QA 551 S656 UNIVERSITY OF CALIFORNIA SAN DIEGO E 3 1822 01276 0328 EL- i1 LIBRARY UNIVERSITY Of= CALIFORNIA SAN DIEGO t ^vl,......--^/'^ '^^ .iU/xU=*=- OA 55! 5656 UNIVERSITY OF "^IFORNIA 5»N DIEGO E 3 1 («,b« S"'oi276 '0328" "^^S^i^i-^ " vr\; BA = 0A-0B=-S-rj = -8 ■ IV. AB=OB-OA = -6-{-2) = -'i;BA = OA-OB=-2-{-6) = +4. The following properties of lengths on a directed line are obvious : (2) AB = ~BA. (3) AB is positive if the direction from A to B agrees with the positive direction on the line, and negative if in the con- trary direction. The phrase "distance between two points" should not be used if these points lie upon a directed line. Instead, we speak of the length AB, remembering that the lengths ABaud BA are not equal, but that AB=-BA. 9. Cartesian* coordinates. Let .Y'A' and F'y be two directed lines intersecting at O, and let P be any point in their plane. Draw lines through P parallel to X'X and Y'Y respectively. Then, if OAf=a, ON = b, * So called after Rene Descartes, 1596-16.50, who first introduced the idea of coordinates into the study of geometry. 10 NEW ANALYTIC GEOMETRY the numbers a, b are called the Cartesian coordinates of P, a the abscissa and h the ordinate. The directed lines X'X and y'Fare called the axes of coordi- nates, X'X the axis of / / P abscissas, Y'Y the axis of ordinates, and their in- tersection O the origin. The coordinates a, b of P are written {q,, b), and the symbol P(a, b) is to be read, " The point P, whose coordinates are a and b." Any point P in the l)lane determines two numbers, the coordinates of P. Con- versely, given two real numbers a' and b', then a point P' in the plane may always be constructed whose coordinates are (a', b'). Por lay off OM' = a', ON' = b', and draw lines parallel to the axes through M' and N'. These lines intersect at P' (a', Z»'). Hence Every point determines a pair of real numbers, and, conversely, a pair of real numbers determines a 2^0 int. The imaginary numbers of algebra have no place in this representation, and for this reason elementary analytic geome- try is concerned only with the real numbers of algebra. 10. Rectangular coordinates. A rectangular system of coordi- nates is determined when the axes A''A and l''Fare perpendicular to each other. This is the usual case, and will be assumed unless otherwise stated. The work of plotting points in a rectangular system is much simplified by the use of coordinate ox plotting pi'^^P'i^'i constructed by ruling off the plane into equal scpiares, the sides being parallel to the axes. CARTESIAN COORDINATES 11 In the figure several points are plotted, the unit of length being assumed equal to one division on each axis. The method is simply this : Count off from O along A' 'A' a number of divisions equal to the given abscissa, and then from the point so determined a A"' f-9,- Kie) (0, (6, (10 0) X Y\ number of divisions up or down equal to the given ordinate, observing the Rule for signs : Abscissas are positive or negative according as tlietj are laid off to the right or left of the origin. Ordinates are positive or negative according as they are laid off above or below the axis of x. Rectangular axes divide the plane into four portions called quadrants ; these are numbered as in the figure, in which the proper signs of the coordinates are also indicated. As distinguished from rectangular coordinates, the term oblique coordinates is employed when the axes are not Second X' Third First (f.+) A' Fourth (f,-) 12 NEW ANALYTIC GEOMETRY perpendicular, as in the figure of Art. 9. The rule of signs given above applies to this case also. Note, however, in plotting, that the ordinate MP is drawn iKirallel to OY. In the following problems assume rectangular coordinates unless the contrary is stated. PROBLEMS 1. Plot accurately the points (3, 2), (3, - 2), (- 4, 3), (6, 0), (- 5, 0), (0, 4). 2. What are the coordinates of the origin ? J.jis. (0, 0). 3. In what quadrants do the following points lie if a and h are posi- tive numbers : (- a, 6) ? (- a, - 6) ? (6, - a) ? (o, h) ? 4. To what quadrants is a point limited if its abscissa is positive ? negative ? if its ordinate is positive ? negative ? 5. Draw the triangle whose vertices are (2, — 1), (— 2, 5), (— 8,-4). 6. Plot the points whose oblique coordinates are as follows, when the angle between the axes is 60°: (2, - 3), (3, - 2), (4, 5), (- 6, -7), (- 8, 0), (9, -5), (-6, 2). 7. Draw the quadrilateral whose vertices are (0, — 2), (4, 2), (0, 6), (— 4, 2), when the angle between the axes is 60°. 8. If a point moves parallel to the axis of x, which of its coordinates i-emains constant ? If parallel to the axis of ?/ ? 9. Can a point move when its abscissa is zero ? Where ? Can it move when its ordinate is zero ? Where ? Can it move if both abscissa and ordinate are zero ? Where will it be ? 10. Where may a point be found if its abscissa is 2 ? if its ordinate is -3? 11. Where do all those points lie whose abscissas and ordinates are equal ? 12. Two sides of a rectangle of lengths a and 6 coincide with the axes of X and y respectively. What are the coordinates of the vertices of the rectangle if it lies in the first quadrant ? in the second quadrant ? in the third quadrant ? in the fourth quadrant ? 13. Construct the quadrilateral whose vertices are (— 3, 6), (— 3, 0), (3, 0), (3, 6). What kind of a quadrilateral is it ? What kind of a quad- rilateral is it when the axes are oblique ? CARTESIAN COORDINATES 13 14. Show that {x, y) and (x, — y) are symmetrical with respect to X'X ; (x, y) and (— x, y) with respect to Y'Y; and (x, y) and (— x, — y) with respect to the origin. 15. A line joining two points is bisected at the origin. If the coordinates of one end are (a, — 6), what will be the coordinates of the other end ? 16. Consider the bisectors of the angles between the coordinate axes. What is the relation between the abscissa and ordinate of any point of the bisector in the first and third quadrants ? second and fourth quadrants ? 17. A square whose side is 2 a has its center at the origin. "What will be the coordinates of its vertices if the sides are parallel to the axes ? if the diagonals coincide with the axes ? Ans. {a, a), (a, —a), (— a, — a), (— a, a) ; (aV2, O), (-aV2, O), (O, aV2), (O, - aV2). 18. An equilateral triangle whose side is a has its base on the axis of X and the opposite vertex above X'X. What are the vertices of the tri- angle if the center of the base is at the origin ? if the lower left-hand vertex is at the origin ? (0, 0), (a, 0), (-, Ans. 11. Lengths. |,o),(-^,o),(o, /a aV3\ \2'"T~/ Then in the figure OM, = x Consider any two given points 2V^2' ^a)" We may now easily prove the important Theorem. The length I of the line Joining two points P.^(x^, i/^), P.^i^^' V-i) is given by the formula ' OM=x.^,M^P^=y^,M^P^=y.^ ro(X2,;?/„) (I) l=^{x^-x,y + {y^-y,)\ Proof. Draw lines through P^ and P^ parallel to the axes to form the right triangle P^SP^. Then P^S = OM, - SP = M P 2 2 2 P^P =V^' + i\s- and hence ^ = V(.r^-a-/+(/A-.y,y^ Q.E. D. 14 NEW ANALYTIC GEOMETRY The same method is used in deriving (I) for any positions of P and P, ; namely, we construct a right triangle by drawing lines parallel to the axes through P^ and P^. The horizontal side of this triangle is equal to the difference of the abscissas of P and Pg, while the vertical side is equal to the difference of the ordinates. The required length is then the square root of the sum of the squares of these sides, which gives (I). A number of different figures should be drawn to make the method clear. EXAMPLE Find the length of the line joining the points (1, 3) and (— 5, 5). Solution. Call (1, 3) P^, and (-5, 5) P„. Then x^ = 1, 2/i = 3, and Xo = — 5, y„ = b; and substituting in (I), we have Y' (-5 5-) r^^ ■•^ (1 3-) .Y' X Y I = V(l + 5)2 + (3 - 5)2 = V40= 2 VlO. It should be noticed that we are simply finding the hypotenuse of a right triangle whose sides are 6 and 2. Remark. The fact that formula (I) is true for all positions of the points P^ and 7^„ is of funda- mental importance. The application of this formula to any given problem is therefore simply a matter of direct substitu- tion. In deriving such general formulas it is most convenient to draw the figure so that the points lie in the first quadrant, or, in general, so that all the quantities assumed as knoirn shall he positive. PROBLEMS 1. Find the lengths of the lines joining the following points : (a) (- 4, - 4) and (1, 3). Ans. Vli. Aivi. VlO. (b) (-V2, V3) and (Vs, V2). (c)(0,0)and(|.^^). (il) {(t + b, c + a) and (c + a, b + c). Ans. V(b — c)2 + (a — 6)2. Ans. a. CARTESIAN COORDINATES 15 2. Find the lengths of the sides of the following triangles : (a) (0, 6), (1, 2), (3, - 5). (b) (1,0), (-1,-5), (-1, -8). (c) («, 6), (6, c), (c, d). (d) (a, - 6), (6, - c), (c, - d). (e) (0,2/), (-X, _y), (-x, 0). 3. Find the lengths of the sides of the triangle whose vertices are (4, 3), (2, - 2), (- 3, 5). 4. Show that the points (1, 4), (4, 1), (6, 5) are the vertices of an isosceles triangle. 5. Show that the points (2, 2), (- 2, - 2), (2 Vs, - 2 Vs) are the vertices of an equilateral triangle. 6. Show that (3, 0), (6, 4), (— 1, 3) are the vertices of a right triangle. What is its area ? 7. Prove that (- 4, - 2), (2, 0), (8, 6), (2, 4) are the vertices of a parallelogram. Also find the lengths of the diagonals. 8. Show that (11, 2), (6, - 10), (- 6, - 5), (- 1, 7) are the vertices of a square. Find its area. 9. Show that the points (1, 3), (2, V6), (2, —Vo) are equidistant from the origin ; that is, show that they lie on a circle with its center at the origin and its radius equal to the VlO. 10. Show that the diagonals of any rectangle are equal. 11. Find the perimeter of the triangle whose vertices are (a, 6), (— a, 6), (-«,-&)• ' . 12. Find the perimeter of the polygon formed by joining the following points two by two in order : (6, 4), (4, - 3), (0, — 1), (— 5, — 4), (— 2, 1). 13. One end of a line whose length is 13 is the point (— 4, 8) ; the ordinate of the other end is 3. What is its abscissa ? Ans. 8 or — 16. 14. What equation must the coordinates of the point (x, y) satisfy if its distance from the point (7, — 2) is equal to 11 ? 15. What equation expresses algebraically the fact that the point (x, y) is equidistant from the points (2, 3) and (4, 5) ? 16. Find the length of the line joining Pi(x,, ?/,) and P^i'f-zi V-z) when the coordinates are oblique. Hint. Use the law of cosines, 44, p. 4. 16 NEW ANALYTIC GEOMETRY '' 12. Inclination and slope. The angle between two intersecting directed lines is defined to be the angle made by their positive directions. In the figures the angle between the directed lines is the angle marked 6. If the directed lines are parallel, then the angle between them is zero or 180°, according as the positive directions agree or do not agfree. Evidently the angle between two directed lines may have any value from to 180° inclusive. Reversing the direction of either directed line changes B to the supplement 180° directions are reversed, the angle is unchanged. When it is desired to assign a positive direction to a line intersecting X'X, we shall always assume the upward direction as positive. * The inclination of a line is the angle be- tween the axis of x and the line when the latter is given the upward direction. e. If both e^Q This amounts to saying that the inclination is the angle above the x-axis and to the right of the given line, as in the The slojje of a line is the tangent of its inclination. The inclination of a line will be denoted by the Greek letter a, ftj, «:.„ a', etc. ("alpha," etc.); its slope by m, m^, m.^, vi', etc., so that VI = tan a, iti^ = tan a^, etc. The inclination may be any angle from to 180° inclusive. The slope may be any real num- ber, since the tangent of an angle in the first two qiiadrants may be any number positive or nega- tive. The slope of a line parallel to A' 'A' is of course zero, since CARTESIAN COORDINATES 17 the inclination is or 180°. For a line parallel to Y'Y the slope is infinite. Theorem. The slojie m of the line passing through two points i\{^\^ 2/i)' ^2(*'2' 2/2) '^ 9^^^'^ h (II) '"=^- -*1 •*2 Proof. In the figure Draw P^S parallel to OX. Then in the right triangle P.2''^P^, since angle P,P„S = a, we have 12 7 S' y- 1 SP, (1) But and 971 = tan a = — — ; • 2 y(px = MJ-^-M^P=y^-y.^; —4s 1 P^S = M^M^ /^ M2 Mix = OM, — OM, = X, — .7- . X Substituting these values in (1) gives (II). Q. e. d. The student should derive (II) when a is obtuse.* We next derive the conditions for parallel lines and for per- pendicular lines in terms of their slopes. Theorem. If two lines are parallel, their slopes are equal ; if perpendicular, the slope of one is the negative reciprocal of tht slope of the other, and conversely. Proof. Let a^ and a^ be the inclinations and m^ and in the slopes of the lines. If the lines are parallel, a^ = a,^. .'. vi^ = m,^. * To construct a line passing through a given point P^ whose slope is a pos- itive fraction - , we mark a point S h units to the right of I\ and a point P^ a units above iS, and draw P^P^- If the slope is a negative fraction, — , then plot S 6 units to the left of P^. ^ 18 NEW ANALYTIC GEOMETRY If the lines are perpendicular, as in the figure, TT «,= 77 + «r . m.^ = tan «, = tan ( — + ^j ) = - cot a^ (by 31, p. 3) (By 26, p. 3) tan a^ J_ m. The converse is proved by retracing the steps with the assumption, in the second part, that a, is greater than a^. PROBLEMS 1. Find the slope of the line joining (1, 3) and (2, 7). Ans. 4. 2. Find the slope of the line joining (2, 7) and (— 4, — 4). Aixs. y. 3. Find the slope of the line joining (V3, V^) and (— V2, Vs). Ans. 2V6- 5. 4. Find the slope of the line joining (a + 6, c + «), (c + a, 6 + c). . b — a Ans. c- b 5. Find tlie slopes of the sides of the triangle whose vertices are (1, 1), (-1, -1), (V3, -V3). ^^^ ^ I+V3 I-V3 'i-Vs'i + Vs ' 6. Prove by means of slopes that (—4, — 2), (2, 0), (8, 6), (2, 4) are the vertices of a parallelogram. ■ 7. Prove by means of slopes that (3, 0), (6, 4), (— 1, 3) are the vertices of a right triapgle. ' 8. Prove by means of slopes that (0, - 2), (4, 2), (0, 6), (- 4, 2) are the vertices of a rectangle, and hence, by (I), of a square. 9. Prove by means of their slopes that the diagonals of the square in Problem 8 are perpendicular. V 10. Prove by means of slopes that (10, 0), (5, 5), (5, — 5), (— 5, 5) are the vertices of a trapezoid. 