simple means of telling whether at any particular point the curve is concave upward or downward. It follows that at any point where the first derivative van- ishes, the ordinate is a minimum if the second derivative is positive ; it is a maximum if the second derivative is negative. 80. Points of Inflexion. A point at which the curve changes from being concave downward to being concave up- ward, or vice versa, is called a point of inflexion. At such a point the second derivative vanishes. Our cubic curve obviously has but one point of inflection, viz. the point whose abscissa is a; = — ai/(3 Oq). 84 PLANE ANALYTIC GEOMETRY [V, § 80 EXERCISES 1. Find the first and second derivatives of y v^hen : (a) y = 6 x8 - 7 a;2 - X + 2. (6) y = 20 -\- ix- Sx^ - oi*. (c) l0y = a^-5x^ + Sx + 9. (d) y = (x - l)(x- 2)(x-3). (e) 2/ = x2(x+3). (/) 7 2/ = 3x-2x(x2-l). 2. Sketch the curve y = (x— 2)(x + 1) (x + 3), observing the sign of y between the intersections with Ox, and determining the minimum, maxi- mum, and point of inflection. 3. In the curve y = aox^ + aix^ + azx + as, what is the meaning of as ? 4. Sketch the curves : (a) 6 y = (X - 1) (X -f iy. (&) y = (x - 3)8. (c) 6 y = 6 + X + x'* - «*. (d) y = x« - 4 X. (e) 8 y = 5 x* - x8. (/) y = x« - 3 x* + 4 x - 5. 6. Draw the curves y = x, y = x^, y = x«, with their tangents at the points whose abscissas are 1 and — 1. 6. Find the equation of the tangent to the curve 14 y = 6 x* — 2 x^ +x — 20 at the point whose abscissa is 2. 7. At what points of the curve y = x' — 6 x^ + 3 are the tangents parallel to the line y=— 3x4-6? 8. Are the following curves concave upward or downward at the indicated points ? Sketch each of them. (a) y = 4x8-6x, atx = 3. (6) 3y = 5x- 3x8, at x =- 2. (c) y = x8-2x2 + 6, atx = ^. (d) 2y = x8 -3x2, at x = 1. Ce) y = 1 -x-x8, atx = 0. (/) 10y=x8+x2-15x-!-6,atx=-|. 9. Show that the parabola y = ox* + 6x -f c is concave upward or concave downward for all values of x according as a is positive or negative. 10. The angle between two curves at a point of intersection is the angle between their tangents. Find the angles between the curves y = x^ and y = x8 at their points of intersection. 11. Find the angle at which the parabola y = 2x2 — 3x — 6 intersects the curve y = x8 + 3 x — 17 at the point (2, — 3). 18. The ordinate of every point of the curve y = x8 + 2 x* is the sum of the ordinates of the curves y = x8 and y = 2 x^. From the latter two curves construct the former. V, § 81] POLYNOMIALS 85 13. From the curve y = x^ construct the following curves : (a)y = 4x8. (6)2/ = ^|y. (c)y = x3-.2. (d)y = 2a:8+4. 14. Draw the curve 2 ^ = ic^ — 3 x^ and its reflection in the line y = x. What is the equation of this reflected curve ? What is the equation of the reflection in the axis Oy ? 15. A piece of cardboard 18 inches square is used to make a box by- cutting equal squares from the four corners and turning up the sides. Draw the curve whose ordinates represent the volume of the box as a function of the side of the square cut out. Find its maximum. 16. The strength of a rectangular beam cut from a log one foot in diameter is proportional to (i.e. a constant times) the width and the square of the depth. Find the dimensions of the strongest beam which can be cut from the log. Draw the curve whose ordinates represent the strength of the beam as a function of the width. 17. Find the equation of the curve in the form y = ax^ + bx^ + cx + d which passes through the following points : (a) (0, 0), (2, - 1), (- 1, 4), (3, 4) ; (6) (1, 1), (3,-1), (0, 5), (-4,1). 18. Show that every cubic curve of the form y = aoofi + aix^ + azx + as is symmetric with respect to its point of inflection. 81. Polynomials. The methods used in studying the quad- ratic and cubic functions and the curves represented by them can readily be extended to the general case of the polynomial, or rational integral function, of the nth degree, y = a^x^ + a^x^-^ -f a^x^~'' + - + (^n-v^ + ^n? where the coefficients a,,, a^, ••• a„ may be any real numbers, while the exponent n, which is called the degree of the poly- nomial, is a positive integer. We shall often denote such a polynomial by the letter y or by the symbol f{x) (read : function of x, or / of a;) ; its value for any particular value of x, say a; = iCi or a; = ^, is then de- noted by f(x^ or f(h), respectively. Thus, for a; = we have /(0) = a„. 86 PLANE ANALYTIC GEOMETRY [V, § 82 82. Calculation of Values of a Polynomial. In plotting the curve y =f(x) by points (§§ 66, 76) we have to calculate a number of ordinates. Unless f(x) is a very simple poly- nomial this is a rather laborious process. To shorten it ob- serve that the value f{x{) of the polynomial f{x) = 0^ + ttiaf*-^ H f- a„ f or a; = a^ can be written in the form /(^) = ( - (((«oa?i + «i)»i + «2)a5i + cis)xi -f ... + a,_,)xi + a„. To calculate this expression begin by finding a^pCi -\- a^ ; mul- tiply by Xi and add aj ; multiply the result by Xi and add a^ ; etc. This is best carried out in the following form : «o «i Oj ..., aoX,+ «! ( Vi + < !ll)Xi {a^i + < 2y)Xi - f o^... instance, if /W = = 2ar «-3a;2- -12a ^ + 5 = ((2 jc - 3)a; - -12); K4-5, to find /(3) write the coefficients in a row and place 2x3 = 6 below the second coefficient ; the sum is 3. Place 3 x 3 = 9 be- low the third coefficient ; the sum is — 3. Place 3x(— 3)=— 9 below the last coefficient; the sum, —4, is =/(3). 2-3-12 6 6 9-9 2 3-3-4 This process is useful in calculating the values of?/ that cor- respond to various values of x, as we have to do in plotting a curve by points. V, § 83] POLYNOMIALS 87 EXERCISES 1. If /(x)=5x8-13x + 2, what is meant by /(a)? hj f{x-\-h)? What is the value of /(O)? of/(2)? of/(-3/5)? of/(-l)? 2. Find the ordinates of the curve y=x^ — oi^ + Sx'^ — 12x-\-S for a; = 3, -9, -^ 3. Find the ordinates ot2y = x^ + Sx^-20x-251otx = 1,2,3,- 1,-2. 4. Suppose the curve y =/(a;)drawn ; how would you sketch : {a)y=f(x-2)? ib)y=f(x+S)? {c)y=f{2x)2 (d)y=f(-x)? (e) y=f(^iy (/) y=fix)+5? (g) y=f(x)-2x? 83. Derivative of the Polynomial. We have seen in the preceding sections how greatly the sketching of a curve and the investigation of a function is facilitated by the use of the derivatives of the function. Thus, in particular, the first derivative y' is the rate of change of the function y with a;, and hence determines the slope, or steepness, of the curve y =f{x). We begin therefore the study of the polynomial by determining its derivative. The method is essentially the same as that used in § § 73, 74 for finding the derivative of a quadratic function. The first derivative y' of any function y of a; is defined, as in § 73, to be the limit of the quotient Ay/^x as Ax approaches zero, Ay being the increment of the function y corresponding to the increment Ax otx; in symbols : Sr' = liin^. Ax=0 Ax Geometrically this means that y' is the slope of the tangent of the V " ^/7 1 1 ^ /\\ \ *. y A / i « .Qi FiQ. 47 curve whose ordinate is y. PP, (Fig. 47) : For, Ay /Ax is the slope of the secant 88 PLANE ANALYTIC GEOMETRY [V, § 83 — ^ = tan «i ; Ax and the limit of this quotient as Ax approaches zero, i.e. as Pi moves along the curve to P, is the slope of the tangent at P : y^ = tan a = lim^. Ax=oAa; If the function y be denoted by /(«), then Ay=f{x + Ax)-f(x)', hence Ax=o Ax 84. Calculation of the Derivative. To find, by means of the last formula, the derivative of the polynomial y =/(aj)= V + ai«""* + - + a.> we should have to form first /(a; -|- Ax), i.e. (x + Axy -\-ai{x-\- Axy-^ + ... + a„, subtract from this the original polynomial, then divide by Aa?, and finally put Ax = 0. This rather cumbersome process can be avoided if we observe that a polynomial is a sum of terms of the form ax* and apply the following fundamental propositions about derivatives : (1) the derivative of a sum of terms is the sum of the derivor tives of the terms ; (2) the derivative of ax'' is a times the derivative ofx""; (3) the derivative of a constant is zero; (4) the derivative of x" is rix"~^ The first three of these propositions can be regarded as obvious ; a fuller discussion of them, based on an exact defi- nition of the limit of a function, is given in the differential V, § 85] POLYNOMIALS 89 calculus. A proof of the fourth proposition is given in the next article. On the basis of these propositions we find at once that the derivative of the polynomial y = a^"" + QiX""-^ -h aiX""-"^ 4- - + a„_ia; + a„ is ?/' = aowx^-i -f ai(n — l)a;"-2 _j_ ^2(72 — 2)x''-^ + — + ««-!• 85. Derivative of x^. By the definition of the derivative (§83) we have for the derivative of 2/ = x" : Ax=0 Ax Now by the binomial theorem we have {x + ^xY = a;" + na;"-iAa; + '*^^^^~^ x''-\/^xy + — +(Aa;)", JL • ^ and hence {x + Aa;)'» — aj** = na;"-^Ax + ^^^ ~ -^ a;"-'^(Aa;)'^ + ... + (Aa;)'*. Dividing by Aa; and then letting Aa; become zero, we find y' = na;""^ EXERCISES 1. Find the derivatives of the following functions of x by means of the fundamental definition (§ 83) and check by § 84 : (a) a;8. (6) x^ + x. (c) x* + 6 x^. (d) - 6 a;8. (c) x* - S x^. (J) mx + b. 2. Find the derivatives of the following functions : (a) 5 a;*- 3x2 + 6 X. (b) 1-x+l x^-^x^ (c) (x - 2)3. (d) (2 X + 3)5. (e) 3(4 x - l)^. (/) x'^ + ax'»-i+ ftx'-s. 3. For the following functions write the derivative indicated : (a) 6 x8 - 3 X, find y'". (6) ax^ + bx + c, find y'". (c) x6, find^. W ax3 + ftx2 + ex + (?, find yiv. (e) ix6,findy". (/) i^x«, find i,vii. (g) x^ - gx8, find y'". {h) (2 x - 3)3, find y'". 90 PLANE ANALYTIC GEOMETRY [V, § 86 86. Properties of the General Polynomial Curve. In plot- ting the curve y = Ooaj* + aiaf~^ -f a.^"'^ -{- ... -{-a^ observe that (Fig. 48) : (a) the intercept OB on the axis Oy is equal to the constant term a„ ; (6) the intercepts OA^ OA2, ••. on the axis Ox are roots of the equation y = 0, i.e. ' FiQ. 48 aoX"H-ai.'B"-iH- ... +a, = 0; (c) the abscissas of the least and greatest ordinates are found by solving the equation y' = 0, i.e. (§ 84) every real root giving a minimum ordinate if for this root y" is positive and a maximum ordinate if y" is negative ; (d) the abscissas of the points of inflection are found by solving the equation y" = 0, i.e. 7i(n-l)aoX''-2+ — +2a„_s = 0, every real root of this equation being the abscissa of a point of inflection provided that y'"=^0. (If y"' were zero, y' might not be a maximum or minimum, and further investigation would be necessary.) 87. Continuity of Poljnaomials. It should also be ob- served that the function y = a^pf" + Ojaf ~^ + — -\- a„ is one- valued, real, and finite for every x ; i.e. to every real and finite abscissa x belongs one and only one ordinate, and this ordinate is real and finite. Moreover, as the first derivative y' = nooa:""^ -{-••• +a„_i is again a polynomial, the slope of the curve is everywhere one-valued and finite. V, § 88] POLYNOMIALS 91 Thus, so-called discontinuities of the ordinate (Fig. 49) or of the slope (Fig. 50) cannot occur : the curve y = a^x"" -\ h a„ is continuous. Strictly defined, the continuity of the function y = Ooic" + ••• 4- a„ means that, for every value of x, the limit of the function is equal to the value of the function. The function y = aoCC* + — + a„ has one and only one value for any value x = Xi, viz. a^j^ + ... + a^. The value of • the function for any other value of X, say for a^ + ^x, is a^ix^ + Aa;)" -}- ••. -f a^ which can be written in the form ao^i" + — + «n + terms containing Ax as factor. Therefore as Ax approaches zero, the function approaches a limit, viz. its value for x = Xi. 88. Intermediate Values. A continuous function, in varying from any value to any other value, must necessarily pass through all intermediate values. Thus, our polynomial y = aoX"" + — -f a„, if it passes from a negative to a positive value (or vice versa) ^ must pass through zero. It follows from this that beticeen any two ordinates of opposite sign the curve y = cTo^" + ••• + <^u ''^ust cross the axis Ox at least once. It also follows from the continuity of the polynomial and its derivatives that between any two intersections with the axis Ox there must lie at least one maximum or minimum^ and be- tween a maximum and a minimum there must lie a point of inflection. Ordinates at particular points can be calculated by the pro- cess of § 82. 92 PLANE ANALYTIC GEOMETRY [V, § 88 EXERCISES 1. Sketch the following curves : (a) y=(a:-l)(x-2)(x-3). (&) iy = x^-l. (c) 10y = x^. {d) 10y = x5 + 5. (e) 4y = (a;+2)2(a;-3). (/) y={x-iy. 8. When is the curve y = aox" + aix^-^ + ••• + an symmetric with respect to Oy? 3. Determine the coefficients so that the curve y = oqx^ + aix* + a^^^ + asx + a4 shall touch Ox at (1, 0) and at ( — 1, 0) and pass through (0, 1), and sketch the curve. 4. Find the coordinates of the maxima, minima, and points of inflec- tion and then sketch the curve 4 y = x* — 2 x^. 5. Are the following curves concave upward or downward at the indi- cated points ? (a) 16y=16x*-8x2 + l, atx=-l, -J, 0, i, 3. (6) y = 4x-x*, atx=:-2, 0, 1, 3. (c) y = x", at any point ; distinguish the cases when n is a positive even or odd integer. 6. What happens to the curves y = ox* and y = aX^ as a changes ? For example, take a =2, 1, ^, 0, — |, — 1, — 2. 7. Find the values of z for which the following relations are true : (a) X*- 6x2 + 9^0. (6) (x - l)2(x2- 4) ^0. 8. Those curves whose ordinates represent the values of the first, second, etc., derivatives of a given polynomial are called the first, second, etc., derived curves. Sketch on the same coordinate axes the following curves and their derived curves : (a) 6y = 2x«-3x2-12x. (6) y =(x - 2)2(x + 1). (c) y=(x-|-l)8. (d) 2y = x* 4x2 + 1. 9. At what point on Ox must the origin be taken in order that the equation of the curve y = 2x'— 3x2 — 12x — 5 shall have no term in x* ? no term in x ? 10. Find, to three significant figures, the roots of the equation x8-3x+l=0. CHAPTER VI THE PARABOLA 89. The Parabola. The parabola can be defined as the locus of a point whose distance from a fixed point is equal to its distance from a fixed line. The fixed point is called the focus, the fixed line the directrix, of the parabola. Let F (Fig. 51) be the fixed point, d the fixed linej then every point P of the parabola must satisfy the condition FP=PQ, Q being the foot of the perpendicular from P to d. Let us take F as origin, or pole, and the perpendicular FD from F to the directrix as polar axis, and let the given distance FD = 2 a. Then FP =r 2ind PQ = 2 a -r cos cf>. The condition PP=PQ becomes therefore ^^ r = 2 a — r cos <^, (1) r= ^^ . ^ ^ 1 + cos This equation, which expresses the radius vector of P as a function of the vectorial angle , is the polar equation of the parabola, when the focus is taken as pole and the perpendicular from the focus to the directrix as polar axis. 90. Polar Construction of Parabolas. By means of the equation (1) the parabola can be plotted by points. Thus, for <^ = we find r = a as intercept on the polar axis. As <^ increases from the value 0, r continually increases, reaching 93 Fig. 51 94 PLANE ANALYTIC GEOMETRY [VI, § 90 the value 2 a for <^ = i7r, and becoming infinite as ap- proaches the value tt. For any negative value of <^ (between and — tt) the radius vector has the same length as for the corresponding positive value of <}> ; this means that the parabola is symmetric with respect to the polar axis. The intersection A of the curve with its axis of symmetry is called the vertex, and the axis of symmetry FA the axis, of the parab- ola. The segment BB' cut off by the parabola on the perpendicular to the axis drawn through the focus is called the latus rectum; its length is 4 a, if 2 a is the distance between focus and directrix. Notice also that the vertex A bisects this distance FD so that the distance between focus and vertex as well as that between vertex and directrix is a. In Fig. 62 the polar axis is taken positive in the sense from the pole toward the directrix. If the sense from the directrix to the pole is taken as positive (Fig. 52), we have again with -Fas pole FP=r, but the distance of P from the directrix is 2 a + ?• cos , So that the polar equation is now (2) r = -^^~ ^ ^ 1 - cos <^ We have assumed a as a positive number, 2 a denoting the absolute value of the distance between the fixed point (focus) and the fixed line (directrix). The radius vector r is then always positive. But the equations (1) and (2) still represent parabolas if a is a negative number, viz. (1) the parabola of Fig. 52, (2) the parabola of Fig. 51, the radius vector r being negative (§ 16). Fig. 52 VI, § 92] THE PARABOLA 95 , r = (6) r = a, r = 1 — cos 1 + cos 8 2 (I (c) r = 4 cos 0, r = (d) r cos = 2 a, r = - • 1 + cos 1 - cos 3. Sketch the following parabolas : (a) (y_2)2 = 8(x~5). (6) (x + 3)2 = 5(3 - y). (c) x2 = 6(y + 1). {d) (2/ + 3)2 = - 3 X. 98 PLANE ANALYTIC GEOMETRY [VI, 96 4. Sketch each of the following parabolajs and find the coordinates of the vertex and focus, and the equations of the directrix and axis : (a) 2/2-2y-3x- 2 = 0. (6) x2 + 4x- 4 y = 0. (c) a;2__4a; + 3y + 1 = 0. (d) Sx^ - 6x - y = 0. (e) 8 y2 - 16 y 4- X + 6 = 0. (/) y^ + y + x = 0. (^) x2~x-3y + 4 = 0. (A) 8y2_3x + 3=0. 6. Sketch the following loci and find their intersections : (a) y = 2x, y =x2. (6) y* = 4ax, x -f y = 3a. (c) y2 = x + 3, y2 = 5-x. (d) y2 4-4x + 4=0, x2+y2=41. 6. Sketch the parabolas with the following lines and points as direc- trices and foci, and find their equations : (a) X - 4 = 0, (6, - 2). (6) y 4- 3 = 0, (0, 0). (c) 2x + 6 = 0, (0, -1). (d) x = 0, (2, -3). (e) 3y-l = 0, (-2,1). (/) x-2a = 0, (a, 6). 7. Find the parabola, with axis parallel to Ox, and passing through the points : (a) (1, 0), (5, 4), (10, -6). (6) (V, -5), (f, 0), (j, -3). 8. Find the parabola, with axis parallel to Oy^ and passing through the points : (a) (0, 0), (-2, 1), (6, 9). (d) (1, 4), (4, -1), (-3, 20). 9. Find the parabola whose directrix is the line 3x — 4y — 10=0 and whose focus is: (a) at the origin; (6) at (6, — 2). Sketch each curve. When does the equation of a parabola contain an xy term ? 10. Find the parabolas with the following points as vertices and foci (two solutions) : (a) (-3, 2), (-3, 6). (6) (2, 5), (-1, 5). (c) (-1, -1), (1, -1). {d) (0, 0), (0, -a). 11. If s denotes the distance (in feet) from a point P in the line of motion of a falling body, at a time < (in seconds), s-so = ig(t-to)% where g is the gravitational constant (32.2 approximately) and Sq is the distance from F at the time ^q, show that this equation can be put in the standard form s = lgT, VI, § 97] THE PARABOLA 99 where s denotes the distance from some other fixed point in the line of motion and 7 is the time since the body was at that point. 12. The melting point t (in degrees Centigrade) of an alloy of lead and zinc is found to be f = 133 + .875a; + .01125a;2, where x is the percentage of lead in the alloy. Reduce the equation to standard form t = kx ; and show that x — x — h^ t= t — k, where h is the percentage of lead that gives the lowest melting point, and k is the temperature at which that alloy melts. 13. Show that the locus of the center of the circle which passes through a fixed point and is tangent to a fixed line is a parabola. 14. Show that the locus of the center of a circle which is tangent to a fixed line and a fixed circle is a parabola. Find the directrix of this parabola. 97. Slope of the Parabola. The slope tan a of the parabola at any point P (x, y) (Fig. 57) can be found (comp. § 72) by first determining the slope tan «! = y^^y of the secant PP^ , and then letting -Pi(2Jij 2/i) move along the curve up to the point P(x, y). Now as P^ conies to coincide with P, x-^ becomes equal to x, and y^ equal to y, so that the expression for tan «! loses its meaning. But observing that P and Pi lie on the parabola, we have y^^^^ax and y^ = 4 aa^i , and hence y^ — y^ — A^a{xi — x). Substituting from this relation the value of x^ — x in the above expression for tan «!, we find for the slope of the secant : tan a^ = ^a -^ — '- = Vi-y^ 2/1 + 2/ Fig. 57 100 PLANE ANALYTIC GEOMETRY [VI, § 97 If we now let Pj come to coincidence with P so that yi becomes = y, we find for the slope of the tangent at P{x, y) : (7) tana = ^. y This slope of the tangent at P is also called the slope of the parabola at P. The ordinate y of the parabola is a function of the abscissa x ; and the slope of the parabola at P (x, y) is the rate at which y increases with increasing x at P; in other words, it is the derivative y' of y with respect to x (compare § 73). As by the equation of the parabola we have y = ± 2 Va^, we find: (8) 2/' = tana = ?^=±J^. y ^x The double sign in the last expression corresponds to the fact that to a given value of x belong two points of the curve with equal and opposite slopes. 98. Equation of the Tangent. As the slope of the parabola 2/2 =4 oa; at the point P(x, y) is 2 a/y (§ 97), the equation of the tangent at this point is Y-y=^{X-x), y where X, Fare the coordinates of any point of the tangent, while Xj y are the coordinates of the point of contact. This equation can be simplified by multiplying both sides by y and observing that y* = 4 aa; ; we thus find (9) yT=^2aix + X). Notice that (as in the case of the circle, § 64) the equation of the tangent is obtained from the equation of the curve, y^ = 4:ax, by replacing y^ hy yT, 2 x hy x -\- X. VI, § 99] THE PARABOLA 101 The segment TP (Fig. 58) of the tangent from its intersec- tion T with the axis of the parabola to the point of contact P is called the length of the tangent at P ; the projection TQ of this segment TP on the axis of the parabola is called the subtangent at P. Now, with Y= 0, equation (9) gives X = ^ x, i.e. TO = OQ; hence the subtangent is bisected by the vertex. This furnishes a simple construction for the tangent at any point P of the parabola if the axis and vertex of the parabola are known. 99. Equation of the Normal. The normal at a point P of any plane curve is defined as the perpendicular to the tan- gent through the point of contact. The slope of the normal is therefore (§ 27) minus the recip- rocal of that of the tangent. Hence the equation of the normal to the parabola is : Y that is : , = -J^(X-.), (10) yX-^2aY=(2a-\-x)y. The segment PN of the normal from the point P{x, y) on the curve to the intersection N of the normal with the axis of the parabola is called the leyigth of the normal at P; the projection QJSf of this segment PN on the axis of the parabola is called the subnormal at P. Now, with F=0, equation (10) gives X = 2 a-{-x, and as x=OQ, it follows that Q]Sf=2a; i.e. the subnormal of the parabola is constant, viz. equal to half the latus rectum. 102 PLANE ANALYTIC GEOMETRY [VI, § 100 100. Intersections of a Line and a Parabola. The inter- sections of the parabola with the straight line y = mx + h are found by substituting the value of y from the latter in the former equation : {mx -I- 6)2 = 4 ax, or, reducing : m V + 2(m6-2a)a; + 62 = 0. The roots of this quadratic in x are the abscissas of the points of intersection ; the ordinates are then found from y = mx -4- h. It thus appears that a straight line cannot intersect a parabola in more than two points. If the roots are imaginary, the line does not meet the parabola ; if they are real and equal, the line has but one point in common with the parabola and is a tangent to the parabola (provided m =^ 0). 101. Slope Equation of the Tangent. The condition for equal roots is (6m-2a)» = 6»m», which reduces to The point that the line of this slope has in common with the parabola is then found to have the coordinates 2a — bm 6* _^ . . ok X = = — , y = ma; -f 6 = J o. m* a As the slope of the parabola at any point (x, y) is (§ 97) 2/ = 2 a/y, the slope at the point just found is 3/ = ajb — m ; i.e. the slope of the parabola is the same as that of the line y = mx-\-b\ this line is therefore a tangent. Thus, the line VI, § 103] THE PARABOLA 103 (11) y=^mx+- y is tangent to the parabola y^ = 4 ax whatever the value of m. This may be called the slope-form of the equation of the tangent. Equation (11) can also be deduced from the equation (9), by putting 2 a/y = m and observing that 2/^ = 4 aaj. 102. Slope Equation of the Normal. The equation (10) of the normal can be written in the form 2a 2a or since by the equation (3) of the parabola x = y^/4: a : 2a ^^^Sa' If we denote by n the slope of this normal, we have : n = -^, y = -2an, ^^-an\ 2 a 8 a^ so that the equation of the normal assumes the form (12) T=nX-2an- an\ This may be called the slopeform of the equation of the normal. 103. Tangents from an Exterior Point. The slope-form (11) of the tangent shows that from any point (x, y) of the plane not more than two tangents can be drawn to the parabola y'^ = ^ax. For, the slopes of these tangents are found by substituting in (11) for X, y the coordinates of the given point and solving the resulting quadratic in m. This quadratic may have real and different, real and equal, or complex roots. Those points of the plane for which the roots are real and different are said to lie outside the parabola ; those points for which the roots are imaginary are said to lie within the parab- 104 PLANE ANALYTIC GEOMETRY [VI, § 103 ola; those points for which the roots are equal lie on the parabola. The quadratic in m can be written xm^ — ym -f a = 0, so that the discriminant is y* — 4 ax. Therefore a point (x^ y) of the plane lies within, on, or outside the parabola according as 2/* — 4 oa; is less than, equal to, or greater than zero. Similarly, the slope-form (12) of the normal shows that not more than three normals can be drawn from any point of the plane to the parabola, since the equation (12) is a cubic for n when the coordinates of any point of the plane are substituted for X, T. As a cubic has always at least on§ real root there always exists one normal through a given point; but there may be two or three. 104. Geometric Properties. Let the tangent and normal at P (Fig. 59) meet the axis at T, N-, let Q be the foot of the perpendicular from P to the axis, D that of the per- pendicular to the directrix d; and let be the vertex, F the focus. As the subtangent TQ is bisected by (§ 98) and the subnormal QN is equal to 2 a (§ 99), while OF = a, it follows that F lies midway between T and N. The triangle TPN being right-angled at P, and F being the midpoint of its hypotenuse, it follows that FP=^FT=FN. VI, § 105] THE PARABOLA 105 Hence, if axis and focus are given, the tangent and the normal at any point P of the parabola are found by describing about F a circle through P which will meet the axis at T and N. As FP=DP, it follows that FPDT is a rhombus; the diagonals PT and FD bisect therefore the angles of the rhombus and intersect at right angles. As TP (like TQ) is bisected by the tangent at the vertex, the intersection of these diagonals lies on this tangent at the vertex. The properties just proved that the tangent at P bisects the angle between the focal radius PF and the parallel PD to the axis and that the perpendicular from the focus to the tangent meets the tangent on the tangent at the vertex are of particular importance. 105. Diameters. It is known from elementary geometry that in a circle all chords parallel to any given direction have their midpoints on a straight line which is a diameter of the circle. Similarly, in a parabola, the locus of the midpoints of all chords parallel to any given direction is a straight line^ and this line which is parallel to the axis is called a diameter of the parabola. To prove this, take the vertex as origin and the axis of the parabola as axis Ox (Fig. 60) so that the equation is 2/^^ = 4 ax. Any line of given slope m has the equation y = mx-\-b, Fig. 60 and with variable b this represents a pencil of parallel lines. Eliminating x we find for y the quadratic 106 PLANE ANALYTIC GEOMETRY [VI, § 105 The roots ^j, y^ are the ordinates of the points Pj, Pg at which the line intersects the parabola. The sum of the roots is 4a 2/1 + ^2 = — ; m hence the ordinate \{yi + 2/2) of the midpoint P between Pj , Pj is constant (i.e. independent of »), viz. = 2 a/m, and independ- ent of 6. The midpoints of all chords of the same slope m lie, therefore, on a parallel to the axis, at the distance 2 a/m from it. The condition for equal roots (§ 101) gives h = a/m. That one of the parallels which passes through the point where the diameter meets the parabola is, therefore, y = mx-Jf-—\ m by § 101 this is a tangent. Thus, the tangent at the end of a diameter is parallel to the chords bisected by the diameter. EXERCISES 1. Find and sketch the tangent and normal of the following parabolas at the given points : (a) 2y2 = 26x, (2, 5). (6) 3y2 = 4x, (3, - 2). (c) y^ = 2x, {\, 1). (d) 5y2=i2x, (i-2). (6) y^ = x,{hl). (/) 46y« = x, (5, i). 2. Show that the secant through the points P (a; , y) and Pi (xi , j/i) of the parabola y^ = iax has the equation 4aX— (y + yi) F+ yyi = 0, and that this reduces to the tangent at P when Pi and P coincide. 3. Find the angle between the tangents to a parabola at the vertex and at the end of the latus rectum. Show that the tangents at the ends of the latus rectum are at right angles. 4. Find the length of the tangent, subtangent, normal, and subnormal of the parabola y^ ^^ 4 ^ at the point (1, 2). 6. Find and sketch the tangents to the parabola y^ = Sx from each of the following points : (a) (- 2, 3). (6) (- 2, 0). (c) (-6, 0). (d) (8, 8). VI, § 105] THE PARABOLA 107 6. Draw the tangents to the parabola ?/2 = 3 x that are inclined to the axis Ox at the angles: (a) 30°, (&) 45°, (c) 135°, (d) 150°; and find their equations. 7. Find and sketch the tangents to the parabola y^ = Ax that pass through the point ( — 2, 2). 8. Find and sketch the normals to the parabola y^ = 6z that pass through the points : («) (1,0). (6) (¥,-3). (c) (-*/,-!). ((^)(f,-|). (e) (0,0). 9. Are the following points inside, outside, or on the parabola Sy^ = x? (a) (3,1). (6) (2, i). (c) (8, 1). (c?) (10, f). 10. Show that any tangent to a parabola intersects the directrix and latus rectum (produced) in points equally distant from the focus. 11. Show that the tangents drawn to a parabola from any point of the directrix are perpendicular. 12. Show that the ordinate of the intersection of any two tangents to the parabola y^ = 'iax is the arithmetic mean of the ordinates of the points of contact, and the abscissa is the geometric mean of the abscissas of the points of contact. 13. Show that the sum of the slopes of any two tangents of the parab- ola 2/2 = 4 aa: is equal to the slope Y/Xoi the radius vector of the point of intersection (X, F) of the tangents ; find the product of the slopes. 14. Find the locus of the intersection of two tangents to the parabola 2/2 = 4 ax^ if the sum of the slopes of the tangents is constant. 15. Find the locus of the intersection of two perpendicular tangents to a parabola ; of two perpendicular normals to a parabola. 16. Show that the angle between any two tangents to a parabola is half the angle between the focal radii of the points of contact. 17. From the vertex of a parabola any two perpendicular lines are drawn ; show that the line joining their other intersections with the parabola cuts the axis at a fixed point. 18. Find and sketch the diameter of the parabola y"^ — Qx that bisects the chords parallel to 3a; — 2y + 5 = 0; give the equation of the focal chord of this system. 19. Find the system of parallel chords of the parabola y2 = 8 x bisected by the line y = 3. 108 PLANE ANALYTIC GEOMETRY [VI, § 105 20. Show that the tangents at the extremities of any chord of a parab- ola intersect on the diameter bisecting this chord. Compare Ex. 12. 21. Find the length of the focal chord of a parabola of given slope m. 22. Find the angles at which the parabolas y^ = 4ax and x^ = 4ay intersect. 23. Two equal confocal parabolas have the same axis but open in op- posite sense ; show that they intersect at right angles. 24. If axis, vertex, and one other point of the parabola are given, ad- ditional points can be constructed as follows : Let be the vertex, P the given point, and Q the foot of the perpendicular from P to the tangent at the vertex ; divide QP into equal parts by the points ^i, J.2, ••• ; and OQ into the same number of equal parts by the points Pi, P2, ••• ; the intersections of 0-4i, 0^2» ••• with the parallels to the axis through Pi, P2, •••are points of the parabola. 26. If two tangents APu AP^ to a parabola with their points of con- tact Pi, P2 are given and ^Pi, ^P2 be divided into the same number of equal parts, the points of division being numbered from Pi to A and from A to P2, the lines joining the points bearing equal numbers are tangents to the parabola. To prove this show that the intersections of any tangent with the lines ^Pi, ^P2 divide the segments Pi^, APi in the same division ratio. 