11. Show that the line joining (a, b) and (c, — d) is parallel to the line joining (— a, — b) and (— c, d). CARTESIAN COORDINATES 19 12. Show that the line joining the origin to (a, b) is perpendicular to the line joining the origin to (— b, a). " 13. What is the inclination of a line parallel to Y'Y? perpendicular to FT? ' 14. What is the slope of a line parallel to Y'Y ? perpendicular to Y'Y? 15. What is the inclination of the line joining (2, 2) and (— 2, — 2) ? Ans. — . 4 16. What is the inclination of the line joining (— 2, 0) and (— 5, 3) ? Stt Ans. 4 17. What is the inclination of the line joining (3, 0) and (4, y/S) ? Ans. -. 18. What is the inclination of the line joining (3, 0) and (2, -\/3) ? Ans. 3 19. Whatistheinclinationof the line joining (0, — 4) and(— a/3,— -5) ? Ans. -■ 6 20. What is the inclination of the line joining (0, 0) and (— -y/s, 1) ? Ans. U ^ 21. Prove by means of slopes that (2, 3), (1, — 3), (3, 9) lie on the same straight line. 22. Prove that the points (a, b + c), (5, c + a), and (e, a + b) lie on the same straight line. 23. Prove that (1, .5) is on the line joining the points (0, 2) and (2, 8) and is equidistant from them. ■ 24. Prove that the line joining (3, — 2) and (.5, 1) is perpendicular to the line joining (10, 0) and (13, - 2). 13. Point of division. Let P^ and P, be two fixed points on a directed line. Any third point on the line, as P or P', is said " to divide the line into — _c . ^ o two segments, and is ' called a point of division. The division is called internal or external according as the point falls within or without PJ^.^ The position of the point of division depends upon the nitio 20 NEW ANALYTIC GEOMETRY of its distances from P^ and P.^. Since, however, the line is directed, some convention must be made as to the manner of reading these distances. We therefore adopt the rule : If P is a point of division on a directed line passing through P^ and P^, then P is said to divide P^P.2 into the segments P^P p p and PP„. The ratio of division is the value of the ratio * — ^ • PP^ We shall denote this ratio by X (G-reek letter "lambda"), that is, p p If the division is internal, PJ^ and PP^ agree in direction and therefore in sign, and X is therefore positive. In external division X is negative. „, . p N J.1 o -10 X=.o _x / 3/i M 31, X 1+A 1 + A « To assist the memory in writing down this ratio, notice that the jtoint of division /' is written last in tlie numerator and, first in tlie denominator. CARTESIAN COORDINATES 21 P,P Proof. Given X = PP 2 Draw the ordinates M^P^, MP, and M^P,^. Then, by geometry, these ordinates will intercept proportional segments on the transversals P^P,^ and OX ; that is,* (1) M,M_P,P But and, by hypothesis, Substituting in (1), — Clearing of fractions and solving for x, X. + Xa^'o X = MM^ PP,^ M^M =0M - OM^ = X — x^, MM.^ = OM.^ - - OM = x.^ — X, PP, ~ ' x-x, _^ 1 + A Similarly, by drawing the abscissas of P.^, P, and P., to the axis or y we may prove ?/ = — ~- Q.e.d. 1+A Corollary. Middle point. T/ie coordinates (x, y) of the middle point of tlte line Joining Pj(a"j, ^j), Po(:''o, y^ are found hy taking the averages of the given abscissas and ordinates; that is, (IV) x=l(x^-\-x^), y = l(y, + y^). P P For if P is the middle point of P^P.,, then \ = -^— = 1, and substituting A = 1 in (III) gives (IV). '^ To apply (III), mark the point of division P, the extremities of the line to be divided P^ and P„ and make sure that the value of X satisfies X = P^P h- PP,. * Care must be taken to read the segments on the transversals (since we are dealing with directed lines) so that they all have positive directions 22 NEW ANALYTIC GEOMETRY EXAMPLES 1. Find tlie point P dividing P^(— 1, - Solution. By the statement, PjP _ 1 PP^- 4 Hence, applying (III), x■^^=—l,y^=— (i, X2 = 3, 2/2 = 0. -1-1.3 -|_ 6), P., (3. 0) in tlie ratio n -6-^.0 3 4 1-i f I — 2i y = Hence P is (- 2i, - 8). ^tw. The result is checked by plotting. The point P lies outside the seg- ment P^P., and the length of PP2 is four times that of PPi- 2. Center of gravity of a triangle. Find the coordinates of the point of intersection of the medians of a triangle whose vertices are (x^, y{), {x^, y^), (Xg, y^). Solution. By plane geometry we have to find the point P on the median AD such that AP — f AD ; that is, AP : PZ) : : 2 : 1, or \ = 2. By the corollary, D is [i(x2 + X3), | (2/3 + 2/3)]. To find P, apply (III), remembering that A corresponds to (x^, y-^) and D to (x„ y„). J (^i.2/1) (x2,yo) This gives x ^ Xi + 2-^(X2 + X,) 1 + 2 = H^l + X2 + X^), _ yi + -^■i{y.2 + Vs) 1 + 2 = 3(2/1 + yo + Vs)- ^ns. Hence the abscissa of the intersection of the medians of a triangle is the average of the abscissas of the vertices, and similarly for the ordinate. The symmetry of these answers is evidence that the particular median chosen is Immaterial, and the formulas therefore prove the fact of the intersection of the medians. The result just found admits of extension to any polygon, and, with formulas (IV), illustrates the fact that the cooi'dinates of centers of gravity are found by taking average values. CARTESIAN COORDINATES 23 PROBLEMS 1. Find the coordinates of the middle point of the line joining (4, — 6) and (- 2, - 4). Ans. (1, - 5). 2. Find the coordinates of the middle point of the line joining (a + b, c + d) and (a — b, d — c). Ans. {a, d). 3. Find the middle points of the sides of the triangle whose vertices are (2, 3), (4, — 5), and (— 3, — 0). Also find the lengths of the medians. 4. Find the coordinates of the point which divides the line joining (— 1, 4) and (— 5, — 8) in the ratio 1 : 3. Ans. (- 2, 1). 5. Find the coordinates of the point which divides the line joining (_ 3, _ 5) and (6, 9) in the ratio 2 : 5. Ans. (- ?, — 1). 6. Find the coordinates of the point which divides the line joining (2, 6) and (-p 4, 8) into segments whose ratio is — |. Ans. (— 22, 14). 7. Find the coordinates of the point which divides the line joining (— 3, — 4) and (5, 2) into segments whose ratio is — |. Ans. (— 19, — 16). 8. Find the coordinates of the points which trisect the line joining the points (— 2, — 1) and (3, 2). Ans. (— |, 0), (J, 1). 9. Prove that the middle point of the hypotennse of a right triangle i is equidistant from the three vertices. 10. Show that the diagonals of the parallelogram whose vertices are (1, 2), (— 5, — 3), (7, — 6), (1, — 11) bisect each other. . 11. Prove that the diagonals of any parallelogram bisect each other. 12. Show that the lines joining the middle points of the opposite sides of the quadrilateral whose vertices are (6, 8), (— 4, 0), (— 2, — 6), (4,— 4) bisect each other. 13. In the quadrilateral of Problem 12 show by means of slopes that the lines joining the middle points of the adjacent sides form a parallelogram. 14. Show that in the trapezoid whose vertices are (— 8, 0), {— 4, — 4), (— 4, 4), and (4, — 4) the length of the line joining the middle points of the nonparallel sides is equal to one half the .sum of the lengths of the parallel sides. Also prove that it is parallel to the parallel sides. 15. In what ratio does the point (— 2, .3) divide the line joining the points (- 3, 5) and (4, - 9) ? Ans. J. 16. In what ratio does the point (16, 3) divide the line joining the points (- 5, 0) and (2, 1) ? Ans. - I. 17. In any triangle show that a line joining the middle points of any ^^ two sides is parallel to the third side and equal to one half of it. 24 NEW ANALYTIC GEOMETRY 18. If (2, 1), (3, 3), (6, 2) are the middle points of the sides of a triangle, what are the cooi'dinates of the vertices of the triangle ? Ans. (-1,2), (5,0), (7, 4). 19. Three vertices of a parallelogram are (1, 2), (— 5, — 3), (7, — 6). What are the coordinates of the fourth vertex ? Ans. (1, - 11), (- 11, 5), or (13, - 1). 20. The middle point of a line is (6, 4), and one end of the line is (5, 7). What are the coordinates of the other end ? Ans. (7, 1). 21. The vertices of a triangle are (2, 3), (4, - 5), (— 3, — 6). Find the coordinates of the point v^here the medians intersect (center of gravity). 14. Areas. In this section the problem of determining the area of any polygon, the coordinates of whose vertices are given, will be solved. We begin with the Theorem. Tke area of a triangle whose vertices are the origin, P^(x^, y^, and P^{x,-^, y^ is given by the formula (V) Area of A OPJ*^ = | {x^y^ - x^y^) . Proof. In the figure' let a = Z. XOP^, P (Greek '' beta") = Z XOP.^, e (Greek " theta ") = Z P^OP^. (1) .-.e^p-a. By 45, p. 4, (2) Area A OP^P^ = ^0P^- OP^ sin = iOP^.OP^sm((3-a) (by (1)) (3) = i OP^ ■ OP^ (sin ^ cos a — cos /3 sin a). But in the figure sma M, X e (By 34, p. 3) M,P-2__ ?/2 OP, OP,' o OM, X, M,Pi_ y, OP, OP, ' OMi Xi ''''''= oprop- Substituting in (3) and reducing, we obtain Area A OPJ^.^ = ^{x^y^ - x^y^). Q.E.D. CARTESIAN COORDINATES 25 (-2;4) Y\ / \ / \ / \ (1,1) y f l\ III L — ■ ■ — '' X (-6, -1) \ EXAMPLE Find the area of the triangle whose vertices are the origin, (—2, 4), and (—5, — 1). Solution. Denote (- 2, 4) by P^, (- 6, - 1) by Fi. Tlien x, =-2, 2/i = 4, ;c., = — 6, i/.,^— 1. Substituting in (V), Area = | [- 2 • - 1 -(- 5) • 4] = 11. Then area =11 tmii squares. If, however, the fonnula (V) is applied by denoting (— 2, 4) by Pg, and (—5, — 1) by P,., the result will be — 11. The two figures for this example are drawn below. The cases ot positive and negative area are distinguished by the Theorem. Passing around the perimeter in the order of the vertices 0, P^, P^, if the area is on the left, as v/i. Fig.l, then (V) gives a posi- tive result ; if the area is on the right, as in Fig. 2, then (V) gives a negative result. Proof. In the formula (4) Area A OPJ\^ = | OP^ ■ 0I\ sin 9 the angle 6 is measured from OP^ to OP,, irithi/i. the triangle. Hence 6 is positive when the area p^ p lies to the left in passing around the perimeter 0, J\, P^, as in Fig. 1, since is then measured counter- clockwise (p. 2). But in Fig. 2, 6 is measured clockwise. Hence 6 is negative and sin in (4) is also negative. q.e.D. We apply (V) to any triangle by regarding its area as made up of trinngles with the origin as a common vertex. Fig. 1 Fig. 2 Fig. 1 26 NEW ANALYTIC GEOMETRY Theorem. The area of a triangle whose vertices are P^(x^, y^, (VI) Area A P^P^P^ = | (x^y^ — x^y^ + x^y^ — x^y^ + x^y^ — x^y^). This formula gives a jwsltlve or negative result according as the area lies to the left or right In 2)asslng around the perimeter in the order P, P„ P,. j> /> Proof Two cases must be distin- [C^^-'/ f^ /\ guished according as the origin is '^V/ t //C I within or Avithout the triangle. ;/ ' / ^^'-'^^ Fig. 1, origin within the triangle. p p By inspection, Fig. 1 Fig. 2 (5) Area A P^P.,P^ = A OP.P^ +' A OP^P^ + A OP^P^, since these areas all have the same sign. Fig. 2, origin without the triangle. By inspection, (6) Area A PjP^^g = A OP^P,^ + A OP,,P^ + A OP^P^, since OP^P.^, OPJ\ have the sr?w-e sign, but OP.^P^ the opposite sign, the algebraic sum giving the desired area. By (V), AOP^P, = i(x,y.3 - ,T^.y^), and AOP^P^ = ^(x^g^-x^g.;). Substituting in (5) and (6), we have (VI). Also in (5) the area is positive, in (6) negative. q. e. d. An easy way to apply (VI) is given by the following "Rule for findmg the area of a triangle. First step. Write down the vertices In two colmnns, abscissas in one, ordlnates In the other, repeating the coordinates of tlie first vei'tex. Second step. 3Iultlpli/ each abscissa by the ordinate of the next row, and add results. This gives x^y^ + 3?oy3'+ ^^Ui- Third step. Multiply each ordinate by the abscissa, of the next row, and add results. This gives y^x,^ + y,^x^ 4- y.^Xy ^x Vi ^2 y^ ^3 ?/8 ^1 y^ CARTESIAN COORDINATES 27 Fourth step. Subtract the result of the third step from that of the second step, and divide by 2. This gives the required area, namely formula (VI). Formula (VI) may be readily memorized by remarking that the right-hand member is a determinant of simple form, namely Area A P^P.f"^ In fact, when this determinant is expanded by the usual rule, the result, when divided by 2, is precisely (VI). It is easy to show that the above rule applies to any polygon if the following caution be observed in the first step : Write down the coordinates of the vertices in an order agreeing with that established bg passing continuously aro^ind the pjerim- eter, and repeat the coordinates of the first vertex. Area 25 + 25 = 26 unit squares. Ans. The result has the positive sign, since the area is on the left. 1 -1 -3 2 1 EXAMPLE Find the area of the quadrilateral whose vertices are (1, 6), (— 3, (2, -2), (-1,3). Solution. Plotting, we have the figure from which we choose the order of the vertices as indicated by the arrows. Following the rule : First step. Write down the vertices in order. Second step. Multiply each abscissa by the ordinate of the next row, and add. This gives 1 X 3 + (- 1 X - 4) + (- 3 X - 2) + 2 X 6 = 2.5. Third step. Multiply each ordinate by tlie abscissa of the next row, and add. This gives G X - 1 + 3 X - 3 + (- 4 X 2) + (- 2 X 1) = - 25. Fourth step. Subtract the result of the third step from the result of the second step, and divide by 2 6 3 - 4 _ 2 6 Y (1, / 1 r i' / I ± ^ / I It / \\ / y ' X ^ -^ (2,- 2) ^ « .-4) 28 NEW ANALYTIC GEOMETRY PROBLEMS 1. Find the area of the triangle whose vertices are (2, 3), (1, 5), ({- 1, -2). Anf<. V-. 2. Find the area of the triangle whose vertices are (2, 3), (4, — 5), ,(_ 3, _ 0). Anfi. 29. 3. Find the area of the triangle whose vertices are (8, 3), (— 2, 3), (4, - 5). Ans. 40. 4. Find the area of the triangle whose vertices are (rt, 0), (— a, 0), i(0, b). Ans. ab. 5. Find the area of the triangle whose vertices are (0, 0), (x,, y^), 2 6. Find the area of the triangle whose vertices are (a, 1), (0, b), (c, 1). Ans. (^LzJ^lfc^i). 2 7. Find the area of the triangle whose vertices are (a, 6), (6, «), (c, -c). Am. i(a2-6'-2). 8. Find the area of the triangle whose vertices are (3, 0), (0, 3\'^), (6, 3V3). Ans. 9 Vs. 9. Prove that the area of the triangle whose vertices are the points (2, 3), (5, 4), (— 4, 1) is zero, and hence that these points all lie on the same straight line. 10. Prove that the area of tlie triangle whose vertices are the points (a, b -\- c), (6, c + a), (c, a + b) is zero, and hence that these points all lie on the same straight line. 11. Prove that the area of the triangle whose vertices are the points (a, c + a), (— c, 0), {— a., c — a) is zero, and hence that these points all lie on the same straight line. 12. Find the area of the quadrilateral whose vertices are (— 2, 3), (_ 3, _ 4), ( 5, - 1), (2, 2). Ans. 31. 13. Find the area of the pentagon whose vertices are (1, 2), (3, — 1), (6, - 2), (2, 5), (4, 4). Ans. 18. 14. Find the area of the parallelogram whose vertices are (10, 5), (-2,5), (-5, -3), (7, -3). Ans. 96. 15. Find the area of the quadrilateral whose vertices are (0, 0), (5, 0), (9,11), (0, 3). ^ns. 41. CARTESIAN COORDINATES 29 16. Find the area of the quadrilateral whose vertices are (7, 0), (11, 9), (0, 5), (0, 0). Ans. 59. 17. Show that the area of the triangle whose vertices are (4, 6), (2, — 4), (— 4, 2) is four times the area of the triangle formed by joining the middle points of the sides. 18. Show that the lines drawn from the vertices (3, — 8), (— 4, 6), (7, 0) to the point of intersection of the medians of the triangle divide it into three triangles of equal area. 19. Given the quadrilateral whose vertices are (0, 0), (6, 8), (10, — 2), (4, — 4); show that the area of the quadrilateral formed by joining the middle points of its adjacent sides is equal to one half the area of the given quadrilateral. CHAPTER III CURVE AND EQUATION 15. Locus of a point satisfying a given condition. The curve* (or group of curves) passing through all points which satisfy a given condition, and through no other points, is called the locus of the point satisfying that condition. For example, in plane geometry, the following results are proved : The perpendicular bisector of the line joining two fixed points is the locus of all points equidistant from these points. The bisectors of the adjacent angles formed by two lines are the locus of all points equidistant from these lines. To solve any locus problem involves two things : 1. To draw the locus by constructing a sufficient number of points satisfying the given condition and therefore lying on the locus. 2. To discuss the nature of the locus; that is, to determine properties of the curve. Analytic geometry is peculiarly adapted to the solution of both parts of a locus problem. 16. Equation of the locus of a point satisfjring a given condition. Let us take up the locus problem, making use of coordinates. We imagine the point P{x, y) vioving in such a manner that the given condition is fulfilled. Then the given condition will lead to an equation involving the variables x and y. The following example illustrates this. *The word "curve" will hereafter signify any continuous line, straight or curved. 30 CURVE AND EQUATION 31 EXAMPLE The point P (x, y) moves so that it is always equidistant from A (— 2, 0) and B{— 3, 8). Find the equation of the locus. Solution. Let P{x, y) be any point on the locus. Then by the given condition (1) PA = PB. But, by formula (I), p. 13, PA = ^(x-\-2y + (y-0y, and PB = -s/(x + Sf + (y - Sf. Substituting in (1), (2) V(:r + 2)^ + (y-0)2 = V(.; + 3)2 + (y-8)l Squaring and reducing, (3) 2a; -16?/ + 69 0. In the equation (3), x and ?/ are variables representing the coordinates of any point on the locus ; that is, of any point on the perpendicular bisector of the line AB. This equation is called the equation of the locus ; that is, it is the equation of the perpendicular bisector CP. It has two important and characteristic properties : 1. The coordinates of any point on the locus may be sub- stituted for X and y in the equation (3), and the result will be true. For let P^{x^, y^) be any point on the locus. Then P^A = P^B, by definition. Hence, by formula (I), p. 13, (4) V(a;^ + 2y + yl = ^(x^ + Sf + (y, - Sf, or, squaring and reducing, (5) 2 a;^- 16//, + 69 = 0. 32 NEW ANALYTIC GEOMETRY But this equation is obtained by substituting x^ and ?/j for x and 1/ respectively in (3). Therefore x^ and i/^ satisfy (3). 2. Conversely, every point whose coordinates satisfy (3) will lie upon the locus. For if P^i^Cp y^ is a point whose coordinates satisfy (3), then (5) is true, and hence also (4) holds. q.e. d. In particular, the coordinates of the middle point C oi A and B, namely, a:=— 2J^, y/ = 4 (IV, p. 21), satisfy (3), since 2(-2i)-16 X 4 + 69 = 0. This discussion leads to "the definition : The equation of the locus of a point satisfying a given condi- tion is an equation in the variables x and y representing coor- dinates such that (1) the coordinates of every point on the locus will satisfy the equation ; and (2) conversely, every point whose coordinates satisfy the equation will lie upon the locus. This definition shows that the equation of the locus must be tested in two ivays after derivation, as illustrated in the exam- ple of this section. The student should supply this test in the examples and problems of Art. 17. From the above definition follows at once the Corollary. A point lies upoii a curve ivhen and only lahen its coordinates satisfy the equation of the curve. 17. First fundamental problem. To find the equation of a curve which is dejined as the locus of a point satisfying a given condition. The following rule will suffice for the solution of this prob- lem in many cases : Rule. First step. Assume that P(x, y) is any jjoint satisfying the given condition, and is therefore on the curve. Second step. Write down the given condition. Third step. Express the given condition in coordinates and simplify the result. The final equation, containing x, y, and the given constants of the problem, will he the required equation. CURVE AND EQUATION 33 EXAMPLES 1. Find the equation of the straight line passing through P^ (4, — 1) and having an inclination of 4 Solution. First step. Assume P (x, y) any point on the line. Second step. The given condition, since the . ,. . . Sir inclination a: is — , may be written (1) slope of P^P =: tan or = — 1. Third step. From (11), p. 17, (2) slope of P^P = tan a = hJlJ^ _ L+i . [By substituting (x, y) for (x^, y-^), and (4, — 1) for (X2, y^)-] y + i Therefore, from (1), (3) — 1, or X — 4 X + y — 3 = 0. Ans. 2. Find the equation of a straight line parallel to the axis of y and at a distance of 6 units to the right. Solution. First step. Assume that P (x, ?/) is any point on the line, and draw NP per- pendicular to OY. Second step. The given condition may be written (4) NP = 6. Third step. Since NP = OM = x, (4) be- comes (5) X = 6. Ans. 3. Find the equation of the locus of a point whose distance from (—1,2) is always equal to 4. Solution. First step. Assume that P (x, y) is any point on the locus. Second step. Denoting (— 1, 2) by C, the given condition is (6) PC = 4. }'i 1. :v' ._. — , __. ._. _.. p (X iv) M X 34 NEW ANALYTIC GEOMETRY Third step. By formula (I), p. 13, PC = V(x+l)2 + {2/-2)2 Substituting in (6), Squaring and reducing, (7) x2^ 2/2 + 2a; -42/ -11 = 0. This is the required equation, namely, the equation of the circle whose center is (— 1, 2) and radius equal to 4. }', k. .^ ^ "^ s. / / > < {Ai> / '' \l C ,' K -1,2) V / N \^ y / X s •^^ ^ / 1 PROBLEMS 1. Find the equation of a line parallel to OY and (a) at a distance of 4 units to the right. (b) at a distance of 7 units to the left. (c ) at a distance of 2 units to the right of (3, 2) . (d) at a distance of 5 units to the left of (2, — 2). 2. Find the equation of a line parallel to OX and (a) at a distance of 3 units above OX. (b) at a distance of 6 units below OX. (c) at a distance of 7 units above (— 2, — 3). (d) at a distance of 5 units below (4, — 2). 3. What is the equation of XX' ? of YY' ? 4. Find the equation of a line parallel to the line x = 4 and 3 units to the right of it ; 8 units to the left of it. 5. Find the equation of a line parallel to the line y = — 2 and 4 units below it ; 5 units above it. 6. What is the equation of the locus of a point which moves always at a distance of 2 units from the axis of x ? from the axis oi y? from the line x = — 5 ? from the line ?/ = 4 ? 7. What is the equation of the locus of a point which moves so as to be equidistant from the lines x = 5 and x = 9 ? equidistant from y = 3 and y =—7? 8. What are the equations of the sides of the rectangle whose vertices are (5, 2), (5, 5), (-2. 2). (-2. F,\ ? CURVE AND EQUATION 35 In Problems 9 and 10, P^ is a given point on the required line, rn is the slope of the line, and a its inclination. 9. What is the equation of a line if (a) Pj is (0, 3) and m=-3? Ans. 3x + y - S = 0. (b) Pj is (- 4, - 2) and 7n = i? Ans. x-Sy-2 = 0. V2 (c) P^ is (- 2, 3) and 7n = ^^ ? j^^s. V2x - 2?/ +6 + 2 V2 = 0. y/3 " (d) Pi is (0, 5) and ?n = -— ? ^ns. V3 x - 2 y + 10 = 0. (e) P^ is (0, 0) and m = - | ? Anfi. 2 x + 3 «/ = 0. (f) P, is (a, b) and 77i = 0? Ans. y = b. (g) P^ is (— a, b) and m = oo ? Ans. z =— a. 10. What is the equation of a line if (a) Pi is (2, 3) and a = 45°? Ans. x - y + 1 = 0. (b) Pj is (- 1, 2) and a = 45^ ? Ans. x — y + 3 = 0. (c) Pj is (— a, — b) and or = 45° ? ^ns. x — y = b — a. (d) Pj is (5, 2) and a = QO°? Ans. VSx - y + 2- bVs = 0. (e) P, is (0, - 7) and a -60°? Ans. VSx- y—7 = 0. (f) Pj is (- 4, 5) and a = 0°? Ans. y = 5. (g) Pj is (2, — 3) a,nd a = 90° ? Ans. x = 2. (h) Pj is (3, - 3 V3) and a = 120° ? Ans. ^3x + y^ 0. (i) Pj is (0, 3) and a= 150° ? Ans. V3x + 3y-9 = 0. (j) P, is (a, 6) and a = 135°? Ans. x + y = a + 6. 11. Find the equation of the strai;:,dit line which passes through the points (a) (2, 3) and {- 4, - 5). Ans. 4x-3!/4-l = 0. Hint. Find the slope by (II), p. 17, and then proceed as in Problem 9. (b) (2, - 5) and (- 1, 9). Ans. 14 x + 3?/ - 13 = 0. (c) (- 1, 6) and (G, - 2), Ans. 8x+72/-84 = 0. (d) (0, - 3) and (4, 0). Ans. 3x-42/-12 = 0. (e) (8, - 4) and (- 1, 2). Ans. 2x + 3?/-4 = 0. 12. Find the equation of the circle with (a) center at (3, 2) and radius = 4. Ans. x^ ■\- y'^ — Qx — 4y — 3 — 0. (b) center at (12, — 6) and /• = 13. Ans. x- + Z/"^ — 24x + lOy = 0. (c) center at (0, 0) and radius — r. Ans. x^ + y- = r-. (d) center at (0, 0) and r = 5. Ans. x^ + y^ =^ 26. (e) center at (3 a, 4 a) and r = 5 a. A ns. x'^ + y^ — 2 a (3 x + 4 y) = 0. (f ) center at (6 4- c, 6 — c) and r = c. Ans. x^ + y^ - 2 {b + c)x — 2 {b — c.)y + 2 h^ + c- = 0. 36 NP]W ANALYTIC GEOMETRY 13. Find the equation of a circle wliose center is (5, — 4) and whose circumference passes througli the point (— 2, 3). 14. Find the equation of a circle having the line joining (.3, — 5) and (— 2, 2) as a diameter. 15. Find the equation of a circle touching each axis at a distance 6 units from the origin. 16. Find the equation of a circle whose center is the middle point of the line joining (— 6, 8) to the origin and whose circumference passes through the point (2, 3). 17. A point moves so that its distances from the two fixed points (2, — 3) and (— 1, 4) are equal. Find the equation of the locus. Ans. 3x—7y + 2 = 0. 18. Find the equation of the perpendicular bisector of the line joining (a) (2, 1), (- 3, - .3). Ans. lOx + 8 y +13 = 0. (b) (3, 1), (2, 4). Ans. x - 3?/ + 5 = 0. (c) (- 1, - 1), (3, 7). Ans. x + 2y-7 = 0. (d) (0, 4), (3, 0). Ans. 6x-8y + 7 = 0. (e) (Xi, ?/i), (Xo, ?y,). Ans. 2 (x^ — X2)x + 2 (y^ — y^y + x^ - x^ + y^J - y^- = 0. 19. Show that in Problem 18 the coordinates of the middle point of the line joining the given points satisfy the equation of the perpendicular bisector. 20. Find the equations of the perpendicular bisectors of the sides of tlie triangle (4, 8), (10, 0), (G, 2). Show that they meet in the point (11, 7). 18. Locus of an equation. The preceding sections have iUus- trated the fact that a locus problem in analytic geometry leads at once to an equation in the variables x and ?/. This equation having been found or being given, the complete solution of the locus problem requires two things, as already noted in Art. IS of this chapter, namely : 1. To draw the locus by plotting a sufficient number of points whose coordinates satisfy the given equation, and through which the locus therefore passes. 2. To discuss the nature of the locus ; that is, to determine properties of the curve. CURVE AND EQUATION 37 These two problems are respectively called : 1. Plotting the locus of an equation (second fundamental problem). 2. Discussing an equation (third fundamental problem). For the present, then, we concentrate our attention upon some given equation in the variables x and y (one or both) and start out with the definition : The locus of an equation in two variables representing coordi- nates is the curve or group of curves passing through all points whose coordinates satisfy that equation,* and through such points only. From this definition the truth of the following theorem is at once apparent : Theorem I. If the form of the given equation he changed in any way {for example, by transposition, by multipUcation by a constant, etc.), the locus is entirely unaffected. We now take up in order the solution of the second and third fundamental problems. 19. Second fundamental problem. Rule to plot the locus of a given equation. First step. Solve the given equation for one of the variables in terms of the other.f * An equation in the variables x and y is not necessarily satisfied by the coordinates of any points. For coordinates are real numbers, and the form of the equation may he such that it is satisfied by no Teal values of x and y. For example, the equation 2.2 _(. ^2 _|. j = q is of this sort, since, when x and y are real numbers, x^ and y^ are necessarily positive (or zero), and consequently X"+y^ + 1 is always a positive number greater than or equal to 1, and therefore 7iot equal to zero. Such an equation therefore has no locus. The expression "the locus of the equation is imagi- nary" is also used. An equation may be satisfied by the coordinates of a, finite number of points only. For example, a;- + ?/'^ = is .satisfied by a; = 0, ?/ = 0, but by no other real values. In this case the group of points, one or more, whose coordinates sat- isfy the equation, is called the locus of the equation. t The form of the given equation will often be such that solving for one vari- able is simpler than solving for the other. Always choose the simpler solution. 38 NEW ANALYTIC GEOMETRY Second step. By this formula compute the values of the vari- able for which the equation has been solved by assuming real values for the other variable. Third step. Plot the points corresponding to the values so determined. Fourth step. If the points are numerous enough to suggest the general shape of the locus, draw a smooth curve through the points. Since there is no limit to the number of points which may be computed in this way, it is evident that the locus may be drawn as accurately as may be desired by simply plotting a sufficiently large number of points. Several examples will now be worked out. The arrangement of the work should be carefully noted. r^ k /, ^ r y ^ / Y (0,2; Y / y r-3,0j \ A[ EXAMPLES 1. Draw the locus of the equation 2a;-3?