26. The shape assumed by a uniform chain or cable suspended between two fixed points Pi, P2 is called a catenary ; its equation is not algebraic and cannot be given here. But when the line P1P2 is nearly horizontal and the depth of the lowest point below P1P2 is small in comparison with P1P2, the catenary agrees very nearly with a parabola. The distance between two telegraph poles is 120 ft. ; P2 lies 2 ft. above the level of Pi ; and the lowest point of the wire is at 1/3 the distance be- tween the poles. Find the equation of the parabola referred to Pi as origin and the horizontal line through Pi as axis Ox ; determine the posi- tion of the lowest point and the ordinates at intervals of 20 ft. 27. The cable of a suspension bridge assumes the shape of a parabola if the weight of the suspended roadbed (together with that of the cables) is uniformly distributed horizontally. Suppose the towers of a bridge 240 ft. long are 60 ft. high and the lowest point of the cables is 20 ft. above the roadway ; find the vertical distances from the roadway to the cables at intervals of 20 ft. VI, § 106] THE PARABOLA 109 28. When a parabola revolves about its axis, it generates a surface called a paraboloid of revolution ; all meridian sections (sections through the axis) are equal parabolas. If the mirror of a reflecting telescope is such a surface (the portion about the vertex), all rays of light falling in parallel to the axis are reflected to the same point ; explain why. 106. Parameter Equations. Instead of using the cartesian or polar equation of a curve it is often more convenient to express x and y (or r and <^) each in terms of a third variable, which is then called the parameter. Thus the parameter equations of a circle of radius a about the origin as center are : x = a cos , y = a sin <^, corresponds a definite x and a definite y, and hence a point of the curve.- The elimination of <^, by squaring and adding the equations, gives the cartesian equation a^-\-y^ = a^. Again, to determine the motion of a projectile we may observe that, if gravity were not acting, the projectile, started with an initial velocity v^ at an angle c to the horizon, would have at the time t the position a; = ^>o cos c • ^, y = VQ sin e • t^ the horizontal as well as the vertical motion being uniform. But, owing to the constant acceleration g of gravity (down- ward), the ordinate y is diminished by ^gt"^ in the time i, so that the coordinates of the projectile at the time t are x = Vq cos c • ^, y = VQmie't — \gt'^. These are the parameter equations of the path, the parameter here being the time t. The elimination of t gives the cartesian equation of the parabola described by the projectile : y=zVQtdiXi€'X- / x\ 2 V cos^ € 110 PLANE ANALYTIC GEOMETRY [VI, § 107 107. Parameter Equations of a Parabola. For any parabola ^ = 4 oa; we can also use as parameter the angle a made by the tangent with the axis Ox\ we have for this angle (§ 97) : 2a tan a = — ; y it follows that y = 2a cot a and hence x = y^/4: a = a cot* «. The equations x = a cot' a, y = 2 a cot a are parameter equations of the parabola ?/' = 4 oic ; the elimina- tion of cot a gives the cartesian equation. 108. Parabola referred to Diameter and Tangent. The equation of the parabola y^ = 4ax preserves this simple form if instead of axis and tangent at the vertex we take as axes any diameter and the tangent at its end. We shall show that the equation in these oblique coordinates is yi* = 4 aiXi , where oi is a new constant determined below. To prove this observe that since the new origin Oi (A, A;) is a point of the parabola y* = 4 aXy we have by § 107 h = acot^a, k = 2a cot a, ^o- 61 where a is the angle at which the tangent at Oi is inclined to the axis. Hence, transferring to parallel axes through Oi, we obtain the equation (y + 2 a cot a)2 = 4 a (x + a cot2 a), which reduces to y^ -\-iacot(t'y = iax. The relation between the rectangular coordinates x, y and the oblique coordinates Xi , yi , both with Oi as origin, is readily seen from the figure to be X = xi + yi cos a, y = yi sin a. Substituting these values we find 2/1^ sin2 a + 4 a cos a • yi = 4 oici + 4 ayi cos a, y / h / \ 1 X /' or, if we put a/sin^ a = ai, y^ — 4 xi = 4 a\X\. VI, § 108] THE PARABOLA 111 The meaning of the constant ai appears by observing that sin2 a tan2 a ai is therefore the distance of the new origin Oi from the directrix, or, what amounts to the same, from the focus F. EXERCISES 1. Show that the parameter equations of a circle with center at (A, k) and radius a are X = h -i- a cos 4>, y = k-{- asiiKp. 2. Sketch the curves whose equations are : (a) x = t,y = t'^; (b) x = t^ -1, y = 3-21"^; (c) x = 2t-l, y = t^-St'; ((?) X = 3 + 2 cos 0, 2/ = 4 + 2 sin ; (e) a; = 4 + 5 cos 0, y = 2 + 5 sin 0. 3. What must be the initial velocity vq of a projectile if, with an eleva- tion of 30°, it is to strike an object 100 ft. above the horizontal plane of starting point at a horizontal distance from the latter of 1200 ft.? 4. What must be the elevation e to strike an object 100 ft. above the horizontal plane of the starting point and 6000 ft. distant, if the initial velocity be 1200 ft. per second ? 6. Prove that a projectile whose elevation is 60° rises three times as high as when its elevation is 30°, the magnitude of the initial velocity being the same in each case. 6. If a golf ball be driven from the tee horizontally with initial speed = 300 ft. /sec, where and when would it land on ground 16 ft. below the tee if resistance of air and rotation of ball could be neglected ? 7. A man standing 15 feet from a pole 150 ft. high aims at the top of the pole. If the bullet just misses the top, where will it strike the ground if vo - 1000 ft. /sec. ? 8. The ends ^, J5 of a straight rod of length 2 a move along two per- pendicular lines ; find the locus of the midpoint of AB. 9. Four rods are jointed so as to form a parallelogram ; if one side is fixed, find the path described by any point rigidly connected with the op- posite side. 112 PLANE ANALYTIC GEOMETRY [VI, § 109 109. Area of Parabolic Segment. A parabola, together with any chord perpendicular to its axis, bounds an area OPF^ (shaded in Fig. 62). It was shown by Archimedes (about 250 B.C.) that this area is two thirds the area of the rectangle PPfQ'Q that has the chord J" P as one side and the tangent at the vertex as opposite side. Fig. 62 This rectangle PJPQ'Q is often called (somewhat improperly) the cir- cumscribed rectangle so that the result can be expressed briefly by saying that the area of the parabola is 2/3 of that of the circumscribed rectangle. This statement is of course equivalent to saying that the (non-shaded) area OQP is 1/3 of the area of the rectangle OQPB. In this form the proposition is proved in the next article. 110. Area by Approximation Process. To obtain first an ap- proximate value {A) for the area OQP (Fig. 63) we may subdivide the area into rectangular strips of equal width, by dividing OQ into, say, n equal parts and drawing the ordiuates j/i , y2, ••• ^n- If the width of these strips is Ax so that OQ = nAx, we have as approximate value of the area : (A) = Ax • yi -\- Ax ' 7/2 + •" + Ax • y„. Now yi is the ordinate corresponding to the abscissa Ax ; yz corresponds to the abscissa 2 Ax, etc. ; yn corresponds to the abscissa nAx = OQ. Hence, if the equation of the curve is a;'* = 4 ay, we have : FiQ. 63 4a 2/2 T^(2Ax)2, 4a yn = ^{nAxy. 4a Substituting these values we find : (^)=(M-'(i + 22 + 32+ ... +n2). 4 a Now, 1 + 22+ ... +n2 = in(n+l)(2n+l)=i(2n8 + 3n2 + n);- hence (^) = ^'(2n« + 3n2 + n) = C^Y2 + § + I). ^ ' 24 a 24 a \ n n^) Now nAx = OQ — Xn, the abscissa of the terminal point P, whatever the number n and length Ax of the subdivisions. Hence, if we let the num- VI, § 111] THE PARABOLA 113 ber n increase indefinitely, we find in the limit the exact expression A for the area OQP: ._^_1. 4a 3 where y„ = Xr?/4, a is the ordinate of the terminal point P. As XnVn is the area of the rectangle OQPB, Archimedes' proposition (§ 109) is proved. 111. Area expressed in Terms of Ordinates. The area (shaded in Fig. 64) between the parabola cc^ = 4 ay, the axis Ox, and the two ordinates yi, y^, whose abscissas differ by y 2 Ax is evidently, by the formula of § 110, A = -^{x^^-Xi^). 12 a Ax 12 a = -=^ (6 xi^ + 12 xiAx 12 a [(xi + 2 Ax)8-xi3] - 8 (Ax)2). Fig. 64 This expression can be given a remarkably simple form by introducing not only the ordinates y\ = a;iV4 a, yz = (ici + 2 Axy/i a, but also the ordinate 2/2 midway between yi and yz, whose abscissa is Xi + Ax. For we have : yi + 4 2/2 + = ~[xi^-hHxi + Axy +{xi + 2 Ax)2] 4a We find therefore : = — [6 Xi2 + 12 XiAx + 8(Ax)2]. 4/z ^ = |Ax(yi + 4?/2 + y3). This formula holds even if the vertex of the parabola is at any point {h, k) , pro- vided the axis of the parabola is parallel to Oy. For (Fig. 65 ) , to find the area under the arc P1P2P3 we have only to add to the doubly shaded area the simply shaded rectangle whose area is 2 kAx. We find therefore for the whole area : I Ax(yi + 4 ?/2 + 2/3) + 2 Mx = ^ Ax(yi + 4 1/2 + 2/3 + 6 A:) = 1 Ax l(yi + A;) + 4 (2/2 + k) +(2/3 + A:)], I Fio. 65 114 PLANE ANALYTIC GEOMETRY [VI, § 111 Fig. 66 where yi,y2, yz are the ordinates of the parabola referred to its vertex, and hence yi + k^ yi + k^ yz-\- k the ordmates for the origin O. We have therefore for any parabola whose axis is parallel to Oy : A = \ ^x{yi + iy2-{-y3). 112. Approximation to any Area. Simpson's Rule. The last formula is sometimes used to find an approximate value for the area under any curve (i.e. the area bounded by the axis Ox, an arc AB of the curve, and the ordinates of ^ and B, Fig. 66). This method is particularly convenient if a number of equidistant ordinates of the curve are known, or can be found graphically. Let Ax be the distance of the ordi- nates, and let yi , 2/2 » ys be any three consecutive ordinates. Then the doubly shaded portion of the required area, between yi and ys, will be (if Ax is sufficiently small) very nearly equal to the area under the parabola that passes through Pi , P3 , P3 and has its axis parallel to Oy. This parabolic area is by § 111 = ^Ax(yi+^yi + y3). The whole area under AB is a sum of such expressions. This method for finding an approximate expression for the area under any curve is known as Simpson's rule (Thomas Simpson, 1743) although the funda- mental idea of replacing an arc of the curve by a parabolic arc had been suggested previously by Newton. 113. Area of any Parabolic Segment. As the equation of a parabola referred to any diameter and the tangent at its end has exactly the same form as when the parabola is referred to its axis and the tangent at the vertex (§ 108) it can easily be shown that the area of any parabolic segment is S/3 of the area of the circumscribed paral- lelogram formed by the chord, the parallel tangent, and the two parallels to the axis through the extremities of the chord (Fig. 67). ^ JjE Qi Ax Q, FlQ. 67 VI, § 113] THE PARABOLA 115 With the aid of this proposition Simpson's rule can be proved very simply. For, the area of the parabolic segment P1P3P2 (Fig. 67) is then equal to 2/3 of the parallelogram formed by the chord P1P2, the tangent at P2, and the ordinates yi, ys (produced if necessary). This parallelo- gram has a height = 2 Ax and a base = ilfP2 = 2/2 — i (yi + Vs) ', hence the area of P1P3P2 is f Ax (2 y^^y^- y^) = 1 Ax [4 2/2 - 2 (2/1 + ys)]. To find the whole shaded area we have only to add to this the area of the trapezoid QiQzPzP\^ which is Ax (1/1+2/3). Hence A = QiQsPsP2Pi = ^ Ax[4 2/2 - 2(2/1 + 2/3) + S(yi + 2/3)] = ^ Ax(2/i +42/2 + 2/3). EXERCISES 1. Show that the area of any parabolic segment is 2/3 of the area of the circumscribed parallelogram. 2. In what ratio does the parabola y^ = 4ax divide the area of the circle (x - a)2 + 2/2 = 4 a^ ? 3. Find the area bounded by the parabola y'^ = 4 ax and a line of slope m through the focus. 4. Find and sketch the curve whose ordinates represent the area bounded by : (a) the line ^ = | x, the axis Ox, and any ordinate, (&) the parabola 2/ = f x^, the axis Ox, and any ordinate. 5. Find an approximation to the areas bounded by the following curves and the axis Ox (divide the interval in each case into eight or more equal parts) : (a) 4 2/ = 16 - x2. (6) 2/ = (x + 3) (x - 2)2. (c) 2/ = a;2 - x^. 6. The cross-sections in square feet of a log at intervals of 6 ft. are 3.25, 4.27, 5.34, 6.02, 6.83 ; find the volume. 7. The cross-sections of a vessel in square feet measured at intervals of 3 ft. are 0, 2250, 5800, 8000, 10200 ; find the volume. Allowing one ton for each 35 cu. ft. , what is the displacement of the vessel ? 8. The half-widths in feet of a launch's deck at intervals of 6 ft. are 0, 1.8, 2.6, 3.2, 3.3, 3.3, 2.7, 2.1, 1 ; find the area. CHAPTER VII ELLIPSE AND HYPERBOLA At Ft :^>^ fiA FiQ. 68 114. Definition of the Ellipse. The ellipse may be defined as the locus of a point whose distances from two fixed points have a constant sum. If Fi , F2 (Fig. 68) are the fixed points, which are called the focif and if P is any point of the ellipse, the condition to be satisfied by P is FiP -h F,P = 2 a. The ellipse can be traced mechan- ically by attaching at F^y Ff the ends of a string of length 2 a and keeping the string taut by means of a pencil. It is obvious that the curve will be symmetric with respect to the line FiF^, and also with respect to the perpendicular bisector of F1F2. These axes of symmetry are called the axes of the ellipse ; their intersection O is called the center of the ellipse. 115. Axes. The points A^, A^y B^, B^ (Figs. 68 and 69) where the ellipse intersects these axes are called vertices. The distance A^A^ of those vertices that lie on the axis containing the foci -Fi, i^2 is = 2 a, the length of the string. For when the point P in describing the ellipse arrives at -4i, the string is doubled along F^A^ so that 116 FiQ. 69 VII, § 116] ELLIPSE AND HYPERBOLA 117 and since, by symmetry, A2F2 = i^i A , we have A2F2 + F^F^ + F^A^ = A^A^ = 2 a. The distance AoA^ = 2 a, which is called the major axiSj must evidently be not less than the distance 1^2-^1 between the foci, which we shall denote by 2 c. The distance B2B1 of the other two vertices is called the minor axis and will be denoted by 2 b. We then have 62 = a2 - c2 ; for when P arrives at Bi , we have B^F^ = B^F^ = a. 116. Equation of the Ellipse. If we take the center as origin and the axis containing the foci as axis Ox, the equation of the ellipse is readily found from the condition F^P + F^P = 2 a, which gives, since the coordinates of the foci are c, and — c, : ■Vix - c)2 + 2/2 + V(aj + c)2 + 2/2 = 2 a. Squaring both members we have a.2 _^ 2/2 -f- c2 + V(a;2 + 2/^ + 02 — 2 cx){x^ + y^ -^ c'^ -{- 2 ex) = 2 a^ -, transferring a;2 -f 2/2 + c2 to the right-hand member and squar- ing again, we find (a;2 + 2/2 -f c2)2 - 4 c2a;2 = 4 a^ - 4 a2(a;2 -}- 2/2 -f- c^) + (x^ + y^-\- c^y, i.e. (a2 - c2)a;2 + aY = a^(a^ - c2). Now for the ellipse (§ 115) a2 — c2 = b^. Hence, dividing both members by a^b% we find as the cartesian equation of the ellipse referred to its axes. This equation shows at a glance : (a) that the curve is sym- metric to Ox as well as to Oy ; (6) that the intercepts on the axes Ox, Oy are ± a, and ± b. The lengths a, b are called the semi-axes. 118 PLANE ANALYTIC GEOMETRY [VII, § 116 Solving tlie equation for y we find (2) 2^ = ±-V^^'=^, ct which shows that the curve does not extend beyond the vertex Ai on the right, nor beyond A2 on the left. If a and h (or, what amounts to the same, a and c) are given numerically, we can calculate from (2) the ordinates of as many points as we please. If, in particular, a = b (and hence c = 0), the ellipse reduces to a circle, EXERCISES 1. Sketch the eUipse of semi-axes a = 4, 6 = 3, by marking the ver- tices, constructing the foci, and determining a few points of the curve from the property FiP -\- F2P= 2 a. Write down the equation of this ellipse, referred to its axes. 2. Sketch the ellipse x^/l6 + y^/9 = 1 by drawing the circumscribed rectangle and finding some points from the equation solved for y. 3. Sketch the ellipses : (a) «« +2 y* = 1. (6) 3 x^ + 12 y* = 6. (c) 3 a;2 + 3 1/2 = 20. (d) x^ + 20 y^ = 1. 4. If in equation (1) a < 6, the equation represents an ellipse whose foci lie on Oy. Sketch the ellipses : (a) T + 7^=l' (^) 20x2 + 2/2 = 1. (c) 10x2 + 9y2=io. 4 lo 6. Find the equation of the ellipse referred to its axes when the foci are midpoints between the center and vertices. 6. Find the product of the slopes of chords joining any point of an ellipse to the ends of the major axis. What value does this product assume when the ellipse becomes a circle ? 7. Derive the equation of the ellipse with foci at (0, c), (0, — c), and major axis 2 a. 8. Write the equations of the following ellipses : (a) with vertices at (5, 0), (- 5, 0), (0, 4), (0, - 4) ; (&) with foci at (2, 0), (- 2, 0), and major axis 6. 9. Find the equation of the ellipse with foci at (1, 1), (—1, — 1), and major axis 6, and sketch the curve. VII, § 118] ELLIPSE AND HYPERBOLA 119 117. Definition of the H3rperbola. The hyperbola can be defined as the locus of a point whose distances from two fixed points have a constant difference. The fixed points F^, F^ are again called the foci; if 2 a is the constant difference, every point P of the hyperbola must satisfy the condition F^P-F^P=^±2a. Kotice that the length 2 a must here be not greater than the distance F^^ = 2 c of the foci. The curve is symmetric to the line -Fa^i and to its perpen- dicular bisector. A mechanism for tracing an arc of a hyperbola consists of a straightedge F2Q (Fig. 70) which turns about one of the foci, F2 ; a string, of length F2Q — 2a, is fastened to the Fig. 70 straightedge at Q and with its other end to the other focus, Fi. As the straightedge turns about F^, the string is kept taut by means of a pencil at P which describes the hyperbolic arc. Of course only a portion of the hyperbola can be traced in this manner. 118. Equation of the Hyperbola. If the line F^F^ be taken as the axis Ox, its perpendicular bisector as the axis Oy, and if F^F^ = 2 c, the condition F^P-F^P^ ±2 a becomes (Fig. 71) ; ■y/(x + cy -{-f -Vix - cy -\-y' = ±2 a. 120 PLANE ANALYTIC GEOMETRY Squaring both members we find [VII, § 118 a^+2/' 4- c^- V(ar^ 4-2^' 4- c^- 2 ca;)(a^H- 2/2 -fc^-f 2 caj) =2 a^ squaring again and reducing as in § 116, we find exactly the same equation as in § 116 : Fig. 71 But in the present case c^ a, while for the ellipse we had c < a. We put, therefore, for the hyperbola c2-a2 = &2; the equation then reduces to the form (3) a2 62 which is the cartesian equation of the hyperbola referred to its axes. 119. Properties of the Hjrperbola. The equation (3) shows at once: (a) that the curve is symmetric to Ox and to Oy\ (6) that the intercepts on the axis Ox are ± a, and that the curve does not intersect the axis Oy. The line F^F^ joining the foci and the perpendicular bisector of F^F^ are called the axes of the hyperbola ; the intersection of these axes of symmetry is called the center. The hyperbola has only two vertices, viz. the intersections Ax , Ai with the axis containing the foci. VII, § 120] ELLIPSE AND HYPERBOLA 121 The shape of the hyperbola is quite different from that of the ellipse. Solving the equation for y we have h (4) y=±-^o?-a?, which shows that the curve extends to infinity from A^ to the right and from A^ to the left, but has no real points between the lines x = a, x = — a. The line F^F^ containing the foci is called the transverse axis; the perpendicular bisector of F^F^ is called the conjugate axis. The lengths a, b are called the transverse and conjugate semi-axes. In the particular case when a=b, the equation (3) reduces to 7^ — y^ — a^f and such a hyperbola is called rectangular or equilateral. 120. As3rmptotes. In sketching the hyperbola (3) or (4) it is best to draw first of all the two straight lines i.e. (5) y^±U a which are called the asymptotes of the hyperbola. Comparing with equation (4) it appears that, for any value of X, the ordinates of the hyperbola (4) are always (in absolute value) less than those of the lines (5); but the difference becomes less as x increases, approaching zero as x increases indefinitely. Thus, the hyperbola approaches its asymptotes more and more closely, the farther we recede from the center on either side, without ever reaching these lines at any finite distance from the center. 122 PLANE ANALYTIC GEOMETRY [VII, § 120 EXERCISES 1. Sketch the hyperbola a;2/16 — yV4 = 1, after drawing the asymp- totes, by determiuing a few points from the equation solved for y ; mark the foci. 2. Sketch the rectangular hyperbola x^ — y"^ — 9. Why the name rectangular ? 3. With respect to the same axes draw the hyperbolas : (a) 20a;2 - ys = 12. (6) x^ - 20 y^ = 12. (c) x^-y'^z= 12. 4. The equation — a^^/a^ + y'^jlP- — 1 represents a hyperbola whose foci lie on the axis Oy. Sketch the curves : (a) - 3x2 + 4j/2 = 24. (6) a;2_ 3^2 -|- 18 = 0. (c) x^-y'^ + 16 = 0. 6. Sketch to the same axes the hyperbolas : 9 ^ ' 9 ^ Two such hyperbolas having the same asymptotes, but with their axes interchanged, are called conjugate. 6. What happens to the hyperbola x^/a'^ — y'^/h'^ = 1 as a varies ? as 6 varies ? 7. The equation x^/aP' — y^/b^ = k represents a family of similar hyperbolas in which k is the parameter. What happens as k changes from 1 to — 1 ? What members of this family are conjugate ? 8. Find the foci of the hyperbolas : (a) 9 a;2 - 16 1/2 = 144. (6) 3 a;2 _ y2 = 12. 9. Find the hyperbola with foci (0, 3), (0, — 3) and transverae axis 4. 10, Find the equation of the hyperbola referred to its axes when the distance between the vertices is one half the distance between the foci. 11. Find the distance from an asymptote to a focus of a hyperbola. 13. Show that the product of the distances from any point of a hyper- bola to its asymptotes is constant. 13. Find the hyperbola through the point (1, 1) with asymptotes y=±2x. 14. Find the equation of the hyperbola whose foci are (1, 1), (—1, — 1), and transverse axis 2, and sketch the curve. VII, § 122] ELLIPSE AND HYPERBOLA 123 121. Ellipse as Projection of Circle. If a circle be turned about a diameter ^2^1 = ^ a through an angle €(<|-7r) and then projected on the original plane, the projection is an ellipse. For, if in the original plane we take the center as origin and OAi as axis Ox (Fig. 72), the ordinate QP of every point P of the projection is the projection of the corresponding ordinate QP^ of the circle; i.e. —j^ q QP=QPi cose. Fig. 72 The equation of the projection is therefore obtained from the equation x' + y^ = a' of the circle by replacing y by y/cos c. The resulting equation x^-\- y' COS^e represents an ellipse whose semi-axes are a, the radius of the circle, and 6 = a cos c, the projection of this radius. 122. Construction of Ellipse from Circle. We have just seen that, if a > 6, the ellipse can be obtained from its circumscribed circle a^ + 2/^ = a^ by re- ducing all the ordinates of this circle in the ratio b/a. This also appears by comparing the ordinates y = ±- V a^ — x"^ a of the ellipse with the ordinates ?/ = ± Va^ — x^ of the circle. 124 PLANE ANALYTIC GEOMETRY [VII, § 122 But the same ellipse can also be obtained from its inscribed circle a^ -{- y^ = b^ hj increasing each abscissa in the ratio a/b, as appears at once by solving for x. It follows that when the semi-axes a, b are given, points of the ellipse can be constructed by drawing concentric circles of radii a, b and a pair of perpendicular diameters (Fig. 73) ; if Fig. 73 any radius meets the circles at Pj , P^ > the intersection P of the parallels through P^ , Pj ^o the diameters is a point of the ellipse. 123. Tangent to Ellipse. It follows from § 121 that if P («, y) is any point of the ellipse and Pj that point of the cir- cumscribed circle which has the same abscissa, the tangents at P to the ellipse and at P^ to the circle must meet at a point T on the major axis (Fig. 74). Q A, Fig. 74 For, as the circle is turned about A^A^ into the position in which P is the projection of Pj , the tangent to the circle at Pi is turned into the position whose projection is PT, the point T on the axis remaining fixed. VII, § 124] ELLIPSE AND HYPERBOLA 125 The tangent a^jX + 2/1 Y= a^ to the circle at Pi {xi , y^) meets the axis Ox at the point T whose abscissa is OT=ayxi = a''/x, Hence the equation of the tangent at P(a;, y) to the ellipse is IT «^ = — , or a2 y X — yX-fx-^^Y-a^^=0: X dividing by a^y/x and observing that, by the equation of the ellipse, x^ — o? = — {a?/lf)y'^ we find (6) xX yY. as equation of the tangent to the ellipse (1) at 4lie point P(x, y). It follows from the equation of the tangent that the slope of the ellipse at any point P(x, y) is ¥x tan a = a^y 124. Eccentricity. For the length of the focal radius F^P of any point P{x, y) of the ellipse (1) we have (Fig. 75), since a^ — 6^ = c^ : F,P^={x-cy+y^=(x-cy-{-—(a''-x^)=^^(a'- 2 a''cx-{-c'x% whence FiP=± {-i^y The ratio c/a of the distance 2 c of the foci to the major axis 2 a is called the (numerical) ec- centricity of the ellipse. Denot- ing it by e we have F^P= ± (a — ex), and similarly we find F^P^ ± (a + ex). Fig. 75 126 PLANE ANALYTIC GEOMETRY [VII, § 124 For the hyperbola (3) we find in the same way, if we again put e = cla, exactly the same expressions for the focal radii F^P^ F^P (in absolute value). But as for the ellipse c^^o?— b^ while for the hyperbola c^^a^ + b- it follows that the eccentric- ity of the ellipse is always a proper fraction becoming zero only for a circle^ while the eccentricity of the hyperbola is always greater than one. 125. Equation of Normal to Ellipse. As the normal to a curve is the perpendicular to its tangent through the point of contact, the equation of the normal to the ellipse (1) at the point P{Xy y) is readily found from the equation (6) of the tangent as y X ^ /I IN c' I.e. ^X--Y=c\ X y The intercept made by this normal on the axis Ox is there- fore ON=z^x=^e'hi. From this result it appears by § 125 that (Fig. 76) FiN= c-\-e^x=i e(a -f-ea;) = e • F^P, FiN= c-e^ = e(a-ex)=e' F^P; hence the normal divides the dis- tance F2F1 in the ratio of the adjacent sides F^P^ F^P of the triangle F^PF.^. It follows that the normal bisects the angle betiveen the focal radii PF^ , PF^ ; in other words, the focal radii are equally inclined to the tangent. Fia. 76 VII, § 127] ELLIPSE AND HYPERBOLA 127 126. Construction of any Hyperbola from Rectangular Hyperbola. The ordinates (4), y = ±- Va;2 a of the hyperbola (3) are b/a times the corresponding ordinates y = ± 'vx^ — a^ of the equilateral hyperbola (end of § 119) having the same transverse axis. When 6 < a, we can put b/a = cos e and re- gard the general hyperbola as the projection of the equilateral hyperbola of equal transverse axis. When 6 > a, we can put a/b = cos € so that the equilateral hyperbola can be regarded as the projection of the general hyperbola. In either case it is clear that the tangents to the general and equilateral hyperbolas at corresponding points (i.e. at points having the same abscissa) must intersect on the axis Ox. 127. Slope of Equilateral Hyperbola. To find the slope of the equilateral hyperbola X'^ — y^ — a^j observe that the slope of any secant joining the point P{x, y) and Pi{xyy 2/i) ^^ (2/i~2/)/(^i~^)> ^-nd that the relations 2/2=aj2-a2, y^ = x^—o? give y^-yi^ = x^-x^\ i.e. (y-yi)(y + yi) =(x-^x{){x-\'x;)f whence .VZL^ _ ^"^ i . x-x^ y+yi Hence, in the limit when Pi comes to coincidence with P, we find for the slope of the tangent at P(x, y) : tan a =x/y. Hence the equation of the tangent to the equilateral hyperbola is T-y = -{X-x),oTxX-yT=a\ y since x^ —y^^ a^. 128 PLANE ANALYTIC GEOMETRY [VII, § 128 128. Tangent to the Hyperbola. It follows as in § 123 that the tangent to the general hyperbola (3) has the equation (7) ^-yl=i, ^ ^ a^ ¥ The slope of the hyperbola (3) is therefore tan a = a^y Notice that the equations (6), (7) of the tangents are obtained from the equations (1), (3) of the curves by replacing x*, y'^ by xXj yY, respectively (compare §§ 54, 98). It is readily shown (compare § 126) that for the hyperbola (3) the tangent meets the axis Ox at the point T that divides the distance of the foci F^Fi proportionally to the focal radii F2P, FiP, so that the tangent to the hyperbola bisects the angle between the focal radii. EXERCISES 1. Show that a right cylinder whose cross-section (i.e. section at right angles to the generators) is an ellipse of semi-axes a, b has two (oblique) circular sections of radius a ; find their inclinations to the cross-section. 2. Derive the equation of the normal to the hyperbola (3). 3. Find the polar equations of the ellipse and hyperbola, with the center as pole and the major (transverse) axis as polar axis. 4. Find the lengths of the tangent, subtangent, normal, and sub- normal in terms of the coordinates at any point of the ellipse. 6, Show that an ellipse and hyperbola with common foci are orthogonal. 6. Show that the eccentricity of a hyperbola is equal to the secant of half the angle between the asymptotes. 7. Express the cosine of the angle between the asymptotes of a hyperbola in terms of its eccentricity. 8. Show that the tangents at the vertices of a hyperbola intersect the asymptotes at points on the circle about the center through the foci. VII, § 128] ELLIPSE AND HYPERBOLA 129 9. Show that the point of contact of a tangent to a hyperbola is the midpoint between its intersections with the asymptotes. 10. Show that the area of the triangle formed by the asymptotes and any tangent to a hyperbola is constant. 11. Show that the product of the distances from the center of a hyper- bola to the intersections of any tangent with the asymptotes is constant. 12. Show that the tangent to a hyperbola at any point bisects the angle between the focal radii of the point. 13. As the sum of the focal radii of every point of an ellipse is con- stant (§ 116) and the normal bisects the angle between the focal radii (§ 125), a sound wave issuing from one focus is reflected by the ellipse to the other focus. This is the explanation of "whispering galleries." Find the semi-axes of an elliptic gallery in which sound is reflected from one focus to the other at a distance of 69 ft. in 1/10 sec. (the velocity of sound is 1090 ft, /sec). 14. Show that the distance from any point of an equilateral hyperbola to its center is a mean proportional to the focal radii of the point. 15. Show that the bisector of the angle formed by joining any point of an equilateral hyperbola to its vertices is parallel to an asymptote. 16. Show that the tangents at the extremities of any diameter (chord through the center) of an ellipse or hyperbola are parallel. 17. Let the normal at any point Pof an ellipse referred to its axes cut the coordinate axes at Q and B ; find the ratio PQ/PB. 18. Show that a tangent at any point of the circle circumscribed about an ellipse is also a tangent to the circle with center at a focus and radius equal to the focal radius of the corresponding point of the ellipse. 19. Show that the product of the y-intercept of the tangent at any point of an ellipse and the ordinate of the point of contact is constant. 20. Find the locus of the center of a circle which touches two fixed non-intersecting circles. 21. Find the locus of a point at which two sounds emitted at an interval of one second at two points 2000 ft. apart are heard simul- taneously. 130 PLANE ANALYTIC GEOMETRY [VII, § 129 129. Intersections of a Straight Line and an Ellipse. The intersections of the ellipse (1) with any straight line are found by solving the simultaneous equations y = mx 4- k. Eliminating y, we find a quadratic equation in x: (m'a} + 62)a^ ^ 2 mka'x + {Jc" - 6')a' = 0. To each of the two roots the corresponding value of y results from the equation y = mx + k. Thus, a straight line can intersect an ellipse in not more than two points. 130. Slope Form of Tangent Equations. If the roots of the quadratic equation are equal, the line has but one point in common with the ellipse and is a tangent. The condition for equal roots is m'lea^ = {m^a^ + &')(*:* - &*), whence k=± Vm^a^ + b\ The two parallel lines (8) y = mx ± VmV+^ are therefore tangents to the ellipse (1), whatever the value of m. This equation is called the slope form of the equation of a tangent to the ellipse. It can be shown in the same way that a straight line cannot intersect a hyperbola in more than two points, and that the two parallel lines y = WM5± -Vm^a^ — 6' have each but one point in common with the hyperbola (3). 