/+6=:0. Solution. First step. Solving for t/, Second step. Assume values for x and com- pute ?/, arranging results in the form of the accompanying table : Thus, if X = 1, 2/ = 5 • 1 + 2 = 2f, X = 2, 2/ = 1 . 2 + 2 = 3i, etc. Third step. Plot the points found. Fourth step. Draw a smooth curve through these points. 2 Plot the locus of the equation 2/ = x2 - 2 X - 3. Solution. First step. The equation as given is solved for y. X y X y 2 2 1 2t - 1 1* 2 H — 2 1 3 4 -3 4 n -4 -1 etc. etc. etc. etc. CURVE AXD EQUATION 39 Second step. Computing y by assuming values of x, we find the table of values below : ., y. X y X y -3 -3 1 -4 - 1 2 -3 -2 5 3 -3 12 4 5 -4 21 5 12 etc. etc. 6 21 etc. etc. Third step. Plot the points. Fourth step. Draw a smooth curve through these points. This gives the curve of the fiirure. 16 = 0. 3. Plot the locus of the equation cc^ + 2/^ + 6 X Solution. First step. Solving for ?/, y -± Vl6-6x-x2. Second step. Compute y by assuming values of x. For this purpose the table of Art. 3 will be found convenient. X y X y ±4 ±4 1 ±3 - 1 ±4.6 2 -2 ±4.9 3 imag. -3 ±5 4 " -4 ±4.9 5 (; - 5 ±4.6 6 (; - 6 ±4 7 (( - 7 ±3 -8 -9 imag. For example, if x = — 1, r/ = ± Vl6 +0 — 1 = V21 = ± 4.6, if X = 3, ?/ = ± Vl6 - 18 - 9 = ± V- 11, an imaginary number. 40 NEW ANALYTIC GEOMETRY Third step. Plot the corresponding points. Fourth step. Draw a smooth curve through these points. The student will doubtless remark that the locus of Example 1, p. 38, appears to be a straight line, and also that the locus of Example 3, p. 39, appears to be a circle. This is, in fact, the case. But the proof must be reserved for later sections. PROBLEMS 1. Plot the locus of each of the following equations : (a) X + 2 2/ = 0. ( i ) X = 2/2 + 2 2/ - 3. (q) x^ + y^ = 25. (h) x + 2y = 3. (j)ix = y^. {r) x^ + y^ + 9x = 0. (c) 3x-2/ + 5 = 0. {k)4x = ?/3-l. (s) x^ + y^ + 4y - 0. (d)z/ = 4x-2. {l)y = x^-l. (t) x2 + 2/2-6x-16 = 0. (e)x2 + 4j/ = 0. {m)y = x^-x. (u) x^ + y^ — 6y -W = 0. (f)2/ = x2-3. (n) ?/ = x3-x2-5. (v)4?/ = x*-8. (g)x2 + 4j/-5 = 0. (o)x2 + ?/2 = 4. (w)4x = y* + 8. (h) ?/ = x2 + X + 1. (p) x2 + 2/2 = 9. (X) 4 2/2 = x3 - 1. The following problem illustrates the Theorem. If an equation can be put in the form of a product of variable factors equal to zero, the locus is found by setting each factor equal to zero and plotting each equation separately. 2. Draw the locus of 4 x2 — 9 2/^ = 0. Solution. Factoring, (1) (2x-32/)(2x + 32/) = 0. Then, by the theorem, the locus consists of the straight lines (p. 59) (2) 2 X - 3 y = 0, and (3) 2 X + 3 2/ = 0. Proof. 1 . The coordinates of any point (x^ , 2/ j ) which satisfy (1) will satisfy eit/i€r{2) or (3). For if (Xj, 2/1) satisfies (1), (4) {2Xi-3 2/i){2x^ + 3 2/i) = 0. This product can vanish only when one of the factors is zero. Hence either 2 Xj- 3 2/^ = 0, and therefore (Xj, y^) satisfies (2) ; or 2Xi + 32/i = 0, and therefore (x^, y^) satisfies (3). CURVE AND EQUATION 41 2. A point (Xj, ?/j) on eitha' of the lines defined by (2) and (3) will also lie on the locus of (l). For if (xi, y^) is on the line 2 x — 3 y = 0, then (Corollary, p. 32) (5) 2x1-32/1 = 0. Hence the product (2 x^ — 3 ?/j) (2 Xj + 3 y.^) also vanishes, since by (5) the first factor is zero, and therefore (x^, ?/,) satisfies (1). Therefore every point on the locus of (1) is also on the locus of (2) and (3), and conversely. This proves the theorem for this example. Q. E. D. 3. Show that the locus of each of the following equations is a pair of straight lines, and plot the lines : (a) xV = 0. {\) x^- y-^ + x + y = 0. (b) x2 = 9 2/2. (m) X- - 3 xy -4y^ = 0. (c) x^ — (/2 = 0. {n ) x'^ — xy + 5 X — 5 y — 0. , (d) ?/- 6?/ = 7. (o) x2-4?/2 + .5x + 10?/ = 0. (e) xy — 2x = 0. (p) x^ + 2 xy + ?/ + x + ?/ = 0. (f ) 9 x2 - ?/2 = 0. {q) x^ + S xy + 2y^_ + X + y = 0. (g) x2- ?jxy = 0. (r) x2 — 2x7/+?/ + Qx- Gy = 0. (h) ?/ + 4x2/ = 0. (s) 3x2 + x?/ — 2?/2 + 6x- 4?/ = 0. (i) x2-4x— 5 = 0. (t) 3x2 — 2x?y — 2/2 + 6x — 52/ = 0. (j) x?/— 2x2 — 3x = 0. (u) x2 — 4x?/- 5?/ + 2x — lOy = 0. (k) 2/2- 5x?/+ G?/ = 0. (v) x2 + 4x?/ + 4 2/2+ 5x+10?/ + 6 = 0. 4. Show that the locus of Ax- + 2Jx + C = is a pair of parallel lines, a single line, or that there is no locus according as A = ^2 _ 4 ^k;; jg positive, zero, or negative. 5. Show that the locus of ^4x2 + Bxy + Cy^ = is a pair of intersect- ing lines, a single line, or a point according as A = B'^ — 4 AC is positive, zero, or negative. 6. Show that the following equations have no locus (footnote, p. 37) : (a) X2 + 2/2 + 1 =0. (e) (x + l)2 + ,/ + 4 = 0. (b) 2x2 + 3y-2 ^_ 8. (f) x2 + 2/2 + 2x + 2y + 3 = 0. (c) x2 4. 4 = 0. (g) 4.x2 + 2/2 + 8x + 5 = 0. (d) x* + 2/2 + 8 = 0. (h) 2/* + 2x2 + 4 = 0. (!) 9x2 + 4 7/2 + I8x + 82/ + 15 = 0. Hint. Write each equation in the form of a sum of squares, and reason as in the footnote on page 37. 42 NEW ANALYTIC GEOMETRY 20. Third fundamental problem. Discussion of an equation. The method exphiined of solving the second fundamental prob- lem gives no knowledge of the required curve except that it passes through all the points whose coordinates are determined as satisfying the given equation. Joining these points gives a curve more or less like the exact locus. Serious errors may be made in this way, however, since tlie nature of the curve between any two successive jyoints j^lotted is not determined. This objec- tion is somewhat obviated by determining before jdotting cer- tain properties of the locus by a discussion of the given equation now to be explained. The nature and properties of a locus depend upon the form of its equation, and hence the steps of any discussion must depend upon the particular problem. In every case, however, certain questions should be answered. These questions will now be presented. * 1. Is the curve symmetrical with respect to either axis of coor- dinates or with respect to the origin? , To answer this question we may proceed as in the following example : EXAMPLE ' ' Discuss the symmetry of the locus of (1) .'c--f4 7/-=16. Solution. The equation contains no odd powers of x ov y \ hence it may be written in any one of the forms (2) (;xf -(- 4 (- 7/)- = 16, replacing {x, y) by {x, - y) ; (3) (- x)"" + 4 {yf = 16, replacing {x, y) by (- x, y) ; (4) (- xf -f 4 (- yf = 16, replacing (x, y) by (- x, - //). The transformation of (1) into (2) corresponds in the figure to replacing each point P(x, y) on the curve by the point CURVE AND EQUATION 4a Q{x, — y). But the points P and Q are symmetrical with respect to XX\ and (1) and (2) have the same locus (Theo- rem I, p. 37). Hence the locus of (1) is unchanged if each point is changed to a second point symmetrical to the first with respect to A' A''. There- fore the loc2cs is si/ynmetrlcal with respect to the axis of x. Similarly, frowi .(3), the locus is symmetrical with respect to the axis of y, and from (4) the locus is symmetricdl with respect to the origin, for the points P{x, y) and S(^—x, — y) are symmetrical with respect to the origin, since OP = OS. In plotting the equation we take advantage of our knowl- edge of the symmetry of the curve by limiting the calcula- tion to points in the first quadrant, as in the table.' We plot these points, mark off the j^oints symmetrical to them with respect to the axes and the origin, and then draw the curve. The . locus is called an ellipse. The facts brought out in the example are stated in Theorem II. Symmetry. If the locus of an equation is un- affected by replacliKj y by — y throughout its equation, the locus is syvfiTTietrical with respect to the axis of x. If the locus is unaffected by changing x to — x tlirough- out its equation, the locus is symmetrical with respect to the axis of y. If the locus is unaffected by changing both x and y to — x and — y throughout its equation, the locus is syviinetrlcal with respect to the origin. These theorems may be made to assume a somewhat differ- ent form if the equation is algebraic in x and y. The locus of X y 4 3.4 2.7 1 ^\ 2 44 NEW ANALYTIC GEOMETRY an algebraic equation in the variables x and y is called an algebraic curve. Then from Theorem II follows Theorem III. Symmetry of an algebraic curve. If no odd jiowers of y occur in an equation, the locus is symmetrical with respect to XX' ; if no odd po^vers of x occur, the locus is sym- inetrical with respect to YY'. If every term is of even* degree, or every term oJ%dd deff7-ee, the loctis is symmetrical xoith respect to the origin. ', The second question arises from the following considerations : Coordinates are real numbers. Hence all values of a; which give imaginary values of y must be excluded in the calculation. Similarly, all values of y which lead to imaginary values of x must be excluded. The second question is, then : 2. IVhat values, if any, of either coordinate will give im,agi- nary values of the other coordinate ? The following examples illustrate the method : ^ EXAMPLES 1. What values of x and y, if any, must be excluded in determining points on the locus of (1) x2 + 4 2/2 = 16 ? X = ± 2 V4 — y'^, y = ± iVl6-x2. Solution. Solving for x in terms of y, and also, for y in terms of x, (2) f — Ji 9 a/4 _ 1/2 (3) From the radical in (2) -we see that all values of y numerically greater than 2 will make 4 — 2/^ negative, and hence make x imaginary. Hence all values of y greater than 2 or less than — 2 must be excluded. Similarly, from the radical in (3), it is clear that values of x greater than 4 or less than — 4 must be excluded. * A constant term is to be regarded as of zero (even) degree, as 16 iu (1), p. 42. CURVE AND EQUATION 45 Therefore in determining points on the locus, we need assume for y values only between and 2, as on page 43, or values of x between and 4 inclusive. A further conclusion is this : The curve lies entirely within the rec- tangle bounded by the four lines a; = 4, x=— 4, y-2, y-—2, and is therefore a closed curve. 2. What values, if any, of the coordinates are to be •(^eluded in deter- mining the locus of '/ (4) 2/^-4x4. 15 = 0? Solution. Solving for x in terms of y, and also for y in terms of x, (5) x= i(1.5 + 2/2), (6) y =. i V4X-15. From (5) any value of y will give a real value of x. Hence no values of y are excluded. K X y 31 4 ±1 4f ±2 6 ±3 7f ±4 10 ±5 12f ±6 etc. etc. From the radical in (0) all values of x for which 4x — 15 is negative mu.st be excluded ; that is, all values of x less than 3f . The locus therefore lies entirely to the right of the line x = 3|. Moreover, since no values of y are excluded, the locus extends to infinity, y increasing as x increases. The locus is, by Theorem III, symmetrical with respect to the x-axis, and is called a parabola. 46 NEW AXALYTIC GEOMETRY - ~ — — - - A X l> / / /^ / v • / V V r / f / r r If t-.f ; 1 _ — r _ _ _ 3. Determine what values of x and ?/, if any, must be excluded in determining the locus of <7) 4 y = x3. Solution. Solving for x in terms of y, and also for y in terms of x (8) X = VTi], (9) y = \x^. From these equations it appears that no values of either coordinate need be excluded. The locus is, by Theorem III, symmetrical with respect to the origin. The coordinates in- crease together ; the curve extends to infinity and is called a cubical parabola. The method illustrated in the examples is summed up in the Rule to determine all values of x and y which must he excluded. Solve the equation for' x in terms of y, and from, this result determine all values of y for which the computed value of x ivould he imaginary. These values of y must he excluded. ^olve the equation f>r y in terms of x, and from this result determine all values of x for which the computed value of y would he imaginary. These values of x mtist he excluded. In determining excluded values of x, and y we obtain also an answer to the question : 3. Is the curve a, closed curve, or does it extend to infinity? The points of intersection of the curve with the coordinate axes should be found. The intercepts of a curve on the axis of x are the abscissas of the points of intersection of the curve and A'A''. The intercepts of a curve on the axis of y are the ordinates of the points of intersection of the curve and YY^. CURVE AND EQUATION 47 Rule to find the intercejjts. Substitute y = and solve for real values of x. This gives the intercepts on the axis of x. Substitute X =■ and solve for real values of y. This gives the intercepts on the axis of y. The proof- of the rule follows at once from the definitions. The rule just given explains how to answer the question : 4. What are the intercepts of the locus? In particular, the locus may pass through the origin, in which case one intercept on each axis will be zero. In this case the coordinates (0, 0) must satisfy the equation. When the equa- tion is algebraic we have Theorem IV. The locus of an algebraic eq^iation passes throvgh the origin when there is no constant term in the equation. The proof is immediate. 21. Directions for discussing an equation. Given an equation, the following questions should be answered in order before plotting the locus. X Is the origin on the locus ? , 3- What are the intercepts ? (D2>. Is the locus symmetrical with respect to the axes or the origin ? 4. What values of x and y must I>e excluded? 5. Is the curve closed, or does it 2^ass off indefinitely far ? Answering these questions constitutes what is called a general discussion of the given equation. The successive results should be immediately transferred to the figure. Thus when the inter- cepts have been determined, ynark them off on th,e axes. Indicate which axes are axes of symmetry. The excluded values of x and y will determine lines parallel to the axes which the locus will not cross. Draw these lines. 48 NEW ANALYTIC GEOMETRY EXAMPLE Give a general discussion of the equation (1) x^-i2j^ + 16y = 0. Draw the locus. Solution. 1. Since the equation contains no constant term, the origin is on the curve. 2. Putting ?/ = 0, we find a; = 0, the intercept on the axis of x. Putting a; = 0, we find y = and 4, the intercepts on the axis of y. Lay off the intercepts on the axes. >J s Y. X k V *«s *»• ^ ">> •v, ^ ^ ^ , — ^, --- .'