131. The condition that a line be a tangent to an ellipse or hyperbola assumes a simple form also when the line is given in the general form Ax-{-By+C=0. VII, § 132] ELLIPSE AND HYPERBOLA 131 Substituting the value of y obtained from this equation in the equation (1) of the ellipse, we find for the abscissas of the points of intersection the quadratic equation : {A^a? + jB262)ic2 + 2 ACo^x + (O^ - I^h'')a^ = 0; the condition for equal roots is A^C^a^ = {AH^ + Wh^C^ - B'b''), which reduces to A^a'-^-B'b'^C'. The line is therefore a tangent whenever this condition is satisfied. When the line is given in the normal form, a; cos )8 + 2/ sin ^ = j9, the condition becomes 132. Tangents from an Exterior Point. By § 130 the line y = mx + V m'-^a'-^ + b^^ is tangent to the ellipse (1) whatever the value of m. The condition that this Une pass through any given point {xi , yi) is 2/1 = mxi + y/m^d^ + 6^ ; transposing the term mxi, and squaring, we find the following quadratic equation for m : mHx^ - 2 mxxyx + y^ = m'^a^ + b^, i.e. (xi^ - a^)m^ - 2 xiyim + yi^ -b'^ = 0. The roots of this equation are the slopes of those lines through (xi , yi) that are tangent to the ellipse (1). Thus, not more than two tangents can be drawn to an ellipse from any point. Moreover, these tangents are real and different, real and coin- cident, or imaginary, according as 132 PLANE ANALYTIC GEOMETRY [VII, § 132 This condition can also be written in the form i.e. ^\yl^i^o. a^ h"^ < Hence, to see whether real tangents can be drawn from a point (xi , yi) to the ellipse (1) we have only to substitute the coordinates of the point for X, y in the expression ^ + 2^-1- if the expression is zero, the point (xi, yi) lies on the ellipse, and only one tangent is possible ; if the expression is positive, two real tangents can be drawn, and the point is said to he outside the eUipse ; if the expres- sion is negative, no real tangents exist, and the point is said to he within the eUipse. These definitions of inside and outside agree with what we would naturally call the inside or outside of the eUipse. But the whole discus- sion applies equally to the hyperbola (3) where the distinction between inside and outside is not so obvious. 133. Sjonmetry. Since the ellipse, as well as the hyperbola, has two rectangular axes of symmetry, the axes of the curve, it has a center ^ the intersection of these axes, i.e. a point of symmetry such that every chord through this point is bisected at this point (compare § 70). Analytically this means that since the equation (1), as well as (3), is not changed by replac- ing a; by — ic, nor by replacing yhy —y, it is not changed by replacing both x and yhy — x and — y, respectively. In other words, if (a;, y) is a point of the curve, so is (— a;, — y). This fact is expressed by saying that the origin is a point of sym- metry, or center. 134. Conjugate Diameters. Any chord through the center of an ellipse or hyperbola is called a diameter of the curve. VII, § 134] ELLIPSE AND HYPERBOLA 133 Just as in the case of the circle, so for the ellipse the locus of the midpoints of any system of parallel chords is a diameter. This follows from the corresponding property of the circle because the ellipse can be regarded as the projection of a circle (§121). But this diameter is in general not perpen- dicular to the parallel chords ; it is said to be conjugate to the diameter that occurs among the parallel chords. Thus, in Fig. 77, P'Q' is conjugate to PQ (and vice versa). Fig. 77 To find the diameter conjugate to a given diameter y = mx of the ellipse (1), let y—mx -\-k be any parallel to the given diameter. If this parallel intersects the ellipse (1) at the real points (a^, 2/1) and (X2, 2/2)? the midpoint has the coordinates ^(ajj + X2), J(yi + 2/2)- The quadratic equation of § 129 gives X=-{Xi-{-X2) = ma^k m^a^ + 62 If instead of eliminating y we eliminate x, we obtain the quad- ratic equation (m2a2+ 62)2/2 _ 2 Wy + Qc" - m' a')W = 0, whence y 1/ , N ^'^ 7^ (2/1 + 2/2) = —TTTTi 2 mV+d^ Eliminating k between these results, we find the equation of the locus of the midpoints of the parallel chords of slope m : 134 PLANE ANALYTIC GEOMETRY [VII, § 134 (9) y = --\x, ^ ^ ma? If m = tan a is the slope of any diameter of the ellipse (1), the slope of the conjugate diameter is mi = tan «! = -• The diameter conjugate to this diameter of slope wij has there- fore the slope mo= — '' - m,a' \ ma^J i.e. it is the original diameter of slope m (Fig. 77). In other words, either one of the diameters of slopes m and m^ is conjugate to the other ; each bisects the chords parallel to the other. 135. Tangents Parallel to Diameters. Among the parallel lines of slope m, y = mx -{- k, there are two tangents to the ellipse, viz. (§ 130) those for which k=±VrrM^f¥, their points of contact lie on (and hence determine) the conju- gate diameter. This is obvious geometrically; it is readily verified analytically by showing that the coordinates of the intersections of the diameter of slope — b^/ma^ with the ellipse (1) satisfy the equations of the tangents of slope m, viz. y = mx ± -y/m^a^ + b^. The tangents at the ends of the diameter of slope m must of course be parallel to the diameter of slope wij. The four tan- gents at the extremities of any two conjugate diameters thus form a circumscribed parallelogram (Fig. 77). The diameter conjugate to either axis of the ellipse is the other axis ; the parallelogram in this case becomes a rectangle. VII, § 136] ELLIPSE AND HYPERBOLA 135 136. Diameters of a Hjrperbola. For the hyperbola the same formulas can be derived except that 6^ is replaced throughout by — fe^. But the geometrical interpretation is somewhat different because a line y = mx meets the hyperbola (3) in real points only when m < b/a. Fig. 78 The solution of the simultaneous equations y = mx, y^x^ — a^y"^ = a'^ft* gives : ah y- mob V6^ — m^a^ V6^ — m^a} These values are real if m < h/a and imaginary if m'^hfa (Fig. 78). In the former case it is evidently proper to call the distance PQ between the real points of intersection a diameter of the hyperbola ; its length is PQ == 2 V^^Tf = 2ab J j + ^'^ - If m>b/a, this quantity is imaginary; but it is customary to speak even in this case of a diameter, its length being defined as the real quantity By this convention the analogy between the properties of the ellipse and hyperbola is preserved. 136 PLANE ANALYTIC GEOMETRY [VII, § 137 137. Conjugate Diameters of a Hyperbola. Two diameters of the hyperbola are called conjugate if their slopes m, mi are such that mmi =— • a* One of these lines evidently meets the curve in jeal points, the other does not. If m < 6/a, the line y = mx, as well as any parallel line, meets the hyperbola (3) in two real points, and the locus of the midpoints of the chords parallel to ^ = ma; is found to be the diameter conjugate to y = mx, viz. y = 77hx= — -X. If m > 6/a, the coordinates x^^ y^ and aja, 2/2 of the intersec- tions of y = mx with the hyperbola are imaginary; but the arithmetic means ^ (xi + x^), ^(^1 + 2/2) ^^^ ^^^^) a-ud the locus of the points having these coordinates is the real line y = rriiX = X. ma} It may finally be noted that what was in § 136 defined as the length of a diameter that does not meet the hyperbola in real points is the length of the real diameter of the hyper- bola a' + 6« ' two such hyperbolas are called conjugate. VII, § 138] ELLIPSE AND HYPERBOLA 137 138. Parameter Equations. Eccentric Angle. Just as the parameter equations of the circle x"^ -{- y^ = a^ are (§ 106) ; x = a cos 9, y = a sin 0, so those of the ellipse (1) are x=:a cos $, y = b sin Oj and those of the hyperbola (3) are x= a sec 6, y = b tan 6. In each case the elimination of the parameter 6 (by squaring and then adding or subtracting) leads to the cartesian equation. The angle 0, in the case of the circle, is simply the polar angle of the point P{x, y). In the case of the ellipse, as appears from Fig. 79 (compare § 122), 6 is the polar angle not of the point P (x, y) of the ellipse, but of that point Pi of the circum- scribed circle which has the same abscissa as P, and also of that point Pg of the inscribed circle which has the same ordinate as P. This angle 6 = xOP^ is called the eccentric angle of the point P (x, y) of the ellipse. In the case of the hyperbola the eccentric angle 6 determines the point P{x, y) as follows (Fig. 80). ' Let a line through inclined at the angle to the trans- verse axis meet the circle of radius a about the center at A, and let the transverse axis meet the circle of radius b about the center at B. Let the tangent at A meet the transverse axis at A' and the tangent at B meet the line OA at B'. Then the parallels to the axes through A and B' meet at P. Fia. 79 y -^N a ^%:-^Y X ■ r b b A' Fia. 80 138 PLANE ANALYTIC GEOMETRY [VII, § 139 139. Area of Ellipse. Since any ellipse of semi-axes a, b can be regarded as the projection of a circle of radius a, inclined to the plane of the ellipse at an angle c such that cos € = 6/a, the area A of the ellipse is ^ = -n-a^ cos e = irdb. EXERCISES 1. Find the tangents to the ellipse x^ + 4 j/^ = 16, which pass through the following points : (a) (2, V3), (6) (-3, iV7), (c) (4,0), {d) (-8,0). 2. Find the tangents to the hyperbola 2 a:^ — 3 ?/2 = 18, which pass through the following points : (a) (-6, 3V2), (6) (-3,0), (c) (4, -V5), {d) (0,0). 3. Find the intersections of the line x — 2 y = 7 and the hyperbola x2 - ?/2 = 5. 4. Find the intersections of the line 3x-j-2/ — 1=0 and the ellipse x2 + 4 y2 = 65. 6. For what value of k will the line i/ = 2x + A;bea tangent to the hyperbola x2-4 2/2_4 = o? 6. For what values of m will the line y= mx + 2 be tangent to the ellipse x2-f4y2_i=o? 7. Find the conditions that the following lines are tangent to the hyperbola x2/a2 - ^2/52 _ 1 . (a) Ax -\- By + C = (i, (&) x cos /3 + «/ sin /3 = p. 8. Are the following points on, outside, or inside the ellipse x2 -f 4 y2 = 4p (a) (!,f), (?>) a, -i)» (c) (-i-l). 9. Are the following points on, outside, or inside the hyperbola 9x2-y2 = 9? (a) (f, - f ), (6) (1.35,2.15), (c) (1.3,2.6). 10. Find the difference of the eccentric angles of points at the extremi- ties of conjugate diameters of an ellipse. 11. Show that conjugate diameters of an equilateral hyperbola are equal. 12. Show that an asymptote is its own conjugate diameter. 13. Show that the segments of any line between a hyperbola and its asymptotes are equal. 14. Find the tangents to an ellipse referred to its axes which have equal intercepts. VII, § 1391 ELLIPSE AND HYPERBOLA 139 15. What is the greatest possible number of normals that can be drawn from a given point to an ellipse or hyperbola ? 16. Show that tangents drawn at the extremities of any chord of an ellipse (or hyperbola) intersect on the diameter conjugate to the chord. 17. Show that lines joining the extremities of the axes of an ellipse are parallel to conjugate diameters. 18. Show that chords drawn from any point of an ellipse to the ex- tremities of a diameter are parallel to conjugate diameters. 19. Find the product of the perpendiculars let fall to any tangent from the foci of an ellipse (or hyperbola). 20. The earth's orbit is an ellipse of eccentricity .01677 with the sun at a focus. The mean distance (major semi-axis) between the sun and earth is 93 million miles. Find the distance from the sun to the center of the orbit. 21. Find the sum of the squares of any two conjugate semi-diametere of an ellipse. Find the difference of the squares of conjugate semi-diam- eters of a hyperbola. 22. Find the area of the parallelogram circumscribed about an ellipse with sides parallel to any two conjugate diameters. 23. Find the angle between conjugate diameters of an ellipse in terms of the semi-diameters and semi-axes. 24. Express the area of a triangle inscribed in an ellipse referred to its axes in terms of the eccentric angles of the vertices. 25. The circle which is the locus of the intersection of two perpendicu- lar tangents to an ellipse or hyperbola is called the director-circle of the conic. Find its equation : (a) For the ellipse. (6) For the hyperbola. 26. Find the locus of a point such that the product of its distances from the asymptotes of a hyperbola is constant. For what value of this constant is the locus the hyperbola itself ? 27. Find the locus of the intersection of normals drawn at correspond- ing points of an ellipse and the circumscribed circle. 28. Two points J., 5 of a line I whose distance is AB = a move along two fixed perpendicular lines ; find the path of any point P oil. CHAPTER VIII CONIC SECTIONS — EQUATION OF SECOND DEGREE PART I. DEFINITION AND CLASSIFICATION 140. Conic Sections. The ellipse, hyperbola, and parabola are together called conic sections, or simply conies, because the curve in which a right circular cone is intersected by any plane (not passing through the vertex) is an ellipse or hyper- bola according as the plane cuts only one of the half-cones or both, and is a parabola when the plane is parallel to a gener- ator of the cone. This will be proved and more fully dis- cussed in §§ 148-152. 141. General Definition. The three conies can also be defined by a common property in the plane : the locus of a point for ivhich the ratio of its distances from a fixed point and from a fixed line is constant is a conic, viz. an ellipse if the constant ratio is less than one, a hyperbola if the ratio is greater than one, and a . parabola if the ratio is equal to one. We shall find that this constant ratio is equal to the eccentricity e = c/a as defined in § 124. Just as in the case of the parabola for which the — above definition agrees with that of Fio. 81 § 89, we shall call the fixed line d^ directrix, and the fixed point Fi focus (Fig. 81). 142. Polar Equation. Taking the focus F^ as pole, the perpendicular from Fi toward the directrix dj as polar axis, 140 VIII, § 143] CONIC SECTIONS 141 and putting the given distance FiD = q, we have FiP=rj PQz=q—r cos , r and being the polar coordinates of any point P of the conic. The condition to be satisfied by the point P, viz. F^P/PQ = e, i.e. F^P— e • PQ, becomes, therefore, r = e(g — rcos<^), or r= ^ 1 + e cos <^ This then is the polar equation of a conic if the focus is taken as pole and the perpendicular from the focus toward the directrix as polar axis. It is assumed that q is not zero; the ratio e may be any positive number. 143. Plotting the Conic. By means of this polar equation the conic can be plotted by points when e and q are given. Thus, for <^ = and <^ = tt, we find eq/{l + e) and eq/{l — e) as the intercepts FiA^ and jF\^2 on the polar axis ; A > ^2 are the vertices. For any negative value of (between and — tt) the radius vector has the same length as for the same positive value of . The segment LL' cut off by the conic on the per- pendicular to the polar axis drawn through the pole is called the latus rectum; its length is 2 eg. Notice that in the ellipse and hyperbola, i.e. when e ^1, the vertex Ai does not bisect the distance FiD (as it does in the parabola), but that F,A,/A,D = e. If in Fig. 81, other things being equal, the sense of the polar axis be reversed, we obtain Fig. 82. We have again FiP=r; but the distance of P from the directrix di is QP = q + r cos , so that the polar equation of the conic is now : r = , '1 . 1 — e cos 1 these equations represent respectively an ellipse and a hyper- bola as defined in §§ 114, 117. To show this we need only introduce cartesian coordi- nates and then transform to the center, i.e. to the midpoint between the intersections Ai, A2 of the curve with the polar axis. 145. Transformation to Cartesian Coordinates. The equa- tion of § 142, r = e(q — r cos <^) becomes in cartesian coordinates, with the pole Fi as origin and the polar axis as axis Ox (Fig. 81) : V?+y = e{q — x)y or, rationalized : (1 - e2)flj2 ^2e^qx-hf = e\^. The midpoint between the vertices A^^ A^ at which the curve meets the axis Ox has, by § 143, the abscissa 2 ^Vl + e \-e) l-e^' this also follows from the cartesian equation, with ?/ = 0. 146. Change of Origin to Center. To transform to paral- lel axes through this point O we have to replace x by X — e^q/(^. — e?) ; the equation in the new coordinates is there- fore and this reduces to (1 - e^^ + / = eV(l + j^) = ^, VIII, § 147] CONIC SECTIONS 143 (1 _ ^y 1 _ e2 If e < 1 this is an ellipse with semi-axes if e > 1 it is a hyperbola with semi-axes 147. Focus and Directrix. The distance c (in absolute value) from the center to the focus F^ is, as shown above, for the ellipse 2 c = — ^ = ae, for the hyperbola c = ^ ^ ^ = ae. The distance (in absolute value) of the directrix from the center is for the ellipse, since g = a(l — e^)/e = a/e — ae : e e and for the hyperbola, since q = ae — a/e : OD = c — q = ae — ae -{-- = -' e e It is clear from the symmetry of the ellipse and hyperbola that each of these curves has two foci, one on each side of the center at the distance ae from the center, and two directrices whose equations are a; = ± a/e. EXERCISES 1. Sketch the following conies : («)'- = ;r-r-| 7' (&) r = ^ .^ , ic)r~ ^ 2 + 3 cos 2 4- cos 1 — 2 cos 144 PLANE ANALYTIC GEOMETRY [VIII, § 147 2. Sketch the following conies and find their foci and directrices : (a) a;2 + 4 2/2 := 4, (b) 4 x2 + y^ = 4, (c) a:2 - 4 2/2 = 4, (d) 4x^ - y' = 4, (e) 16 a;2 -\-2by^ = 400, (/) 9 a;2 _ 16 y2 = 144^ {g) 9 ic2 - 16 2/2 + 144 = 0, (h) x^-y^ = 2. 3. Show that the following equations represent ellipses or hyperbolas and find their centers, foci, and directrices : (a) a;2 + 32/2_2x+ 62/4- 1 = 0, (6) 12x2- 42/2 - 12x - 9 = 0, (c) 5x2 + 2/2 + 20x + 15 = 0, (d) 5x2-42/2 + 82/4-16 = 0. 4. Find the length of the latus rectum of an ellipse and a hyperbola in terms of the serai-axes. 5. Show that when tangents to an ellipse or hyperbola are drawn from any point of a directrix the line joining the points of contact passes through a focus. 6. From the definition (§ 141) of an ellipse and hyperbola, show that the sum and difference respectively of the focal radii of any point of the conic is constant. 7. Find the locus of the midpoints of chords drawn from one end of : (a) the major axis of an ellipse ; (6) the minor axis. 8. Find the locus of § 141 when the fixed point lies on the fixed line. 148. The Conies as Sections of a Cone. As indicated by their name the conic sections, i.e. the parabola, ellipse, and hyperbola, can be defined as the curves in which a right circu- lar cone is cut by a jjlane (§ 140). In Figs. 83, 84, 85, Fis the vertex of the cone, ^ CVC=2 a the angle at its vertex ; OQ indicates the cutting plane, CVC that plane through the axis of the cone which is perpen- dicular to the cutting plane. The intersection OQ of these two planes is evidently an axis of symmetry for the conic. The conic is a parabola, ellipse, or hyperbola, according as OQ is parallel to the generator VC of the cone (Fig. 83), meets VC at a point (7 belonging to the same half-cone VIII, § 149] CONIC SECTIONS 145 as does (Fig. 84), or meets FC at a point (X of the other half-cone (Fig. 85). If the angle GOQ be called /8, the conic is a parabola if y8 = 2 a (Fig. 83), an ellipse \i p>2a (Fig. 84), a hyperbola if y8 < 2 « (Fig. 85). In each of the three figures CC represents the diameter 2 r of any cross-section of the cone (i.e. of any section at right angles to its axis). We take as origin, OQ as axis Fig. 83 Oa;, so that (Fig. 83) OQ = x, QP=y are the coordinates of any point P of the conic. As QP is the ordinate of the circular cross-section CPQ'P* we have in each of the three cases y^ = QP^ = CQ, • QC 149. Parabola. In the first case (Fig. 83), when ^ = 2 a so that OQ is parallel to VQ\ the expression X OQ oq ^ is constant, i.e. the same at whatever distance from the vertex we may take the cross-section CPQ'P' . For, QC" is equal to the diameter OB — 2r^ of the cross-section through 0, and CQ/OQ = CO/ VC= 2 r/r esc « = 2 sin a. Hence, denoting the constant r^ sin a by j) we have ||.QC' = 4rosin«=4p. The equation of the conic in this case, referred to its axis OQ and vertex 0, is therefore y^ = 4:px. Notice that asp = ^o sin a the focus is found as the foot of the perpendicular from the midpoint of OB on OQ. 146 PLANE ANALYTIC GEOMETRY [VIII, § 150 150. Ellipse. In the second case (Fig. 84), i.e. when yS > 2 a, if we put '00' = 2 a, it can be shown that x{2a-x) OQ'QO' is constant. For we have QI^ = CQ • QC and from the triangles CQOj QOO'j observing that -^ QaC =l3-2a: whence C'(^ _ sm /d / ]>. ^ OQ sin(i,r-a)' / '" -^ ->\ Q(7_sin(^-2a) Q(y sin(i,r + a)' Fig. 84 QP^ _sin^sin()8-2a) OQ . Q(y --2 ' cos' a an expression independent of the position of the cross-section CC Denoting this positive constant by Zc*, we find the equation y^ = fe(2a-a;), or (^~^)' which is an ellipse, with semi-axes a, ka, and center (a, 0). 151. Hyperbola. In the third case (Fig. 85), proceeding as in the second and merely observing that now QC/ = — (2 a -\- x), we find the equation y^ = k'^x (2 a -\- x) y I.e. a" (kay ' which represents a hyperbola, with semi-axes a, ka and center (—a, 0). VIII, § 152] CONIC SECTIONS 147 152. Limiting Cases. The conic is an ellipse, hyperbola, or parabola according as ^8 > 2 a, < 2 a, or = 2 a. Hence the parabola can be regarded as the limiting case of either an ellipse or a hyperbola whose center is removed to infinity. If when /8 > 2 a (Fig. 84), we let /8 approach tt, or if when /8 < 2 a (Fig. 85), we let /? approach 0, the cutting plane be- comes in the limit a tangent plan^ to the cone. It then has in common with the cone the points of the generator VC, and these only. A single straight line can thus appear as a limiting case of an ellipse or hyperbola. Finally we obtain another class of limiting cases, or cases of degeneration, of the conies if, in any one of the three cases, we let the cutting plane pass through the vertex V of the cone. In the first case, /8 = 2 a, the cutting plane is then tan- gent to the cone so that the parabola also may degenerate into a single straight line. In the second case, ^ > 2 a, if /3=^ ir, the ellipse degenerates into a single point, the vertex V of the cone. In the third case, /8 < 2 a, if /? ^ 0, the hyperbola de- generates into two intersecting lines. The terra conic section, or conic, is often used as including these limiting cases. EXERCISES 1. For what value of /3 in the preceding discussion does the conic be- come a circle ? 2. Show that the spheres inscribed in a right circular cone so as to touch the cutting plane (Figs. 83, 84, 86) touch this plane at the foci of the conic. 3. The conic sections were originally defined (by the older Greek mathematicians, in the time of Plato, about 400 b.c.) as sections of a cone by a plane at right angles to a generator of the cone ; show that the section is a parabola, ellipse, or hyperbola according as the angle 2 a at the vertex of the cone is = | tt, < ^ tt, > ^ tt. 148 PLANE ANALYTIC GEOMETRY [VIII, § 153 PART XL REDUCTION OF GENERAL EQUATION 153. Equations of Conies. We have seen in the two pre- ceding chapters that by selecting the coordinate system in a con- venient way the equation of a parabola can be obtained in the simple form that of an ellipse in the form a^^b^-^' and that of a hyperbola in the form a" b-' ' When the coordinate system is taken arbitrarily, the carte- sian equations of these curves will in general not have this simple form ; but they will always be of the second degree. To show this let us take the common definition of these curves (§ 141) as the locus of a point whose distances from a fixed point and a fixed line are in a constant ratio. With respect to any rectangular axes, let x^ , y^ be the coordinates of the fixed point, ax + 6?/ + c = the equation of the fixed line, and e the given ratio. Then by §§ 9 and 42 the equation of the locus is V(^-.0^ + (,-,0^ = e . «^±M±^, or, rationalized: (X - x,y + {y-y,f = ^^ (ax-\-by-^cy. a -j- It is readily seen that this equation is always of the second degree; i.e. that the coefficients of x^, y^, and xy cannot all three vanish. VIII, §155] EQUATION OF SECOND DEGREE 149 154. Equation of Second Degree. Conversely, every equa- tion of the second degree, i.e. every equation of the form (§47) (1) Ax^ -{-2 Hxy -\- By"" + 2 Qx + 2Fy -^ C = 0, where A, H, B are not all three zero, in general represents a conic. More precisely, the equation (1) may represent an ellipse, a hyperbola, or a parabola; it may represent two straight lines, different or coincident ; it may be satisfied by the coordinates of only a single point; and it may not be satisfied by any real point. Thus each of the equations ar^ _ 3 2/2 = 0, x?/ = evidently represents two real different lines ; the equation aj2_2a; + l=0 represents a single line, or, as it is customary to say, two coin- cident lines ; the equation a;2 + 2/' = represents a single point, while is satisfied by no real point and is sometimes said to represent an "imaginary ellipse." The term conic is often used in a broader sense (compare § 152) so as to include all these cases ; it is then equivalent to the expression "locus of an equation of the second degree.'"' It will be shown in the present chapter how to determine the locus of any equation of the form (1) with real coefficients. The method consists in selecting the axes of coordinates so as to reduce the given equation to its most simple form. 155. Translation of Axes. The transformation of the equation (1) to its most simple form is very easy in the par- ticular case when (1) contains no term in xy, i.e. when 11=0. Indeed it suffices in this case to complete the squares in x and y and transform to pO/TQ/lhl axes. 150 PLANE ANALYTIC GEOMETRY [VIII, § 155 Two cases may be distinguished: (a) 11=: 0, A=^ Oj B ^0, so that the equation has the form (2) Ax''-\-By^-{-2Gx-\-2Fy-{-C= 0. Completing the squares in x and y (§ 48), we obtain an equation of the form A{x-hy + B(y-ky = K, where ^ is a constant ; upon taking parallel axes through the point (hy k) it is seen that the locus is an ellipse, or a hyper- bola, or two straight lines, or a point, or no real locus, accord- ing to the values of A^ B, K. (h) H=0, and either jB= or ^4=0, so that the equation is (3) Aa!' + 2Gx-\-2Fy-]-C=0,oT By^ + 2Ox-{-2Fy+C=0. Completing the square in x or y, we obtain (x-hy=p(y-k), or (y -ky = q{x-h); with (h, k) as new origin we have a parabola referred to vertex and axis, or two parallel lines, real and different, coincident, or imaginary. It follows from this discussion that the absence of the term in xy indicates that, in the case of the ellipse or hyperbola, its axes, in the case of the parabola, its axis and tangent at the vertex^ are parallel to the aaxs of coordinates. EXERCISES 1. Beduce the following equations to standard forms and sketch the loci : (a) 2 1/2 - 3 a; + 8 y + 11 = 0, (6) ic^ + 4 y2 _ g -,. + 4 y + 6 = 0, (c) 6 x2 + 3 2,2 _ 4 a; ^ 2 y + 1 = 0, {d) a;2 _ 9 y2 _ 6 x + 18 y = 0, (c) 9 x2 + 9 y2 _ 36 x+6 y + 10=0, (/) 2 x2 - 4 y2 -f 4 x + 4 y - 1 = 0, {g) x2 + y2 - 2 X + 2 y + 3 = 0, ih) 3 x2 - 6 x + y + 6 = 0, (0 x2 - y2 _ 4aj - 2 y + 3 = 0, (j) 2 x2 - 5 x + 12 = 0, (A:) 2x2 -5 a; + 2 = 0, (?) y2-4y + 4 = 0. VIII, §156] EQUATION OF SECOND DEGREE 151 2. Find the equation of each of the following conies, determine the axis perpendicular to the given directrix, the vertices on this axis (by division-ratio), the lengths of the semi-axes, and make a rough sketch in each case : (a) with x — 2 = as directrix, focus at (6, 3), eccentricity | ; (&) with 3a; + 4y — 6 = 0as directrix, focus at (5, 4), eccentricity \ ; (c) with x — y — 2 = as directrix, focus at (4, 0), eccentricity |. 3. Find the axis, vertex, latus rectum, and sketch the parabola with focus at (2, — 2) and 2a: — 3?/— 5 = as directrix (see Ex. 2). 4. Prove the statement at the end of § 166. 5. Find the equation of the ellipse of major axis 5 with foci at (0, 0) and (3, 1). 156. Rotation of Axes. If the right angle xOy formed by the axes Ox, Oy be turned about the origin through an angle 6 so as to take the new position x^Oyi (Fig. 86), the w (40 relation between the old coordinates OQ = x, QP = y of any point P and the new coordinates OQ^^x^, QiP=yi of the same point P are seen from the figure to be x = Xi cos — yi sin 9, y = x^ sin -\- yi cos 6. By solving for x^, y^, or again from Fig. 86, we find Xi = X cos $ -\-ysmOj yi = — X sin 6 -{-y cos 6. If the cartesian equation of any curve referred to the axes 152 PLANE ANALYTIC GEOMETRY [VIII, § 156 Ox, Oy is given, the equation of the same curve referred to the new axes Ox^ , Oy^ is found by substituting the values (4) for X, y in the given equation. 157. Translation and Rotation. To transform from any- rectangular axes Ox, Oy (Fig. 87) to any other rectangular y 1 k h ^ i . ji Fia. 87 axes OiiCi, Oi2/i, we have to combine the translation 00^ (§ 13) with the rotation through an angle B (§ 156). This can be done by first transforming from Ox, Oy to the parallel axes Oix\ Oiy' by means of the translation (§ 13) x = x' -{- h, y = y'-\-k, and then turning the right angle x'Oiy' through the angle $ = x'OiXi, which is done by the transformation (§ 156) x' = Xi cos 6 — yi sin 0, y' == Xi sin 0-^yi cos 6. Eliminating x', y', we find x = Xi cos 6 (5) 2/i sin 0-\-hy ,y = Xi sin 6 -\- yi cos 6 -\- k. The same result would have been obtained by performing first the rotation and then the translation. It has been assumed that the right angles xOy and x^Oy^ are superposable ; if this were not the case, it would be necessary to invert ultimately one of the axes. VIII, § 157] EQUATION OF SECOND DEGREE 153 EXERCISES 1. Find the coordinates of each of the following points after the axes have been rotated about the origin through the indicated angle : (a) (3, 4), iT. (6) (0, 5),-|7r. (c) (-3, 2), e? = tan-i|. (d) (4, -3), ^tt. 2. If the origin is moved to the point (2, — 1) and the axes then rotated through 30°, what will be the new coordinates of the following points ? (a) (0,0). (6) (2,3). (c) (6,-1). 3. Find the new equation of the parabola y^ = 4 ax after the axes have been rotated through : (a) ^tt , (&) Jtt , (c) ir . 4. Show that the equation x"^ + y'^ = «^ is not changed by any rotation of the axes about the origin. Why is this true ? 5. Find the center of the circle (x— a)^ + y'^ =a^ after the axes have been turned about the origin through the angle d. What is the new equation ? 6. For each of the following loci rotate the axes about the origin through the indicated angle and find the new equation : (a) x2-y2 + 2 = o, Itt. (b) x^-y^ = aMTr. (c) y = mx + 6, ^ = tan-i m. (d) 12a;2 - 7xy - 12i/2 = 0, ^ = tan-i|. (^) -2 + !^=^'^'^- ^^^ x2_2/2 = 0,i,r. a2 0^ 7. Through what angle must the axes be turned about the origin so that the circle a;2 + i/2_3x + 4?/ — 5 = will not contain a linear term in x? 8.- Suppose the right angle XiOyi (Fig. 89) turns about the origin at a uniform rate making one complete revolution in two seconds. The coordinates of a point with respect to the moving axes being (2, 1), what are its coordinates with respect to the fixed axes xOy at the end of ; (a) i sec. ? (&) I sec. ? (c) 1 sec. ? (d) 1| sec. ? 9. In Fig. 89, draw the line OP, and denote Z QOP by 0. If G and F are not both zero, the equation (15) can be inter- preted as meaning that the square of the distance of the point {x, y) from the line (16) ax + by = is proportional to the distance of (x, y) from the line (17) 2Gx + 2Fy-^C=0. Hence if these lines (16), (17) happen to be at right angles, the 160 PLANE ANALYTIC GEOMETRY [VIII, § 164 locus of (15) is a parabola, having the line (16) as axis and the line (17) as tangent at the vertex. But even when the lines (16) and (17) are not at right angles the equation (15) can be shown to represent a parabola. For if we add a constant k within the parenthesis and compensate the right-hand member by adding the terms 2 akx + 2 bky + k^, the locus of (15) is not changed ; and in the resulting equation (18) (ax + by + ky = 2{ak - G)x + 2{bk -F)y + k^-G we can determine k so as to make the two lines (19) ax-\-by-^k = 0, (20) 2(ak - G)x + 2{bk -F)y + k^-C=0 perpendicular. The condition for perpendicularity is a{ak-G) + b{bk-F)=0, whence (21) k = ^^±^. With this value of k, then, the lines (19), (20) are at right angles ; and if (19) is taken as new axis Ox and (20) as new axis Oy, the equation (18) reduces to the simple form y^ = px. The constant p, i.e. the latus rectum of the parabola, is found by writing (18) in the form / ax -{-by-\- fe y 2 V(afc - Gy + (bk - Fy 2(ak - G)x + 2(bk-F)y-^k^- C , a^+b^ 2V(ak-Gy-\-{bk-Fy hence ^ = ^^^ V(aA: -GO* + (bk - Fy. Substituting for k its value (21) we can reduce it to ^_ 2(aF-bCf) (a2 + 62)i VIII, § 164] EQUATION OF SECOND DEGREE 161 EXERCISES 1. Find the equation of each of the following loci after transforming to parallel axes through the center : (a) Sx'^-ixy-y^-Sx-iy+7 = 0. (6) 5 x2 + 6 x?/ + ?/^ + 6 X - 4 y — 5 = 0. (c) 2 x2 + xy - 6 y2 _ 7 a; — 7 ?/ + 5 = 0. (d) x2 - 2 xy - 2/2 + 4 X - 2 ?/ - 8 = 0. 2. Find that diameter of the conic 3x^ — 2xy—4:y^-{-6x—^y ■{■2=0 (a) which passes through the origin, (&) which is parallel to each co- ordinate axis. 3. For what values of k do the following equations represent straight lines ? Find their intersections. (a) 2x^-xy-Sy^-6x + l9y -\-k = 0. (&) kx^ + 2 xy + y"^ - X - y - 6 = 0. (c) 3 x2 - 4 xi/ + ^•^/2 + 8 2/ - 3 = 0. (d) X* + 2 ?/2 + 6 X - 4 y + A; = 0. 4. Show that the equations of conjugate hyperbolas x^/a^—y^/b^=±l and their asymptotes x^/a'^—y^/b^ = 0, even after a translation and rota- tion of the axes, will differ only in the constant terms and that the con- stant term of the asymptotes is the arithmetic mean between the constant terms of the conjugate hyperbolas. 6. Find the asymptotes and the hyperbola conjugate to 2 x2 - xy - 15 2/2 + X + 19 y + 16 = 0. 6. Find the hyperbola through the point (—2, 1) which has the lines 2x — y+l=0, 3x + 2?/ — 6 = as asymptotes. Find the conjugate hyperbola. 7. Show that the hyperbola xy = a^ is referred to its asymptotes as coordinate axes. Find the semi-axes and sketch the curve. Find and sketch the conjugate hyperbola. 8. The volume of a gas under constant temperature varies inversely as the pressure (Boyle's law), i.e. vp = c. Sketch the curve whose ordi- nates represent the pressure as a function of the volume for different values of c ; e.g. take c = 1, 2, 3. 9. Sketch the hyperbola (x — a)(y — b) = c^ and its asymptotes. In- terpret the constants a, 6, c geometrically. 162 PLANE ANALYTIC GEOMETRY [VIII, § 164 10. Sketch the hyperbola xy-\-3y — 6 = and its asymptotes. 11. Find the center and semi-axes of the following conies, write their equations in the most simple form, and sketch the curves : (a) 6 x^ - 6 xy + 6 y^ + 12V2 z - W2y + 8 = 0. (6) a;2 - 6 V3 xy - 6 y2 _ 16 = 0. (c) x^ + xy + y^ _ 3 y + 6 = 0. (d) 13 x2 - 6V3xy + 7 y^ _ 64 = 0. (e) 2 x2 - 4 xy + y2 _|_ 2 x - 4 2/ - f = 0. (/) 3x2 + 2xy + y2 + 6x + 4y + ^ = 0. IS. Sketch the following parabolas : (a) x2 - 2VSxy -\-Sy^- eVSx - 6 y = 0. (6) x2 - 6 xy + 9 y2 _ 3 X + 4y - 1 = 0. IS. Show that the following combinations of the coeflBcients of the general equation of the second degree are invariants {i.e. remain un- changed) under any transformation from rectangular to rectangular axes : (a) A-hB. (6) AB - H\ (c) {A -5)2 + 4 m. 14. Show that x2 + y 2 = a^ represents a parabola. Sketch the locus. 15. Find the parabola with x + y = as directrix and (^ a, J a) as focus. 16. Let five points A^ J5, C, D^ E be taken at equal intervals on a line. Show that the locus of a point P such that AP • EP = BP • DP is an equilateral hyperbola. (Take C as origin.) 17. The variable triaYigle AQB is isosceles with a fixed base AB. Show that the locus of the intersection of the line AQ with the perpen- dicular to QB through B is an equilateral hyperbola. 18. Let ^ be a fixed point and let Q describe a fixed line. Find the locus of the intersection of a line through Q perpendicular to the fixed line and a line through A perpendicular to A Q. 19. Find the locus of the intersection of lines drawn from the extrem- ities of a fixed diameter of a circle to the ends of the perpendicular chords. 20. Show by (14'), §163, that if the equation of the second degree represents an ellipse, parabola, hyperbola, we have, respectively, AB-H^ >0, = 0, < p. CHAPTER IX HIGHER PLANE CURVES PART I. ALGEBRAIC CURVES 165. Cubics. It has been shown (§ 30) that every equation of the first degree, Oo + a^x + 6^2/ = 0, represents a straight line; and (§ 154) that every equation of the second degree, ao + a^x + h^y + a^^ + ^^y + C22/2 = 0, either represents a conic or is not satisfied by any real points. The locus represented by an equation of the third degree, ao + a^x -f h^ 4- ttgoj^ + h^y + Cay* -I- a^ + h^^y + c^xy'^ + d^^= 0, I.e. the aggregate of all real points whose coordinates x, y satisfy this equation, is called a cubic curve. Similarly, the locus of all points that satisfy any equation of the fourth degree is called a quartic ctirve; and the terms quintiCj sextic, etc., are applied to curves whose equations are of the Jifthf sixth, etc., degrees. Even the cubics present a large variety of shapes; still more so is this true of higher curves. We shall not discuss such curves in detail, but we shall study some of their properties. 163 164 PLANE ANALYTIC GEOMETRY [IX, § 166 166. Algebraic Curves. The general form of an algebraic equation of the nth degree in x and y is «o 4- a^x + h^ (1) +a^-\- b^y -f C22/2 ■^a^-\-b^y-\-c^'^-\-d^ + a^a;" + b^x^'-^y -\ \-Kxy''~^+ Z^" = 0. The coefficients are supposed to be any real numbers, those in the last line being not all zero. The number of terms is not more than 1 -f 2 -f 3 + ... -^(n + 1) = i(n + l)(n + 2). If the cartesian equation of a curve can be reduced to this form by rationalizing and clearing of fractions, the curve is called an algebraic curve of degree n. An algebraic curve of degree n can be intersected by a straight line, Ax + By-{-C=0, in not more than n points. For, the substitution in (1) of the value of y (or of x) derived from the linear equation gives an equation in x (or in y) of a degree not greater than n ; this equation can therefore have not more than n roots, and these roots are the abscissas (or ordinates) of the points of intersec- tion. We have already studied the curves that represent the poly- nomial function y=aQ-{-a^x-\-a^-\ \- a^iC* ; such a curve is an algebraic curve, but it is readily seen by comparison with the preceding equation that this equation is of a very special type, since it contains no term of higher de- gree than one in y. Such a curve is often called a parabolic curve of the nth degree. IX, § 169] ALGEBRAIC CURVES 165 167. Transformation to Polar Coordinates. The cartesian equation (1) is readily transformed to polar coordinates by sub- stituting x = r cos <^, 2/ = ^ sin <^ ; it then assumes the form : + (ai cos <^ + &i sin ^)r (2) + (as cos^ <^ + 62 cos sin^ <}!> + c^g sin' c^)/^ 4- (a„ cos" <^ + 6„ cos''"^ <^ sin <^ -f- +A;„cos sin"-^ <^+Z„sin'»<^)r'" = 0. If any particular value be assigned to the polar angle <^, this becomes an equation in r of a degree not greater than n. Its roots ri, • represent the in- tercepts OPi, OP2, ••• (Fig. 89) made by the curve (2) on the line y = tan - x. Some of these roots may of course be imaginary, and there may be equal roots. Fig. 89 168. Curve through the Origin. The equation in r has at least one of its roots equal to zero if, and only if, the constant term Uq is zero. Thus, the necessary and sufficient condition that the origin be a point of the curve is ao = 0. This is of course also apparent from the equation (1) which is satisfied by aj = 0, 2/ = if , and only if, Oq = 0. 169. Tangent Line at Origin. The equation (2) has at least two of its roots equal to zero if ao=0 and a^ cos <^ + 61 sin = 0. If tti and 61 are not both zero, the latter condition 166 PLANE ANALYTIC GEOMETRY [IX, § 169 can be satiefied by selecting the angle <^ properly, viz. so that tan<^ = -^. The line through the origin inclined at this angle <^ to the polar axis is the tangent to the curve at the origin (Fig. 90) . Its cartesian equation is y = tan <^ • a; = — (ai/bi)x, i.e. (3) a,x + b,y = 0. Thus, if ao = while Oj , bi are not both zero, the curve has at the origin a single tangent ; the origin is therefore called a simple, or ordinary, point of the curve. In other words, if the lowest terms in the equation (1) of- an algebraic curve are of the first degree, the origin is a simple point of the curve, and the equor tion of the tangent at the origin is ob- tained by equating to zero the terms of the first degree. 170. Double Point. The condition ajcos <^+6i sin <^ = necessary for two zero roots is also satisfied if aj = and 6i = 0; indeed, it is then satisfied whatever the value of . Hence, if ao = 0, Oj = 0, 6i = 0, the equation (2) has at least two zero roots for any value of <\>. If in this case the terms of the second degree in (1) do not all vanish, the curve is said to have a double point at the origin. Thus, the origin is a double point if, and only ify the lowest terms in the equation (1) are of the second degree. 171. Tangents at a Double Point. The equation (2) will have at least three of its roots equal to zero if we have a© = 0, Oi = 0, 6i = and Oj cos* <^ + 62 cos sin <^ -h Cj sin' <^ = 0. IX, § 172] ALGEBRAIC CURVES 167 If tta, 62J C2 are not all zero, we can find two angles satisfying this equation which may be real and different, or real and equal, or imaginary. The lines drawn at these angles (if real) through the origin are the tangents at the double point. Multiplying the last equation by r^ and reintroducing carte- sian coordinates we obtain for these tangents the equation w a^"^ + h^y + C22/^ = 0. Fig. 91 Thus, if the loivest terms in the equation (1) are of the second degree^ the origin is a double point, and these terms of the second degree equated to zero represent the tangents at the origin. 172. Types of Double Point, (a) If the two lines (4) are real and different, the double point is called a node or crunode; the curve then has two branches passing through the origin, each with a different tangent (Fig. 91). J- (6) If the lines (4) are coincident, i.e. if a-fi? + bcfcy -f- c^y^ is a complete square, the double point is called a cusp, or spinode; the curve then has ordinarily two real branches tangent to one and the same line at the origin (Fig. 92 represents the most simple case). (c) If the lines (4) are imaginary, the double point is called an isolated point, or an acnode; in this case, while the coordi- nates 0, of the origin satisfy the equation of the curve, there exists about the origin a region containing no other point of the curve, so that no tangents can be drawn through the origin (Fig. 93). Fia. 92 Fig. 93 168 PLANE ANALYTIC GEOMETRY [IX, § 172 It should be observed that, for curves of a degree above the third, the origin in case (b) may be an isolated point ; this will be revealed by investigating the higher terms (viz. those above the second degree). 173. Multiple Points. It is readily seen how the reasoning of the last articles can be continued although the investigation of higher multiple points would require further discussion. The result is this : If in the equation of an algebraic curve, when rationalized and cleared of fractions, the lowest terms are of degree k, the origiri is a k-tuple point of the curve, and the tan- gents at this point are given by the terms of degree k, equated to zero. To investigate whether any given point (Xi , 2/j) of an alge- braic curve is simple or multiple it is only necessary to trans- fer the origin to the point, by replacing xhy x-\-Xi and y by y -{■ yi, and then to apply this rule. EXERCISES 1. Determine the nature of the origin and sketch the curves : (a) y = x'^-2x. (6) x'^ = iy-y^. (c) (x + a){y, + a) = a'^. (d) y2 = a;2(4 - x). (c) y^ = 3i^. (/) x^ + ^2 = ^^j. ig)y^ = x^ + 3fi. (A) x8 - 3 aa;y + 1/8 = 0. (i) x*-y*-^Qxy^ = 0. 8. Determine the nature of the origin and sketch the curve (y—x^y=x**, for: (a) n = l. (6) n = 2. (c) n = 3. (d) w = 4. 3. Locate the multiple points, determine their nature, and sketch the curves : (a) y^ = x{x + 3)2. (6) (y - 3)2 = xK (c) (y + iy = (x- 3)8. (d)y8=(x + l)(x-l)2. 4. Sketch the curve y^ =:{x — a')(x — b)(x — c) and discuss the multi- ple points when : Ca) 0 -\- b. 170 PLANE ANALYTIC GEOMETRY [IX, § 175 If 6 = 2 a, the curve is called the cardioid; the equation then becomes r = 4 a cos^ ^ . Sketch the limaQons for 6 = 3a, 2a, a; transform to car- tesian coordinates and determine the character of the origin. 176. Cissoid. OC/ = a being a diameter of a circle, let any radius vector drawn from meet the circle and its tangent at O' at the points Q, D, respectively; if on this radius vector we lay- off OR = QD, the locus of R is called the cissoid of Diodes. With as pole and OCy as polar axis, we have OD = a/cos , OQ = a cos <^ ; the equation is therefore /I A sin2 r = af cos )= a \cos J cos or in cartesian coordinates a? a—x Fio. »4 If instead of taking the difference of the radii vectores of the circle and its tangent we take their sum, we obtain the so-called companion of the cissoid, r = a(cos <^ + sec <^), .2a-ar I.e. r x — a Sketch this curve. 177. Versiera. With the data of § 176, let us draw through Q a parallel to the tangent, through D a parallel to the diameter ; the locus of the point of intersection P of these parallels is called the versiera (wrongly called the " witch of Agnesi "). IX, § 178] SPECIAL CURVES 171 We have evidently with as origin and 00^ as axis Ox: x = a cos^ , y = a tan , whence eliminating <^ : x= 2/2 + a2 If we replace the tangent at 0' by any perpendicular to 00' (Fig. 95), at the distance b from 0, we obtain the curve x = a cos^ (f>, y = h tan ^y _ a¥ which reduces to the versiera for h = a. Sketch the versiera, and the last curve for 6 = i a. 178. Cassinian Ovals. Lemniscate. Two fixed points Fj, F2 being given it is known that the locus of a point P is : Fig. 96 (a) a circle if i^iP/i^^^P = const. (Ex. 7, p. 54); (6) an ellipse if F^P + F^P^ const. (§ 114) ; (c) a hyperbola if PiP- ^2^= const. (§ 119). The locus is called a Cassinian oval if FiP > F2P = const. If 172 PLANE ANALYTIC GEOMETRY [IX, § 178 we put F1F2 = 2 a, the equation, referred to the midpoint between Fi and F2 as origin and OF2 as axis Oxj is [(x + ay + /] l{x - ay + y2] = k^. In the particular case when k — a^ the curve passes through the origin and is called a lemniscate. The equation then re- duces to the form ^x^ + y^y = 2a%x'-y% which becomes in polar coordinates r^ = 2 a^ cos 2 . Trace the lemniscate from the last equation. (Ito/ Cycloid. The common cycloid is the path described by any point Pofa circle rolling over a straight line (Fig. 97). If A be the point of contact of the rolling circle in any posi- tion, the point of the given line that coincided with the point P of the circle when P was point of contact, it is clear that the length OA must equal the arc AP=a$, where a is the radius of the circle, and 6= "^ACP, the angle through which the circle has turned since P was at 0. The figure then shows that, with as origin and OA as axis Ox : X = OQ = aO — a sin 6, y = a — a cos 6. These are the parameter equations of the cycloid. The curve has an infinite number of equal arches, each with an axis of sym- metry (in Fig. 97, the line x = wa) and with a cusp at each end. Write down the cartesian equation. IX, § 181] SPECIAL CURVES 173 180. Trochoid. The path described by any point P rigidly connected with the rolling circle is called a trochoid. If the Fig. 98. —The Trochoids distance of P from the center C of the circle is 6, the equations of the trochoid are x= aO— b sin 6, y = a — b cos 0. Draw the trochoid for b = ^a and for 6 = | a. 181. Epicycloid. The path described by any point P of a circle rolling on the outside of a fixed circle is called an epicy- cloid (Fig. 99). Let be the center, b the radius, of the fixed circle, Cthe center, a the radius, of the rolling circle; and let Aq be that point of the fixed circle at which the describing point P is the point of contact. Put A^OA = , ACP = 6. As the arcs AAq and AP are equal, we have 6(^ = a6. ^^^- ^ With as origin and OAq as axis of x we have X = (a -\- b) cos -i- a sin [^ — (i tt — <^)], y := {a -\- b) sin (f> — a cos [^ — (|- tt — <^)], I.e. X = (a -\- b) cos (f> — a cos 9, y=: (a -\- b) sin - 174 PLANE ANALYTIC GEOMETRY [IX, § 182 182. H3rpocycloid. If the circle rolls on the inside of the fixed circle, the path of any point of the rolling circle is called a hypocycloid. The equations are obtained in the same way ; they differ from those of the epicycloid (§ 181) merely in hav- ing a replaced by — a. Write down these equations. Show that : (a) for b = 2a the hypocycloid reduces to a straight line, and illustrate this graphically ; (b) for 6 = 4 a the curve, called the four-cusped hypocycloid, has the equations x = Sa cos -\-a cos 3 <^ = a cos' <^, y=Sa sin — asmS = a sin' , whence x^ -\-y^ = a^, EXERCISES 1. Sketch the following curves : (a) Spiral of Archimedes r = ai>; (6) Hyperbolic spiral r = a ; (c) Lituus r'^tp = a^. 2. Sketch the following curves : (a) r = a sin ; (6) r = a cos ; (c) r = a8in2 ; (d) r = acos2; (e) r= acos30 ; (/) r = asin30; (g) r = a cos 4 ; {h) r = a sin 4 . 3. Sketch with respect to the same axes the Cassinian ovals (§ 178) for a = 1 and k = 2, 1.5, 1.1, 1, .75, .5, 0. 4. Let two perpendicular lines AB and CD intersect at O. Through a fixed point Q of AB draw any line intersecting CD at R. On this line lay off in both directions from B segments BP of length OB. The locus of P is called the strophoid. Find the equation and sketch the curve. 6. Show that the lemniscate (§ 178) is the inverse curve of an equi- lateral hyperbola with respect to a circle about its center. 6. Show that the strophoid (Ex. 4) is the curve inverse to an equilat- eral hyi)erbola with respect to a circle about a vertex with radius equal to the transverse axis. 7. Show that the cissoid (§ 176) is the curve inverse to a parabola with respect to a circle about its vertex. 8. Find the curve inverse to the cardioid (§175) with respect to a circle about the origin. IX, § 183] SPECIAL CURVES 175 9. Transform the equation a{x^ 4- y'^) = y? to polar coordinates, in- dicate a geometrical construction, and draw the curve. 10. A tangent to a circle of radius 2 a about the origin intersects the axes at T and T. Find and sketch the locus of the midpoint P of TT'. 11. From any point ^ of the line x — a draw a line parallel to the axis Ox intersecting the axis Oy at Q. Find and sketch the locus of the foot of the perpendicular from (7 on 0§. 12. The center of a circle of radius a moves along the axis Ox. Find and sketch the locas of the intersections of this circle with lines joining the origin to its highest point. 13. The center of a circle of radius a moves along the axis Ox. Find and sketch the locus of its points of contact with the lines through the origin. 183. The Sine Curve. The simple sine curvCj y=8m x, is best constructed by means of an auxiliary circle of radius one. In Fig. 100, OQ is made equal to the length of the arc OA = X ; the ordinate at Q is then equal to the ordinate BA of the circle. y Fig. 100 Construct one whole period of the sine curve, i.e. the portion corresponding to the whole circumference of the auxiliary circle ; the width 2 tt of this portion is called the period of the function sin x. The simple cosine curve, y = GOSx, is the same as the sine curve except that the origin is taken at the point (Jir, 0). The simple tangent curve, y = tan x, is derived like the sine curve from a unit circle. Its period is tt. 176 PLANE ANALYTIC GEOMETRY [IX, § 184 184. The Inverse Trigonometric Curves. The equation y = sin X can also be written in the form X = sin~^ y, ox x — arc sin y. The curve represented by this equation is of course the same as that represented by the equation y = sin x. But if X and y be interchanged, the resulting equation x = sin y, OT y = sin"* ^, y = arc sin x, represents the curve obtained from the simple sine curve by reflection in the line y=x(^ 70). Notice that the trigonometric functions sina;, cos x, tan x, etc., are one-valued, i.e. to every value of x belongs only one value of the function, while the inverse trigonometric functions sin~* x, COS"* a;, tan"* a;, etc., are many-valued; indeed, to every value of X, at least in a certain interval, belongs an infinite number of values of the function. 185. Transcendental Curves. The trigonometric and in- verse trigonometric curves, as well as, in general, the cycloids and trochoids, are transcen- dental curves, so called because the relation between the carte- sian coordinates x, y cannot be expressed in finite form (i.e. without using infinite series) by means of the algebraic opera- tions of addition, subtraction, multiplication, division, and raising to a power with a con- Fio. lOX stant exponent. IX, §186] SPECIAL CURVES 177 186. Logarithmic and Exponential Curves. Another very important transcendental curve is the exponential curve y = a% and its inverse, the logarithmic curve y = loga X, where a is any positive constant (Fig. 101). A full discussion of these curves can only be given in the calculus. EXERCISES 1. From a table of trigonometric functions, plot the curve y = sinoj. 2. Plot the curve y = sinx geometrically, as in § 183. 3. Plot the curve y = cos x (a) from a table ; (6) by a geometric con- struction similar to that of § 183. 4. Plot the curve y = tanaj from a table. 5. Plot each of the curves (a) y = sm2 x. (6) y = 2 cos Sx. (c) y = S tan (x/2). (df) y = sec x. (e) y = cot 2 sc. (f)y = 2 tan 4 x. 6. Plot each of the curves (a) y = sin-i X. (6) y = cos~i x. (c) y = tan-i x, 7. By adding the ordinates of the two curves y = s'mx and y = cos x, construct the graph of y = sin x + cos x. 8. Draw each of the curves (a) y = smx + 2 cos x. (c) y = sec x + tan x. (6) ?/ = 2 sin a; + cos(x/2). (d) 2/ = sin x -f 2 sin 2 a; + 3 sin 3x. 9. The equation aj = sin t, where t means the time and x means the distance of a body from its central position, represents a Simple Harmonic Motion. From the graph, describe the nature of the motion. 10. From a table of logarithms of numbers, draw the curve y—\ogiox. 11. By multiplying the ordinates of the curve of Ex. 10 by 3, construct the curve y = S logio x. 12. From the figure of Ex. 10, construct the curve y = 10* by reflec- tion of the curve of Ex. 10 in the line y = x. 13. Draw the curve y = ^ logio^: by the process of Ex. 11. Show that it represents the equation y = logiooaJ, since y = logioo X = logioo 10 X logio x = l logio x. N 178 PLANE ANALYTIC GEOMETRY [IX, § 187 PART III. EMPIRICAL EQUATIONS 187. Empirical Formulas. In scientific studies, the rela- tions between quantities are usually not known in advance, but are to be found, if possible, from pairs of numerical values of the quantities discovered by experiment. Simple cases of this kind have already been given in §§ 15, 29. In particular, the values of a and b in formulas of the type y = a-\-hx were found from two pairs of values of x and y. Compare also § 34. Likewise, if two quantities y and x are known to be connected by a relation of the form ?/ = a -f 6a; + cx^, the values of a, 6, c can be found from any three pairs of values of x and y. For, if any pair of values of x and y are substituted for x and y in this equation, we obtain a linear equation for a, 6, and c. Three such equations usually determine a, 6, and c. In general the coefficients a, h, c, •••, Z in an equation of the ^^® y:=a-\-hx-\-cy?+ ... -\-lxr can be found from any n + 1 pairs of values of x and y. 188. Approximate Nature of Results. Since the measure- ments made in any experiment are liable to at least small errors, it is not to be expected that the calculated values of such coefficients as a, 6, c, .-. of § 187 will be absolutely accu- rate, nor that the points that represent the pairs of values of X and y will all lie absolutely on the curve represented by the final formula. To increase the accuracy, a large number of pairs of values of X and y are usually measured experimentally, and various pairs are used to determine such constants as a,b,Cf .•• of § 187. The average of all the computed values of any one such con- stant is often taken as a fair approximation to its true value. IX, § 189] EMPIRICAL EQUATIONS 179 189. Illustrative Examples. Example 1. A wire under tension is found by experiment to stretch an amount I, in thousandths of an inch, under a tension T, in pounds, as follows : — T in pounds . " 10 15 20 25 30 I in thousandths of an inch . 8 12.5 15.5 20 23 Find a relation of the form I = kT (Hookers Law) which approx- imately represents these results. First plot the given data on squared paper, as in the adjoining figure. 30 25 20 15 10 ZZIZZZy^LZZZZZZZZIZZZZZ 10 15 20 25 30 35 Fig. 102 Substituting I = 8, T = 10 in I = kT, we find k = .8. From I = 12.5, T = 15, we find k = .833. Likewise, the other pairs of values of I and T give, respectively, k = .775, k = .S, k= .767. The average of all these values of A; is A; = .795 ; hence we may write, approximately, I = .795 T. 180 PLANE ANALYTIC GEOMETRY [IX, § 189 This equation is represented by the line in Fig. 102 ; this line does not pass through even one of the given points, but it is a fair compromise be- tween all of them, in view of the fact that each of them is itself probably slightly inaccurate. Example 2. In an experiment with a Weston Differential Pulley Block, the effort E^ in pounds, required to raise a load IF, in pounds, was found to be as follows : w 10 20 30 40 50 60 70 80 90 100 E 3i 4| 6i n 9 lOi ]2i 13| 16 16i Find a relation of the form E = aW -\- h that approximately agrees with these data. [Gibson] These values may be plotted in the usual manner on squared paper. They will be found to lie very 20 10 nearly on a straight line. If E is plotted vertically, 6 is the in- tercept on the vertical axis, and a is the slope of the line ; both can be measured directly in the figure. To determine a and h more exactly, we may take various points that lie nearly on the line. Thus {E = Q\, 11'= 30) and {E = 16J, W = 100) lie nearly on a line that passes close to all the points. Substituting in the equation E m JW 20 40 Fig. GO 103 80 100. 6 J = 30 a + 6, 16J aW-hhwe obtain 100 a + 6, whence a = 0.146, 6 = 1. !6. Hence we may take ^=0.146 TF+ 1.86, approximately. Other pairs of values of E and W may be used in like manner to find values for a and b, and all the values of each quantity may be averaged. 50 60 70 80 90 205° 226° 250° ' ' 276° 304* IX, § 189] EMPIRICAL EQUATIONS 181 Example 3. If 6 denotes the melting point (Centigrade) of an alloy of lead and zinc containing x per cent of lead, it is found that X = % lead 40 e = melting point .... 186° Find a relation of the form 6 = a -{- bx -\- cx^ that approximately expresses these facts. [Saxelby] Taking any three pairs of values, say (40, 186), (70, 250), (90, 304), and substituting in d = a + bx + cx^ we find 186 = a + 40 & 4- 1600 c, 260 = a + 70 b + 4900 c, 304 = a + 90 & + 8100 c, whence a = 132, b = .92, c = .0011, approximately ; whence ^ = 132 + .92 X + .0011 x^. Other sets of three pairs of values of x and y may be used in a similar manner to determine a, &, c ; and the resulting values averaged, as above. EXERCISES 1. In experiments on an iron rod, the amount of elongation I (in thou- sandths of an inch) and the stretching force p (in thousands of pounds) were found to be {p = 10, I =8 ), (p = 20, I = 15), (p = 40,1 = 31). Find a formula of the type I = Jc-p which approximately expresses these data. Ans. k = .715. 2. The values 1 in. =2,5 cm. and 1 ft. = 30.5 cm. are frequently quoted, but they do not agree precisely. The number of centimeters, c, in i inches is surely given by a formula of the type c = ki. Find k ap- proximately from the preceding data. 3. The readings of a standard gas-meter S and those of a meter T being tested on the same pipe-line were found to be (aS^=3000, T'=0), (/S'=3510, T = 500), {S = 4022, T = 1000). Find a formula of the type T = aS+b which approximately represents these data. 4. An alloy of tin and lead containing x per cent of lead melts at the temperature 9 (Fahrenheit) given by the values (a! = 25%, ^ = 482°), (x = 50 %, d = 370°), (x = 75 %, 6 = 356°) . Determine a formula of the type d = a + bx + cx'^ which approximately represents these values. 182 PLANE ANALYTIC GEOMETRY [IX, § 189 5. The temperatures (Centigrade) at a depth d (feet) below the sur- face of the earth in a mine were found to be d = 100, d = 15.7° ; d = 200, ^=16.5 ; J =300, ^=17.4. Find a relation of the form e=a-\-bd between d and d. 6. Determine a line that passes reasonably near each of the three points (2, 4), (C, 7), (10, 9). Determine a quadratic expression y=a-\-bx-\-cx^ that represents a parabola through the same three points. 7. Determine a parabola whose equation is of the form y=a+bx+cx^ that passes through each of the points (0, 2.5;, (1.5, 1.5), and (3.0, 2.8). Are the values of rt, &, c changed materially if the point (2.0, 1.7) is substituted for the point (1.5, 1,5) ? 8. If the curve y = sin a; is drawn with one unit space on the x-axis representing 60^, the points (0, 0), (^, J), (1|, 1) lie on the curve. Find a parabola of the form y = a+bx-\-cx^ through these three points, and draw the two curves on the same sheet of paper to compare them. 190. Substitutions. It is particularly easy to test whether points that are given by an experiment really lie on a straight line ; that is, whether the quantities measured satisfy an equa- tion of the form y = a-{-bx. This is done by means of a trans- parent ruler or a stretched rubber band. For this reason, if it is suspected that two quantities x and y satisfy an equation of the form it is advantageous to substitute a new letter, say u, for a? : u = x^, y = a-{-hu and then plot the values of y and u. If the new figure does agree reasonably well with some straight line, it is easy to find a and &, as in § 189. Likewise, if it is suspected that two quantities x and y are connected by a relation of the form y = a-\-h--OTxy = ax-\-hf X it is advantageous to make the substitution u = 1/x. IX, § 191] EMPIRICAL EQUATIONS 183 Other substitutions of the same general nature are often useful. In any case, the given values of x and y should be plotted first unchanged, in order to see what substitution might be useful. 191. Illustrative Example. If a body slides down an inclined plane, the distance s that it moves is connected with the time t after it starts by an equation of the form s = kt'^. Find a value of k that agrees reasonably with the following data : s, in feet 2.6 10.1 23.0 40.8 63.7 t, in seconds 1 2 3 4 5 In this case, it is not necessary to plot the values of s and t themselves, because the nature of the equation, s = kt^, is known from physics. Hence we make the substitution t^ = u, and write down the supple- mentary table : 8, in feet 2.6 10.1 23.0 40.8 63.7 w (or t2) 1 4 9 16 25 These values will be found to give points very nearly on a straight line whose equation is of the form s — ku. To find A;, we divide each value of s by the corresponding value of u ; this gives several values of k : k 2.6 2.525 2.556 2.55 2.548 The average of these values of k is approximately 2.556 ; hence we may write s = 2.556 M, or s = 2.556 t^. EXERCISES 1. Find a formula of the type u = kv^ that represents approximately the following values : u 3.9 15.1 34.5 61.2 95.5 137.7 187.4 V 1 2 3 4 5 6 7 184 PLANE ANALYTIC GEOMETRY [IX, § 191 2. A body starts from rest and moves s feet in t seconds according to the following measured values : s, in feet 3.1 13.0 30.6 50.1 79.5 116.4 t, in seconds 5 1 1.5 2 2.5 3 Find approximately the relation between s and t. 3. The pressure p, measured in centimeters of mercury, and the volume V, measured in cubic centimeters, of a gas kept at constant temperature, were found to be : 191 V 146 155 165 178 p 117.2 109.4 102.4 95.0 Substitute u for 1/v, compute the values of m, and determine a relation of the form p = ku; that is, p = k/v. 4. Determine a relation of the form y = a + bx^ that approximately represents the values : X 1 2 3 4 5 6 7 y 14.1 25.2 44.7 71.4 105.6 147.9 197.7 192. Logarithmic Plotting. In case the quantities y and x are connected by a relation of the form it is advantageous to take logarithms (to the base 10) on both sides : log y = log kaf" = log k -\-n log x, and then substitute new letters for log x and log y : u = log X, v = log y. For, if we do so, the equation becomes v = l -\- nuj where I = log k. IX, § 192] EMPIRICAL EQUATIONS 185 If the values of x and ?/ are given by an experiment, and if w = log £c and v = log y are computed, the values of u and v should correspond to points that lie on a straight line, and the values of I and n can be found as in § 189. The value of k may be found from that of Z, since log k=l. Example 1. The amount of water J., in cu. ft., that will flow per minute through 100 feet of pipe of diameter d, in inches, with an initial pressure of 50 lb. per sq. in. , is as follows : 1 1.5 13.4.3 2 27.50 3 75.13 4 152.51 6 409.54 Find a relation between A and d. Let u = \ogd, V = log A ; then the values of u and v are \ogd . losA . 0.000 0.688 0.176 1.128 0.301 1.439 0.477 1.876 0.602 2.183 0.778 2.612 " J .2 .3 4 .5 .6 .7 .8 Fig. 104 These values give points in the (w, v) plane that are very nearly on a straight line ; hence we may write, approximately, where a and b can be determined directly by measurement in the figure, 186 PLANE ANALYTIC GEOMETRY [IX, § 192 or as in § 189. If we take the first and last pairs of values of u and v, we find .688 = a + 0, 2.612 = a +.778 ft. Solving these equations, we find approximately, a = .688, b = 2.473, and we may write V = .688 + 2.473 u or log A = .688 + 2.473 log d. Since .688 = log 4.88, the last equation may be written in the form log ^ = log 4. 88 + 2. 