/= 4 ro> i) ,'/= 2 X ' 6 "^ — ■ X ^ ^ -*' ">». ^ y^ ^ y ^ "^ * ^\ r' ^ 3. The equation contains no odd powers of x ; hence the locus is symmetrical with respect to YY' . 4. Solvino; for x, (2) X = ± 2 V;/- - 4 y. All values of y must be excluded which make the expression beneath the radical sign negative. Now the roots of ?/2 — 4?/ = are y = and ?/ = 4. For any value of y between these roots, y- — 4y is negative. For example, y = 2 gives 4 — 8 = — 4. Hence all values of y between and 4 must be excluded. Draw the lines y = and y = 4. The locus does not come between these lines. Solving for ?/, (3) y = 2±i Vx^ + 16. Hence no value of x is excluded, since x^ + 16 is positive for all values of x. CURVE AND EQUATION 49 5. From (3), y increases as x increases, and tlie curve exte^ids out indefinitely far from botli axes. Plotting the locus, using (2), the curve is found to be as in the figure. The curve is a hyperbola. Sign of a quadratic. In the preceding example it became necessary to dfterniine for what values of y the quadratic expression y'^ — iy in (2) was positive. The fact made use of is this : If the sign of a quadratic expression is negative (or positive) for any one value of the unknown taken betv?een the roots, it is also negative (or positive) for every value of the unknown between the roots. This is easily seen graphically. For take any quadratic (4) Ax- + Bx+ C. Plot the equation " ■ (.5) y = Ax" + Bx+ C. The locus of ' (5) will be a parabola turned upward if A is positive, downward if A is negative (see Example 2, p. 38). The intercepts on the ,£-axis will be the . roots of (4). The values of y from (5) will clearly all have one sign for all values of a> be- tween the intei'- cepts, and the op- posite sign for all other values of x. We see, then, that the values of the quadratic (4) will have one sign for all values of x taken between the roots, and the opposite sign for all other values. To apply this, consider the locus of ( 1. 50 NEW ANALYTIC GEOMETRY PROBLEMS 1. Give a general discussion of each of the following equations and draw the locus. Make sure that the discussion and the figure agree. i(a)x2-4y = 0. ' {n) 9y- - x" = 0. -(b) ?/-4x + 3 = 0. (o) 9(/'- + r' = 0. (c) x'^ + 4?/- l(i = 0. (p) 2x2/ + 3j'- 4 = 0. (d) 9x2 _,. y-z _ 18 = 0. (q) x^ + 4x2/ + S?/^ + 8 = 0. , (e) x2 — 42/2_iG = 0. (r) x^ + xy + y^ — 4 = 0. ( f ) x^ - 4 2/2 + 10 = 0. ( s ) x2 + 2 xy - 3 i/2 = 4. (g) x2-?/2 + 4 = 0. (t) 2xy-i>^ + 4x = 0. ( h ) x2 — ?/ + X = 0. ( u ) 3 X- — y + X = 0. 1 ( i ) xy - 4 = 0. ( V ) 4 (/•- - 2 X - y = 0. ( j ) Oy + x^ = 0. (w) x2 - 2/2 + Gx = 0. \ (k) 4x- 2/3 = 0. (X) x2 + 4 2/2 + 8z/ = 0. ( 1 ) 6x - 2/^ = 0. (y) 9x2 + y- + I8x-6y = 0. .(m) 5x — 2/ + 2/" = 0. (z) 9x2 _ ^^2 ^_ jgx + 62/ = 0. 2. Determine the general nature of the locus In each of the following equations. In plotting, assume particular values for the arbitrary con- stants, but not special values ; that is, values which give the equation an added peculiarity.* = 2?-x. ' = 2ry. (a) 2/- = 2 )/(.(•. (f) x2- y (b) x2 — 2 my = vi'^. (g) x2 + y' lc\ ^ + ^ - 1 (^^) ^"^ + ^' ^ ' «2^ 62 • (i) x" + y (d) 2x2/ = a2. ' (J) "^^ = , . ^2 ^ _ (k) a22/ = x3. a^ h- The loci of the equations (a) to (i) in Problem 2 are all of the class known as conies, or conic sections, — curves following straight lines and circles in the matter of their simplicity. These curves are obtained when cross sections are taken of a right circular cone. Various definitions and properties will be given later. A definition often used is the following : A conic section is the locus of a point whose distances from a fixed point and a fixed line are in a constant ratio. * For example, in (a) and (b) )» = i.s a special value. In fact, in all these examples zero is a special value for any constant. CURVE AND EQUATION 51 3. Show that every conic is represented by an equation of tlie second degree in z and y. Hint. Take 71" to coincide with the fixed line, and di-aw XX' through the fixed point. Denote the fixed point hy (p, 0) and the constant ratio by e. Ans. (1 — e2) ^2 + y-2 _ 2px + p^ = 0. 4. Discuss and plot the locus of the equation of Problem 3 : (a) when e = 1. The conic is now called a. parabola (see p. 45). (b) when e < 1. The conic is now called an ellipse (see p. 43). (c) when e > 1. The conic is now called a hyperbola (see p. 48). 5. A point moves so that the sum of its distances from the two fixed points (3, 0) and (— 3, 0) is constant and equal to 10. What is the locus ? Ans. Ellipse 16 x^ + 25 1/^ = 400. 6. A point moves so that the difference of its distances from the two fixed points (5, 0) and (— 5, 0) is constant and equal to 8. What is the locus ? Ans. Hyperbola 9z- — 16y^ = 144. 7. Find the equations of the following loci, and discuss and plot them. (a) The distance of a point from the fixed point (0, 2) is equal to its distance from the a;-axis increased by 2. (b) The distance of a point from the fixed point (0, — 2) is equal to its distance from the y-axis increased by 2. (c) The distance of a point from the origin is equal to its distance from tlie ?/-axis increased by 2. (d) The distance of a point from the fixed point (2, — 4) is equal to its distance from the a;-axis increased by 5. Ans. 2y = x^ — 4x — 5. (e) The distance of a point from the point (3, 0) is equal to half its distance from the point (6, 0). (f ) The distance of a point from the point (8, — 4) is twice its distance from the point (2, — 1). (g) One third of the distance of a point from the point (0, 3) is equal to its distance from the z-axis increased by unity. Ans. x- — 8 y- — 24?/ = 0. (h) The distances of a point to the fixed point (— 1, 0) and to the line 4x — 5 = are in the ratio f. Ans. Qx"^ + 25 y" + 90a; = 0. 8. Prove the statement : If an equation is unaltered when x and y are interchanged, the locus is symmetrical with respect to the line y — x. Make use of this result in drawing the loci of : {^)xy = 4. {h) x^- + xy + y-^ = 9. {c) x^ + ij'^ := 1 . {d)xi + y^ = 1. 22. Asymptotes. The following examples elueidatt^ difficul- ties arising frequently in drawing the locus of an equation. 52 NEW ANALYTIC GEOMETRY EXAMPLES 1. Plot the locus of the equation (1) xy-2y-4 = 0. Solution. Solving for y, 4 (2) y X AVe observe at once, if x i2/ This is interpreted thus : The curve ap- proaches the line x = 2 as it passes off to infinity. In fact, if we solve (1) for x and write the result in the form 4 x = 2 + -' y it is evident that x approaches 2 a,s y increases indefinitely. Hence the locus extends both upward and downward indefinitely far, approaching in each case the line x = 2. The vertical line x = 2 is called a vertical asymptote. X y X y _ 2 _ 2 1 - 4 -1 ■t 3 H -8 -2 -1 l| - 16 -4 — I 2 00 -5 4 2i 16 2^ 8 -10 -i 3 4 etc. etc. 4 2 5 1 6 1 12 0.4 etc. etc. Va 1 h = \ \ V > s ^ -^ ^ — J — ^ — ;^ oo oo y tj , ^ O f2,o; *Y > N S V \ — 1 1 " 1 In plotting, it is necessary to assume values of x differing from 2, both less and greater, as in the table. slightly CURVE AND EQUATION 53 From (2) it appears that y diminishes and approaches zero as x in- creases indefinitely. The curve tlaerefore extends indefinitely far to the right and left, approaching constantly the axis of x. The axis of x is therefore a horizontal asymptote.* This curve is called a hyperbola. In the problem just discussed it was necessaiy to learn what value x approached when y became very large, and also what value y approached when X became very large. These questions, when important, are usually readily answered, as in the following examples : 2. Plot the locus of V 2x + 3 (Fig- 1) 3x-4 When X is very great, we may neglect the .3 in the numerator (2x4-3) and the — 4 in the denominator (3x — 4). That is, when x is very large, V = — = - • Hence y = - is a horizontal asymptote. ^ 3x 3 3 ^ ^ The equation shows directly that 3x — 4 = or x = f is a vertical asymptote. Or we may solve the equation for x, which gives 4^ + 3 Hence, when y is very large, x Fig. 2 3. The locus of y 2x + 3 x2_3x + 2 is shown in Fig. 2. There are two vertical asymptotes, a: = 1 and x = 2, since the denominator x^ — 3x + 2 = (x — 1) (x — 2). A branch of the * For oblique asymptotes, that is, asymptotes not parallel to either axis, see Art. 66. 54 NEW ANALYTIC GEOMETRY curve- lies between these lines. Furthermore, when x becomes large, we 2 X 2 may write the equation y = — = - = 0. Hence the x-axis X- X is a horizontal asymptote. A few points of the locus are given in the table. Note that different scales are used for ordinates and abscissas. X y 3 3 2 5 — 5 3 2 -24 1 3 12 1 B The determination of the vertical and hori- zontal asymptotes of a curve should be added to the discussion of the equation as outlined in Art. 21. PROBLEMS riot each of the following, and determine the horizontal and vertical asymptotes : 1. (a) xy + y-S = 0. (b) xy + X + 3 = 0. (c) 2x(/ + 2x + 3?/ (d) X- + xy + 8 = 0. 2. (a) x^y -5 = 0. (b) x^y — y -\-2x (e) y = (f) y = (g) y = (h) y = (i) y = 5 x2- -3x 4X--2 X-- -4 X - -3 x+1 X2- -4 X- - -1 (X- -2)(x + 3) (x + 2)(x-.3) (e) 2x;/ + 4x- 6?/ + 3 = (f) ?/2 + 2xi/-4 = 0. = 0. ). (g) X2/ + X + 2 y - 3 = 0. (h) X2/ + 2/-x2 + 2x = 0. (c) X2/2- 4x+ 6 = 0. = 0. (d) x^y - 2/ + 8 = 0. (J) y x2-4 X^ -f X ?/2 (0) 4x = ^^. 2/ — 9 (k)x 2/2 (P)i2x= ^y . ^ ' 3-2/2 (l)x y-2 2/-3 fn) v-Z'-'-^V ^'^^ ^ - I x j (m) y _x2-l 4-x2 x"- X— 1 (n) y x2-3x x2 + 3x + 2 + 2 X2 ^-^y X2-3X + 2 PROBLEMS FOR INDIVEDUAL STUDY Discuss fully and draw carefully the following loci : 1. 2/2 - 4 X2/ + x3 = 0. 5. {y - x)2 - x- {a- - x^) = 0. 2. 2/2 - 2 X2/ - 2 x2 + x3 = 0. 6. (y - x2)2 _ (a2 - x^) = 0. 3. 2/2 _ 3-2 4. a.4 _ 0. 7. (2/ - x2)2 - x2 {(l- - X"") = 0. 4. (y _ a;)2 _ (a2 _ 3.2) _ q. 8. ?/ (« _ a;) _ ^3 = q (the cissoid). 9. 2/2 (a — x) — x2 (a + x) = (the strophoid). 10. x* + 2 ax^y - ay^ = 0. CURVE AND EQUATION 55 11. X* — axy'^ +(/4 = 0. 12. a*2/2 — a^x* + x^ = o. 13. 02/2 _ bz* - x6 = 0. 14. a^i/2 _ 2 a6x^?/ — x^ = 0. 15. 2/2 - (a2 - x2) (62 - x2)2 = 0. 16. x8y2 _ (fij.2 ^ aij* = 0. 17. X (2/ - x)2 - h^y = 0. 18. (x2 + 2/2)2 _ ^2 (j;2 _ y2) ^ (the lemniscate). 19. (x2 - a2)2 = 02/2 (3 a + 2 y). 20. (x2 + ,/-2 _ 1) y _ ax = 0. 21. 2/2 - x2 — X (x — 4)2 = 0. 22. (x2 + 2/2-2 a2/)2 = a2 (x2 + y^) (the limacon). 23. (X* + x22/2 +y*) = x (ax2 — 4 ay^). 24. (X2 + 2/2 ^ 4 oty _ (j2) (j.2 _ q2) ^ 4 g[2y2 = Q. 25. (2/2 - x2) (x - 1) (x - I) = 2 (2/2 + x2 - 2 x)2. 26. (x2 + ,/2 -I- 4 a2/ -a^) (x2 _ a2) + 4 ahj- = (the cocked hat). 23. Points of intersection. If two curves whose equations are given intersect, the coordinates of each point of intersection must satisfy both equations when substituted in them for the variables. In algebra it is shown that all values satisfying two equations in two unknowns may be found by regarding these equations as simultaneous in the unknowns and solving. Hence the Rule to find the points of intersection of tivo curves whose eqiiatlons are given. Consider the equations, as simultaneous in tlie coordinates and solve ms in algebra. Arrange the real solutions in corresponding jJf^i^'s. These will be the coordinates of all tlie points of intersection. Notice that only real solutions correspond to common points of the two curves, since coordinates are always real numbers. 56 NEW ANALYTIC GEOMETRY EXAMPLES 1. Find the points of intersection of <1) X- 72/ + 25 = 0, (2) x2 + ?/2 = 25, Solution. Solving (1) for x, (3) x=72/-25. Substituting in (2), {7y- 25)-^ + y^ = 25. Reducing, 2/2-72/4-12 = 0. .-. y — S and 4. Substituting in (3) [not in (2)], X = — 4 and + 3. Arranging, the points of intersection are (— 4, 3) and (3, 4). Ans. In the figure the straight line (1) is the locus of equation (1), and the circle the locus of (2). 2. Find the points of intersection of the loci of (4) 2x2 + 32/2 = 35, (5) 3x2-42/ = 0. Solution. Solving (5) for x^, (6) x2 = |2/. Substituting in (4) and reducing, 92/2 + 8 2/ -105 = 0. .-. 2/ = 3 and — ^. Substituting in (6) and solving, X = ± 2 and ± i V- 210. Arranging the real values, we find the points of intersection are (4- 2, 3), (- 2, 3). Ans. In the figure the ellipse (4) is the locus of (4), and the parabola (5) the locus of (-5). CURVE AND EQUATION 67 PROBLEMS Find the points of intersection of the following loci : 7x-Uy + l=0^ x^ + y^ = 41 y = 3x + 21 •x2 = 2p2/j- ^- x^ + 2/2 = 4i" ^"^- (0, 0), (2p, 2j^). ^rw. (0, 2), (- I, - f). 9. ^^' + ^' = ^1.. 4.pJ!'o}' ^«..(J,2),(!.-2,. ^ns. (0, 0), (16, 16). x^ + y^ = 100^ x2 + ,/ = a2 ^ 10- ^2^^ [• • S, + y + a = 0j- 2 J /.ji^-. (0, -a),(- — ,^). ^7W. (8, 6), (8, - 6). 2/2 = 16"! jj_ a'2+ y2 = 5„2| x^ = »?/ x^ = iay J ylTis. (± 4 V2, 4). .•l?is. (2r(. a), (— 2a, a). Find the area of the triangles and polygons whose sides are the loci of the following equations : 12. 3x + y + 4 = 0, 3x - 5 2/ + 34 = 0, 3x - 2?/ + 1 = 0. Ans. 36. 13. X + 2?/ = 5, 2x + ?/ = 7, ?/ = X + 1. Ans. |. 14. x + y = «, X — 2 ij = 4 a, y — x + 7 a = 0. Ans. 12 a^. 15. X = 0, ?/ = 0, X = 4, ?/ = - 6. Ans. 24. 16. X — y = 0, X + y = 0. X — ?/ = a, X + 1/ = 6. ^ns. -^ ■ 17. 1/ = 3x - 9, ?/ = 3x + 5, 2 2/ = X - 6, 2 2/ = X + 14. ^ns. 56. 18. Find the distance between the points of intersection of the curves 3x - 2 ?/ + 6 = 0, x2 + v/2 = 9. Ans. jf VlS. 19. Does the locus of ?/2 = 4 x intersect the locus of2x + 3r/ + 2 = 0? Ans. Yes. 20. For what value of a will the three lines 3x+?/ — 2 = 0, f'x + 2 ?/ — 3 = 0, 2 X — ?/ — 3 = meet in a i^oint ? Aiw. a = 5. 21. Find the length of the common chord of j-'^ + t/^ = IS and y^ = Sx + S. Ans. (i. 22. If the equations of the sides of a triangle are x + 7y + ll = 0, 3x + ?/ — 7 = 0, X — 3?/ + l = 0, find the length of each of the medians. Ans. 2\/5, |\/2, J VlTO. CHAPTER IV THE STRAIGHT LIPfE V 24. The degree of the equation of any straight line. It will now be shown that any straight line is represented by an equa- tion of the first degree in the variable coordinates x and y. Theorem. The eqiiation of the straight line passing through a point B (0, l>) on the axis of y and having its slope equal to m u< (I) y=mx-\-b. Proof Assume that P (x, ?/) is any point on the line. The given condition may be written slope of PB = m. Since by (II), p. 17, ^ slope of PB = ^^ ~ ^ , X — [Substituting {x, y) for (j-^, y^) and (0, h) for (Xg, y^-l 1 y — b , then ' = m, or y = 7nx: -{■ b. Q. E. D. X In equation (I), m and b may have any values, positive, negative, or zero. Equation (I) will represent any straight line which inter- sects the ?/-axis. But the equation of any line 2)araUel to the ?/-axis has the form a; = a constant, since the abscissas of all points on such a line are equal. The two forms, y = mx -f- b and X = constant, will therefore represent all lines. Each of these equations being of the first degree in x and y, we have the Theorem. The equation of any straight line is of the first degree in the coordinates x and y. 58 THE STRAIGHT LINE 5& r 25. Locus of any equation of the first degree. The question now arises : Given an equation of the first degree in the coordinates x and y, is the locus a straight line ? Consider, for example, the equation (1) 3.r- 2^ + 8 = 0. Let us solve this equation for y. This gives (2) Z/=t-'' + ^- Comparing (2) with the formula (I), y = vix + b, we see that (2) is obtained from (I) if we set m = |, Z» = 4. Now in (I) m and h may have any values. The locus of (I) is, for all values of ??i and h, a straight line. Hence (2), or (1), is the equation of a straight line through (0, 4) with the slope equal to |. This discussion prepares the way for the general theorem. The equation (3) Ax+By + C = 0, where A, B, and C are arbitrary constants, is called the general equation of the first degree in x and y because every equation of the first degree may be reduced to that form. Equation (3) represents all straight lines. For the equation y — mx + b may be written mx — y + b = 0, which is of the form {S) it A = m, B = — 1, C = b ; and the e(}uation x = con- stant may be written x — constant = 0, which is of the form (3) if A = 1, B = 0, C =— constant. Theorem. The locns of the general equation of t lie first degree Ax -{-By + C = is a straight line. Proof. Solving (3) for y, we obtain (4) , y=-T--B 60 NEW ANALYTIC GEOMETRY Comparison with (I) shows that the locus of (4) is the straisfht line for which A C ')n = — —I 6=— — . B B If, however, B = 0, the reasoning fails. But if 5 = 0, (3) becomes Ax-\-C = Q, C or X = — —• A The locus of this equation is a straight line parallel to the ?/-axis. Hence in all cases the locus of (3) is a straight line. Q.E.D. Corollary. The slope of the line Ax +% + C = Is VI = — — ; that Is, the coefficient of x with its sign changed B divided hy the coefficient of y. 26. Plotting straight lines. If the line does not pass through the origin (constant term not zero, p. 47), find the intercepts (p. 47), mark them off on the axes, and draw the line. If the line passes through the origin, lind a second point whose coordinates satisfy the equation, and draw a line through this point and tlie origin. EXAMPLE Plot the locus of 3 X — 2/ + G = 0. Find the slope. Solution. Letting ?/ = and solving for x, X = — 2 = intercept on x-axis. Letting x = and solving for y, 2/ = 6 = intercept on ?/-axis. The required line passes through the points (-2, 0) and (0, 6). To find the slope : Comparison with the general equation (3) shows that yl = 3, B = — 1, (7 = 6. Hence in=.— — — ^. Otherwise thus : Reduce the given equation to the form y = mx + b by solving it for y. This gives ?/ = 3x + 6. Hence m = 3, 6 = 6, as before. THE STRAIGHT LINE 61 - 5, 2i ; m = i -1, 3; m = 3. i, 3; m=-|. .\C'' PROBLEMS 1. Find the intercepts and the slope of the following lines, and plot the lines : (a) 2 X + 3 2/ = 6. Ans. 3, 2 ; m = - |. (b) x — 2y+5 = 0. Ans. >'(c) 3x — ?/ + 3 = 0. Ans. ^ (d) 5x + 2?/ — 6 = 0. Ans. 2. Plot the following lines and find the slope : (a)2x-37/ = 0. (c)3x + 22/ = 0. (b) y -4x = 0. {d) x-3y = 0. K 3. Find the equations, and reduce them to the general form, of the lines for which (a)m = 2,6=-3. Ans. 2x — y — S = 0. (b) jn=-i, 6= 5. Ans. x + 2y — 3 = 0. ^ (c) m = |,6= — f. ^?xs. 4x-10?/- 25 = 0. Ans. X — y — 2 = 0. l{d) a = -,b=-2. (e) a = — J> = S. ^ ' 4 Ans. X + y — 3 = 0. Hint. Substitute imj = mx + b and transpose. ^ 4. Select pairs of parallel and perpendicular lines from the following : ' L^:y = 2x-3. L^:y=-3x-\-2. L,:y = 2x+ 7. (a) Ans. L^ II ig ; ^2 -^ -^4- ^L^:y = ix + i. ^^L^:x + 3y = 0. (b) -! ig : 8x + 2/ + 1 = 0. Ans. L^ ± ig. [Lg:Qx-3y + 2 = 0. fij :2x- 5?/ = 8. (c) < L^: 5y + 2x = 8. Ans. L^ ± ig. [ig:85x-14?/ = 8. t^ 5. Show that the quadrilateral whose sides are 2x — 32/+4 = 0, 3x — ?/ — 2 = 0, 4x — 6^ — 9 = 0, and 6x — 2y + 4 = is a paral- lelogram. 6. Find the equation of the line whose slope is — 2, which passes through the point of intersection of y = 3x + 4 and y =— x + 4. Ans. 2x + y — 4 = 0. 62 NEW ANALYTIC GEOMETRY ^ 7. Write an equation which will represent all lines parallel to the line (a) 2/ = 2a; +7. (c) y _ 3x - 4 = 0. {b);/=-a; + 9. (d) 2?/ - 4x + 3 = 0. 8. Find the equation of the line parallel to 2x — ^y = Q whose intercept on the F-axis is — 2. Ans. 2x — 3j/ — 6 = 0. 9. Show that the following loci are straight lines and plot them : (a) The locus of a point whose distances from the axes XX' and YY' are in a constant ratio equal to |. Ans. 2x — 3?/ = 0. (b) The locus of a point the sum of whose distances from the axes of coordinates is always equal to 10. Ans. a; + ?/ — 10 = 0. (c) A point moves so as to be always equidistant from the axes of coordinates. Ans. x — y = 0. (d) A point moves so that the difference of the squares of its distances from (3, 0) and (0, — 2) is always equal to 8. Ans. The parallel straight lines 6x + 42/ + 3=:0, 6x + 42/ — 13 = 0. (e) A point moves so as to be always equidistant from the straight lines X — 4 = and 2/ + 5 = 0. Ans. The perpendicular straight lines x — y — 9 = 0, x + y + 1— 0. 10. A point moves so that the sum of its distances from two perpen- dicular lines is constant. Show that the locus is a straight line. Hitit. Choosing the axes of coordinates to coincide with the given lines, the equation \sx + y — constant. 11. A point moves so that the difference of the squares of its distances from two fixed points is constant. Show that the locus is a pair of straight lines. Hint. Draw A'A'' through the fixed points, and YY' through their middle point. Then the fixed points may be written (a,0), (- a,0), and if the" constant difference " be denoted by k, we find for the locus 4 ax = k and 4 ax =- k. 12. A point moves so that the difference of the squares of its distances from two perpendicular lines is zero. Show that the locus is a pair of perpendicular lines. 13. A point moves so that its distance from a fixed line is in a constant ratio to its distance from a fixed point on the line. For what values of the ratio is the locus real ? What is the locus ? 27. Point-slope form. If it is required that a straight line shall pass through a given point in a given direction, the line is determined. THE STRAIGHT LINE 63 The following problem is therefore definite : To find the equation of the straight line passing throxigh a given -point P^(x^, y,) and having a given slope m. Solution. Let P(x, y) be any other point on the line. By the hypothesis, ^^^^ pp^ ^ ^^ (1) •'•^^^ = ^^- (n,p.i7) Clearing of fractions gives the formula (II) y-yi = m{x-x^). 28. Two-point form. A straight line is determined by two of its points. Let us then solve the problem : To find the equation of the line passing through two given points P^(x^, ?/j), P^(x^, y.^. Solution. The slope of the given line is slopeP,P, = ^5^^- Let P (x, y) be any other point on the line P^P^- Then slope pp, = y ~y^ . ^ a* — Xj Since P, Pj, and P„ are on one line, slope PP^ = slope PJ*^- Hence we have the formula (III) y-Vx Vx-Vi, M X^ — — Xn Equation (III) may be written in the determinant form X y \ (2) Xi z/i 1 =0. X y 1 Xi Vi 1 ^1 2/2 1 For the determinant, when expanded, is of the first degree in x and y. Hence (2) is the equation of a line. But (2) is satisfied when x = x^, y = ?/j, and also when x = x^, y = yn, for then two rows become identical and the determinant vanishes. Otherwise thus : Comparison of (2) with the formula at the close of Art. 14 shows that the area of the triangle PP^P,, is zero. Hence these three points lie on a line. 64 NEW ANALYTIC GEOMETRY 3, EXAMPLES 1. Find the equation of the line passing througli P^ (3, — 2) whose slope is — J. Solution. Use the point-slope equation (II), substituting x^ = ?/j = — 2, m =— I. This gives y + 2=-^{X-S). Clearing and reducing, X + 4y + 5 = 0. 2. Find the equation of the line through the two points P^ (5, — 1) and ^2(2,-2). Solution. Use the two-point equation (III), substituting Xj = 5, ?/j = — 1, P ( 3,-2> = 2, 2/2 =-2 This gives y + i -1 + 2 1 x-5 5-2 3 CI earing and reducing. X — Sy-8^0. The answer should be checked. To do this, we must prove that the coor- dinates of the given points satisfy the answer. Thus for Pj, substituting X = 5, 2/ =— 1, the answer holds. Similarly for P^. The student should supply checks for Examples 1, 3, and 4. 3. Find the equation of the line through the point P^ (3, — 2) parallel to the line L^:2x-Sy-i = 0. Solution. The slope of the given line ij equals |. Hence the slope of the re- quired line also equals | (Theorem, p. 17), and it passes through Pj(3, — 2). Using the point-slope equation (II), we have y + 2 = 1 (a;-3), or 2X-32/- 12 = 0. 4. Find the equation of the line through the point Pi(— 1, 3) perpen- dicular to the line i^:5x — 22/ + 3 = 0. Solution. The slope of the given line L^ equals |. Hence the slope of the required line equals — | (Theorem, p. 17). Since we know a point Pi(— 1, 3) on the line, we use the point-slope equation (II), and obtain y - 3 = - Kx + 1), or 2 X 4- 5 ?/ - 13 =: 0. THE STRAIGHT LINE 65 PROBLEMS 1. Find the equation of the line satisfying the following conditions, and plot the line. Check the answers : (a) Passing through (0, 0) and (8, 2). Ans. x — 4y = 0. (b) Passing through (— 1, 1) and (— 3, 1). Ans. tj — 1 = 0. (c) Passing through (— 3, 1) and slope = 2. Ans. 2x — y + 7 = 0. (d) Having the intercepts* a = 3 and 6 =— 2. Ans. 2x — Sy— 6 = 0. (e) Slope =—3, intercept on cc-axis = 4. Ans. 3x + v/ — 12 = 0. (f) Intercepts a =—3 and 6 =— 4. Ans. Ax + Sy + 12 = 0. (g) Passing through (2, 3) and (— 2, - 3). Ans. 3x — 2y = 0. (h) Passing through (3, 4) and (— 4, — 3). Ans. x — y + 1 - 0. (1) Passing through (2, 3) and slope =—2. Ans. 2x + y — 7 = 0. 2. Find the equation of the line passing through the origin parallel to the line 2a; — 3?/ = 4. Ans. 2x — 3y = 0. 3. Find the equation of the line passing through the origin perpen- dicular to the line 5x + y — 2 — 0. Ans. x — ^y = Q. 4. Find the equation of the line passing through the point (3, 2) par- allel to the line 4x — ?/ — 3 = 0. Ans. 4x — y — 10 = 0. 5. Find the equation of the line passing through the point (3,0) per- pendicular to the line 2x + y — o = 0. Ans. x — 2 y — 3 = 0. 6. Find the equation of the line whose intercept on the ?/-axis is 5, which passes through the point (6, 3). Ans. x + Sy — 15 = 0. 7. Find the equation of the line whose intercept on the a; -axis is 3, which is parallel to the line x — Ay + 2 = 0. Ans. x — 4?/ — 3 = 0. 8. Find the equation of the line passing through the origin and through the intersection of the lines x — 2y + S = and x + 2y — 9 = 0. Ans. x — y = 0. 9. Find the equations of the sides of the triangle whose vertices are (- 3,2), (3, -2), and(0, -1). Ans. 2x-l-32/ = 0, x-f-32/-l-3 = 0, and x + y + 1 =0. 10. Find the equations of the medians of the triangle in Problem 9, and show that they meet in a point. Ans. X = 0, 7 X + 9 y + 3 = 0, and fjx + dy + 3 = 0. Hint. To show that three lines meet in a point, tind the point of intersection of two of them and prove that it lies on the third. * Intercept on avaxis = a, intercept on y-axis = b. The given points are (3, 0) and (0, - 2) . 66 NEW ANALYTIC GEOMETRY 11. Determine whether or not the following sets of points lie on a straight line : (a) (0, 0), (1, 1), (7, 7). Ana. Yes. (b) (2, 3), (- 4, - 6), (8, 12). Ans. Yes. (c) (3,4), (1,2), (.5,1). Ans. No. (d) (3, - 1), (- 6, 2), (- I 1). Ans. No. (e) (5,6), (i,l), (-1,-5). Ans. Yes. (f ) (7, 6), (2, 1), (0, - 2). Ans. No. (g) (3, - 2), (6, - 4), (- 5, 4). (h) (1, 0), (0, 1), (7, - 8). (i) (-3, -1), (6, 2), (8,3). 12. Find the equations of the lines joining the middle points of the sides of the triangle in Problem 9, and show that they are parallel to the sides. Ans. 4x + 6y+ 3 = 0, j + 3y = 0, and x + y = 0. 13. Find the equation of the line passing through the origin and through the intersection of the lines x + 2 y = 1 and 2x — -iy — S = 0. Ans. X + 10 y -0. 14. Show that the diagonals of a square are perpendicular. Hint. Take two sides for the axes and let the length of a side be a. 15. Show that the line joining the middle points of two sides of a tri- angle is parallel to the thiixl. Hint. Choose the axes so that the vertices are (0,0), ( Pi Mi D Fia. 1 Thus in Fig. 1 the closing line is P^P^ ; in Fig. 2 the closing line is P^P^- With reference to such broken lines, the following theorem, of frequent application, holds. Second Theorem of PROJECxioisr. If each segment of a hroTien line he given the direction determined injxissing continu- oti sly from one extremity to the other, then the algebraic sum of the 2}roJectio?is of the segments ujion any directed line equals the projection of the closing line. Proof The proof results immediately. For, in Fig. 1, M^il/^ = projection of Pf'.-^; M^M^ — projection of P.f'^ ; M^M^ = projection of closing line P-^P^. But obviously M^M^ + MJVI^ = M^M^, and the theorem follows. Similarly in Fig. 2. Q. e. d. 70 NEW ANALYTIC GEOMETRY Corollary. If the sides of a closed polygon he given the direc- tion established by passing contimtously around the perimeter, the sum of the p/rojections of the sides upon any directed line is zero. For the closing line is now zero. 32. The normal equation of the straight line. In the preceding sections the lines considered were determined by two points or by a point and a direction. Both of these methods of determin- ing a line are frequently used in elementary geometry, but we have now to consider a line determined by two conditions which belong essentially to analytic geometry. Let AB be any line, and let ON be drawn from the origin perpendicular to Ali at C. Let the j^ositive divcrtion on ON be from O toward N, that is, from the origin toward the line, and denote the positive directed length OC by /i, and the posi- tive angle A'OiV, measured, as in trigonometry, from OX as initial line to ON as ter- minal line,tjf o) (Greek letter "omega"). Then it is evident from the figures that the position of any line is deterTnined by a jniir of values ofp and w, both j' and to being pjositive and w < 3C0°. On the other hand, every line which does not pass through the origin determines a single positive value of ^ and a single positive value of w which is less than 360°. The problem now is this : Given for the line AB of the figure the perpendicular distance OC (=7>) from the origin and the angle XOC (= w); to find the equation of AB. Solution. Let P(x,y) be any point on the given line AB. Then since yl/> is perpendicular to ON, the projection of OP on ON is equal to pj. Consider the broken line ODP. The THE STRAIGHT LINE 71 closing line is OP. By the second theorem of projection (p. 69), the projection of OP on OX is equal to the sum of the projec- tions of OD and DP on ON. Then (1) projection of OD on OX + projection of DP on ON =^p. By the first theorem of projec- tion (p. 68), (2) projection of OD on ON = OD cos iii =x cos o), and (3) projection of DP on OX = DP coslZ - = 0. Q. E.D. .v\ \ s^ z^-- ^x* ^ X '^ \x' This equation is known as the normal eq nation. ^Yhen^; = 0, however, AB passes through the origin, and the rule given above for the posi- tive direction on OX becomes meaningless. From the fig- ures we see that we can choose for w either of the an gles A' OX or A' 6* A' '. IVh en J/ =z ive sJuill alirosl- tlve (Jlrcctum. on <>N is the iijucard directum. 