473 log d = log(4.88 (P««) whence ^ = 4.88 ^2.478. Slightly different values of the constants may be found by using other pairs of values of u and v. 193. Logarithmic Paper. Paper called logarithmic paper may be bought that is ruled in lines whose distances, horizon- tally and vertically, from one point (Fig. 105) are propor- tional to the logarithms of the numbers 1, 2, 3, etc. Such paper may be used advantageously instead of actually looking up the logarithms in a table, as was done in § 192. For if the given values be plotted on this new paper, the result- ing figure is identically the same as that obtained by plotting the logarithms of the given values on ordinary squared paper. Example. A strong rubber band stretched under a pull of p kg. shows an elongation of E cm. The following values were found in an ex- periment : p 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0 E 0.1 0.3 0.6 0.9 1.3 1.7 2.2 2.7 3.3 3.9 5.3 6.9 [RiGGS] If these values are plotted on logarithmic paper as in Fig. 105, it is evi- dent that they lie reasonably near a straight line, such as that drawn. IX, § 193] EMPIRICAL EQUATIONS 187 By measurement in the figure, the slope of this line is found to be 1.6, approximately. Hence if u = logp and v = log E, we have where Z is a constant not yet determined ; whence log E=l + 1.6logp or E = kp^-^ l(p .IJ - - ■ ■ 7 Q / / ^ r _J / / / ■ j& = elongation in c p — pull in kg. / m. / / Z z-zz TTfTTI — I — '"'"""^ /. — ^ t-: EE 15 — " J / ■ - _ s i. / i it 'fi ^/ - — fe _ / / ~t 2 *i _ __ .. ^'-^.Z J ^ ----- A 4= = = :: Z'l^'^. ■ _^ EE 45 2 — .■ ^ -I "-I2 1 :^ 1 y -- P ^4 . 15 .2 .2 i ,4 i,j i .7.ajSi 16 2 ' 2 I 4 5 € » 7 8 9i0 Fig. 105. — Elongation of a Rubber Band where Z = log A:. If p = 1, ^ = A: ; from the figure, if p = 1, E = .B hence A: = .3, and E = .3i>i-6. The use of logarithmic paper is however not at all essential ; the same results may b© obtained by the method of § 192. 188 PLANE ANALYTIC GEOMETRY [IX, § 193 EXERCISES 1. In testing a gas engine corresponding values of the pressure p, meas- ured in pounds per square foot, and the volume r, in cubic feet, were obtained as follows : tJ = 7.14, p = 54.6 ; 7.73, 60.7 ; 8.59, 45.9. Find the relation between |) and v (use logarithmic plotting). Ans. p = 387.6 v-^, or pv^ = 387.6. 2. Expansion or contraction of a gas is said to be adiabatic when no heat escapes or enters. Determine the adiabatic relation between pressure p and volume v (Ex. 1) for air from the following observed values: p = 20.64, V = Q.'27 ; 26.79, 6.34 ; 64.25, 3.15. Ans. pri« = 273.5. 3. The intercollegiate track records for foot-races are as follows, where d means the distance run, and t means the record time : d 100 yd. 220 yd. 440 yd. 880 yd. 1 mi. 2 mi. t 0:09^ 0:21^ 0:48 1:54| 4:15§ 9:24^ Plot the logarithms of these values on squared paper (or plot the given values themselves on logarithmic paper). Find a relation of the form t = kd^. What should be the record time for a race of 1320 yd. ? [See Kennelly, Popular Science Monthly, Nov. 1908.] 4. Solve the Example of § 193 by the method of § 192. 6. Each of the following sets of quantities was found by experiment. Find in each case an equation connecting the two quantities, by §§ 192- 193. («) V p 1 137.4 2 62.6 3 39.6 4 28.6 5 22.6 (b) u V 12.9 63.0 17.1 27.0 23.1 13.8 28.6 8.6 3.0 6.9 (c) c 82° 2.09 212° 2.69 390° 2.90 570° 2.98 750° 3.09 1100° 3.28 SOLID ANALYTIC GEOMETRY CHAPTER X COORDINATES 194. Location of a Point. The position of a point in three- dimensional space can be assigned without ambiguity by giv- ing its distances from three mutually rectangular planes, pro- vided these distances are taken with proper signs according as the point lies on one or the other side of each plane. The three planes, each perpendicular to the other two, are called the coordinate planes ; their common point (Fig. 106) is called the origin. The three ^ mutually rectangular lines Ox, q^ ^pf Oy, Oz in which the planes in- tersect are called the axes of coordinates; on each of them a positive sense is selected arbitrarily, by affixing the letter x, y, z, respectively, y^ The three coordinate planes, Oyz, Ozx, Oxy, divide the whole of space into eight compartments called octants. The first octant in which all three coordinates are positive is also called the coordinate trihedral. If P', P", P"' are the projections of any point P on the coordinate planes Oyz, Ozx, Oxy, respectively, then P'P = x, P"P = y, P"'P=z are the rectangular cartesian coordinates of 189 ?<- ^Q" Fig. 106 190 SOLID ANALYTIC GEOMETRY [X, § 194 P. If the planes through P parallel to Oyz, Ozx, Oxy intersect the axes Ox, Oy, Oz in Q\ Q", Q'", the point P is found from its coordinates x, y, z by. passing along the axis Ox through the distance OQ'=x, parallel to Oy through the distance Q'P"=y, and parallel to Oz through the distance P"P=z, each of these distances being taken with the proper sense. Every point in space has three definite real numbers as coordi- nates; conversely, to every set of three real numbers corresponds one and only one point. Locate the points : (2, 3, 4), (- 3, 2, 0), (5, 0, - 3), (0, 0, 4), (0, - 6, 0), (- 5, - 8, -2). 195. Distance of a Point from the Origin. For the distance OP=r (Fig. 106) of the point P(x, y, z) from the origin we have, since OP is the diagonal of a rectangular parallelepiped with edges OQ' = x, OQ" = y, OQ"' = z: r= Vic^ -f 2/^ + 196. Distance between two Points, the t>vo points Pi (x^ , ?/i , z^) and Pj (X2,y2, Z2) can be found if the coordi- nates of the two points are given. For (Fig. 107), the planes through P^ and those through Pj parallel to the coordinate planes bound a rectangular parallelepiped with P1P2 = d as di- agonal ; and as its edges are P^Q=zX2-x,, P,R = 7j2-yi., we find The distance between M ••* Pn-i^n on the axis Ox is : (x2-x^)-{-(xs-X2)-{ \-(x^ 00 „ 1 ) — Vu„ "^ Out where the right-hand member is the projection of the closing side or resultant PiP„ on Ox. Any line can of course be taken as axis Ox. 200. Division Ratio. Two points Pi(xi, y^, z^ and Pi (^2 J 2/2 ? ^2) beirig given by their coordinates, the coordinates x, y, z of any point P of the line P^P^ can he found if the division ratio PyP/P^P^ = k is known in luhich the point P divides the segment P,P, (Fig 109). Let Qi, Q, ^2^36 the projections of Pi, P, P2 on the axis Ox\ as Q divides Q^Q^ in the same ratio k in which P divides PiPj, we have as in § 3 : x=x^-\- k{x2 — x{). Similarly we find by projecting on Oy, Oz : y = yi + ^Cv2 - Vi), z = z,-{- k{z. - z^. If k is positive, P lies on the same side of Pi as does P2; if k is negative, P lies on the opposite side of Px (§ 3). o Fig. 109 194 SOLID ANALYTIC GEOMETRY [X, § 201 201. Direction Cosines. Instead of using the cartesian coordinates a, y^ z to locate a point P (Fig. 110) we can also use its radius vector r = OP, i.e. the length of the vector drawn from the origin to the point, and its direction cosines, i.e. the cosines of the angles a, p, y, made by the vector OP with the axes Ox, Oy, Oz. We have evidently 05 = r cos a, y = r cos^, z = r cosy. As a line has two opposite senses we can take as direction cosines j of any line parallel to OP either cos a, cos p, cos y, or — cos a, — cos p, — cos y. The direction cosines cos a, cos )8, cos y of a vector OP are often denoted briefly by the letters I, m, n, respectively, so that the coordinates of P are x = lr, y = mr, z = nr. The direction cosines of any parallel line are then Z, m, n or — ?, — m, — n. 202. Pythagorean Relation. Tlie sum of the squares of the direction cosines of any line is equal to one. For the equations of § 201 give upon squaring and adding, since a^ -\- y'^ -\- z^ = r^ : cos'^a + cos* P + cos* 7 = 1* or P + m2 + n2 = 1 ; and this still holds when I, m, n are replaced by —l,—m,— n. Since this result is derived directly from the Pythagorean Theorem of geometry, it may be called the Pythagorean Pela- tion between the direction cosines. Notice that I, m, n can be regarded as the coordinates of the extremity of a vector of unit length drawn from the origin parallel to the line. X, § 202] COORDINATES 195 EXERCISES 1. Find the length of the radius vector and its direction cosines for each of the following points : (5, — 3, 2); (— 3, — 2, 1); (—4, 0, 8). 2. The direction cosines of a line are proportional to 1, 2, 3 ; find their values. 3. A straight line makes an angle of 30° with the axis Ox and an angle of 60° with the axis Oy ; what is the third direction angle ? 4. What is the direction of a line when Z = ? when Z = m = ? 5. What are the direction cosines of that line whose direction angles are equal ? 6. What are the direction cosines of the line bisecting the angle between two intersecting lines whose direction cosines are Z, m, n and Z', wi', n'^ respectively ? 7. Find the direction cosines of the line which bisects the angle between the radii vectoresof the points (3, — 4, 2) and (— 1, 2, 3). 8. Three vertices of a parallelogram are (4, 3, —2), (7, —1, 4), (— 2, 1, — 4); find the coordinates of the fourth vertex (three solutions). 9. In what ratio is the line drawn from the point (2, — 5, 8) to the point (4, 6, — 2) divided by the plane Ozx ? by the plane Oxy ? At what points does this line pierce these coordinate planes ? 10. In what ratio is the line drawn from the point (0, 5, 0) to the point (8, 0, 0) divided by the line in the plane Oxy which bisects the angle between the axes ? 11. Find the coordinates of the midpoint of the line joining the points (4, —3, 8) and (6, 5, — 9). Find the points which trisect the same segment. 12. If we add to the segment joining the points (4, 1, 2) and (— 2, 5, 7) a segment of twice its length in each direction, what are the coordi- nates of the end points ? 13. Find the coordinates of the intersection of the medians of the tri- angle whose vertices are Pi(a;i, yi , ^i), P2{X2 , 2/2, Zi) ^ Pz{xz , 2/3 , ^z)- 14. Show that the lines joining the midpoints of the opposite edges of a tetrahedron intersect and are bisected by their common point. 15. Show that the projection of the radius vector of the point P(a;, ?/, z) on a line whose direction cosines are V, m', n' is Vx-\-m'y-\-n'z. 196 SOLID ANALYTIC GEOMETRY [X, § 203 M (T Fig. Ill 203. Projections. Components of a Vector. If two points -f\(^ij 2/i > ^\) and Piix^^ y^, z^ are given by their coordinates, the projections of the vector, P^P^ on the axes, or what amounts to the same, on parallels to the axes drawn through Pi (Fig. Ill), are evidently (§ 198) : P^q = x^-x^, PiR^Vi — yi, P^S = z^-z,. These projections, or also the vectors P\Q, QNfNPz, are called the rectangular components of the vector P1P2 , or its components along the axes. If d is the length of the segment P^P^ , its direction cosines Z, m, n are, since P2Q is perpendicular to PiQ, P^R to P^R, P^S to P,S: J Xn — X, ?/, — Vl 2;., — Z, a a a These relations can also be written in the form : ^2 - a?i ^ .V2 - Vi ^ Zj-Zi ^ ^ I m n flt,mt.iit) 204. Angle between Two Lines. If the directions of two lines are given by their direction cosines l^ ?Wi, ?^l ajid h, Wo, tio, the angle ij/ between the two lines is given by the formula cos <|/ = lih + inxnii + nin2 . For, drawing through the origin two lines of direction cosines ^ , wij , Til ^-rid I2, m^, W2 3-nd taking on the former a vector OPi of unit length, Fia. 112 the projection OP of OPi on the other line is equal to the X, § 206] COORDINATES 197 cosine of the required angle if/. On the other hand, OPi has li, rrii , rii as components along the axes ; hence, by § 199 : cos k{/ = Zi?2 4- wi^mg + riiWa • Two intersecting lines (or any two parallels to them) make two angles, say ij/ and -rr—ip. But if the direction cosines of each line are given, a definite sense has been assigned to each line, and the angle between the lines is understood to be the angle between these senses. 205. Conditions for Parallelism and for Perpendicularity. If, in particular, the lines are parallel, we have either li^h, mi = m2 , ?ii = ^2 , or Z^ = — ?,, m^ = — m2 , w i = — Wg ; hence in either case ^i _ '^i _ ^ I2 m2 ^2 This then is the condition of parallelism of two lines whose direction cosines are li, rrii, % and I2, mg, %2- If the lines are perpendicular, i.e. ii i{/ = -|- tt, we have cos j/^ = ; hence the condition of perpendicularity of two lines whose direction cosines are l^, m^, n^ and I2 , m2 , ^2 is hh + Wim2 + nin2 = 0. 206. The formula of § 204 gives sin2 ^ = 1 — cos2 1// = 1 —(hl2 + Wim2 + nin2)^. As (§ 202) (Zi2 + mi2 + ni^)(ih^ + m-P- + n-?^^ 1, we can write this ex- pression in the form Zi2 + m^ + n^ hh + mirm + fiiUi hh + mim2 + win2 h^ + m2^+ n-^ sin^i// which can also be expressed as follows sin^ i// = mo 712 h wi 711 h «2 h The direction (I, m, n) perpendicular to two given different directions (li , mi , ni) and (^2, «i2, n^) is found by solving the equations (§ 205) hi + mim + Win = 0, ZgZ + 7W2WI + n2n = 0, 198 SOLID ANALYTIC GEOMETRY [X, § 206 I m n Hi h whence If we denote by k the common value of these ratios, we have 1 = wii ni k, m Til h 112 h w= \k; 12 Wl2 I substituting these values in the relation (§ 202) l^ -\- m^ -{■ n^ = 1, and observing the preceding value of sin^, we find: TOi ni m2 7121 n2 li n=± h wii I2 wtg sin^ sin ^ sin yj/ where V is the angle between the given directions. 207. Three directions (h , w»i , ni), (Z2, WI2, W2), (^3 , ^3 , ws) are com- planar, i.e. parallel to the same plane, if there exists a direction (/, m, n) perpendicular to all three. This will be the case if the equations hi + fnim + wiw = 0, hi + wi2m 4- 712/1 = 0, hi + w»3W» + nzn = have solutions not all zero ; hence the condition of complanarity h wii Wi h WI2 W2 I3 WI3 ns = 0. EXERCISES 1. Find the length and direction cosines of the vector drawn from the point (5, —2, 1) to the point (4, 8, — 6) ; from the point (a, b, c) to the point (—a, —b, —c) ; from ( — a, —6, — c) to (a, 6, c). 2. Show that when two lines with direction cosines Z, w, n and Z', w*', n', respectively, are parallel, IV + mm' + nn' = ± 1. 3. Show that when two lines with direction cosines proportional to a, 6, c and a', 6', c' are perpendicular aa' + 66'+ cc' = ; and when the lines are parallel a/a' = 6/6' = c/c'. 4. Show that the points (5, 2, -3), (6, 1, 4), (-2, -3, 6), (—1, — 4, 13) are the vertices of a parallelogram. X, § 208] COORDINATES 199 6. Show by direction cosines that the pomts (6, -3, 5), (8, 2, 2), (4, —8, 8) lie in a line. 6. Find the angle between the vectors from (5, 8, — 2) to (—2, 6,-1) and from (8, 3, 5) to (1, 1, -6). 7. Find the angles of the triangle whose vertices are (5, 2, 1), (0,3, -1),(2, -1,7). 8. Find the direction cosines of a line which is perpendicular to two lines whose direction cosines are proportional to 2, —3, 4, and 6, 2, —1, respectively. 9. Derive the formula of § 204 by taking on each line a vector of unit length, OPi and OP2, and expressing the distance P1P2 first by the cosine law of trigonometry, then by § 196, and equating these expressions. 10. Find the rectangular components of a force of 12 lb. acting along a line inclined at 60° to Ox and at 45° to Oy. 11. Find the resultant of the forces OPi , OP2 , OP3 , OP4 if the co- ordinates of Pi, P2, P3, P4, with Oas origin, are (3, —1, 2), (2, 2,-1), (-1,2, 1), (-2, 3, -4). 12. If any number of vectors, applied at the origin, are given by the coordinates x, y, z of their extremities, the length of the resultant B is V'(Sx)2 + (Sy)'-^ + (S2:)2 (see Ex. 9, p. 20), and its direction cosines are 2x/i?, Sy/iJ, ^zjB. 13. A particle at one vertex of a cube is acted upon by seven forces represented by the vectors from the particle to the other seven vertices ; find the magnitude (length) and direction of the resultant. 14. If four forces acting on a particle are parallel and proportional to the sides of a quadrilateral, the forces are in equilibrium, i.e. their resultant is zero. Similarly for any closed polygon. 208. Translation of Coordinate Trihedral. Let x, y, z be the coordinates of any point P with respect to the trihedral formed by the axes Ox, Oy, Oz (Fig. 113). If parallel axes Oi^i, Oi?yi, O^Zi be drawn through any point Oi(a, b, c), and if ^ij Vi, ^1 are the coordinates of P with respect to the new tri- 200 SOLID ANALYTIC GEOMETRY [X, § 208 hedral OiXiy^Zi, then the relations between the old coordinates X, y, z, and the new coordinates «i , 2/1, ^i of one and the same point P are evidently x = a-{-x^, y = b-\-yi, z = c-{-Zi. The coordinate trihedral has thus been given a translation, represented by the vector 00^. This operation I ^1 ic ! is also called a transformation to X^ — •«- > I I parallel axes through Oj. Fio. 113 209. Area of a Triangle. Any two vectors OPi , OP2 drawn from the origin determine a triangle OPxPi, whose area A can easily be ex- pressed if the lengths ri , r2 and direction cosines of the vectors are given. For, denoting the angle P1OP2 by ^, we have for the area A : A = \ ViTi sin \{/^ where sin ^ can be expressed in terms of the direc- tion cosines by § 206. % FiQ. 114 210. Moment of a Force. Such areas are used in mechanics to represent the moments of forces. The moment of a force about a point O is defined as the product of the force into the perpendicular distance of from the line of action of the force. Thus, if the vector P1P2 (Fig. 115) represent a force (in magnitude, direction, and sense) the moment of this force ^L.,,,,^^^^ about the origin O is equal to twice the area of the triangle OP1P2, i.e. to the area of the parallelogram OP1P2P3 , where OP3 is a vector equal to the vector P1P2. Fig. 116 It is often more convenient to represent this moment not by such an area, but by a vector OQ^ drawn from O at right angles to the triangle, and of a length equal to the number that represents the moment. If the body on which the force acts could turn freely about this perpendicular, the moment would represent the turning effect of the force P1P2. X, § 211] COORDINATES 201 The sense of this vector that represents the moment is taken so as to make the vector point toward that side of the plane of the triangle from which the force P1P2 is seen to turn counterclockwise. 211. If we square the expression found in § 209 for the area of the triangle OP1P2 and substitute for sin^ xp its value from § 206, we find : ^2 = iriW( mi Hi + 712 h + Zi Wi 1 Hence A^ is the sum of the squares of the three quantities Ax = \ ri?'2 mi Wi mi 712 , Ay = I rir2 7ii h 7l2 h Inri h «i2 which have a simple geometrical and mechanical interpretation. For, as the coordinates of Pi , P2 are we have, e.g., Xi = hri, yx = wiri, zi = win, X2 = hr2, y2 = W2r2, Z2 = n2^2» A, = l hn min ?2^2 W2»*2 = i xi yi X2 2/2 and as Xi , 2/1 and X2 , 2/2 are the coordinates of the projections Qi , Q2 of Pi , P2 on the plane Oxy, Ag represents (§ 12) the area of the triangle 0QiQ2, ie. the projectio7i on the plane Oxy of the area OP1P2. Sim- ilarly, Ax and Ay are the projections of the area OP1P2 on the planes Oyz and Ozx, respectively. As any three mutually rectangular planes can be taken as coordinate trihedrals, our formula A^ = A^ -\- A^ + A^ means that the square of the area of any triangle is equal to the sum of the squares of its projections on any three mutually rectangular planes. In mechanics, 2Ag is the moment of the projection ^1^2 of the force P1P2 about 0, or what is by definition the same thing, the moment of P1P2 about the axis Oz. Similarly, for 2A^, 2 Ay . The proposition means, therefore, that the moments of P1P2 about the axes Ox, Oy, Oz laid off as vectors along these axes can be regarded as the rectangular components of the moment of Pi P2 about the point ; in other words, 2Agy 2 Ay, 2 A, are the components along Ox, Oy, Oz of that vector 2 ^ (§ 210) which represents the moment of P1P2 about 0. 202 SOLID ANALYTIC GEOMETRY [X, § 212 Fio. 116 212. Polar Coordinates. The position of any point P (Fig. 116) can also be assigned by its radius vector 0P= r, i.e. the dis- tance of P from a fixed origin or pole 0, and two angles : the colati- tude $, i.e. the angle NOP made by OP with a fixed axis ON, the X>olar axis, and the longitude , i.e. the angle AOP made by the plane of $ with a fixed plane NOA through the polar axis, the initial meridian plane. A given radius vector r confines the point P to the sphere of radius r about the pole 0. The angles 6 and <^ serve to determine the position of P on this sphere. This is done as on the earth's surface except that instead of the latitude, which is the angle made by the radius vector with the plane of the equator AP', we use the colatitude or polar distance 6 = NOP. The quantities r, 6, and <^ are the j^olar or spherical coordi- nates of P. After assuming a point as pole, a line ON through 0, with a definite sense, as polar axis, and a (half-) plane through this axis as initial meridian plane, every point P has a definite radius vector r (varying from zero to infinity), colatitude $ (varying from to -n), and a definite longitude <^ (varying from to 2 it). The counterclockwise sense of rotation about the polar axis is taken as the positive sense of <^. 213. Transformation from Cartesian to Polar Coordinates. The relations between the cartesian coordinates x, y, z and the polar coordinates r, 6, of any point P appear directly from Fig. 117. If the axis Oz coincides with the polar axis, the plane Oxy with the equatorial plane, i.e. the plane through the X, § 2131 COORDINATES 203 pole at right angles to the polar axis, while the plane Ozx is taken as initial meridian plane, the pro- ' jections of OP=r on the axis Oz and * on the equatorial plane are OIi = rGose, OQ = r sine. Projecting OQ on the axes Ox, Oy,we -/^ y H& find Fig, 117 a; = r sin ^ cos <^, ?/ = r sin ^ sin , z = rcos0. Also r = Va^ + 2/^ + 2-, cos (9 = ^ tan<^ = ^. Vaj2 + y'^-\-z^ » EXERCISES 1. Find the area of the triangle whose vertices are (a, 0, 0), (0, 6, 0), (0, 0, c). 2. Find the area of the triangle whose vertices are the origin and the points (3, 4, 7), (- 1, 2, 4). 3. Find the area of the triangle whose vertices are (4, — 3, 2), (6,4,4), (-5, -2, 8). 4. The cartesian coordinates of a point are 1, V3, 2 V3 ; what are its polar coordinates ? 5. If r = 5, fl = i TT, = I TT, what are the cartesian coordinates ? 6. The earth being taken as a sphere of radius 3962 miles, what are the polar and cartesian coordinates of a point on the surface in lat. 42° 17' N. and long. 83° 44' W. of Greenwich, the north polar axis being the axis Oz and the initial meridian passing through Greenwich ? What is the distance of this point from the earth's axis ? 7. Find the area of the triangle whose vertices are (0, 0, 0) , (ri , ^1 , 0i) , (r2, ^2, 02). 8. Express the distance between any two points in polar coordinates. 9. Find the area of any triangle when the cartesian coordinates of the vertices are given. 10. Find the rectangular components of the moment about the origin of the vector drawn from (1, — 2, 3) to (3, 1, — 1). CHAPTER XI THE PLANE AND THE STRAIGHT LINE PART I. THE PLANE 214. Locus of One Equation. In plane analytic geometry any equation between the coordinates x, y ov r, of a. point in general represents a plane curve. In particular, an equation of the first degree in x and y represents a straight line (§ 30) ; an equation of the second degree in x and y in general repre- sents a conic section (§154). In solid analytic geometry any equation between the coordi- nates X, y, z or r, 0, 4> of a. point in general represents a surface. Thus, if any equation in Xj y, z, F{x,y,z)^0, be imagined solved for z so as to take the form 2!=/(a;,y), we can find from this equation to every point (a?, y) in the plane Oxy one or more ordinates z (which may of course be real or imaginary), and the locus formed by the extremities of the real ordinates will in general form a surface. It may how- ever happen in particidar cases that the locus of the equation F(Xy y, z) = 0, i.e. the totality of all those points whose coordi- nates X, y, z when substituted in the equation satisfy it, con- sists only of isolated points, or forms a curve, or that there are no real points satisfying the equation. Similar considerations apply to an equation in polar coordinates F(r, $,<(>) =0, 204 XI, § 216] THE PLANE 205 215. Locus of Two Simultaneous Equations. Two simulta- neous equations in cc, y, z (or in the polar coordinates r, 6, ) will in general represent a curve in space, namely, the inter- section of the two surfaces represented by the two equations separately. Thus, in the present chapter, we shall see that an equation of the first degree in x, y, z represents a plane and that therefore two such equations represent a straight line, the intersection or the two planes. In chapters XII and XIII we shall discuss loci represented by equations of the second degree, which are called quadric surfaces. 216. Equation of a Plane. Every equation of the first degree in X, y, z represents a plane. The plane is defined as a surface such that the line joining any two of its points lies completely in the surface. We have therefore to show that if the general equation of the first degree (1) Ax-{-By-^Cz + D=0 is satisfied by the coordinates of any two points Pi(aJi, ?/i, Zj) and P^ix.^ , 2/2 J %)? i-^- if Ax^ -f Byi + (7% + J5 = 0, ^ -^ ^ Ax2 + By2-\-Cz2 + D=:0, then (1) is satisfied by the coordinates of every point P(x, y, z) of the line PiP2- Now, by § 200, the coordinates of every point of the line P1P2 can be expressed in the form x = Xi-\- k(x2 — a^i), y = yi-\- Kv^ — Vi), z = ^i + K^2 — %)» where k is the ratio in which P divides PiP^, i.e. h = P,P/P^P,. We have therefore to show that A[x^ + K^2 - a^)] + ^[2/1 +^fe-2/i)] + C[_z^+K^2-Zi)'] +^=0, 206 SOLID ANALYTIC GEOMETRY [XI, § 216 ■whatever the value of k. Adding and subtracting kD, we can write this equation in the form {l-k){Ax^ + By^+Cz^-\-D)-^k{Ax^-\-By^+Cz^ + U)=zO', and this is evidently true for any k, owing to the conditions (2). 217. Essential Constants. The equation (1) will still rep- resent the same plane when multiplied by any constant differ- ent from zero. Since A, B, C cannot all three be zero, we can divide (1) by one of these constants ; it will then contain not more than three arbitrary constants. We say therefore that the general equation of a plane contains tJwee essential constants. This corresponds to the geometrical fact that a plane can, in a variety of ways, be determined by three condi- tions, such as the conditions of passing through three points. 218. Special Cases. If, in equation (1), Z> = 0, the plane evidently passes through the origin. If, in equation (1), C=0, so that the equation is of the form Ax -[- By -\- D = 0, this equation represents the plane perpendicular to the plane Oxy and passing through the line whose equation in the plane Oxy is Ax -\- By -\- D = 0. For, the equation Ax + J5?/ -|- 2) = is satisfied by the coordinates of all points («, ?/, z) whose x and y are connected by the re- lation Ax -\- By -{- D = and whose z is arbitrary, but it is not satisfied by the coordinates of any other points. Similarly, if 5 = in (1), the plane is perpendicular to Ozx; if ^ = 0, the plane is perpendicular to Oyz. It B = and C = in (1), the equation obviously represents a plane perpendicular to the axis Ox ; and similarly when and A, or A and B are zero. Notice that the line of intersection of (1) with the plane Oxy, for instance, is represented by the simultaneous equations Ax + By+Cz + D=0, 2 = 0. XI, § 220] THE PLANE 207 219. Intercept Form. If D =^ 0, the equation (1) can be divided by Z) ; it then assumes the form If Aj Bf G are all different from zero, this equation can be written + 1, -D/A ' -D/B ' -D/G or, putting - D/A = a, - D/B = b, - D/G = c : (3) a c In this equation, called the intercept form of the equation of a plane, the constants a, h, c are the intercepts made by the plane on the axes Ox, Oy, Oz respectively. For, putting, for instance, 2/ = and 2: = 0, we find x = a\ etc. 220. Plane through Three Points. If the plane Ax + By + Gz-\-D = is to pass through the three points Pi(xi, y^, z^, ^2(^2) 2^2 ? ^2)) A (^3 J 2/3? ^z)j the three conditions Ax, + By, + Gz,-\-D = 0, Ax^-^By^-\-Gz^+D = 0, Ax^ + By^ + Gz^ + D = (i must be satisfied. Eliminating A, B, G, D between the /owr preceding equations, as in § 55, we find the equation of the plane passing through the three points in the form X y z 1 X, 2/1 ^l 1 x^ 2/2 2^2 1 X3 2/3 2=3 1 208 SOLID ANALYTIC GEOMETRY [XI, § 220 EXERCISES 1. Find the intercepts made by the following planes : (a) 4a; + 12y + 3^ = 12; (&) 15a:- 6y + 10« -f 30 = ; Qc) x-y ■}-z-l=0; (rf)x+2y + 30 + 4 = O. 2. Interpret the following equations : (a)a; + y + « = l; {b)5y-3z = 12, (c) x + y=0; (d) 5 y + 12 = 0. 3. Find the plane determined by the points (2, 1, 3), (1, — 5, 0), (4,6, -1). 4. Write down the equation of the plane whose intercepts are 3, 2, — 5. 6. Find the intercepts of the plane passing through the points (3, -1,4), (6, 2, -3), (-1,-2, -3). 6. If planes are parallel to and a distance a from the coordinate planes, what are their intercepts ? What are their equations ? 7. Show that the four points (4, 3, 3), (4,-3,-9), (0, 0, 3), (2, 1, 2) lie in a plane and find its equation. 221. Normal Form. The position of a plane in space is fully determined by the length p = ON (Fig. 118) of the per- pendicular let fall from the origin on the plane and the direction co- sines I, m, n of this perpendicular regarded as a vector ON. Let Pbe any point of the plane and OQ =x, QR = y, EP= z its coordinates ; as the projection of the open polygon OQRP on ON is equal to ON (§ 199) we have (4) lx + my-\- nz =p. This equation is called the normal form of the equation of a plane. Observe that the number p is always positive, being the distance of the plane from the origin, or the length of the vector ON. Hence Ix + my -{- nz is always positive. FiQ. 118 XI, §222] THE PLANE 209 222. Reduction to the Normal Form. The equation Ax + By + Cz -\- D = is in general not of the form ly-{-my-\-nz—p since in the latter equation the coefficients of x, y, z, being the direction cosines of a vector, have the property that the sum of their squares is equal to 1, while A^ + B^-\- C^ is in general not equal to 1. But the general equation can be reduced to the normal form by multiplying it by a constant factor Ic properly chosen. The equation TcAx + l<^By + hCz -\-kD=0 evidently represents the same plane as does the equation Ax -i- By + Cz -\- D = 0; and we can select Jc so that {JcAy + (JcBy + (kCy = l, viz. k. ^ ±VA^ + B^+C^ As in the normal form the right-hand member p is positive (§ 221) the sign of the square root should be selected so that kD becomes negative. TTie normal form is therefore obtained by dividing the equation Ax + By -\- Cz+D = Oby ± VA^ + B^ + C^ according as D is negative or positive. It follows at the same time that the direction cosines of any normal to the plane Ax -\- By -{- Cz -{- D = are proportional to A, B, C, viz. I z= , m = C and that the distance of the plane from the origin is -D P = ±VAr-hB'-^0' the upper sign of the square root to be used when D is nega- tive, the lower when D is positive. 210 SOLID ANALYTIC GEOMETRY [XI, § 223 223. Distance of Point from Plane. Let Ix -\- my + nz = p be the equation of a plane in the normal form, Pi(iCi, yi, z^ any point not on this plane (Fig. 119). The projection OS of the vector OPi on the normal to the plane being equal to the sum of the projections of its components 0Q = Xi , QR = 2/i, RPi = 2!i, we have OS = Za^ + myx + nzj . Hence the distance d of Pi from the plane, which is equal to NS, will be d= OS — 0N= Ixi + myi + 7iZi — ;>. fig. ii9 If this expression is negative, the point Pj lies on the same side of the plane as does the origin ; if it is positive, the point Pj lies on the opposite side of the plane. Any plane thus di- vides space into two regions, in one of which the distance of every point from the plane is positive, while in the other the distance is negative. If the plane does not pass through the origin, the region containing the origin is the negative region ; if it does, either side can be taken as the positive side. To find the distance of a point Pi(xi, ?/i, Zi) from a plane given in the general form Ax-^By + Cz-hD = 0, we have only to reduce the equation to the normal form (§ 223) and then to substitute for x, y, z the coordinates a^i, 3/1, Zi of Pi ; thus ^^ Axi + Byi-{- Czi 4- D the square root being taken with + or — according as Z) is negative or positive. Notice that d is the distance from the plane to the point Pi, not from P^ to the plane. XI, §225] THE PLANE 211 224. Angle between Two Planes. As two intersecting planes make two angles whose sum = ir, we shall, to avoid any ambiguity, define the angle between the planes as the angle between the perpendiculars (regarded as vectors) drawn from the origin to the two planes. If the equations of the planes are given in the normal form, liX-{-miy + niZ=pi, l^-{-m^ + n.;Z =P2i we have, by § 204, for the angle i/^ between the planes : cos \p = l-J.2 -f mim2 + n^a^ . If the equations of the planes are in the general form, we find by reducing to the normal form (§ 222) : cosj/^ = A^A2-^B,B,+ C,C2 ± V A' + B,' + 0^2 . ± ^A,' + B,' + C,^ 225. Bisecting Planes. To find the equations of the two planes that bisect the angles formed by two intersecting planes given in the normal form, liX-{-miy-{-niZ—pi = 0, l2X-{-m^ -{-n^z—pz^O, observe that for any point in either bisecting plane its distances from the two given planes must be equal in absolute value. Hence the equations of the required planes are liX + rriiy -f- n^z —pi=z± (Icpc -\- m^y -\-n^— p^. To distinguish the two planes, observe that for the plane that bisects that pair of vertical angles which contains the origin the perpendicular distances are in the one angle both positive, in the other both negative; hence the plus sign gives this bisecting plane. 