33. Reduction to the normal form. In Art. 25 it appeared that the slope of any line could be found after its equation was reduced to the form y = mx + h. If now the equation of any line can be reduced to the normal form (V), we shall be able to find the perpendicular distance p from the origin to the line, as well as the ingle w which this perpendicular makes with OX. 72 NEW ANALYTIC GEOMETRY To reduce a given equation (1) Ax + lU/ + C = to the normal form, it is necessary to determine w and^ so that the locus of (1) is identical with the locus of (2) X cos w + y s^i^ 0) — 2^ = 0. This is the case when corresponding coefficients are propor- tional.* Hence we must have cos w _ sin < 180° (p. 71) ; then sin (ri is positive, and from (4) r and B must have the same signs. The advantages of the normal form of the equation of the straight line over the other forms are twofold. In the first place, every line may have its equation put in the normal form ; whether it is parallel to one of the axes or passes through the origin is immaterial. In the second place, as will be seen in the following section, it enables us to find immediately the perpen- dicular distance from a line to a point. PROBLEMS 1. In what quadrant will ON (see figure on page 70) lie if sin w and cos w are both positive ? both negative ? if sin w is positive and cos a> negative ? if sin w is negative and cos oj positive ? 2. Find the equations and plot the lines for which • (a) w = 0, p = 5. An&. x = 6. (b)a; = ^,p = 3. Ann. ^ + 3 = 0. \.(c) o) = -,p = 3. Am. \/2x + V27/ - 6 = 0. 4 74 NEW ANALYTIC GEOMETRY 2 ir r- v^/ (d) w = — , P = 2. Ans. X — V3 2/ + 4 = 0. O (e) oj = ■^,p = 4. ^TW. V2x — V2i/ — 8 = 0. 4 3. Reduce the following equations to the normal form and find p and w : '(a) 3x + 4(/ — 2 = 0. ^ns. p = f, w = cos-i \ = sin-i|. (b) 3x - 4?/ — 2 = 0. ^ns. p = |, w = cos-i | = sin-i (— |). , (c) 12x — 52/ = 0. -4ns. p = 0, w = cos-^ (— 1|)= sin-i j^^. (d) 2 X + 5 2/ + 7 = 0. ^7W. p = ^- — , u> = cos~ 1 / — - — ) = sin-i + V29 \-V29/ \-\''29 (e) 4a; — 3 // + 1 = 0. A-m. p = I, u = cos-i(— |) = sin-i | (f) 4x- 5y + 6 = 0. Ans. p + V41 \-V41/ \+^'41 (g)x-4 = 0. (h)2/-3 = 0. (i)x + 2 = 0. (j)2/ + 4 = 0. 4. Find the perpendicular distance from the origin to each of the following lines : (a) 12x+ 52/-26 = 0. Ans. 2. (b) x + y + 1 = 0. Ans. i V2. (c) 3x-2?/-l = 0. Ans. ^"3 Vl3. (d) X + 4 = 0. (e) y-b = 0. 5. Derive (V) when (a) - < w < - . \ 6. For what values of p and w will the locus of (V) be parallel to the X-axis ? the y-axis ? pass through the origin ? 7. Find the equations of the lines whose slopes equal — 2, which are at a distance of 5 from the origin. Ans. 2V5x + v'Sy — 25 = and 2\/5x + VS?/ + 25 = 0. 8. Find the lines whose distance from the origin is 10, which pass through the point (5, 10). --Ins. y = 10 and 4x + 3?/ = 50. 9. Write an equation representing all lines whose perpendicular dis- tance from the origin is 5. 34. The perpendicular distance from a line to a point. The positive direction on the line OX drawn through the origin perpendicular to AB is from O to AB (Art. 32). The positive THE STRAIGHT LINE 75 direction on ON will now be assumed to determine the positive direction on all lines perpendicular to AB. Hence the perpen- dicular distance from the line AB to the point P^ is positive if Pj and the origin are on opposite sides of AB, and negative if /'j and the origin are on the same side (if AB. Thus in the figure the distance from AB to P^ is positive, and from AB to 1\^ is negative. Let us now solve this problem : Given the equation of any line AB and a point P^ ; to find the perpendicular distance from A B to /\. Solution. Assume that the equation of AB is in the normal form (1) X cos oi + y sin w -^. = 0. Let the coordinates of 7'j be (x^, y^ and denote the perpendicular dis- tance from A B to P^ by d. In the figure project the broken line 01)]^^ upon the normal ON. Then since OP^ is the closing line, by the second theorem of projection (p. 69), projection of OPj on 0N= projection of OD on OiV+ projection of DP^ on ON. From the iigure, projection of OP^ on ON = OE =p + d. By the first theorem of projection (p. GS), projection of ()]> on ON = OD cos w = x^ cos \ = 7/j sin \y = 2x-3. -* ^°' \y.= -3x + i. Ans. V5 Ans. 3 VTo Ans. 1 2\/T3 Ans. 6 ■^(c) |2x-32/ + 4 = 0, ^ ' \4x-6i/4-9=:0. / 2/ = mx + 3, \y = mx-3. ""'■• viT^ 11. Find the equations of the bisectors of the angles of the following triangles, and prove that these bisectors meet in a common point : C? (a) X + 2 ?/ - 5 = 0, 2 X - 2/ - 5 = 0, 2 x + y + 5 = 0. ^(b) 3x+ 2/-1 = 0, x-3y-3=:0, x+3y+n=0. ^(c) 3x + 4?/-22 = 0, 4x-3?/ + 29 = 0, y - 5 - 0. (d)x + 2 = 0, 2/- 3 = 0, x + y = 0. (e) X = 0, 2/ = 0, x + y + 3 = 0. 12. Find the bisectors of the angles formed by the lines 4x — 32/ — 1 = and 3x — 4?/-f2 = 0, and show that they are perpendicular. Ans. 7 X — 7y + 1 = and x + y — 3 = 0. 13. Find the equations of the bisectors of the angles formed by the lines 5x- 12y + 10 = and 12x- 5y + 1.5 = 0. 14. Find the locus of a point the ratio of whose distances from the lines 4x — 3?/ '+ 4 = OandSx + 12)/- 8 = 0isl3to5. Ans. 9x + 9//-4 = 0. 15. Find the bisectors of the interior angles of the triangle formed by the#iiies 4x-3y = 12, 5x-12?/-4 = 0, and 12x — 5?/ - 13 = 0. Show that they meet m a point. Ans. 7x- 9y- 16 = 0, 7x + 7?/- 9 = 0, 112x- 64y- 221 = 0. 16. Find the bisectors of the interior angles of the triangle formed by the lines 5x-12y = 0, 5 x + 12 2/ + 60 = 0, and 12 x - 5 y - 60 = 0. Show that they meet in a point. Ans. 2 2/ + -5 = 0, 1 7 X + 7 2/ = 0, 17 X - 17 // - 60 = 0. 80 NEW ANALYTIC GEOMETRY 17. The sides of a triangle are S x + 4 y — 12 = 0, 3x — 4y = 0, and 4x + 3y + 24 = 0. Show that the bisector of the interior angle at the vertex formed by the first two lines and the bisectors of the exterior angles at the other vertices meet in a point. 18. Find the equations of the lines parallel to 3x -f 4 ?/ — 10 = 0, and at a distance from it equal numerically to 3 units. -4ns. oj + 4?/ = 2.5 or — .5. : 35. The angle which a line makes with a second line. The angle between two directed lines has Ix-rn detiiied (Art. 12) as the angle between their positive directions. When a line is given by means of its equation, no positive direction along the line is fixed. In order to distinguish between the two pairs of equal angles which two intersecting lines make with each other, we' define the angle which a line makes with a second line to be the positive angle (p. 2) from the second line to the first line. Thus the angle which L^ makes with L,^ is the angle 6. We speak always of the " angle which one line makes with a second line," and the use of the phrase " the angle between two lines " should be avoided if those lines are not directed lines. Theorem. If vi^ and m^ ai^e the slopes of tivo lines, then the (ingle which the first line makes ivith the second is given hy (VI) tan B = 1 + m^m^ Proof Let ttj and a^ be the inclinations of L^ and Z., respec- tively. Then, since the exterior angle of a triangle equals the sum of the two opposite interior angles, we have (Fig. 1) a^ = $-\- a,^, or e = n^- a^, (Fig. 2) a^ = Tr-e + a^, or = tt + (a^ - a^. ■ THE STRAIGHT LINE 81 And since (30, p. 3), tan (tt + ^') = tana;, we have, in either case, tan 6 = tan (n^ — a^ tan aj — tan a.^ 1 + tan «! tan ag Y /l^ / X Fig. 1 Fig. 2 But tan a^ is the slope of Z^, and tan or., is the slope of L- hence, writing tan a^ = m^, tan a, = m.^, we have (VI). In applying (VI) we remember that m.^ = slope of the line from which B is measured in t\e, positive direction. (The Greek letter B used here is named " theta.") EXAMPLES 1. Find the angles of the triangle formed by the lines whose equa- tions are L:2x-3?/-6 = 0, M-.dx-y -Q = 0, JV:6x + 42/-25 = 0. Solution. To see which angles formed by the given lines are the angles of the triangle, we plot the lines, obtaining the triangle ABC. Let us find the angle A. In the figure, A is measured from the line L. Hence in (VI), m, = slope of i = I, m, = slope of M = 6. .-. tan A = _ 16 1 + 4~ 15 , and A = tan-i||. 82 NEW ANALYTIC GEOMETRY Next find the angle at B. In the figure, B is measured from N. Hence in., = slope of iV = — L nj, = slope of L = |. Hence m., = , and B= -■ Finally, the angle at C is measured from the line M. Hence in (VI) m^ = slope of M = 6, m^ = slope of iV = — ?. _ 3 _ g 15 •. tan C = —^ — ^ = — , and C = tan- 1 if. 1-9 16 We may vsrify these results. For if B ■ and - , then A = 1-C hence (31, p. 3 ; and 26, p. 3) tan ^ = cotC = r,, which is true for the values found. 2. Find the equation of the line through (3, 5) wliich makes an angle of - with the line x — y + Q 0. Solution. Let m.^ be the slope of the required line. Then its equation is by (II), Art. 27, (1) y-5 = m^{x-3). The slope of the given line is m.-, = 1, and .since the angle which (1) makes with the given line is ~, we have by (VI), since = ~ = 60°, 3 o .-1 3 1 + )«^ ??!, — 1 - 7'/| / /A A 3 \ / -- _\ r3. '>) < I - . \ \ 1 \ \ X \ Whence (2 + V3). tan — 3 1 + J"l 1 + \^'3 m. = -: 1 - \3 Substituting in (1), we obtain 2/-5=-(2 + V3)(x-3), or (2 + V 3) X + 2/ - (11 + 3 Vs) = 0. In plane geometry there would be two solutions of this problem, — the line just obtained and the dotted line of the figure. Why must the latter be excluded here ? In working out the following problems the student should first draw the figure and mark by an arc the angle desired, remembering that this angle is measured from the second line to the first in the counterclock- wise direction. THE STRAIGHT LINE 83 PROBLEMS 1. Find the angle which the line Sx — y + 2 =0 makes with 2x + y — 2 = 0; also the angle which the second line makes with the first, and show that these angles are supplementary. . Sir ir Ans. — -1 — . 4 4 2. Find the angle which tlie line (a) 2x— 5y + 1 = makes with the line x — 2y + S = 0. (b) X + y + I = makes with the line x — y + 1 =0. J (c) 3x — 4y + 2 = makes with the line x + Sy — 7 = 0. (d) 6x — Sy + S = makes with the line x = 6. (e) X — 7 ?/ + 1 = makes with the line x + 2?/ — 4 = 0. In each case plot the lines and mark the angle found by a small arc. Ans. (a) tan-i(- Jj); (b) | ; (c) tan-i(V^) ; (d) tan-i (- ^) ; (e) tan- 1(3^5). ^ 3. Find the angles of the triangle whose sides are z + 3?/ — 4 = 0, 3x — 2y + 1 = 0, andx — y + 3 = 0. Ans. tan-i(- y), tan-i{^), tan-i(2). Hint. Plot the triangle to see which angles formed by the given lines are the angles of the triangle. 4. Find the exterior angles of the triangle formed by the lines 5x-j/ + 3 = 0, 2/ = 2, x-42/ + 3 = 0. Ans. tan- 1(5), tan-i(- |), tan-i(- V). 6. Find one exterior angle and the two opposite interior angles of the triangle formed by the lines 2x — 3y — 6 = 0, 3x + 4j/ — 12 = 0, x — Sy + 6 = 0. Verify the results by formula 37, p. 3. i,' 6. Find the angles of the triangle formed by 3x+2j/ — 4 = 0, X - 3 2/ + 6 = 0, and 4 x - 3 ?/ — 10 = 0. 7. Find the equation of the line passing through the given point and making the given angle with the given line, (a) (2, 1),^, 2x-3?/ + 2 = 0. Ans. 5x-y-9 = 0. 4 (b) (1, - 3), -^, X + 22/ + 4 = 0. Ans. Sx + y = 0. / ^ , V , ^ fn + tan , (c) (Xi, Vi), — and through the intersection of 2x — v + 3 = and x - 3 // 4- 2 = 0. .1 ij.s. l<)x 4- 3 // 4- 20 = 0. //////. Th« syntems of lint^s ^Kissing thr(mi;li tht- |)) = 0, and 2 X - ?/ + 3 + A:' ^ - 3 2/ + 2) = 0. 90 NEW ANALYTIC GEOMf:TRY These lines will coincide if the coefficients are proportional ; that is, if 2 + k' ^ -\--Sk'~ 3 + 2*' ' Letting r be the common value of these ratios, we obtain 1 + A; = 2 r + rk', \-k = -r-Zr¥, and -2 + Q>k='ir + 2-rkf. From these equations we can eliminate the terras in rk' and ;•, and thus fiud the value of k which gives that line of the first system which also belongs to the second system. 