212 SOLID ANALYTIC GEOMETRY [XI, § 225 If the equations of the planes are given in the general form, first reduce the equations to the normal form (§ 222). EXERCISES 1. A line is drawn from the origin perpendicular to the plane X — y — 62 — 10 = 0; what are the direction cosines of this line ? 2. Find the distance from the origin to the plane 2x-\-2y — z=6. 3. Find the distances of the following planes from the origin : (a) 3x-Ay-\-5z-S = 0, ^b)x + y+z = 0, (c) 2y-50 = 3, (d) Sx-4y-\-6 = 0. 4. Find the distances from the following planes to the point (2,1, -3): (a) 3 X + 6 y - 6 2 = 8, (6) 2x-Sy-z = 0, (c) x + y -[- z = 0. 6. Find the plane through the point (4, 8, 1) which is perpendicular to the radius vector of this point ; also the parallel plane whose distance from the origin is 10 and in the same sense. 6. Find the plane through the point (—1,2, — 4) that is parallel to the plane 4x — 3y + 22 = 8; what is the distance between these planes ? 7. Find the distance between the planes 4x — 6y — 2z = 6, 4x — 5y - 2 « + 8 = 0. 8. Are the points (6, 1, — 4) and (4, — 2, 3) on the same side of the plane 2x + Sy - 6z -^ 1 =0? 9. Write down the equation of the plane equally inclined to the axes and at the distance p from the origin. 10. Show that the relation between the distance p from the origin to a plane and the intercepts a, 6, c is 1/a^ + l/b^ + l/c^ = 1/p/^. 11. Show that the locus of the points equally distant from the points -Pi(a;i , yi , 2i) and P^ixo , j/a , 22) is a plane that bisects P1P2 at right angles. 12. Find the equations of the planes bisecting the angles : (a) between the planes a: + 2/ + «-3=0, 2«-3y + 42 + 3 = 0; (6) between the planes 2x-2y-z = 8,x + 2y-2z = 6. XI, § 226] THE PLANE 213 = 0. 226. Volume of a Tetrahedron. The volume of the tetrahe- dron whose vertices are the points Pi(xi, y^, z{), F^ixz, y^, z^, Pzi^zi Vzi ^z)i A(^4> 2/4 > ^a) can be expressed in terms of the coordinates of the points. The equation of the plane deter- mined by the points P2, P^, P4 is (§ 220) X y z 1 •^2 2/2 ^2 1 •^3 2/3 % 1 ^•4 2/4 2:4 1 Now the altitude d of the tetrahedron is the distance from this plane to the point P^ (x^ , 2/1 , ^i), i-e. (§ 223) x^ 2/1 2^1 1 X2 2/2 ^2 -*- ''^S Vz ^3 1 X, Va z. 1 2/2 ^2 1 2 Z2 X2 1 2 a;2 2/2 1 2/3 2:3 1 + 2^3 Xs 1 + Xz 2/3 1 2/4 2^4 1 24 a;4 1 a?4 2^4 1 d = But the denominator is seen immediately to represent twice the area of the triangle with vertices P2, P3, P4 (Ex. 9, p. 203), i.e. twice the base of the tetrahedron. Denoting the base by J5, we then have 2Bd X, Vi 2=1 X2 y2- 2^2 X, 2/3 2^3 X, 2/4 2^4 The volume of the tetrahedron is V=^Bd, and therefore Xi 2/1 2^1 1 X2 2/2 2;, 1 Xs 2/3 2:3 1 2^4 2/4 2:4 1 ^ 214 SOLID ANALYTIC GEOMETRY [XI, § 227 227. Simultaneous Linear Equations. Two simultaneous equations of the first degree, represent in general the line of intersection of the two planes represented by the two equations separately. For, the coordi- nates of every point of this line, and those of no other point, satisfy both equations. See § 215 and §§ 231-232. Three simultaneous equations of the first degree, A,x + B,y+C,z + D, = 0, A^x + ^2^ -h Co2; + A = 0, A,x + ^32/ + C32; -f A = 0, determine in general the point of intersection of the three planes. The coordinates of this point are found by solving the three equations for x, y, z. But it may happen that the three planes have no common point, as when the three lines of intersection are parallel, or when the three planes are parallel ; and it may happen that the planes have an infinite number of points in common, as when two of the planes, or all three, coincide, or when the three planes pass through . one and the same line. Four planes will in general have no point in common. If they do, i.e. if there exists a point (xi ,^1,^1) satisfying the four equations Aixi + BiVi + Cizi + Di = 0, ^23^1 + Biyi + CiZi + Z>2 = 0, ^33:1 + ^3^1 + CiZi + Da = 0, AaXx + -84^1 + CaZx + Z>4 = 0, we can eliminate a;i, j/i, 01, 1 between these equations so that we find the condition = 0. Ax Bx Cx Dx A2 B2 C2 i>2 As Bz Cz Bz A4 Bi c. B, XI, §228] THE PLANE 215 EXERCISES 1. Find the volume of the tetrahedron whose vertices are (0, 0, 0), (o, 0, 0), (0, 6, 0), (0, 0, c). 2. Find the volumes of the tetrahedra whose vertices are the following points : (a) (7, 0, 6), (3, 2, 1), (- 1, 0, 4), (3, 0, -2). (6) (3, 0, 1), (0, - 8, 2), (4, 2, 0), (0, 0, 10). (c) (2, 1,-3), (4, -2, 1), (3, -7, -4), (5, 1, 8). 3. Find the coordinates of the points in which the following planes intersect : (a) 2x + 6y + z-2 = 0, xi-6y-{-z = 0, Sx-Sy -\-2 z - 12 = 0, (6) 2x+y+z=a + b + G, ^x-2 y-\-z=2 a-2b-\-c, 6x-y=3a-6. 4. Show that the four planes 6x — Sy—z = 0, 4:X — 2y + z = Sy Sx + 2y — 6z = 6, x + y -\- z = 6 pass through the same point. What are the coordinates of this point ? 6. Show that the four planes 4x + y + 2! + 4 = 0, a; + 2y — + 3 = 0, y — 50 + 14=0, x + y -\- z — 2 = have a common point. 6. Show that the locus of a point the sura of whose distances from any number of fixed planes is constant is a plane. 228. Pencil of Planes. All the planes that pass through one and the same line are said to form a pencil of planes, and their common line is called the axis of the pencil. If the equations of any two non-parallel planes are given, say A,x + B,y + C,z + A = 0, then the equation of any other plane of the pencil having their intersection as axis can be written in the form (2) {A,x + B,y + 0,z + A) + K^^ + ^22/ + 0,z + A) = 0, where A; is a constant whose value determines the position of the plane in the pencil. For, this equation (2) being of the first degree in x, y, z certainly represents a plane ; and the coordinates of the points 216 SOLID ANALYTIC GEOMETRY [XI, § 228 of the line of intersection of the two given planes (1), since they satisfy each of the equations (1), must satisfy the equa- tion (2) so that the plane (2) passes through the axis of the pencil. 229. Sheaf of Planes. All the planes that pass through one and the same point are said to form a sheaf of planes, and their common point is called the center of the sheaf. If the equations of any three planes, not of the same pencil, are given, say A;,x + B^+C,z-^D^ = 0, then the equation of any other plane of the sheaf having their point of intersection as center can be written in the form {A^x + B,y + C,z 4- A) + K{Ax + Biy-irC^^- A) + k^{A^x-\- B^ 4- C^z + A) = 0, where Ic^ and k^ are constants whose values determine the position of the plane in the sheaf. The proof is similar to that of § 228. 230. Non-linear Equations Representing Several Planes. When two planes are given, say Aa' + Ay + Ci2 + A = 0, A^ + B^-\-C^ + D^ = 0, then the equation {A^x 4- B,y -f C,z + D,){A^x + 5^2/ + CjZ -f A) = 0, obtained by equating to zero the product of the left-hand mem- bers (the right-hand members being reduced to zero), is satis- fied by all the points of the first given plane as well as all the points of the second given plane, and by no other points. The product equation is therefore said to represent the two given planes. The equation is of the second degree. XI, § 230] THE PLANE 217 Similarly, by equating to zero the product of the left-hand members of the equations of three or more planes (the right- hand members being zero) we obtain a single equation repre- senting all these planes. An equation of the nth degree may, therefore, represent n planes ; it will do so if its left-hand mem- ber can be resolved into n linear factors with real coefficients. EXERCISES 1. Find the plane that passes through the Ime of mtersection of the planes 5x — 32/+4^ — 35=0, x + y —z = and through (4, — 3, 2) . 2. Show that the planes 3cc — 2?/ + 5^ + 2=0, x + y — — 5 = 0, 6a; + ?/4-2;3— 13 = belong to the same pencil: 3. Show that the following planes belong to the same sheaf and find the coordinates of the center of the sheaf : 6x + y — 4;2 = 0, x + 2/ + i2 = 5, 2a;-4y-2 = 10, 2a; + 3?/+;s = 4. 4. What planes are represented by the following equations ? (a) iK2-6x + 8 = 0, (6) ?/2-9 = 0, (c) x'^-z^ = 0, {d) x'^-4xy = 0. 5. Find the cosine of the angle between the following pairs of planes : (a) 4:x-3y^z=6, x-\-y-z=S ; (b) 2x+7 y-{-4z=2, x-9y-2z=12. 6. Show that the following pairs of planes are either parallel or perpendicular : (a) Sx-2y-\-5z=0,2x+Sy=8; (b) 6x+2y-z=6, lOx+iy-2 z=3; (c) x + y-2z = S, x-{-y+z=n; (d) x- 2y - z = S, 3x -Qy-S z=6. 7. Find the plane that is perpendicular to the segment joining the points (3, — 4, 6) and (2, 1, — 3) at its midpoint. 8. Show that the planes Aix + Biy + Ciz + Di = 0, A2X + B2y + C2Z -f-i>2 = are parallel (on the same or opposite sides of the origin) if A1A2 + B1B2 + C1C2 ^^1^^ V^i2 -f- Bi^ + Ci2' VA2' + B2^ + CV 9. A cube whose edges have the length a is referred to a coordinate trihedral, the origin being taken at the center of a face and the axes par- allel to the edges of the cube. Find the equations of the faces. 218 SOLID ANALYTIC GEOMETRY [XI, § 230 a: y z 1 Xl y\ Z\ 1 X2 yi 22 1 A B c 10. Show that the plane through the points Pi (xi , yi , Z\) and P2fx2, ?/2, 2^2) and perpendicular to the plane Ax -\- By ■\- Cz -^^ D — ^ can be represented by the equation = 0. 11. Find those planes of the pencil 4a; — 3y + 52 = 8, 2x + 3y — « = 4 which are perpendicular to the coordinate planes. 12. Find the plane that is perpendicular to the plane 2x + 3y — « = 1 and passes through the points (1, 1, — 1), (3, 4, 2). 13. Find the plane that is perpendicular to the planes 4x — 3y + 2 = 6, 2a; + 3y — 60 = 4 and passes through the point (4, — 1, 6). 14. Show that the conditions that three planes A\X + Bxy -|- C\Z + Di = 0, A^x + Biy + C^z + Z>2 = 0, Azx + Bzy + Ca^ + -D3 = belong to the same pencil, are A{-\- kAt _ Bi + kB2 As Bz Ci -t- kCj Cz Di±kD2. Dz ' or, putting these fractions equal to s and eliminating k and a, Ci Di Ci Di C2 Di = C2 Dt Cz Dz Cz Dz A, A2 Az Di D2 Dz Ai Ao Az Ai A2 Az By Cx B2 C2 Bz Cz = 0. (Verify Ex. 2 by using these conditions.) 15. Find the equations of the faces of a right pyramid, with square base of side 2 a and with altitude h, the origin being taken at the center of the base, the axis Oz through the opposite vertex, and the axes Ox, Oy parallel to the sides of the base. 16. Homogeneous substances passing from a liquid to a solid state tend to form crystals ; e.g. an ideal specimen of ammonium alum has the form of a regular octahedron. Find the equations of the faces of such a crystal of edge a if the origin is taken at the center and the axes through the vertices, and determine the angle between two faces. 17. Find the angles between the lateral faces of a right pyramid whose base is a regular hexagon of side a and whose altitude is h. XI, § 232] THE STRAIGHT LINE 219 PART 11. THE STRAIGHT LINE 231. Determination of Direction Cosines. Two simulta- neous linear equations (§ 227), (1) Ax+By + Cz-{-n=0, A'x-{-B'y-{-C'z-\-I>'=0, represent a line, namely, the intersection of the two planes represented by the two equations separately, provided the two planes are not parallel. To obtain the direction cosines ?, m, n of this line observe that the line, since it lies in each of the two planes, is perpen- dicular to the normal of each plane. Now, by § 222 the direc- tion cosines of these normals are proportional to A, B, and A' J B\ C\ respectively. We have therefore Al + Bm-\-Cn = 0, An + B'm -h C"n = 0, whence l:m:n BO B'C' OA C'A' AB A'B' The direction cosines themselves are then found by dividing each of these determinants by the square root of the sum of their squares. 232. Intersecting Lines. The two lines A^x-{-B,y-\- Ci^-f A = 0, ) r A^-^B^y + G^^ + D^^O, A^x -\- B^y + C^z 4- A' = J ^^ 1 Ao}x^B^y-\- C^z+D^ = will intersect if, and only if, the four planes represented by these equations have a common point. By § 227, the condition for this is A A C'l A A^ B,' C/ A A2 Jj2 C2 -^2 A2 -O2 ^2 -^2 = 0. 220 SOLID ANALYTIC GEOMETRY [XI, § 233 233. Special Forms of Equations. For many purposes it is couvenieiit to represent a line by means of one of its points and its direction cosines, or by means of two of its points. Let the line be called A. If (^j ^ij 2;i) is a given point of X and I, m, n are the direc- tion cosines of X, then every point (a;, y, z) of A must satisfy the relations (§ 203) : In these equations, I, m, n, can evidently be replaced by any three numbers proportional to Z, m, n. Thus, if («2, ya? ^Jg) be any point of \ different from (a^, ^u ^j), we have the continued proportion X2 — Xi : y2 — yi : z^ — Zi = l : m : n; hence the equations of the line through the two points (xi , yi , Zi) and (x2 , 2/2 f ^2) 21 re : (S) x-x^ ^ y-Vi ^ g-gj ^ X^-QC^ 2/2-2/1 «2-«i* If, for the sake of brevity, we put x^— x^ = a, y2 — yi = ^> Z2 — Zi = c, we can write the equations of the line in the form (4) ag-a?i ^ 2/-2/i _ g-gi a be* where a, b, c, are proportional to I, m, n, and can be regarded as the components of a vector parallel to the line. The equations (3) also follow directly by eliminating k be- tween the equations of § 200, namely, (5) x=xi+k{x2-x{), 2/=2/i+A:(2/2-2/i), z=Zj^+k(z2-Zj). These equations which, with a variable k, represent any point of the line through (x^, y^, z^ and (ajj, 2/2? ^2) are called the parameter equations of the line. XI, § 234] THE STRAIGHT LINE 221 Fig. 120 234. Projecting Planes of a Line. Each of the forms (2), (3), (4), which are not essentially different, furnishes three linear equations ; thus (4) gives : 6 c G a ah but these three equations are equivalent to only two, since from any two the third follows immediately. The first of these equations, which can be written in the form represents, since it does not contain x (§ 218), a plane perpendicular to the plane Oyz] and as this plane must con- tain the line X it is the plane CCA that projects \ on the plane Oyz (Fig. 120). Similarly the other two equations represent the planes that project A. on the co- ordinate planes Ozx and Oxy. Any two of these equations represent the line X as the intersection of two of these pro- jecting planes. At the same time the equation y-?/i ^ g-gi h c can be interpreted as representing a line in the plane Oyz, viz. the intersection of the projecting plane with the plane x — 0. This line {AC in Fig. 120) is the projection X^ of X on the plane Oyz. As the other two equations (4) can be inter- preted similarly it appears that the equations (2), (3), or (4) represent the line A. by means of its projections A^, Xy, X^ on the three coordinate planes, just as is done in descriptive geometry. Any two of the projections are of course sufficient to determine the line. 222 SOLID ANALYTIC GEOMETRY [XI, § 235 235. Determination of Projecting Planes. To reduce the equations of a line A given in the form (1) to the form (4) we have only to eliminate between the equations (1) first one of the variables x, y, z, then another, so as to obtain two equa- tions, each in only two variables (not the same in both). The process will best be understood from an example. The line being given as the intfersection of the planes (a) 2x-Sy-{-z + S = 0y (5) x + y + z-2 = 0, eliminate z by subtracting (6) from (a) and eliminate x by subtracting (6), multiplied by 2, from (a) ; this gives the line as the intersection of the planes x — 4:y-\-5 = 0, which are the projecting planes parallel to Oz and Ox, i.e. the planes that project the line on Oxy and Oyz. Solving for y and equating the two values of y we find : x-\-5 __y _ z — 7 4 ^i~35"- The line passes therefore through the point (—5, 0, 7) and has direction cosines proportional to 4, 1, — 5, viz. ,4 1 5 1 = — ^, m = — =, n = — . V42 V42 V42 EXERCISES 1. Write the equations of the line through the point (— 3, 1, 6) whose direction cosines are proportional to 3, 5, 7. 2. Write the equations of the line through the point (3, 2, — 4) whose direction cosines are proportional to 6, — 1, 3. 3. Find the line through the point (a, 6, c) that is equally inclined to the axes of coordinates. XI, § 236] THE STRAIGHT LINE 223 4. Find the lines that pass through the following pairs of points : (a) (4, - 3, 1), (2, 3, 2), (6) (- 1, 2, 3), (8, 7, 1), (c) (- 2, 3, - 4), (0, 2, 0), (d) (- 1, - 5, - 2), (- 3, 0, -1), and determine the direction cosines of each of these lines. 6. Find the traces of the plane 2aj — 3?/ — 4^ = 6 in the coordinate planes. 6. Write the equations of the\me2x—Sy-\-6 z—6=0,x—y-\-2z — S=0 in the form (4) and determine the direction cosines. 7. Put the line 4x — Sy — 6 = 0^x — y — z — 4: = in the form (4) and determine the direction cosines. 8. Find the line through the point (2, 1, — 3) that is parallel to the line 2x-Sy + 4z — 6 = 0, 6x + y-2z — S = 0. 9. What are the projections of the line 6x — Sy — lz — 10 = 0, x + y — Sz+6 = on the coordinate planes ? - 10. Obtain the equations of the line through two given points by equating the values of k obtained from § 200. 11. By § 222, the direction cosines of any line are proportional to the coeflBcients of x, y, and z in the equation of a plane perpendicular to the line. Find a line through the point (3, 5, 8) that is perpendicular to the plane 2x-\-y + Sz=:b. 236. Angle between Two Lines. The cosine of the angle ^ be- tween two lines whose direction cosines are Zi , mi , wi and Z2 > wi2 , n2 is, by § 204, cos \j/ = hh + Wim2 + W1W2 . Hence if the lines are given in the form (4) , say x-xi^y — yi_z - zi x -Xi _y — yi _z — Zj we have COS^ : hi ci ai hi C2 aiGi 4- hihi + C1C2 ±V^ ■ + 5l2 + Ci2 . ±Va22 + &2^ + C22 If the lines are parallel^ then ai_&i. ai hi — > Ci if they are perpendicular, then and mce versa. didi + &1&2 + C1C2 = ; 224 SOLID ANALYTIC GEOMETRY [XI, § 237 Let the line and plane plana 237. Angle between Line and Plane. be given by the equations x — x\ _ y— y\ _ z — zi a b c ' Ax-{-By+Cz + D = 0. The plane of Fig, 121 represents the plane through the given line perpendicular to the given plane. The angle /3 between the given line and plane is the complement of the angle a between the line and any perpen- dicular PN to the plane. Hence 8in^= aA + bB + cC Fig. 121 ± Va'^ + 6'-« + c2 . ± y/A^ + 52 + C^ The (necessary and sufficient) condition for parallelism of line and plane is aA -\- bB -\- cC = i the condition of perpendicularity is a _b_± A~ B C 238. Line and Plane Perpendicular at Given Point. If the plane Ax -^ By -^ Cz + D = passes through the point Pi(xi , yi , «i), we must have Axi + Byi -{- Czi -\- D = 0. Subtracting from the preceding equation, we have as the equation of any plane through the point Pi(xi , yi, zi) : A{x - xi) + B{y - yi) + C(z - zi) = 0. The equations of any line through the same point are x — xi __ y — y\ _ z —zx a b c If this line is perpendicular to the plane, we must have (§ 237) : a/ A = b/B = c/C. Hence the equations x — xi _ y—y\ _ z — zi A ~ B ~ C represent the line through Pi(a;i, yi, zi) perpendicular to the plane A(x - xi) + B{y - yi) + C{z - zi) = 0. XI, § 240] THE STRAIGHT LINE 225 239. Distance of a Point from a Line, if the equations of the line X are given in the form z — Z\ I y-yi m where (cci , yi , zi) is a point Pi of X (Fig. 122), the distance d = QP2 of the point P2ix2, 2/2, Zi) from X can be found from the right-angled triangle P1QP2 which gives d^ = P1P2' - PlQ^ by observing that P1P22 = (X2 - Xi)2 + (y2 -yiy + (Z2 - ZiY, while PiQ is the projection of P1P2 on X. This projection is found (§ 199) as the sum of the projections of the components X2 — xi, 2/2 — 2/1, Z2 — z\ of P1P2 on X : PiQ = 1{X2 - xi) + w(2/2 - y\) + n{Z2 - Zl). Hence cP=(^X2-Xiy+ (2/2-2/i)H(^2-2i)2-[Z(x2-xi)4-w(2/2-2/i) +n(z2-z{)Y. 240. Shortest Distance between Two Lines. Two lines Xi , X2 whose equations are given in the form x — xi_y — yi_z — zi x~ X2 _y — y2 _z - Z2 TOl Wl m2 W2 xoill intersect if their directions {l\ , «ii , wi), (Z2, WI2 , W2), and the direc- tion of the line joining the points (xi , 2/1 , ^1), (^2 , 2/2 » ^2) are complanar (§207), i.e. if X2 — Xi 2/2 - 2/1 «2 - «1 Zl wi n\ = 0. Z2 W2 W2 If the lines Xi , X2 do not intersect, their shortest distance d is the dis- tance of P2(X2, 2/2, ^2) from the plane through Xi parallel to X2. As this plane contains the directions of Xi and X2 , the direction cosines of its nor- mal are (§ 206) proportional to mi wi m h h mi m2 W2 112 h ' h mz 226 SOLID ANALYTIC GEOMETRY [XI, § 240 and as it passes through Pi (xi , yi , zi) its equation can be written in the form x — xi y — yi z — zx l\ mi m = 0. I2 Wl2 712 Hence the shortest distance of the lines \\ , X2 is : Xi - xi 2^2 - yi Zi -zi l\ TWi n\ h WI2 rii V r/ii n\ W2 W2 h n»i h n»2 As the denominator of this expression is equal to sin^ (§ 206), we have d sin ^ = X2 — xi yi — yi Zi - zi h mi ni h rrii m EXERCISES 1. Find the cosine of the angle between the lines x-S _ y-5 ^ z + l ^^^ x+l^ y-3 ^ g + 3 2 3 4 -1 2 3 * ' 2. Find the angle between the lines 3x — 2y + 4«— 1 = 0, 2x + y-30 + 10 = 0, a,nd x + y + z = 6, 2x-hSy -5z = S. 3. Find the angle between the lines that pass through the points (4, 2, 6), (- 2, 4, 3) and (- 1, 4, 2), (4, - 2, - 6). 4. Find the angle between the line x + l _ y — 2 __ g + 10 3 -6 3 and a perpendicular to the plane 4x — 3y — 22 = 8. 5. In what ratio does the plane 3x — 4y + 60 — 8 = divide the segment drawn from the origin to the point (10, — 8, 4). 6. Find the plane through the point (2, — 1, 3) perpendicular to the line x—S _ y+2 _ z—7 4 3 -1 * XI, § 240] THE STRAIGHT LINE 227 7. Find the plane that is perpendicular to the line 4x + y — z=6, 'dx + iy-{-Sz + 10 = and passes through the point (4, — 1,3). 8. Find the plane through the origin perpendicular to the line 5x-2y + z=6, Sx + y-4z = S. 9. Find the plane through the point (4, — 3, 1) perpendicular to the line joining the points (3, 1, — 6), (— 2, 4, 7). 10. Find the line through the point (2, — 1, 4) perpendicular to the plane ic — 2^ + 42 = 6. 11. Show that the lines x/S = y/ -- 1 = z/-2 and x/i = y/6 = z/3 are perpendicular. 12. Show that the lines ^L=Ll = y±l = ^:zl and a;-2^y-3^ z 1-2 3 _2 4 -6 are parallel. 13. Find the angle between the line Sx — 2y — z = 4, 4x + 3y — 3^ = 6 and the plane x -i-y -\- z = S. 14. Find the lines bisecting the angles between the lines x — a _ y - & _ ^ — c ^j^j x — a _ y — b _ z — c 15. Find the plane perpendicular to the plane Sx — iy — z = 6 and passing through the points (1, 3, — 2), (2, 1, 4) . 16. Find the plane through the point (3, — 1, 2) perpendicular to the line 2 a; — 3!/ — 42; = 7, ic+y— 2^=4. 17. Find the plane through the point (a, 6, c) perpendicular to the line Aix + Biy + Ciz + Di = 0, Azx + B^y + dz + D2 = 0. 18. Find the projection of the vector from (3, 4, 5) to (2, — 1, 4) on the line that makes equal angles with the axes ; and on the plane 2x-3y + 45!=6. 19. Find the distances from the following lines to the points indicated : (6) 2a; + y-0 = 6, a;- y + 4^ = 8, (3, 1,4); (c) 2x + 3?/ + 60 = l, 3x-6y + 32=0, (4, 1, -2). 228 SOLID ANALYTIC GEOMETRY [XI, § 240 20. Show that the equation of the plane determined by the line x-xi y-yi_z-zi a b c d the point P2(a;2, 2/2 , Z2) can be written in the form X -xi 2/ -2/1 z -zi X2 — Xi 2/2 — Vl Z2 — Zi = 0. a b c 21. Find the plane determined by the intersecting lines a:-3_y-6_0 + l .^. x - S _y - 5 _z + 1 4 3 2 12 3 22. Find the plane determined by the line x-xi_y-yi_z~zi a b c and its parallel through the point P2{x2 , 2/2 , ^2)- 23. Given two non-intersecting lines x — xi __ y — yi _ z- zx x — X2 _ y -y2 _ z — Zj , a\ b\ c\ a2 62 C2 find the plane passing through the first line and a parallel to the second; and the plane passing through the second line and a parallel to the first. 24. What is the condition that the two lines of Ex. 23 intersect ? 26. Find the distance from the diagonal of a cube to a vertex not on the diagonal. 26. Find the distance between the lines given in Ex. 23. 27. Show that the locus of the points whose distances from two fixed planes are in constant ratio is a plane. 28. Show that the plane (m — n)x +(n — l)y +(l — m)z = contains the line x/l = y/m = z/n and is perpendicular to the plane determined by the lines x/m = y/n = z/l and x/n = y/l = z/m. CHAPTER XII THE SPHERE 241. Spheres. A sphere is defined as the locus of all those points that have the same distance from a fixed point. Let C{h, j, h) denote the center, and r the radius, of a sphere ; the necessary and sufficient condition that any point P{Xj ?/, z) has the distance r from C{U, j, k) is (1) (^ - 7^)2 + (y -jY -Y{z- IcY = r^. This then is the cartesian equation of the sphere of center C(h, j, k) arid radius r. If the center of the sphere lies in the plane Oosy, the equa- tion becomes (x-hy + {y-jy + z'=r\ If the center lies on the axis Ox, the equation is (x-hy-\-y^-\-z^ = r\ The equation of a sphere about the origin as center is : 242. Expanded Form. Expanding the squares in the equa- tion (1), we find the equation of the sphere in the form x'^-\.y^-{.z'^-2hx- 2jy -2kz-^h'' +f-\-k'^-r^ = 0. This is an equation of the second degree in x, y, z ; but it is of a particular form. The general equation of the second degree in x, y, z is (2) Ax"^ + By^ -\-Cz^^2 Dyz + 2Ezx-[-2 Fxy -i-2Gx-\-2Hy-{-2Iz-i-J=0', 229 230 SOLID ANALYTIC GEOMETRY [XII, § 242 i.e. it contains a constant term J; three terms of the first degree, one in x, one in y, and one in z ; and six terms of the second degree, one each in x^, y^, z^j yz, zx, and on/. If in (2) we have D = E = F= 0, A = B=C^O,\t reduces, upon division by A, to the form a;2-f.7/2 + 2;2-f_-x+— 2/-I-— Z + — = 0, which agrees with the above form of the equation of a sphere, apart from the notation for the coefficients. 243. Determination of Center and Radius. To determine the locus represented by the equation (3) Ax^ -\- Ay" + Az^ ■\-2 Gx^-2 Hy -\-2 Iz -^r J^Q, where A^ G, II, /, J are any real numbers while ^ =^ 0, we divide by A and complete the squares in x, y, z; this gives The left side represents the square of the distance of the point {x, y, z) from the point (— G/A, —H/A, — I/A)] the right side is constant. Hence, if the right side is positive, the equa- tion represents the sphere whose center has the coordinates (— G/A, — H/A, — I/A), and whose radius is r = -WG^ + H^-\-P-AJ. A If, however, G^ + H^ + I^< AJ, the equation is not satisfied by any point with real coordinates. If G^ -|- H^ -{-/* = AJ, the equation is satisfied only by the coordinates of the point (-G/A,-H/A,-I/A). Thus the equation of the second degree Ax^ + By^-j-Cz^ + 2 Dyz -\-2Ezx-{-2 Fxy + 2Gx-{-2Hy + 2Iz-i-J=0, represents a sphere if, and only if, A=B=C^O, D=:E = F=0, G^ + IP-\-P>AJ. XII, § 244] THE SPHERE 231 244. Essential Constants. The equation (1) of the sphere contains four constants : li, j^ k, r. The equation (2) contains five constants of which, however, only four are essential since we can divide out by one of these constants. Thus dividing by A and putting 2 0/A = a, 2 H/A = 6, 2 I/A = c, J/ A = d, the general equation (2) assumes the form x^ + y'^ -\- z^ + ax -[-hy -\- cz -\- d = 0, with only the four essential constants a, 6, c, d. This fact corresponds to the possibility of determining a sphere geometrically, in a variety of ways, by four conditions. EXERCISES 1. Find the spheres with the following points as centers and with the indicated radii : (a) (4, -1,2), 4; (6) (0,0, 4), 4; (c) (2,-2, 1), 3 ; {d) (3, 4, 1), 7. 2. Find the following spheres : («) with the points (4, 2, 1) and (3, — 7, 4) as ends of a diameter ; (6) tangent to the coordinate planes and of radius a ; (c) with center at the point (4, 1, 5) and passing through (8, 3, — 5). 3. Find the centers and the radii of the following spheres : (a) a;2 + 2/2 +-^2 _3a; + 5y_6^ + 2 = 0. (6) a:2 + y2 _|. 2;2 - 2 6a; + 2 c;? - 62 _ c2 = 0. (c) 2 x2 + 2 1/2 + 2 2;2 -^ 3 a; - y + 5 - 11 = 0. {d) x^-{-y'^ + z^-x-y -z = Q. 4. Show that the equation A{x'^ ■\- y"^ + z"^) + 2 Ox + 2 Hy -^ 2 Iz + J = 0, in which J is variable, represents a family of concentric spheres. 5. Find the spheres that pass through the following points : (a) (1, 1, 1), (3, - 1, 4), (- 1, 2, 1), (0, 1, 0). (6) (0, 0, 0), (a, 0, 0), (0, 6, 0), (0, 0, c). (c) (0, 0, 0), (- 1, 1, 0), (1, 0, 2), (0, 1, - 1). (d) (0, 0, 0), (0, 0, 4), (3, 3, 3), (0, 4, 0). 6. Find the center and radius of the sphere that is the locus of the points three times as far from the point (a, 6, c) as from the origin. 7. Show that the locus of the points, the ratio of whose distances from two given points is constant, is a sphere except when the ratio is unity. 232 SOLID ANALYTIC GEOMETRY [XII, § 244 8. Find the positions of tlie following points relative to the sphere x^ + y^ + z'^-4x-\-^y-2z = 0] (a) the origin, (6) (2, -2, 1), (c) (1,1,1), (d) (3, -2,1). 9. Find the positions of the following planes relative to the sphere (a)4x-\-2y + z-\-2 = 0, (b)8x-y-iz-\-5 = 0. 10. Find the positions of the following lines relative to the sphere of Ex. 9: (a)2x-y + 2z + 7=0, Zx- y-z-lOz=0. (6)3x + 8y + 5r-9=0, a;-8y+«+ll = 0. 245. Equations of a Circle. In solid analytic geometry a curve is represented, by two simultaneous equations (§ 221), that is, by the equations of any two surfaces intersecting in the curve. Thus two linear equations represent together the line of intersection of the two planes represented by the two equations taken separately (§§ 233, 237). A linear equation together with the equation of a sphere, ^ ^ x^ -\- y^ -\- z^ -{- ax -j- by + cz -{- d = 0, represents the locus of all those points, and only those points, which the plane and sphere have in common. Thus, if the plane intersects the sphere, these simultaneous equations rep- resent the circle in which the plane cuts the sphere; if the plane is tangent to the sphere, the equations represent the point of contact; if the plane does not intersect or touch the sphere, the equations are not satisfied simultaneously by any real point. 246. Sections Perpendicular to Axes. Projecting Cylinders. In particular, the simultaneous equations (6) z — k, a^ -\- y^ -\- z^ = r^ represent, if A: < r, a circle about the axis Oz (i.e. a circle whose center lies on Oz and whose plane is perpendicular to Oz). If the value of z obtained from the linear equation be XII, § 247] THE SPHERE 233 substituted in the equation of the sphere, we obtain an equation in X and ?/, oc^ -{- 1/ — r^ — Tc^, which represents (since z is arbitrary) the circular cylinder, about Oz as axis, which pro- jects the circle (5) on the plane Oxy. Interpreted in the plane Oxy^ i.e. taken together with z = 0, this equation represents the projection of the circle (5) on the plane Oxy. Similarly if we eliminate x ov y or 2 between the equations (4), we obtain an equation in y and z, z and x, or x and y, rep- resenting the cylinder that projects the circle (4) on the plane OyZy Ozx, or Oxy, respectively. 247. Tangent Plane. The tangent plane to a sphere at any point Pi of the sphere is the plane through Pj, at right angles to the radius through Pj . ¥oY a sphere whose center is at the origin, a;^ + 2/^ -f 2^ _ ^^ the equation of the tangent plane at Pi(x^, yi, z^) is found by observing that its distance from the origin is r and that the direction cosines of its normal are those of OPi, viz. Xi/r, yi/r, Zi/r. Hence the equation (6) x^x + y{y + 212; = r\ If the equation of the sphere is given in the general form ^(a;2 +2/2+ 22)4. 2 Gx + 2 Hy +2Iz -\-J= 0, we obtain by transforming to parallel axes through the center the equation the tangent plane at Fi(xi, yi, Zi) then is . x,x + 2/12/ + ^i2 = — + — + — - -. Transforming back to the original axes, we have : A^ A^ A^ a' 234 SOLID ANALYTIC GEOMETRY [XII, § 247 Multiplying out and rearranging, we find that the equation of the tangent plane to the sphere Aix" + y"" + z") + 2 Gx + 2 Hy + 2 Iz + J = at the point Pi(a^, yi, Zi) is A{x^xJtyiy^-Ziz) + 0(x^^x) 4-fi-(2/i+y)4- 7(21+2)+ J= 0. 248. Intersection of Line and Sphere. The intersections of a sphere about the origin, x^ -^y^ + z^ = r\ with a line determined by two of its points Pi(xi, yi, Zi) and PiiXi, 2/2> ^2)} a-iid given in the parameter form [(6), § 239] x = x^ + k{x2-x{), y = yi + k(y2-yi)j 2 = % + ^(za - Zi), are found by substituting these values of a:, y, z in the equation of the sphere and solving the resulting quadratic equation in k : Ix, + k(x, - x{)Y + [2/1 + k(y, - 2/0]^ + [^1 + k{z, - z,)Y = r», which takes the form l(x, - x,y + (y, - y,y + (z, - z,y^ k' + 2 [x,{x, - x,)+ y,(y, - y,) + Zi (22 - Zi)] k + W + 2/1' + z,' - r2) = 0. The line P1P2 will intersect the sphere in two different points, be tangent to the sphere, or not meet it at all, according as ^C^f^ ^^ the roots of this equation in k are real and ^x^ different, real and equal, or imaginary ; i.e. according as ^^^' ^^ lx,{x,-x,)+y,(y,-y,)+z,(z,-z,)J-(P{x,^-^y,^+z,^)+(Pr''^0, where d denotes the distance of the points Pi and Pj- Divid- ing by cP, we can write this condition in the form where by § 239 the quantity in square brackets is the square of the distance S from the line P^Pz to the origin (Fig. 123). XII, § 249] THE SPHERE 235 Our condition means therefore that the line P^P^ meets the sphere in two different points, touches it, or does not meet it at all according as r > 8, r = 8, r < 8, which is obvious geomet- rically. 249. Tangent Cone. The condition for the line P^P^ to be tangent to the sphere is (§ 244) : W+ yi'-hzi'-r%)i{-x,y H- (y, - y,y + (22 - z^yy To give this expression a more symmetric form, let us put, to abbreviate, ^1^2 + 2/i2/2 + 2=122 = P, ^i + yi + 21^ = gi, x^ -h 2/2' -f z^ = ^2, so that the condition is i.e. p^-2 r^p = q^q^ - r^q^ - r% ; adding r* in both members, we have i.e. (x,x, + 2/12/2 + 21^2 - r^ = (o^i^ + 2/1' + 2i^ - r')(x2' + 2/2' + z,^ - r»). Now keeping the sphere and the point Pj fixed, let Pj vary subject only to this condition, i.e. to the condition that P1P2 shall be tangent to the sphere; the point Pj, which we shall now call P(x, y, z) is then any point of the cone of vertex P^ tangent to the sphere. Hence the equation of the cone of vertex Pi (^1 ) 2/1 J 2i) tangent to the sphere x^-\-y'^-\- z^ (a^i' + 2/1' + z^ - r2)(a^ + y2 + 2;^ _ ^) ^ (^^^ ^ 2/i2/ ^-z^z - rj. If, in particular, the point Pi is taken on the sphere so that x^ + yx + z^ — r\ the equation of the tangent cone reduces to the form x^x + y^y + z^z = r^, which represents the tangent plane at Pi. 236 SOLID ANALYTIC GEOMETRY [XII, §250 250. Inversion. A sphere of center and radius a being given, we can find to every point P of space (excepting 0) one and only one point P' on OP (produced if necessary) such that OP ■ OP' = a^. The points P, P' are said to be inverse to each other with respect to the sphere (compare § 56). Taking rectangular axes through O, we find as the relations between the coordinates of the two inverse points P(x, y, z) and P'(x', y', z') if we put OP = r = Vx^ + y^ + z'\ OP' = r' = Vx'=^ + y'-^ + z'^ : X _y' _z' _r' _rr' _a'^ , X y z r r^ r^ ' hence z' - "^"^ v'- ^ z' - — • hence x-^^^^^^^^, ^'-^2 + ^2 + ^2' '-^^^y2^^2' and similarly a^x' .. _ aY „ _ a^z' X = x'-i ^ y'i + z'i' " x'2 + y'i + z'^' x'-^ + y'^ + z'^ These equations enable us to find to any surface whose equation is given the equation of the inverse surface, by simply substituting for x, y, z their values. Thus it can be shown, that by inversion every sphere is transformed into a sphere or a plane. The proof is similar to the corresponding propo- sition in plane analytic geometry (§57) and is left as an exercise. EXERCISES 1. Find the radius of the circle which is the intersection : (a) of the plane y = 6 with the sphere x"^ -\- y^ -\- z^ — 6y = ] (6) of the plane 2x — Sy + z-2 = with the sphere x"^ + y^ + z^ -6x + 2y - 16 = 0. 2. A line perpendicular to the plane of a circle through its center is called the axis of the circle. Find the circle : (a) which lies in the plane « = 4, has a radius 3 and Oz as axis ; (6) which lies in the plane y = 6, has a radius 2 and the line x — 3 = 0, — 4=0as axis. 3. Find the circles of radius 3 on the sphere of radius 4 about the origin whose common axis is equally inclined to the coordinate axes. 4. Does the Une joining the points (2,-1,-6), (-1, 2, 3) intersect the sphere x^ -\-y^ + z^ = 10? Find the points of intersection. XII, §251] THE SPHERE 237 5. Find the planes tangent to the following spheres at the given points : (a) a;2 + 2/2 + ^2 _ 3 ^ _ 5 ^ _ 2 = 0, at (2, - 1, 3) ; (&) x2 + y2 ^ 02 ^ 2 X - 6 ?/ + 2 -1 = 0, at (0, 1, - 3) ; (c) 3ix^ + y^-\-z^)-5x + 2y -z = 0, at the origin; (d) x^ -^ y^ + z'^— ax - bij - cz - 0, at (a, b, c). 6. Find the tangent cone : (a) from (4, 1, — 2) to a;2 + y2 ^ ^2 = 9 ; (6) from (2 a, 0, 0) to x^ + y^ + z"^ = a^ ; (c) from (4, 4, 4) to a;2 + y2 + ^^2 = 16 ; (d) from (1, - 5, 3) to x'^ + y^ + z^ = 9. 7. Find the cone with vertex at the origin tangent to the sphere (x-2 a)2+ y'i + z2 = a\ 8. Show that, by inversion with respect to the sphere x^ + y'^ -^ z^ = a^, every plane (except one through the center) is transformed into a sphere passing through the origin. 9. With respect to the sphere x^ -i- y^ + z^ = 25, find the surfaces in- verse to (a) x = 6, (6) x-y = 0, (c) 4 (a;2 + y2 4. ^2) _ 20 a; — 25 = 0. 10. Show that by inversion with respect to the sphere x'^ -\- y^ -\- z^ = a^ every line through the origin is transformed into itself. 11. With respect to the sphere x^ -\-y^ -\- z^ = a^^ find the surface in- verse to the plane tangent at the point Pi {xi ,^yi , zi). 12. Show that all spheres with center at the center of inversion are transformed into concentric spheres by inversion. 13. What is the curve inverse to the circle x^ + y^ + z^ = 25, = 4, with respect to the sphere x^ + y^ -^ z^ = 16? 251. Poles and Polars. Let P and P' be inverse points with respect to a given sphere ; then the plane w through P', at right angles to OP ( being the center of the sphere) , is called the polar plane of the point P, and P is called the pole of the plane tt, with respect to the sphere. With respect to a sphere of radius a, with center at the origin^ x2 + y2 _|_ 2-2 = a2, the equation of the polar plane of any point Pi(iCi, 2/1, z{) is readily found by observing that its distance from the origin is a2/ri, and that the 238 SOLID ANALYTIC GEOMETRY [XII, §251 direction cosines of its normal are equal to Xi/ri, yi/n, ^i/n, where n^ = xi^ -\- y\^ + z^ ; the equation is therefore xix + Viy + ziz = a^- If, in particular, the point Pi lies on the sphere, this equation, by § 254 (6), represents the tangent plane at Pi. Hence the polar plane of any point of the sphere is the tangent plane at that point ; this also follows from the definition of the polar plane. 262. With respect to the same sphere the polar planes of any two points Pi(a;i , yi , zi) and P2(X2 , yz , ^2) are Xix + yiy + ziz = a^ and xix + yzy + z^z = a*. Now the condition for the polar plane of Pi to pass through P2 is v.xXi + yi2/2 + z^zi = a?- ; but this is also the condition for the polar plane of P2 to pass through Pi. Hence the polar planes of all the points of any plane w (not passing through the origin O) pass through a common pointy namely, the pole of the plane ir ; and conversely, the poles of all the planes through a com- mon point P lie in a plane^ namely, the polar plane of P. 263. The polar plane of any point P of the line determined by two given points Pi(xi , j/i , zi) and P2(X2 , 2/2 , 22) (always with respect to the same sphere x^ -\- y^ + z^ = a^) is Ixi + k{x2 - xi)]x + [yi + k{y2 - t/i)]y + \.zi-\-k{z2 - zi)^z = a^. This equation can be written in the form Xix + yiy + ziz — a^ + — — • (xzx + y2y + Z2Z — a^) = 0, which for a variable k represents the planes of the pencil whose axis is the intersection of the polar planes of Pi and P2. Hence the polar planes of all the points of a line X pass through a common line ; and conversely, the poles of all the planes of a pencil lie on a line. Two lines related in this way are called conjugate lines (or conjugate axes, reciprocal polars). Thus the line P1P2 XII, § 256] THE SPHERE 239 and the line xix + yiy + ziz = a^, X2X + y2y + Z2Z = a^ are conjugate with respect to the sphere x^ •{■ y^ -\- s^ = a*. As the direction cosines of these lines are proportional to X2 — X1, y2- yi, Z2- z\ and y\ zx Z\ Xx xi y\ y2 Z2 02 X2 X2 y2 respectively, the two conjugate lines are at right angles (§236). 254. By the method used in the corresponding problem in the plane (§60) it can be shown that the polar plane of any point Pi(cci, yi, zi) with respect to any sphere A(pfi + 2/2 + 02) ^. 2 G^a; + 2 ^y + 2 70 + J'= is A{xix + y\y + 010) + G(xi + x) + H{yi + y) + /(01 4- 0) + J'= 0. 255. Power of a Point, if in the left-hand member of the equation of the sphere (X- /i)2 + {y - j)2 + (^z-kY-r^ = we substitute for x, y, z\ the coordinates xi , yi , 01 of any point not on the sphere, we obtain an expression (xi — hy + (yi — J)^+ {z\ — ky — r^ different from zero which is called the power of the point P\(x\ ^ yi, 01) with respect to the sphere. As {xi — hy + {y\ — j)'^ + (01 — A:)2 is the square of the distance d be- tween the point Pi and the center C of the sphere, we can write the power of Pi briefly ^2 - r2 ; the power of Pi is positive or negative according as Pi lies outside or within the sphere. For a point Pi outside, the power is evidently the square of the length of a tangent drawn from Pi to the sphere. 256. Radical Plane, Axis, Center. The locus of a point whose powers with respect to the two spheres x2 + y2 + ^2 ^_ a-iX + hiy + ci0 + di-O, x^+y^ + z'^ + a2x + b2y + C20 + c?2 = are equal is evidently the plane (ai — a2)x + (61 — 62)2/ + (ci - 02)0 + di — d2 = 0, which is called the radical plane of the two spheres. It always exists un- less the two spheres are concentric. 240 SOLID ANALYTIC GEOMETRY [XII, § 256 It is easily proved that the three radical planes of any three spheres (no two of which are concentric) are planes of the same pencil (§ 228) ; and hence that the locus of the points of equal power with respect to three spheres is a straight line. This line is called the radical axis of the three spheres ; it exists unless the centers lie in a straight line. The six radical planes of four spheres, taken in pairs, are in general planes of a sheaf (§229). Hence there is in general but one point of equal power with respect to four spheres. This point, the radical center of the four spheres, exists unless the four centers lie in a plane. 257. Family of Spheres. The equation (x^-\-y^-\-z^+ aix -\-biy-\-ciz-\-di) -{-k(ix^+y^+z^-+a2X-\-b2y-\-C2Z-{-d2) =0 represents a family, or pencil, of spheres, provided k =^—1. If the two spheres x^ + y^ + z^ + aix + biy + ciz -|- di = 0, «2 + y2 + «2 ^a2pc + b2y + C2Z + d2 = intersect, every sphere of the pencil passes through the common circle of these two spheres. It k =— 1, the equation represents the radical plane of the two spheres. EXERCISES 1. Find the radius of the circle in which the polar plane of the point (4, 3,-1) with respect to x^-{-y^+Z' = 16 cuts the sphere. 2. Find the radius of the circle in which the polar plane of the point (6, — 1, 2) with respect to x^ + y^ -\- z^ — 2x -\- 4y = cuts the sphere. 3. Show that the plane 3a; + y — 4^ = 19 is tangent to the sphere x^ + y^ + z^ — 2x — iy — 6z—l2 = 0, and find the point of contact. 4. If a point describes the plane 4a; — 5y — 3a: = 16, find the coordi- nates of that point about which the polar plane of the point turns with respect to the sphere x^ +y^ + z'^ = 16. 5. If a point describes the plane 2x-\-Sy-\-z = i, find that point about which the polar plane of the point turns with respect to the sphere X2 + y2 + 2!2 = 8. 6. If a point describes the line ^ ~ = ^ = ^ ~ , find the equa- 3 5-2' tions of that line about which the polar plane of the point turns with XII, § 257] THE SPHERE 241 respect to the sphere x^ -\-y'^ + z^ = 25. Show that the two lines are perpendicular. 7. If a point describe the line 2x - Sy -j- iz = 2, x + y -{■ z = 3, find the equations of that line about which the polar plane of the point turns with respect to the sphere x^ + y^ + z^ = 16. Show that the two lines are perpendicular. 8. Find the sphere through the origin that passes through the circle of intersection of the spheres x"^ -\- y'^ -\- z^ — 3x + 4 y — 6z — S = 0^ x^ -^ y^ + z^ - 2 X -\- y - z - 10 = 0. 9. Show that the locus of a point whose powers with respect to two given spheres have a constant ratio is a sphere except when the ratio is unity. 10. Show that the radical plane of two spheres is perpendicular to the line joining their'centers. 11. Show that the radical plane of two spheres tangent internally or externally is their common tangent plane. 12. Find the equations of the radical axis of the spheres x^ -\- y^-{- z^ -3x-2y -z-4: = 0, x^-\-y^ + z^-\-5x~Sy-2z-8 = 0, x'^ -\- y^ -\-z^-16 = 0. 13. Find the radical center of the spheres x^ -i-y'^ -^ z^ — 6x ■j-2y - z + e = 0, x^ -\- y^ + z^ - 10 = 0, x^ + y^ + z^ + 2x - Sy + 5 z - 6 = 0, ^2 + ?/2 + 02 _ 2 X + 4 ?/ - 12 = 0. 14. Show that the three radical planes of three spheres are planes of the same pencil. 15. Two spheres are said to be orthogonal when their tangent planes at every point of their circle of intersection are perpendicular. Show that the two spheres x"^ -}- y^ -{- z"^ + a^x + b^y + c^z + di = 0, x^ 4. ^2 _|_ ^2 + a^x + h^y + C2« + 0^2 = are orthogonal when a\ai + 6162 + C1C2 = 2{di + d2). 16. Write the equation of the cone tangent to the sphere x^ + y^ + ^2 — fi with vertex (0, 0, zi). Divide this equation by zi^ and let the vertex recede indefinitely, i.e. let z\ increase indefinitely. The equation 3.2 _|_ ^2 _ ^2^ thus obtained, represents the cylinder with axis along the axis Oz and tangent to the sphere x^ + i/^ 4. ^2 _ ,.2^ B CHAPTER XIII QUADRIC SURFACES 258. The Ellipsoid. The surface represented by the equation is called an ellipsoid. Its shape is best investigated by tak- ing cross-sections at right angles to the axes of coordinates. Thus the coordinate plane Oyz whose equation is x = in- tersects the ellipsoid in the ellipse Any other plane perpendicular to the axis Ox (Fig. 125) at FiQ. 125 the distance h < a from the plane Oyz intersects the ellipsoid in an ellipse whose equation is i.e. y^ , 2' _j '■('-5)"<'-S) 242 XIII, §259] QUADRIC SURFACES 243 Strictly speaking this is the equation of the cylinder that pro- jects the cross-section on the plane Oyz. But it can also be interpreted as the equation of the cross-section itself, referred to the point (/i, 0, 0) as origin and axes in the cross-section parallel to Oy and Oz. Notice that as ^ < a, W^/o?, and hence also 1 — h^/a}, is a posi- tive proper fraction. The semi-axes 6Vl — h^/o?, c VI — h^/a} of the cross-section are therefore less than h and c, respec- tively. As h increases from to a, these semi-axes gradually diminish from h, c to 0. 259. Cross-Sections. Cross-sections on the opposite side of the plane Oyz give the same results; the ellipsoid is evi- dently symmetric with respect to the plane Oyz. By the same method we find that cross-sections perpendicu- lar to the axes Oy and Oz give ellipses with semi-axes dimin- ishing as we recede from the origin. The surface is evidently symmetric to each of the coordinate planes. It follows that the origin is a center, i.e. every chord through that point is bisected at that point. In other words, if (x, y, z) is a point of the surface, so is (—a?, —2/, —z). Indeed, it is clear from the equation that if (.t, y, z) lies on the ellipsoid, so do the seven other points {x, y, -z), {x, —y, z), (-x, y, z), (x, -y, -z), (- x, y, - 2), (- a;, - y, z), {-x, -y, -z). A chord through the center is called a diameter. It follows that it suffices to study the shape of the portion of the surface contained in one octant, say that contained in the tri- hedral formed by the positive axes Ox, Oy, Oz ; the remaining portions are then obtained by reflection in the coordinate planes. The ellipsoid is a dosed surface; it does not extend to in- finity ; indeed it is completely contained within the parallel- epiped with center at the origin and edges 2 a, 2 6, 2 c, parallel to Ox, Oy, Oz, respectively. 244 SOLID ANALYTIC GEOMETRY- [XIII, § 260 260. Special Cases. In general, the semi-axes a, b, c of the ellipsoid, i.e. the intercepts made by it on the axes of coordi- nates, are different. But it may happen that two of them, or even all three, are equal. In the latter case, i.e. if a = b = c, the ellipsoid evidently reduces to a sphere. If two of the axes are equal, e.g. if 6 = c, the surface ^ + ^ + 51 = 1 a" b^ b^ is called an ellipsoid of revolution because it can be generated by revolving the ellipse n ' 62 1 Fig. 126 about the axis Ox (Fig. 126). Any cross-section at right angles to Ox, the axis of revolution, is a circle, while the cross-sections at right angles to Oy and Oz are ellipses. The circular cross-section in the plane Oyz is called the equator ; the intersections of the surface with the axis of revolution are the poles. li a ^b {a being the intercept on the axis of revolution), the ellipsoid of revolution is called prolate ; if a < b, it is called oblate. In astronomy the ellipsoid of revolution is often called spheroid, the surfaces of the planets which are approximately ellipsoids of revolution being nearly spherical. Thus for the surface of the earth the major semi-axis, i.e. the radius of the equator, is 3962.8 miles while the minor semi- axis, i.e. the distance from the center to the north or south pole, is 3949.6 miles. Xm, §261] QUADRIC SURFACES 245 261. Surfaces of Revolution. A surface that cau be gen- erated by the revolution of a plane curve about a line in the plane of the curve is called a surface of revolution. Any such surface is fully determined by the generating curve and the position of the axis of revolution with respect to the curve. Let us take the axis of revolution as axis Ox, and let the equation of the generating curve be As this curve revolves about Ox, any point P of the curve (Fig. 127) de- scribes a circle about Ox as axis, "et rTTisr , with a radius equal to the ordinate / f i / f(x) of the generating curve. For ^ \ '/ any position of P we have therefore Fig. 127 f +''■'= If i^m and this is the equation of the surface of revolution. Thus if the ellipse revolves about the axis Ox, we find since y = ± {h/a)^a^ — o? for the ellipsoid of revolution so generated the equation 2/2 + 22_5!(a2-a;2), a which agrees with that of § 260. Any section of a surface of revolution at right angles to the axis of revolution is of course a circle ; these sections are called parallel circles, or simply parallels (as on the earth's surface). Any section of a surface of revolution by a plane passing through the axis of revolution is called a meridian section ; it consists of the generating curve and its reflection in the axis of revolution. 246 SOLID ANALYTIC GEOMETRY [XIII, § 261 EXERCISES 1. An ellipsoid has six/oci, viz. the foci of the three ellipses in which the ellipsoid is intersected by its planes of symmetry. Determine the coordinates of these foci : (a) for an ellipsoid with semi-axes 1, 2, 3 ; (6) for the earth (see § 260) ; (c) for an ellipsoid of semi-axes 10, 8, 1 ; (d) for an ellipsoid of semi-axes 1, 1, 5. 2. Find the equations of the surfaces of revolution generated by re- volving the following curves about the given lines : (a) y = x", about the axis Ox. (b) y- = X, about the latus rectum. (c) a;2 _j_ y2 _ 2 ac = 0, about the axis Oy. (d) x2 _ y2 _ 1^ about the axis Ox. 3. Find the equation of the paraboloid of revolution generated by the revolution of t-he parabola y^ = 4 ox about Ox. 4. Find the equation of a torus, or anchor-ring, i.e. the surface generated by the revolution of a circle of radius a about a line in its plane at the distance & > a from its center. 5. Find the equation of the surface generated by the revolution of a circle of radius a about a line in its plane at the distance 6 < a from its center. Is the appearance of this surface noticeably different from the surface of Ex. 4 ? What happens to this surface when 6 = 0; when b = a? 6. Find the equation of the surface generated by the revolution of the parabola y^ = iax about : (a) the tangent at the vertex ; (6) the latus rectum. 7. Find the equation of the surface generated by the revolution of the hyperbola xy = a^ about an asymptote. 8. Find the cone generated by the revolution of the line y = mx -\- b about: (a) Ox, (6) Oy. 9. How are the following surfaces of revolution generated ? (a) y2+22=x*. (6) 2x2-f-2y2-32=0. (c) x^-\-y^-z^-2x+i=0. 10. Find the equation of the surface generated by the revolution of the ellipse x^ 4- 4 y2 _ 4 a; = o : (a) about the major axis ; (b) about the minor axis ; (c) about the tangent at the origin. XIII, § 263] QUADRIC SURFACES 247 262. Hyperboloid of One Sheet. The surface represented by the equation is called a hyperboloid of one sheet (Fig. 128). The intercepts Fig. 128 on the axes Ox, Oy are ± a, ± 6 ; the axis Oz does not intersect the surface. 263. Cross-Sections. The plane Oxy intersects the surface in the ellipse cross-sections perpendicular to Oz give ellipses with ever- increasing semi-axes. The planes Oyz and Ozx intersect the surface in the hyperbolas ^_?. — 1 ^__ — 1 62 ^2~ ' a2 c2~ Any plane perpendicular to Ox, at the distance h from the origin, intersects the hyperboloid in a hyperbola, viz. f "(•-i) <'-3 1. 248 SOLID ANALYTIC GEOMETRY . [XIII, § 263 As long as ^ < a this hyperbola has its transverse axis parallel to Oy while for 1i'>a the transverse axis is parallel to Oz ; for h = a the equation reduces to y''-/h'^ — z^l& = and represents t"wo straight lines, viz. the parallels through (a, 0, 0) to the asymptotes of the hyperbola yV^* ~ ^V^'^ = ^ which is the intersection of the surface with the plane Oyz. Similar considerations apply to the cross-sections perpen- dicular to Oy. The hyperboloid has the same properties of symmetry as the ellipsoid (§ 259) ; the origin is a center , and it suffices to inves- tigate the shape of the surface in one octant. 264. Hyperboloid of Revolution of One Sheet. If in the hyperboloid of one sheet we have a = b, the cross-sections per- pendicular to the axis Oz are all circles so that the surface can be generated by the revolution of the hyperbola about Oz. Such a surface is called a hyperboloid of revolution of one sheet. 265. Other Forms. The equations a* 6» c2 ' a* b^ c^ also represent hyperboloids of one sheet which can be investi- gated as in §§ 262-264. In the former of these the axis Oy, in the latter the axis Ox, does not meet the surface. Every hyperboloid of one sheet extends to infinity. 266. Hyperboloid of Two Sheets. The surface represented by the equation a" b^ c2 is called a hyperboloid of two sheets (Fig. 129). XIII, § 269] QUADRIC SURFACES 249 The intercepts on Ox are ± a ; the axes Oy, Oz do not meet the surface. 267. Cross-Sections. The cross-sections at right angles to Ox, at the distance h from the origin are '■('-i)"<'-S)" these are imaginary as long as h < a; for h>a they are ellipses with ever- increasing semi-axes as we recede from the origin. The cross-sections at right angles to Oy and Oz are hyperbolas. The hyperboloid of two sheets, like that of one sheet and like the ellipsoid, has three mutually rectangular planes of symmetry whose intersection is therefore a center. The surfaces z^ -, x^ Fig. 129 ^2 ^ Z^_ ^ ^=1 are hyperboloids of two sheets, the former being met by Oy, the latter by Oz, in real points. The hyperboloid of two sheets extends to infinity. 268. Hyperboloid of Revolution of Two Sheets. If 6 = c in the equation of § 266, the cross-sections at right angles to Ox are circles and the surface becomes a hyperboloid of revolution of two sheets. 269. Imaginary Ellipsoid. The equation = 1 a2 b^ c2 is not satisfied by any point with real coordinates. It is some- times said to represent an imaginary ellipsoid. 250 SOLID ANALYTIC GEOMETRY [XIII, §270 270. The Paraboloids. — 4- ^ = 2 cz. '2^62 -^ = 2c2, The surfaces a-' 0== a2 which are called the elliptic paraboloid (Fig. 130) and hyper- bolic paraboloid (Fig. 131), respectively, have each only two planes of symmetry, viz. the planes Oyz and Ozx. We here assume that c=^0. The cross-sections at right angles to the Fig. 130 axis Oz are evidently ellipses in the case of the elliptic parab- oloid, and hyperbolas in the case of the hyperbolic paraboloid. The plane Oxy itself has only the origin in common with the elliptic paraboloid ; it intersects the hyperbolic paraboloid in the two lines x^/a?' — y^jW- — 0, i.e. y = ± hx/a. The intersections of the elliptic paraboloid (Fig. 130) with the planes Oyz and Ozx are parabolas with Oz as axis and as vertex, opening in the sense of positive 2; if c is positive, in the sense of negative z if c is negative. Planes parallel to these coordinate planes intersect the elliptic paraboloid in parabolas with axes parallel to Oz, but with vertices not on the axes Oa;, Ot/, respectively. For the hyperbolic paraboloid (Fig. 131), which is saddle- shaped at the origin, the intersections with the planes Oyz and XIII, § 273] QUADRIC SURFACES 251 Ozx are also parabolas with Oz as axis ; if c is positive the parabola in the plane Oyz opens in the sense of negative z, that in the plane Ozx opens in the sense of positives. Similarly for the parallel sections. 271. Paraboloid of Revolution. If in the equation of the elliptic paraboloid we have a = 6, it reduces to the form x^-\-y^ = 2pz. This represents a surface of revolution, called the paraboloid of revolution. This surface can be regarded as generated by the revolution of the parabola y^ = 2pz about the axis Oz. 272. Elliptic Cone. The surface represented by the equation ^ f a^ b^ = is an elliptic cone, with the origin as vertex and the axis Oz as axis (Fig. 132). The plane Oxy has only the origin in common with the surface. Every parallel plane z = k, whether Ic be positive or negative, intersects the surface in an ellipse, with semi-axes increasing proportionally to k. The plane Oyz, as well as the plane Ozx, intersects the surface in two straight lines through the origin. Every plane parallel to Oyx or to Ozx intersects the surface in a hyperbola. Fia. 132 273. Circular Cone. If in the equation of the elliptic cone we have a = b, the cross-sections at right angles to the axis Oz become circles. The cone is then an ordinary circular cone, or 252 SOLID ANALYTIC GEOMETRY [XIII, § 273 cone of revolution, which can be generated by the revolution of the line y = (a/c)z about the axis Oz. Putting a/c = m we can write the equation of a cone of revolution about Oz, with vertex at 0, in the form 274. Quadric Surfaces. The ellipsoid, the two hyper- boloids, the two paraboloids, and the elliptic cone are called quadric surfaces because their cartesian equations are all of the second degree. Let us now try to determine, conversely, all the various loci that can be represented by the general equation of the second degree Ax^ + By^ +Cz'^ + 2 Dyz -h 2 Ezx -h 2 Fxy 'Jt2Gx-\-2Hy + 2Iz + J=0. In studying the equation of the second degree in x and y (§ 253) it was shown that the term in xy can always be removed by turning the axes about the origin through a cer- tain angle. Similarly, it can be shown in the case of three variables that by a properly selected rotation of the coordinate trihedral about the origin the terms in yx, zx, xy can in general all be removed so that the equation reduces to the form (1) Ax^ + By^ + C^^ + 2 Gx + 2 JJy +2 J» -H «/■= 0. This transformation being somewhat long will not be given here. We shall proceed to classify the surfaces represented by equations of the form (1). 275. Classification. The equation (1) can be further sim- plified by completing the squares. TJiree cases may be distin- guished according as the coefficients A, By C are all three differ- ent from zero, one only is zero, or two are zero. XIII, §275] QUADRIC SURFACES 253 Case (a): A =^ 0, B ::^0, C^ 0. Completing the squares in X, y, z we find Referred to parallel axes through the point (— G/A, — H/B, — I/C) this equation becomes (2) Ax''-\-By^-\-Cz^ = J,. Case (6) : A=^0, B=^0, 0=0. Completing the squares in x and y we find (-i) ^^^ B A B ' If 1=^0, we can transform to parallel axes through the point (—G/A, — H/B, J2/2 I) so that the equation becomes (3) Ax" + By''-i-2Iz = 0. If, however, 7=0, we obtain by transforming to the point (-G/A,-II/B,0) (3') Ax'-\-By^=J,. Case (c) : A^O, B = 0, C = 0. Completing the square in X we have ^2 6?2 -(-SJ + 2 Hy + 2 Iz = ^ -J=J^. If H and / are not both zero, we can transform to parallel axes through the point (— G/A, J^/2 H, 0) or through (— G/A, 0, J3/2 /) and find (4) . Ax' + 2Hy + 2Iz = 0. If Zr= and 7= 0, we transform to the point (— G/A, 0, 0) so that we find (4') Ax''=J,. 254 SOLID ANALYTIC GEOMETRY [XIII, § 276 276. Squared Terms all Present, Case (a). We proceed to discuss the loci represented by (2). If J^ ^ 0, we can divide (2) by t/i and obtain : (a) if A/Ji , B/J^ , C/Ji are positive, an ellipsoid (§ 258) ; (fi) if two of these coefficients are positive while the third is negative, a Jiyperholoid of one sheet (§ 262) ; (y) if one coefficient is positive while two are negative, a hyperboloid of tivo sheets (§ 266) ; (8) if all three coefficients are negative, the equation is not satisfied by any real point (§ 269) ; . If J^ = 0, the equation (2) represents an elliptic cone (§ 272) unless A, B, C all have the same sign, in which case the origin is the only point represented. 277. Case (b). The equation (3) of §275 evidently fur- nishes the two paraboloids (§ 270) ; the paraboloid is elliptic if A and B have the same sign; it is hyperbolic if A and B are of opposite sign. The equation (S*), since it does not contain z and hence leaves z arbitrary, represents the cylinder, with generators parallel to Oz, passing through the conic Ax^ + By"^ =^ J^, As ^ and B are assumed different from zero, this conic is an ellipse if AfJ^ and 5/J2 are both positive, a hyperbola if AjJ^ and ^/c/, are of opposite sign, and it is imaginary if AjJ^ and -B/Jg are both negative. This assumes J^ 4^ 0. If J^ — 0, the conic degen- erates into two straight lines, real or imaginary ; the cylinder degenerates into two planes if the lines are real. 278. Case (c). There remain equations (4) and (4'). To sim- plify (4) we may turn the coordinate trihedral about Ox through an angle whose tangent is — HII\ this is done by putting -^H'+P ^/H' + P XIII, § 278] QUADRIC SURFACES 255 our equation then becomes It evidently represents a parabolic cylinder, with generators parallel to Oy. Finally, the equation (4') is readily seen to represent two planes perpendicular to Ox, real or imaginary, unless J3 = 0, in which case it represents the plane Oyz. EXERCISES 1. Name and locate the following surfaces : (a) a;2 + 2 ?/2 + 3^2 = 4. (Jb) x"^ + y'^ - hz -Q = 0. (c) x^ - y'^ + z"^ = 4.. (d) x2-y^ + z^-\-Sz + 6 = 0. (e) 2?/2 -4:^2 _ 5=0. (/) 2a;2 + y2_|.3^2 + 5_0. (g) 6;s2 + 2x2 = 10. ih) z^-9 = 0. (i) x2-y + l = 0. 0*) x^-y^-z^ + 6z = 9. (k) x^ + Sy'^ + z"^ -j- 4 z -\- 4 = 0. {I) z'^-hy -9 = 0. 2. The cone x2/a2 + yyb^ - 02/c2 = is called the asymptotic cone of the hyperboloid of one sheet a;2/a2 + ^2/52 _ ^2/c2 = 1. Show that as z increases the two surfaces approach each other, i.e. they bear a relation similar to a hyperbola and its asymptotes. 3. What is the asymptotic cone of the hyperboloid of two sheets ? 4. Show that the intersection of a hyperboloid of two sheets with any plane actually cutting the surface is an ellipse, parabola, or hyperbola. Determine the position of the plane for each conic. 5. Show that in general nine points deterinine a quadric surface and that the equation may be written as a determinant of the tenth order equated to zero. 6. Show that the surface inverse to the cylinder x^ -\- y^ = a^, with respect to the sphere ^2 4- y2 + ^2 _ ^52^ ig the torus generated by the rev- olution of the circle (y — a/2)2 + z^ = a^ about the axis Ox. 7. Determine the nature of the surface xyz = a^ by means of cross- sections. 256 SOLID ANALYTIC GEOMETRY [XIII, § 279 279. Tangent Plane to the Ellipsoid. The plane tangent to the ellipsoid a" h^ c^ can be found as follows (compare §§ 255, 256). The equa- tions of the line joining any two given points {x^ y^ z^ and {^ , 2/2 , 22) are x=^x^-{-k{Xi — x^\ 2/ = 2/i + %2-yi), z = 2!i + A:(22-2;i). This line will be tangent to the ellipsoid if the quadratic in k [_x, + k{x^-x,)J [yi-^k(y,-y,)J [z, + k(z,-z,)y ^^ a« "^ 6« "^ c« has equal roots. Writing this quadratic in the form nx^-x,y _^ (y,-y,y _^ (z,-z,y\ t \_ 0} b^ c* J o^ xiix^-x^) yijy^-y,) ■ z^jz^-z,) !. fx,^ y,^ z,^ ^\ ^ \_ a^ 6* o" \ ^^a^^h^^ c" J ' we find the condition a2 "^ 62 "^ c2 ; \o? 62 c2 )\ J {x,-x,y (y,-y,y , fe-^On W I yx' , z,^ .\ L a2 6* c2 JVa^ 62 c2 y If now we keep the point {x^ , 2/1 , ^i) fixed, but let the point (^2) .V2) 2:2) vary subject to this condition, it will describe the cone, with vertex (oq , 2/1 , 2i), tangent to the ellipsoid ; to indi- cate this we shall drop the subscripts of x^, y^y z^. If, in particular, the point {x^ , y^ , Zi) be chosen on the ellipsoid, we have ^' + .^' + !l = i, a2 62 c2 XIII, §281] QUADRIC SURFACES 257 and the cone becomes the tangent plane. The equation of the tangent plane to the ellipsoid at the point {x^^ , y^ , z^ is, therefore : a2 &2 "1" c2 280. Tangent Planes to Hj^erboloids. In the same way it can be shown that the tangent planes to the hyperboloids a2 62 c2 ' a2 52 (.2 at (a;i,2/i,2i) are a2 62 c2 > ^2 52 ^2 By an equally elementary, but somewhat longer, calculation it can be shown that the tangent plane to the quadric surface Ax'' + By^ + C0' + 2 D?jz + 2Ezx + 2 Fxy + 2Gx-\-2Hy + 2Iz-\-J=0 at (.Tj , 2/1 J 2;i) is : AxyX 4- %i2/ + C2i2 + Z)(yi2; + ^iV) + ^(^^i.^ + a?i2;) + F{x,y + ^/lO;) + 6?(x, + ir) + ^(2/1 + 2/) + /(^i + ^) + J^= 0, In particular, the tangent planes to the paraboloids tj^yl^2cz, t-t=.2cz a2 62 a" 62 are 281. Ruled Surfaces. A surface that can be generated by the motion of a straight line is called a ruled surface; the line is called the generator. The plane is a ruled surface. Among the quadric surfaces not only the cylinders and cones but also the hyperboloid of one sheet and the hyperbolic paraboloid are ruled surfaces. 258 SOLID ANALYTIC GEOMETRY [XIII, § 282 282. Rulings on a Hyperboloid of One Sheet. To show this for the hyperboloid a" b^ c2 ' we write the equation in the form and factor both members : Ci-t){l-H^!X'-l} It is then apparent that any point whose coordinates satisfy the two equations be \ aj b a k\ aj where k is an arbitrary parameter, lies on the hyperboloid. These two equa- tions represent for every value of A; (:^ 0) a straight line. The hyperboloid of one sheet contains therefore the family of lines represented by the last two equa- tions with variable A:. In exactly the same way it is shown that the same hyper- boloid also contains the family of lines be V aJ b c k'\ aJ Fig. 133 Thus every hyperboloid of one sheet contains two sets of recti- linear generators (Fig. 133). XIII, § 283] QUADRIC SURFACES 259 283. Rulings on a Hyperbolic Paraboloid. The hyperbolic paraboloid (Fig. 134) x^ y^ = 2C2 also contains two sets of recti- linear generators^ namely, a b a b k and a b a b k' Fig. 134 EXERCISES 1. Derive the equation of the tangent plane to : (a) the elliptic paraboloid ; (b) the hyperbolic paraboloid ; (c) the elliptic cone. 2. The line perpendicular to a tangent plane at a point of contact is called the normal line. Write the equations of the tangent planes and normal lines to the following quadric surfaces at the points indicated : (a) a;V9 + y^i - ^716 = 1, at (3, - 1, 2) ; (6) a;2 + 2 2/2 + 2^2^10, at (2,1, -2); (c) a;2 + 2 y2 _ 2 ;22 ^ 0, at (4, 1, 3) ; (d) x^-Sy^-z = 0, at the origin. 3. Show that the cylinder whose axis has the direction cosines I, rn, n and which is tangent to the ellipsoid x^/a^ + y^/b^ + z^/c^ = 1, is w b^ cy U'" &2 c2;U' &■' c2 / • 4. Show that the plane Ix -^ my + nz = Vl'-^a^ + m^b'^ + n^c^ is tangent to the ellipsoid x^/a"^ + y'^/b'^ + z^/c"^ = 1. 5. Show that the locus of the intersection of three mutually perpen- dicular tangent planes to the ellipsoid a:2/a2 -f 2/2/52 _f. ^2/02 = 1, is the sphere (called director sphere) x^ -{■ y"^ +z^ = a^ + b^ + c2. 260 SOLID ANALYTIC GEOMETRY [XIII, § 283 6. Show that the elliptic cone is a ruled surface. 7. Show that any two linear equations which contain a parameter represent the generating line of a ruled surface. What surfaces are gen- erated by the following lines ? (a) x-y-\-kz = 0,x-\-y-z/k = Q; (6) 3 a; - 4 y = A;, (3 a;+4 y)k=l ; (c) X - y + 3 A-2 = 3 k, k{x + y)— 2 = 3. 8. Show that every generating line of the hyperbolic paraboloid a;2/a2 _ y'ljifi = 2 C2 is parallel to one of the planes x^/ct^ — y'^/h'^ = 0. 284. Surfaces in General When it is required to deter- mine the shape of a surface from its cartesian equation the most effective methods, apart from the calculus, are the transformation of coordinates and the taking of cross-sections, generally (though not necessarily always) at right angles to the axes of coordinates. Both these methods have been ap- plied repeatedly to the quadric surfaces in the preceding articles. 285. Cross-Sections. The method of cross-sections is ex- tensively used in the applications. The railroad engineer de- termines thus the shape of a railroad dam ; the naval architect uses it in laying out his ship ; even the biologist uses it in con- structing enlarged models of small organs of plants or animals. 286. Parallel Planes. When the given equation contains only one of the variables x, y, z, it represents of course a set of parallel planes (real or imaginary), at right angles to one of the axes. Thus any equation of the form F(x)=0 represents planes at right angles to Ox, of which as many are real as the equation has real roots. XIII, § 290] QUADRIC SURFACES 261 287. Cylinders. When the given equation contains only two variables it represents a cylinder at right angles to one of the coordinate planes. Thus any equation of the form F(x,y)=0 represents a cylinder passing through the curve F{x, y) = in the plane Oxy, with generators parallel to Oz. If, in particular, F(x, y) is homogeneous in x and y, i.e. if all terms are of the same degree, the cylinder breaks up into planes. 288. Cones. When the given equation F(x, y^ z)=0 is homogeneous in x, y, and z, i.e. if all terms are of the same degree, the equation represents a general cone, with vertex at the origin. For in this case, if (x, y, z) is a point of the sur- face, so is the point (kx, ky, kz), where k is any constant; in other words, if P is a point of the surface, then every point of the line OP belongs to the surface ; the surface can therefore be generated by the motion of a line passing through the origin 289. Functions of Two Variables. Just as plane curves are used to represent functions of a single variable, so surfaces can be used to represent functions of two variables. Thus to obtain an intuitive picture of a given function f(x, y) we may con- struct a model of the surface such as the relief map of a mountainous country. The ordi- nate z of the surface represents the function. 290. Contour Lines. To obtain some idea of such a surface by means of a plane drawing the method of contour lines or level lines can be used. This is done, e.g., in topographical maps. The method consists in taking horizontal cross-sections at equal intervals and projecting these cross-sections on the hori- zontal plane. Where the level lines crowd together the surface is steep ; where they are relatively far apart the surface is flat. 262 SOLID ANALYTIC GEOMETRY [XIII, § 290 EXERCISES 1. What surfaces are represented by the following equations ? (a) Ax + By-{- C = 0. (c) y^ + z'^^a^. (e) zx-a^, (g) x«- 3x2-3;+ 3 = 0. (0 y = x2-x-6. {k) x2 + 2 1/2 = 0. (m) x2 - J/2 = 2-2. (0) (x-l)(y-2)(;^-3) = 0. (6) xcos/3 + ysin j3=p. (/);?2 = 4ay. (h) xyz = 0. U) yz^-9y = 0. (0 x^ = yz. (n) y^-{-2z^-\-izx = 0. 2. Determine the nature of the following surfaces by sketching the contour lines : (a) z=x + y. (6) z = xy. (c) z = y/x. (d) z =x2 + y, (e) 2=x2-y2+4. (/) 2 = x2. (g) z = x^+y^-ix. (h) z = xy~x. (i) z = 2'. (j) y=«2_4x. (^-) y = 3 ^2 + x2. (Z) «=3 x+y2. 3. The Cassinian ovals (§ 178) are contour lines of what surface ? 4. What can be said about the nature of the contour lines of a sur- face z=f(x) ? Discuss in particular : (a) 2 = x2 — 9 ; (b) z = x^ — 8; (c) y = z^-{-2z. 291. Rotation of Coordinate Trihedral. To transform the equation of a surface from one coordinate trihedral Oxyz to another Ox'y'z', with the same origin O, we must find expressions for the old co- ordinates X, y, z of any point P in terms of the new coordinates x', y', z'. We here confine ourselves to the case when each trihedral is trirectangular ; this is the case of orthogonal transformation^ or orthogonal substitution. Let ^1 , mi , ni be the direction cosines of the new axis Ox' with respect to the old axes Ox, Oy, Oz (Fig. 136) ; similarly h, m2 , W2 those of Oy', and ^3 , m^ , na those of Oz'. This is indicated by the scheme Fig. 135 XIII, §293] QUADRIC SURFACES 263 X' h^ + mi2 + ni2 = 1, h^ + mi^ + 7122 = 1, Zs^ + m32 + W32 = 1, Zi^ + 22^ + Za^ = 1, mi2 + m22 + m32 = ] Zl Z2 Z3 Wll Wi2 WI3 wi n2 ns which shows at the same time that then the direction cosines of the old axis Ox with respect to the new axes Ox', Oy' , Oz' are Zi , ^2 , Z3 , etc. 292. The nine direction cosines Zi , Z2 , ••• W3 are sufficient to determine the position of the new trihedral Ox'y'z' with respect to the old. But these nine quantities cannot be selected arbitrarily ; they are connected by six independent relations which can be written in either of the equivalent forms Z2Z3 + m2m3 + 712^3 = 0, ' (1) Z22 + m22 + 7122 = 1, Z3Z1 + mzlUi + WsWi = 0, Z1Z2 + WliW2 4- WiW2 = 0, wiini + miUi + WI3W3 = 0, (1') mi2 + m22 + m32 = 1, riih + n2Z2 + W3Z3 = 0, Wi2 + W22 4- W32 = 1 , limi + Z2TO2 + hmz = 0. The meaning of these equations follows from §§ 202 and 205. Thus the first of the equations (1) expresses the fact that Zi, mi, wi are the direction cosines of a line, viz. Ox' ; the last of the equations (1') ex- presses the perpendicularity of the axes Ox and Oy ; and so on. 293. If X, y, z are the old, x', y\ z' the new coordinates of one and the same point, we find by observing that the projection on Ox of the radius vector of P is equal to the sum of the projections on Ox of its components x', j/', z' (§ 199), and similarly for the projections on Oy and Oz : X = Zix' + hyi + hz', (2) y = Ttiix' + miy' + mzz', z = mx' + n2y' + nzz'. Indeed, these relations can be directly read off from the scheme of direction cosines in § 291. Likewise, projecting on Ox', Oy', Oz', we find x' = Zix + m\y + n\z, (2') y' = hx + nny + n2Z, z' = I3X + mzy + nzz. 264 SOLID ANALYTIC GEOMETRY [XIII, § 293 As the equations (2), by means of which we can transform the equation of any surface from one rectangular system of coordinates to any other with the same origin, give x, y, z as linear functions of x\ y\ «', it follows that 8uch a transformation cannot change the degree of the equation of the surface. 294. The equation (2') must of course result also by solving the equa- tions (2) for x', 2/', z\ and vice versa. Putting l\ h h wii »n2 Wl3 = A ni n2 W3 solving (2) for x', j/', «', and comparing the coefficients of x, y, z with those in (2') we find the following relations : Dll = 7712713 — 7713712 , DMi = 7l2?3 — W3Z2 » DtIi = litn^ — Z3WI2 1 ©tC. Squaring and adding the first three equations and applying the re- lations (1) we find : Z)2 = 1. By § 226, D can be interpreted as six times the volume of the tetrahe- dron whose vertices are the origin and the points x', y', z' in Fig. 135, i.e. the intei-sections of the new axes with the unit sphere about the origin. The determinant gives this volume with the sign + or — according as the trihedral Ox'y'z' is superposable or not (in direction and sense) to the trihedral Oxyz (see § 295). It follows that D=±l and li=± (m27i3 — W3W2), wii = i (712I3 - 713/2), ni=± {hmz — hmi)^ h — ± (w»37ii — ?7»in3), «i2 = ± {nzh — nih), 7i2 = ± Chrni — hmz)^ ^3 =± (77ii7i2 — wi27ii), ^3 = ± (7ii?2 — Wa/i), n^ = ± (hm^ — hmi) , the upper or lower signs to be used according as the trihedrals are super- posable or not. 295. A rectangular trihedral Oxyz is called right-handed if the rotation that turns Oy through 90° into Oz appears counterclockwise as seen from Ox ; otherwise it is called left-handed. In the present work right-handed sets of axes have been used throughout. Two right-handed as well as two left-handed rectangular trihedrals are superposable ; a right-handed and a left-handed trihedral are not super- posable. The difference is of the same kind as that between the gloves of the right and left hand. Two non-superposable rectangular trihedrals become superposable upon reversing one (or all three) of the axes of either one. XIII, § 297] QUADRIC SURFACES 265 296. The fact that the nine direction cosines are connected by six rela- tions (§ 292) suggests that it must be possible to determine the position of the new trihedral with respect to the old by only three angles. As such we may take, in the case of superposable trihedrals, the angles 6, 0, ^, marked in Fig. 135, which are known as Eulefs angles. The figure shows the intersections of the two trihedrals with a sphere of radius 1 described about the origin as center. If OiV is the intersection of the planes Oxy and Ox'y', Euler's angles are defined as d = zOz\ (t> = NOx', xfy = xON. The line ON is called the line of nodes, or the nodal line. Imagine the new trihedral Ox'y'z' initially coincident with the old trihedral Oxyz, in direction and sense. Now turn the new trihedral about Oz in the positive (counterclockwise) sense until Ox' coincides with the assumed positive sense of the nodal line ON; the amount of this rotation gives the angle \p. Next turn the new trihedral about ON in the positive sense until the plane Ox'y' assumes its final position ; this gives the angle d as the angle between the planes Oxy and Ox'y', or the angle zOz' between their normals. Finally a rotation of the new trihedral about the axis Oz', which has reached its final position, in the positive sense until Ox' assumes its final position, determines the angle 0. 297. The relations between the nine direction cosines and the three angles of Euler are readily found from Fig. 135 by applying the fundamen tal formula of spherical trigonometry cos c = cos a cos b + sin a sin b cos y successively to the spherical triangles xNx', xNy', xNz\ yNx', yNy', yNz', zNx', zNy', zNz'. We find in this way : l\ = cos ^p cos — sin i/' sin cos 0, mi = sin ^ cos -f cos V' sin cos 6, Wi = sin sin d, I2— — cos sin — sin cos cos d, I3 = sin sin 0, m2 =— sin sin + cos cos cos 0, mz =— cos sin 0, n2 = cos sin 0, m = cos 0. 266 Four Place Logarithms N 1 2 3 .|. 6 7 8 9 12 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0000 0414 0792 1139 1461 1761 2041 2304 2553 2788 0043 0453 0828 1173 1492 1790 206« ^330 2577 2810 0086 0492 0864 1206 1523 1818 2095 2356 2(X)1 2833 0128 0531 0899 1239 1553 1847 2122 2380 2625 2856 0170 0212 0253 0294 0334 0374 4 8 12 1 17 21 25 29 33 37 05(59 0934 1271 1584 isTr. 2148 2405 2648 2878 0607 0969 1303 1614 1903 2175 2430 2672 2900 0(545 1004 1335 1644 v.m 2201 2455 2695 2923 0682 1038 1367 1673 1959 2227 2480 2718 2^>45 0719 1072 1399 1703 1987 ??53 2504 2742 2967 0755 1106 1430 1732 2014 2279 252<) 2765 2989 4 8 11 3 7 10 3 6 10 3 6 9 3 6 8 3 5 8 2 5 7 2 5 7 2 4 7 15 19 23 14 17 21 13 16 19 12 15 18 11 14 17 11 13 16 10 12 15 9 1214 9 1113 26 30S4 24 28 31 23 26 29 2124 27 20 22 25 18 21 24 17 20 22 16 19 21 16 18 20 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 6 8 1113 16 17 19 21 22 23 24 25 26 27 28 29 30 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 3243 3444 3636 3820 3997 4166 4330 4487 4639 3263 34(>4 3655 3838 4014 4183 4346 4502 4654 3284 3483 3674 3856 4031 4200 4362 4518 m>9 3304 3502 3692 3874 4048 4216 4378 4533 4683 3324 3522 3711 3892 4065 4232 4393 4548 4698 3346 3541 3729 3909 4082 4249 4409 4564 4713 a365 3560 3747 3927 4099 4265 4425 4579 4728 3385 3579 3766 3945 4116 4281 4440 451 »4 4742 »404 3598 3784 3962 413,3 4298 445(5 4(509 4757 2 4 6 2 4 6 2 4 6 2 4 6 2 4 6 2 3 5 2 3 6 2 3 5 13 4 8 10 12 8 10 12 7 9 11 7 9 11 7 9 10 7 8 10 6 8 9 6 8 9 6 7 9 14 16 18 14 16 17 13 15 17 12 14 10 12 14 16 11 13 15 11 12 14 11 12 14 10 12 13 4786 4800 4814 4829 4843 4857 4871 4886 4900 13 4 6 7 9 10 11 13 31 32 33 34 35 36 37 38 39 4914 5051 5185 6315 5441 556;^ 5682 5798 6911 4928 5066 5198 5328 5453 5575 5694 5809 6922 4942 5079 5211 5340 6465 5587 5706 5821 6933 4965 5092 5224 5353 5478 5599 5717 6832 5944 4969 4983 5105 5119 5237 5250 5366 5378 5490 5502 5611 5623 5729 5740 5843 5855 5955 59(3() 4997 5132 5263 5391 5514 5635 5752 r>m) 5977 5011 5145 5276 5403 5527 5&47 5763 5877 5988 5024 5159 5289 5416 5539 6658 6775 5888 5999 5038 5172 5302 5428 5551 5670 5786 6899 6010 1 3 4 13 4 13 4 1 2 4 1 2 4 12 4 1 2 4 1 2 3 1 2 3 5 7 8 5 7 8 5 7 8 5 6 8 5 6 7 5 6 7 5 6 7 5 6 7 4 6 7 10 11 12 911 12 91112 910 11 910 11 810 11 8 9 11 8 9 10 8 910 40 6021 6031 6042 6053 6064 6075 6086 6096 6107 6117 12 3 4 5 6 8 9 10 41 42 43 44 45 46 47 48 49 6128 6232 6335 6435 6632 6628 6721 6812 6902 6138 6243 6345 6444 6542 6637 6730 6821 6911 6149 6253 6355 6454 (;551 6646 6739 6830 6920 6160 6263 6365 6464 6656 6749 6839 6928 6170 6274 6375 6474 6571 6665 6758 6*48 6937 6180 62S4 a385 6484 6580 6675 6767 6857 6W6 6191 6294 6395 &493 6;-)90 6684 6776 6866 6955 6201 6304 6405 6503 6599 6693 6785 6875 6964 6212 6314 6415 ()513 6609 6702 6794 6884 6972 6222 6325 6425 6522 6618 6712 6803 6893 6981 1 2 3 12 3 12 3 12 3 1 2 3 12 3 12 3 12 3 12 3 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 4 6 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 7 8 7 7 8 7 7 8 6 7 8 50 6990 6998 7007 7016 7024 7033 7042 7060 7059 7067 12 3 3 4 6 6 7 8 51 52 53 54 7076 7160 7243 7324 70&4 7168 7251 7332 7093 7177 7269 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 7152 7235 7316 Tim 12 3 1 2 3 1 2 2 1 2 2 3 4 5 3 4 5 3 4 5 3 4 5 6 7 8 6 7 7 6 6 7 6 6 7 N 1 2 8 4 6 6 7 8 9 1 2 2 4 5 6 7 8 9 The proportional parts are stated in full for every tenth at the right-hand side. The logarithm of any number of four significant figures can bo read directly by add- Four Place Logarithms 267 N 1 2 3 4 5 6 7 8 9 12 3 4 5 6 7 8 9 55 56 57 58 59 60 7404 7482 7559 7634 7709 7782 7412 7490 7566 7642 7716 7789 7419 7497 7574 7649 7723 7427 7505 7582 7657 7731 7435 7513 7589 7664 7738 7443 7520 7597 7672 7745 7451 7528 7604 7679 7752 7459 7536 7612 7686 7760 7466 7543 7619 7694 7767 7474 7551 7(527 7701 7774 12 2 12 2 112 112 112 3 4 5 3, 4 5 3 4 5 3 4 4 3 4 4 5 6 7 5 6 7 5 6 7 5 6 7 5 6 7 7796 7803 7810 7818 7825 7832 7839 7846 112 3 4 4 5 6 6 61 62 63 64 65 66 67 68 69 7853 7924 7993 8062 8129 8195 8261 8325 8388 7860 7931 8000 8069 8136 8202 8267 8331 8395 7868 7938 8007 8075 8142 8209 8274 8338 8401 7875 7945 8014 8082 8149 8215 8280 8344 8407 7882 7952 8021 8089 8156 8222 8287 8351 8414 7889 7959 8028 8096 8162 8228 8293 8357 8420 7896 7966 8035 8102 8169 8235 8299 8363 8426 7903 7973 8041 8109 8176 8241 8306 8370 8432 7910 7980 8048 8116 8182 8248 8312 8376 8439 7917 7987 8055 8122 8189 8254 8319 8382 8445 112 1 1 2 112 112 112 112 112 112 112 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 5 6 6 5 5 6 5 5 6 5 5 6 5 5 6 5 5 6 6 5 6 4 5 6 4 5 6 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 112 3 3 4 4 5 6 71 72 73 74 75 76 77 78 79 8513 8573 8633 8692 8751 8808 8865 8921 8976 8519 8579 8639 8698 8756 8814 8871 8927 8982 8525 8585 8645 8704 8762 8820 8876 8932 8987 8531 8591 8651 8710 8768 8825 8882 8938 8993 8537 8597 8657 8716 8774 8831 8887 8943 8998 8543 8603 8663 8722 8779 8837 8893 8949 9004 8549 8609 8669 8727 8785 8842 8899 8954 9009 8555 8615 8675 8733 8791 8848 8904 8960 9015 8561 8621 8681 8739 8797 8854 8910 8965 9020 8567 8627 8686 8745 8802 8859 8915 8971 9025 112 112 112 112 112 112 1 1 2 112 112 3 3 4 3 3 4 2 3 4 2 3 4 2 3 3 2 3 3 2 3 3 2 3 3 2 3 3 4 5 6 4 5 6 4 5 6 4 6 6 4 5 5 4 4 5 4 4 5 4 4 5 4 4 5 80 ^)031 9036 9042 9047 9053 9058 9063 9069 9074 9079 112 2 3 3 4 4 5 81 82 83 84 85 86 87 88 89 90a5 9138 9191 9243 9294 9345 9395 9445 9494 fX)90 9143 9196 9248 9299 9350 9400 9450 9499 9096 9149 9201 9253 9304 9355 9405 9455 9504 9101 9154 9206 9258 9309 9360 9410 9460 9509 910fj 9159 9212 9263 9315 9365 9415 9465 9513 9112 9165 9217 9269 9320 9370 9420 9469 9518 9117 9170 9222 9274 9325 9375 9425 9474 9523 9122 9175 9227 9279 9330 9380 9430 9479 9528 9128 9180 9232 9284 9335 9385 9435 9484 9533 9133 9186 9238 9289 9340 9390 9440 9489 9538 112 1 1 2 1 1 2 112 1 1 2 112 112 1 1 1 1 2 3 3 2 3 3 2 3 3 2 3 3 2 3 3 2 3 3 2 3 3 2 2 3 2 2 3 4 4 5 4 4 5 44 5 4 4 5 4 4 5 4 4 5 4 4 5 3 4 4 3 4 4 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 1 1 2 2 3 3 4 4 91 92 93 94 95 96 97 98 99 9590 9638 9685 9731 9777 9823 9868 9912 9956 9595 9643 9689 9736 9782 9827 9872 9917 9961 9600 9647 9694 9741 9786 9832 9877 9<)21 99(i5 9605 9652 9699 9745 9791 9836 9881 9926 9969 9609 9657 9703 9750 9795 9841 9886 9930 9974 9614 9661 9708 9754 9800 9845 9890 9934 9978 9619 9666 9713 9759 9805 9850 9894 9939 9983 9624 mil 9717 9763 9809 9854 9899 9943 9987 9628 9675 9722 9768 9814 9859 t)903 9948 9991 9633 9(i80 9727 9773 9818 9863 9908 9952 9996 Oil Oil Oil Oil 1 1 1 1 Oil Oil Oil 2 2 3 2 2 3 2 2 3 2 2 3 2 2 3 2 2 3 2 2 3 2 2 3 2 2 3 3 4 4 3 4 4 3 4 4 3 4 4 3 4 4 3 4 4 3 4 4 3 3 4 3 3 4 N 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 ing the proportional part corresponding to the fourth figure to the tabular number corresponding to tbo first three figures. Thero may be an error of 1 in the last place. 268 Four Place Trigonometric Functions [Charactoristics of Lof?arithins orn tted- determine by the usual i -ule from the value] Uadians 1 )egrees Si ME Tangext C!oTANGENT Cosi NE Value Lo?io Value Logio Value Logio Value Lopio .0000 0°00' .0000 .0000 l.OCKX) .0000 .0000 90° 00' 50 1.5708 1.5079 !0029 10 .0029 .4637 !0029 .4637 343.77 .5363 I'.OOOO .0058 20 .0058 .7(>48 .0058 .7648 171.89 .2352 1.0000 .0000 40 1.5650 .0087 30 .0087 .9408 .0087 .i)409 114.69 .0591 1.0000 .0000 30 1.5(521 .0116 40 .0116 .0(558 .0116 .0658 85.940 .9^42 .9995) .0000 20 1.5592 .0U5 50 .0145 .1627 .0145 .1627 68.750 .8373 .9999 .0000 10 1.5563 .0175 1°00' .0175 .2419 .0175 .2419 57.290 .7581 .9998 .99^)0 89° 00' 1.5533 .0204 10 .0204 .3088 .0204 .3089 49.104 .6911 .9998 .fn>9<) 50 1.5504 .0233 20 .0233 .3668 .0233 .36(59 42.964 .6331 .9i)97 .99*.><) 40 1.5475 .0202 30 .0262 .4179 .02(52 .4181 38.188 .5819 .99i)7 .91HHJ 30 1.5446 .0291 40 .0291 .4637 .0291 .4638 34.368 .5362 .9996 .9<)98 20 1.5417 .0320 50 .0320 .5050 .0320 .6053 31.242 .4947 .9995 .ims 10 1.5388 .0349 2° 00' .0^49 .5428 .0349 .5431 28.636 .4569 .9994 .9997 88° 00' 1.5359 .0378 10 .0378 .5776 .0378 .5779 26.4.32 .4221 .9993 .99<)7 50 1.53.30 .0407 20 .0407 .6097 .0407 .6101 24.542 .3899 .9{m .9^>96 40 1.5301 .0436 30 .0436 Ami .0437 .6401 22.904 .3599 .9990 .^y^m 30 1.5272 .0465 40 .0465 .(5677 .0466 .6(582 21.470 .3318 .9989 .9995 20 1.5243 .0495 50 .0494 .(5940 .0495 .6945 20.20(5 .3055 .9988 .9995 10 1.5213 .0524 3° 00' .0523 .7188 .0524 .7194 19.081 .2806 .9986 .9994 87° 00' 1.5184 .0553 10 .0552 .7423 .0553 .7429 18.075 .2571 .9985 .V.W3 50 1.5155 .0582 20 .0581 .7645 .0582 .7(552 17.1(59 .2^48 .9i)83 .9993 40 1.5126 .0611 30 .0(510 .7a')7 .0612 .78(55 16..350 .2i;i5 .9981 .m\2 30 1.5097 .0(U0 40 .0640 .8059 .0(541 .80(57 15.(505 .193:i .9980 .mn 20 1.5068 .0669 50 .0<}<39 .8251 .0(570 .8261 14.924 .1739 .9978 .9f»90 10 1.5039 .0698 4° 00' .0(598 .8436 .0699 .8446 14..301 .1554 .9976 .9989 86° 00' 1.5010 .0727 10 .0727 .8613 .0729 .8(524 13.727 .1376 .9974 .9!>S9 50 1.4981 .0756 20 .0756 .8783 .0758 .8795 13.197 .1205 .9971 .9988 40 1.4952 .0785 30 .0785 .8946 .0787 .8960 12.706 .1040 .9969 .9987 30 1.4923 .0814 40 .0814 .9104 .0816 .9118 12.251 .0882 .9967 .9986 20 1.4893 .0844 50 .0M3 .9256 .0*46 .9272 11.826 .0728 .9964 .9985 10 1.4864 .0873 6° 00' .0872 .9403 .0875 .9420 11.430 .0580 .9962 SY.)H-A 85° 00' 1.4835 .0902 10 .mK)l .9545 .0904 .9m:i 11.059 .0437 .{)959 A¥M'2 50 1.4806 .0931 20 .0929 sm2 .09:i4 .9701 10.712 .02^9 .9957 .92 .99(56 10.078 .0034 .9951 .9«>79 20 1.4719 .1018 50 .1016 .0070 .1022 .0093 9.7882 .9907 .9948 .i>977 10 1.4690 .1047 6° 00' .1045 .0192 .ia5i .0216 9.5144 .9784 .9946 .9976 84° 00' 1.4661 .1076 10 .1074 .0311 .1080 .o:»6 9.2553 .9(5(54 .9942 .9<)75 50 1.4632 .1105 20 .1103 .0426 .1110 .0453 9.0098 .9547 .9939 .9973 40 1.4603 .1134 30 .1132 .0539 .1139 .0567 8.77(59 .94:« .9936 .9972 30 1.4573 .1164 40 .11(51 .0648 .11(59 .0678 8.5555 .9322 .9i)32 .9»»71 20 1.4544 .1193 50 .1190 .0755 .1198 .0786 8.3450 .9214 .9929 .9969 10 1.4516 .1222 7° 00' .1219 .0859 .1228 .0891 8.1443 .9109 .9925 .99(58 83° 00' 1.4486 .1251 10 .1248 .0961 .1257 .0995 7.9530 .9005 .9i>22 .9966 60 1.4457 .1280 20 .1276 .10(50 .1287 .1096 7.7704 .8904 .9918 .9964 40 1.4428 .1309 30 .1305 .1157 .1517 .1194 7.5958 .8806 .9914 .9963 30 1.4399 .1338 40 .1334 .1252 .1346 .1291 7.4287 .8709 .9911 .9961 20 1.4370 .13(>7 50 .1363 .1345 .1376 .1385 7.2687 .8615 .9907 .9iW 10 1.4»41 .1396 8° 00' .1392 .1436 .1405 .1478 7.1154 .8522 .9903 .9958 82° 00' 1.4312 .1425 10 .1421 .152,5 .1435 .1569 6.9682 .8431 .9899 .995(5 50 1.4283 .1454 20 .1449 .1612 .1465 .1(558 6.8269 .8342 .aS94 .9954 40 1.4254 .148-4 30 .1478 .1697 .1495 .1745 6.6912 .82.55 .9890 .{K)52 30 1.4224 .1513 40 .1507 .1781 .1524 .ia3i 6.5(506 .8169 .988(5 .99.50 20 1.4195 .1542 50 .1536 .1863 .1554 .1915 6.4348 .8085 .9881 .9948 10 1.4166 .1571 9° 00' .1564 .1943 .1584 .1997 6.3138 .8003 .9877 .9946 81° 00' 1.4137 Value Log,o Value Lopio Value Logio Value Logio Degrees Radians Cosine Cotangent Tangent Sink Four Place Trigonometric Functions 269 [Characteristics of Logarith ms orai tted — leterraine by the usual rule from the value] Radians Degeees Sine Tangent Cotangent Cosine V^alue Logjo Value i-Ogio Value Logio Value Logio .1571 9° 00' .1564 .1943 .1584 .1997 6.3138 .8003 .9877 .9946 81° 00' 1.413? .1600 10 .1593 .2022 .1614 .2078 6.1970 .7922 .9872 .9944 50 1.4108 .1(529 20 .1622 .2100 .1644 .2158 6.0844 .7842 .98(58 .9942 40 1.4079 .1658 30 .1650 .217() .1673 .2236 5.9758 .7764 .9863 .9940 30 1.4050 .1687 40 .1679 .2251 .1703 .2313 5.8708 .7687 .9858 .9938 20 1.4021 .1716 50 .1708 .2324 .1733 .2389 5.7694 .7611 .9853 .t)936 10 1.3992 .1745 10° 00' .1736 .2397 .1763 .2463 5.6713 .7537 .9848 .9934 80° 00' 1.3963 .1774 10 .1765 .2468 .1793 .2536 5.5764 .7464 .9843 .9931 50 1.3934 .1804 20 .1794 .2538 .1823 .2609 5.4845 .7391 .9838 .9929 40 1.3904 .1833 30 .1822 .2606 .1853 .2680 5.3955 .7320 .9833 .9927 30 1.3875 .1862 40 .J851 .2674 .1883 .2750 5.3093 .7250 .9827 .9924 20 1.3846 .1891 50 .1880 .2740 .1914 .2819 5.2257 .7181 .9822 .9922 10 1.3817 .1920 11°00' .1908 .2806 .1944 .2887 5.1446 .7113 .9816 .9919 79° 00' 1.3788 .1949 10 .1937 .2870 .1974 .2953 5.06.58 .7047 .9811 .9917 50 1..3759 .1978 20 .1965 .2934 .2004 .3020 4.98<)4 .6980 .9805 .9914 40 1.3730 .2007 30 .1994 .25)97 .2035 .3085 4.9152 .6915 .9799 .9912 30 1..3701 .2036 40 .2022 .3058 .2065 .3149 4.8430 .6851 .9793 .9909 20 1.3672 .2065 50 .2051 .3119 .2095 .3212 4.7729 .6788 .9787 .9907 10 1.3(343 .2094 12<'00' .2079 .3179 .2126 .3275 4.7046 .6725 .9781 .9904 78° 00' 1.3614 .2123 10 .2108 .3238 .2156 .3336 4.6382 .6664 .9775 .9901 50 1.3584 .2153 20 .2136 .32% .2186 .3397 4.5736 .6603 .9769 .9899 40 1..3555 .2182 30 .2164 .3353 .2217 .3458 4.5107 .6542 .9763 .9896 30 1.3526 .2211 40 .2193 .3410 .2247 .3517 4.4494 .6483 .9757 .9893 20 1.3497 .2240 50 .2221 .3466 .2278 .3576 4.3897 .6424 .9750 .9890 10 1.3468 .2269 13° 00' .2250 .3521 .2309 .3(534 4.3315 .6366 .9744 .9887 77° 00' 1.3439 .2298 10 .2278 .3575 .2339 .3691 4.2747 .6309 .9737 .9884 50 1.3410 .2327 20 .2306 .3629 .2370 .3748 4.2193 .6252 .9730 .9881 40 1.3381 .2356 30 .2334 .3682 .2401 .3804 4.1653 .619(5 .9724 .9878 30 1.3352 .2385 40 .2;^)3 .3734 .24.32 .3859 4.1126 .6141 .9717 .9875 20 1.3323 .2414 50 .2391 .3786 .2462 .3914 4.04 .9635 50 1.1665 .4072 20 .3961 .5978 .4314 .o:j48 2.3183 .3652 .9182 .9()2f) 40 1.163(5 .4102 30 .3987 .6007 .4348 .6383 2.2^)98 .3617 .9171 .9624 30 1.1606 .4131 40 .4014 .6036 .4383 .6417 2.2817 .3583 .9159 .9618 20 1.1577 .4160 50 .4041 .6065 .4417 .6452 2.2637 .3548 .9147 .9613 10 1.1548 .4189 24° 00' .4067 .6093 .4452 .6486 2.2460 .3514 .9135 .9607 66° 00' 1.1519 .4218 10 .4094 .6121 .4487 .6520 2.2286 .3480 .9124 .9602 50 1.1490 .4247 20 .4120 .6149 .45-22 .()553 2.2113 .3447 .9112 .9596 40 1.1461 .4276 30 .4147 .6177 .4557 .6587 2.1943 .3413 .9100 .9590 30 1.1432 .4305 40 .4173 .6206 .45i)2 .6620 2.1775 .3380 .9088 .9584 20 1.1403 .4334 50 .4200 .6232 .4628 .6654 2.1609 .3346 .9075 .9579 10 1.1374 .4363 26° 00' .4226 .6259 .4663 .6687 2.1445 .3313 .9063 .9573 66° 00' 1.1345 .4392 10 .4253 .628(5 .4699 .6720 2.1283 .3280 .tK)51 .95<57 50 1.1316 .4422 20 .4279 .6313 AIM .67r,2 2.1123 .3248 .9038 .9561 40 1.1286 .4451 30 .4305 .6340 .4770 .6785 2.05 .3215 .9026 .9555 30 1.1267 .4480 40 .4331 .6.36279 .6458 87° 00' .6018 .7795 .7536 .8771 1.3270 .1229 .7986 .9023 68° 00' .9250 .6487 10 .6041 .7811 .7581 .8797 1.3190 .1203 .7969 .9014 50 .9221 .6516 20 .(i065 .7828 .7627 .8824 1.3111 .1176 .7951 .9004 40 .9192 .6545 30 .6088 .7»i4 .7673 .8850 1.3032 .1150 .79;i4 .8995 30 .9163 .6574 40 .6111 .7861 .7720 .8876 1.2954 .1124 .7916 .8985 20 .9134 .6603 50 .6134 .7877 .7766 .8902 1.2876 .1098 .7898 .8975 10 .9105 .6632 88° 00' .6157 .7893 .7813 .8928 1.2799 .1072 .7880 .8965 62° 00' .9076 .6661 10 .6180 .7910 .7860 .89.54 1.2723 .1046 .7862 .8955 50 .9047 .6(>90 20 .6202 .7926 .7907 .8980 1.2W7 .1020 .7844 .8945 40 .9018 .6720 30 .6225 .7911 .7954 .9006 1.2572 .0994 .7826 .8935 30 .8988 .6749 40 .6248 .7957 .8002 .9032 1.2497 .0968 .7808 .8925 20 .8959 .6778 50 .6271 .7973 .8050 .9058 1.2423 .0942 .7790 .8915 10 .8930 .6807 89° 00' .6293 .7989 .8098 .9084 1.2349 .0916 .7771 .8905 61°00' .8901 .6836 10 .6316 .8004 .8146 .9110 1.2276 .0890 .7753 .8895 50 .8872 .6865 20 .6338 .8020 .8195 .9135 1.2203 .0865 .7735 .88^ 40 .8843 .689i 30 .6361 .8035 .8243 .9161 1.2131 .0839 .7716 .8874 30 .8814 .6923 40 .6383 .8050 .8292 .9187 1.2059 .0813 .7698 .8864 20 .878^) .6952 50 .6406 .806(> .8342 .9212 1.1988 .0788 .7679 .8853 10 .8756 .6981 40° 00' .6428 .8081 .8391 .9238 1.1918 .0762 .7660 .8843 60° 00' .8727 .7010 10 .6450 .8096 .8441 .9264 1.1847 .0736 .7642 .8832 50 .8698 .7039 20 .6472 .8111 .8491 .9289 1.1778 .0711 .7623 .8821 40 .86()8 .7069 30 .(>494 .8125 .8541 .9315 1.1708 .0685 .7604 .8810 30 .8639 .7098 40 .(i517 .8140 .8591 .9341 1.1640 .0659 .7585 .8800 20 .8610 .7127 50 .6539 .8155 .8642 .9366 1.1571 .06U .7566 .8789 10 .8581 .7156 41° 00' .6561 .8169 .8693 .9392 1.1504 .0608 .7547 .8778 49° 00' .85,52 .7185 10 .6583 .8184 .8744 .9417 1.1436 .0583 .7528 .8767 50 .8523 .7214 20 .6604 .811)8 .87i)6 .9443 1.1369 .0557 .7509 .8756 40 .8494 .7243 30 .6626 .8213 .8847 .9468 1.1303 .a532 .7490 .8745 30 .8465 .7272 40 .6648 .8227 .8899 .9494 1.1237 .0506 .7470 .8733 20 .8436 .7301 50 .6670 .8241 .8952 .9519 1.1171 .0481 .7451 .8722 10 .8407 .7330 42° 00' .6691 .82.55 .iXXH .9544 1.1106 .0456 .7431 .8711 48° 00' .8378 .7359 10 .6713 .8269 .9057 .9570 1.1041 .0430 .7412 .8699 50 .8*48 .7389 20 .6734 .8283 .9110 .9595 1.0977 .0405 .7392 .8G8S 40 .8319 .7418 30 .6756 .8297 .91(>3 .9()21 1.0913 .0379 .7373 .8676 30 .8290 .7447 40 .6777 .8311 .9217 .9646 1.0850 .0354 .7353 .8665 20 .8261 .7476 50 .6799 .8324 .9271 .9671 1.0786 .0329 .7333 .8653 10 .8232 .7505 48° 00' .6820 .8338 .9325 .9697 1.0724 .0303 .7314 .8641 47° 00' .8203 .7534 10 .6841 .8351 .9380 .9722 1.0661 .0278 .72a; ^aV6. 13. If the vertices of the triangle are (a, 0), (—a, 0), (0, aV3) and A;2 is the constant, the locus is 3 x2 + 3 y2 _ 2 VS ay + 3 a2 — 2 A;2 = 0. Pages 74-75. 10. (a) 2y=3x2 + 5x; (6) 12y =- 5x2 + 29x - 18. 11. 300y =-x2 + 230x; 44.1 ft. above the ground; 230 ft. from the starting point. Page 81. 6. East, East 33° 41' North, East 63° 8' North, East 18° 26' South. 10. 100/(7r + 4). Pages 84-85. 10. 0, 8° 8'. 11. 7° 29'. 15. When the side of the square is 3 in. 17. (a) 6y = x8 + 6x2-19x; (6) 7y = 2x« - x2 - 29a; + 35. Page 92. 10. - 1.88, 1.53, .347. Pages 97-98. 2. (a) (4, jir), (4, |t); (6) (a, ^tt), (a, | tt) ; (c) (4,0); (d) (4 a, JT), (4a, fx). 7. (a) y2_4a;_|-4 = 0; (6) 14y2_ 45 a; + 62y + 60 = 0. 8. (6) x2-10x-3y + 21=0; (c) a;2 + 2x + y - 1 =0. 9. The equation of a parabola contains an xy term when its axis is oblique to a coordinate axis. Pages 106-108. 8. (a) y = ; (6) 2x + 2y - 9=0, 2 x-y-18 = 0; (c) 2x + 2y-9 = 0, 8x+ 16y-27 =0, 24x-16y- 153 = 0. id) 8x-16y-27 = 0. 14. y = kx. 16. Directrix; y2 = a(x - 3 a). 21. — (1 + w2). *n.2 ANSWERS 275 26. x2-80x-2400?/ = 0; 0, - i, - |, - J, 0, |, 2. 29. x2 = 360(y-20). 3 Page 115. 2. (3 7r-4)/6 7r. 3. §a^ (^+f')^ . 3 m** 6. (a) 64/3 ; (6) 625/12 ; (c) 1/12. 7. 123.84 ft^. 8. 1794 J tons. 9. 199.4 ft2. Page 118. 9. 8x2 -2 a;?/ + 8 2/2 -63 = 0. Page 122. 10. 3x2 - ^2 = 3 ^^^ n. 5. 14. 2xy = 1. Pages 128-129. 2. ^X+-r=:c2. 13. 54.5 ft., 42.2 ft. 17.62/^2. X y 20. An ellipse or hyperbola according as one circle lies within or without the other circle. Pages 138-139. 7. (a) A^a^-&h'^=^ C^ ; (&) aP- cos2 ^- &2 sin2 ^=p2. 19. 62. 21. a2-f 62; a2_52. 22. 4 a6. 23. sin-i (a6/a'6')- 25. (a) x2 + y2 ^ Qj2 + 52 . (6) x2 + ^2 ^ (j2 _ 52. Page 144. 3. (a) (1, - 1), (1 ± V2, - 1), x = 1 ± | V2 ; (&) a,0), (f,0), (-f,0),X = 0,X=:l. 4. 2 62/a. 7. (a) a2i/2 = h'h:,{a - x); (6) 52^2 = cfiyQy _ yy 8. Two straight lines. Page 151. 2. (a); Vertices (5, 3), (8, 3); semi-axes 3/2, V2. (6) Vertices (4, 8/3), (8, 8) ; semi-axes 10/3. 5\/3/3. (c) Vertices (17/5, 7/5), (1, 3) ; semi-axes V65/5, ^13/2. 3. 3x +2y- 2 = 0; 21/13, -37/26, 10/ \/l3. Page 153. 5. (acos ^, — asin ^), x2 -|- y2 _ 2 a(xcos0 — ?/ sin ^)= 0. Pages 161-162. 2. (a) 3 x - 14 1/ = ; (6) y =- 3/13, x = - 14/13. 5. 2 x2 - x?/ - 15 ?/2 H- X -f- 19 ?/ - 6 = 0, 2 x2 - xy - 15 ?/2 + x -f 19 ?/ - 28 = 0. 6. 6x2 -f xy- 2 2/2 -9x 4-82/ -46 = 0, 6x^ + xy-2y^ -9x + 8y + Si =0. 11. (a) x2/4 + ?/2 = 1 ; (b) x2/4 - 2/2/2 = 1 ; (c) 3x2 -f y2 + 6 = 0; (d) x2/16 + 2/2/4 = 1 . (e) (3 -h Vl7)x2 -f (3 - V17)?/2 = 4 ; (/) (2-hV2)x2-H(2-V2)2/2 = l. 15. x^-\-y^= aK 19. Equilateral hyperbola. 276 ANSWERS Page 168. 2. (a) Simple point ; (6) node ; (c) cusp ; (d) cusp. 4. (a) None ; (b) node at {b, 0) ; (c) isolated point at (a, 0) ; (d) cusp at (a, 0). Pages 174-175. 4. r=a(sec ± tan 0) or (x - a)y^-{- x'^(x-^ a) =0. 10. a;2?/2 = a2(a;2 ^ yiy ^ Cissoid (a - x)y^ = x^. 12. y(x2 4- y2) := rt(^2 _ ^2). i3_ y _ ^ ctn