4. Find the equation of the line passing through the intersection j-iif = r. Squaring, we have (I). q.e.d. Corollary. Tlis equation of the circle whose center is the origin (0, 0) and whose radius is r is ■ x^ -if-y^ z= f-. If (I) is expanded and ti-ansj^osed, we obtain (1) a^ + / - 2 ^i:a; - 2 )8y + a' + y8' - v' = 0. The form of this equation is clearly 2^" 4- 2/^ + tei-ms of lower degree = 0. In words : Any circle is defined by an equation of the second degree in the variables x and y, in which the terms of the second degree consist of the stim of the squares of x and y. Equation (1) is of the form (2) a-^ -f y' + Dx + By + F = 0, where (3) D=-2a,E=-2p, and F = a^ + 1^ - r\ 92 THE CIRCLE 93 Can we infer, conversely, that the locus of every equation of the form (2) is a circle ? To decide this question transform (2) into the form of (I) as follows: Rewrite (2) by collecting the terms in x and the terms in y thus : (4) x^ -{- Dx + y"" + Ey =- F. Complete the square of the terms in x by adding (^-D)'^ to both sides of (4), and do the same for the terms in y by adding (i E)' to both members. Then (4) may be written (5) (a- + \Df + {y + \Ef = \{D^+l^ - 4F). In (5) we distinguish three cases : If Z)^ + £'- — 4Fis positive, (5) is in the form (I), and hence the locus of (2) is a circle whose center is (— ^D, — \E) and whose radius is r = ^ Vz*'^ + is^ — 4 F. If i)^ + 21^ — 4 F = 0, the only real values satisfying (5) are x——\D,y=—\E (footnote, p. 37). The locus, therefore, is the single point (— \D, — \E). In this case the locus of (2) is often called a point circle, or a circle whose radius is zero. If ly^ + F^ — 4F is negative, no real values satisfy (o), and hence (2) has no locus. The expression If -\- E^ — 4, F \^ called the discriminant of (2), and is denoted by © (Greek letter " Theta "). The result is given by the Theorem. TJte locus of the equation (II) x' + y^-\.Dx + Ey + F=0, whose discriminant is ® = Lf-\- E^ — A F, is determined as foliates : When is positive, the locus is the circle whose center is (— 1. £>, — ^ £•) and tvhose radius is r = ^ Vy/-+7i'^— 4F=^ V®. When is zero, the locris is the point circle (— ^D, — ^-E). When is negative, there is no locus. Corollary. When E=0, the center o/(II) is on the x-axLs, and when D=0, the center is on the y-axis. 94 NEW ANALYTIC GEOMETRY EXAMPLE Find the locus of the equation x^ + y^ — ix + 8y — 5 — 0. First solution. The given equation is of the form (II), where D = -4, ^=8, F = -5, and hence e = 16 + 64 + 20 = 100 > 0. The locus is therefore a circle ■whose center is the point (2, — 4) and whose radius is | VlOO = 5. Second solution. The problem may be solved without applying the theorem if we follow the method by which the theorem was established. Collecting terms, (x2-4x) + (2/2 + 8y) = 5. Completing the squares, (x2 _ 4x + 4) + (y2 + 8 y + 16) = 25. Or, also, (x - 2)2 + {y + 4)2 = 25. Comparing with (I), a: = 2, /3 = — 4, r = 5. The equation Ax!^ + Z>,r?/ + Q/ + Dx -{- Et/ -\- F = is called the general equation of the second degree in x and y because it con- tains all possible terms in x and y of the second and lower de- grees. This equation can be reduced to the form (II) when and only when A = C and iJ = 0. Hence the locus of an equation of the second degree is a circle only when the coefficients of x^ and ?/ are equal and the xy-tenn is lacking. 39. Circles determined by three conditions. The equation of any circle may be written in either one of the forms (x-ay^(y-/3y=r^, or x^-\-y^ + Dx -i- Ey + F=0. Each equation has three arbitrary constants. To determine these constants three equations are necessary. Such an equa- tion means that the circle satisfies some geometrical condition. Hence a circle may be determined to satisfy three conditions. THE CIRCLE 95 Rule to determine the equation of a circle satisfying tltree conditions. First step. Let the required equation be (1) (^^ay + {y-fif=r^, or (2) x'' + y^+Dx + Ey + F=0, as may be more convenient. Second step. Find three equations betiveen the constants a, y3, and r \_or D, E, and F] ivhich express that the circle (1) [o/'(2)] satisfies the three given conditions. Third step. Solve the equations found in the second step for a, /3, end r [_or D, E, and F^. Fourth step. Substitute the resxdts of the third step in (1) [o/> (2) ]. The result is the required equation. In some problems, however, a more direct method results by constructing the center of the required circle from the given conditions and then finding the equations and points of inter- section of the lines of the figrure. EXAMPLES 1. Find the equation of the circle passing through the three points Pi(0, 1), PjCO, 6), and P^(^, 0). First solution. First step. Let the re- quired equation be (3) x^ + y^ + Dx + Ey+ F=0. Second step. Since Pj, P^, and P^ lie on (3), their coordinates must satisfy (3). Hence we have (4) 1 + E + F = 0, (5) 36 + 6 £■ + P = 0, and (6) 9 + 3D+P=0. Third step. Solving (4), (5), and (6), we obtain E =-7, F=6, D = - 5. 96 NEW ANALYTIC GEOMETRY Fourth step. Substituting in (3), the required equation is a;2 + ?/- — 5 X — 7 y + 6 = 0. The center is (|, |), and the radius is f V2 = 3.5. Second solution. A second method which follows the geometrical con- struction for the circumscribed circle is the following. Find the equations of the perpendicular bisectors of P^P<, and P^P^. The point of intersec- tion is the center. Then tind the radius by the length formula. 2. Find the equation of the circle passing through the points P^ (0,— 3y and P„(4, 0) which has its center on the line x + 2y = 0. First solution. First step. Let the required equation be (7) x2 + if- + Dx + Ey + F=0. Second step. Since Pj and P.^ lie on the locus of (7), we have (8) 9 - 3 7? + P = 0, and (9) 16 + 4 Z; + F = 0. , j, and since it lies on the given line, -f-(-D-. or (10) D + 2E = 0. Third step. Solving (8), (9), and (10), U=-V, F = l P = -V- Fourth step. Substituting in (7), we obtain the required equation, X2 + 2/2 _ y X + 7 y _ 2^4 = 0, or 5x2 + .5 2/2-14X + 7?/-24 = 0. The center is the point (|, — y'^), and the radius is ^ V29. Second solution. A second solution is suggested by geometry, as follows r Find the equation of the perpendicular bisector of PiPg. The point of intersection of this line and the given line is the center of the required circle. The radius is then found by the length formula. Vi K / ■^ -vj S ^ \ \ R, Ji "^ (4,( )X \ 4'^ & ^ -4 \ (0.-3) / ^ ~^ P^ "^ 1" THE CIRCLE 9T 3. Find the equation of the circle inscribed in the triangle whose sides are (11) AB:3x-4y-l9 = 0, BC:ix + 3y-n = 0, CA:x+ 7 = 0. Solution. The center is the point of intersection of the bisectors of the anu'les of the triangle. We therefore find the equations of the bisectors of the angles A and C. Reducing equations (11) to \C the normal form, (12) AB: BC Sx-iy-19 4x + 3?/ — 17 = 0: 0; -1 Then, by Example 2, Art. 34, the bisectors ai-e 3x-4?/-19_x+7 2 .r — 2/ + 4 = 0, (13) ^D:- CE 4x + 3y — 17 X + 7 5 -1 or 3x + y + Q = 0. The point of intersection of AI) and CE is (- 2, 0). This is therefore the center of the inscribed circle. The radius is the perpendicular distance from any of the lines (11) to (—2, 0). Taking the side /I/', then, from (12), ^^ 3(-2)-4(0)-19 _ ^ Hence, by (I), tlie equation of the required circle is (x + 2)2 + {!/- 0)2 = 25, or x'^ + ?/ + 4x - 21 = 0. Ans. 98 NEW ANALYTIC GEOMETRY PROBLEMS 1. Find the equation of the circle whose center is (a) (0, 1) and whose radius is 3. Ans. x^ + y^ — 2y — 8 = 0. (b) (— 2, 0) and whose radius is 2. Ans. x^ + y^ +4x = 0. (c) (— 3, 4) and whose radius is 5. Ans. x- + y^ + 6x — 8y = (d) (a, 0) and whose radius is a. Ans. x- ■\- y- — 2ax = 0. (e) (0, /3) and whose radius is /3. Ans. x" + ?/- — 2j8y = 0. (f ) (0, — j8) and whose radius is /3. Ans. x- + y^ + 2 ;8?/ = 0. 2. Draw the locus of the following equations : (a) x2 + ?/2-6x-]6 = 0. {i)x' + y'^-Qx + 4?/ -5 = 0. (b) 3x2 + 3 ;/2 - lOx - 24 2/ = q. (g) (x + 1)2 + {y - 2)2 = 0. (c) x2 + ?/ = 8 X. (h) 7 x2 + 7 7/2 _ 4 X — 2/ = 3. (d) x2 + ?/- 8x-Qy -\- 25 = 0. (i) x2 + 1/2 + 2ax + 26?/ 4-a^ + ^'^ = 0. (e) x2 + 2/2 - 2 X + 2 2/ + 5 = 0. (j ) x2 + ?/2 + 10 x + 100 = 0. 3. Show that the following loci are circles, and find the radius and the coordinates of the center in each case : (a) A point moves so that the s'.m of the squares of its distances from (3, 0) and {- 3, 0) always equals 68. Ans. x'^ -\- y^ = 25. (b) A point moves so that its distances from (8, 0) and (2, 0) are always in a constant ratio equal to 2. Ans. x'^ + 2/2 = 16. (c) A point moves so that the ratio of its distances from (2, 1) and ( — 4. 2) is always equal to ^. Ans. 3x^ -\- 3y" — 24:X — 4y = 0. (d) The distance of a moving point from the fixed point (— 1, 2) is twice its distance from the origin. . , ^ „ 2 Vs Ans. a = i,3 = —2r = (e) The distance of a moving point from the fixed point (2, — 4) is half its distance from the fixed point (0, 3). (f) The square of the distance of a moving point from the origin is proportional to the sum of its distances from the coordinate axes. (g) The square of the distance of a moving point from the fixed point (— 4, 3) is proportional to its distance from the line 3x — 4?/— 5 = 0. (h) The sum of the squares of the distances of a point from the two lines X — 22/ = 0, 2x + 2/ — 10 = 0, is unity. 4. Find the equation of a circle passing through any three of the fol- lowing points : (0,2) (3,3) (6,2) (7,1) (8,-2) (7,-5) (6,-6) (3,-7) (0,-6) (-1,-5) (-2.-2) (-1,1) Ans. x2 + ?/2 _ 6 X + 4 2/ — 12 = 0. THE CIRCLE 99 5. Find the equation of tlie circle which (a) has the center (2, 3) and passes through (3, — 2). Ans. x^ + y"^ — ix — 6y — 13 = 0. (b) has the line joining (3, 2) and (— 7, 4) as a diameter. Ans. ^2 + 2/2 + 4x — 6?/ — 13 = 0. (c) passes through the points (0, 0), (8, 0), (0, — 6). A ns. a;2 + 2/2 — 8 X + 6 2/ = 0. (d) passes through (0, 1), (5, 1), (2, - 3). Ans. 2 x2 + 2 2/2 — 10 x + 2/ - 3 = 0. (e) circumscribes the triangle (4, 5), (3, — 2), (1, — 4). (f) has the center (— 1, — 5) and is tangent to the x-axis. Ans. x2 + 2/2 + 2 X + 10 2/ + 1 = 0. (g) has the center (3, — 5) and is tangent to the line x — 7y + 2 = 0. Ans. x2 + 2/2 — 6x + lOy + 2 = 0. (h) passes through the points (3, 5) and (— 3, 7) and has its center on the X-axis. Ans. x- + y" + ix — 46 = 0. (i) passes through ^the points (4, 2) and (— 6, — 2) and has its center on the 2/-axis. Ans. x^ + y^ + Sy — SO = 0. (j) passes through the points (5, — 3) and (0, 6) and has its center on the line 2 X — 3 2/ — 6 = 0. Ans. 3x2 + 3 //- - 114x - 64?/ + 27G = 0. (k) passes through the points (0, 2), (— 1, 1) and has its center in the line 3 2/ + 2 X = 0. Ans. x2 + 2/2 — 6x + 4 2/ — 12 = 0. (1) circumscribes the triangle x — 6 = 0, x + 22/ = 0, x — 22/ = 8. Ans. 2 x2 + 2 2/2 - 21 X + 8 y + 60 = 0. (m) is inscribed in the triangle (0, 6), (8, 6), (0, 0). Ai^s. x2 + 2/2 — 4x — 82/ + 16 = 0. (n) passes through (1, 0) and (5, 0) and is tangent to the 2/-axis. Ans. x2 + 2/2 — 6 X ± 2 Vs 2/ + 5 = 0. (o) passes through the points (— 3, — 1), (1, 1) and is tangent to the line 4x + 3 2/ + 25 = 0. - , ^ / ,iiA_Lr- — yi 6. Find the eijuations of the inscribed circles of the following triangles : (a) x + 2 2/ - 5 = 0, 2x-2/ — 5 = 0, 2X + 2/+ 5 = 0. (b) 3X + 2/-1 = 0, X - 3 2/ - 3 = 0, x + 32/ + 11 = 0. (c) 3x + 4?/- 22 = 0, 4 X - 3 2/ + 29 = 0, 2/ - 5 = 0. (d) X + 2 = 0, 2/ - 3 = 0, X + 2/ = 0. (e) X = 0, y = o, X + 2/ + 3 = 0. 7. What is the equation of a circle whose radius is 10, if it is tangent to the line 4x + 32/— 70 = 0at the point whose abscissa is 10 ? 100 NEW ANALYTIC GEOMETRY In the proofs of the following theoieins the choice of the axes of coordinates is left to the student, since no mention is made of either coordinates or equations in the problem. In such cases always choose the axes in the most convenient manner possible. 8. A point moves so that the sum of the squares of its distances from two fixed points is constant. Prove that the locus is a cirde. 9. A point moves so that the sum of the squares of its distances from two fixed perpendicular line^ is constant. Prove that the loous is a circle. 10. A point moves so that the ratio of its distances from two fixed points is constant. Determine the nature of the locus. Ans. A circle if the constant ratio is not equal to unity, and a straight line if it is. 11. A point moves so that the square of its distance from a fixed point is proportional to its distance from a fixed line. Show that the locus is a circle. CHAPTER VI TRANSCENDENTAL CURVES AND EQUATIONS In the preceding chapters the emphasis has been laid chiefly on algebraic equatioais ; that is, equations involving only powers of the coordinates. ^Ve now turn our attention to equations such as y _ iQg ^^ y = 2^, x — sin ?/, which are called transcendental equations, and their loci, trany scendental (•■urves. 40. Natural logarithms. The common logarithm of a given number N is the exponent x of the base 10 in the equation (1) 10' = N ; that is, x = log^^ .V. A. second system of logarithms, known as the natural si/stew, is of fundamental importance in mathematics. The base of this system is denoted by e, and is called the natural base. Numer- ically to three decimal places, the natural base is always (2) e = 2.718. The natural logarithm of a given number N is the exponent y in the equation (3) ey = N; that is, y = log, N. To find the equation connecting the coinmon and natural logarithms of a given number, we may take the logarithms of both members of (3) to the base 10, which gives (4) logj^ e« = logj„ N, or y log^^ e = log^„ N. (16, p. 1) (5) .". logjij .V = log^g e ■ logg N (using the value of y in (3)) 101 102 NEW ANALYTIC GEOMETRY 0.434; also - = 2.302. (A) The equation shows that the common logarithm of any number equals the product of the natural logarithm by the constant log e. This constant is called the modulus (= il/) of the com- mon system. That is (Table, Art. 2), <6) M=\og^^ We may summarize in the equations, Common log = natural log times 0.434, Natural log = common log times 2.302. These equations show us how to tind the natural logarithm from the common logarithm, or vice versa. Exponential and logarithmic curves. The locus of the equation (7) // = e^ is called an exponential curve. From the preceding we may write (7) also in the form (8) a. = log,y = 2.3021og^„y. The locus of (7) is therefore the curve whose abscissas are the natural logarithms of the ordinates. Let us now discuss and plot (7). (Figure, p. 103.) Discussion. Since negative num- bers and zero have no logarithms, y is necessarily positive. More- over, X increases as y increases. The coordinates of a few points on the locus are set dov?n in the table. The discussion and figure illustrate the fact that logeO = — 00. For clearly, as y approaches zero, a; becomes negatively larger and larger, without limit. Hence the x-axis is a horizontal asymptote. If the curve is carefully drawn, natural logarithms may be measured off. Thus, by measurement in the figure, if 2/ = 4, x = 1.38 = loge4. .r y X y 1 1 1 e = 2.7 -1 '- = .31 e 2 i'-^ =7.4 -2 e^ etc. etc. etc. etc. TRANSCENDENTAL CURVES AND EQUATIONS 103 More generally, the locus of (9) y = e^, where A; is a given constant, is an exponential curve. The dis- cussion of the difference of this locus from that in the figure is left to the reader. The locus of the equation (10) y = log,„ X, which is called a lo(j 4 2 X for X = 1, are as follows : sin 2x = sin 2 radians = sin 114°.59 = 0.909, X tan — — = 1 • tan | - radians ) = tan 45° = 1, irx /TT ■ cos — cos I — radians cos 30° 0.433. 2.r 2 2 Let us now draw the locus of the equation (3) 7/ = sin;r, in which, as just remarked, x is tlie circular measure of an angle V X Solution. In making the calculation for plotting, it is convenient to ■choose angles at intervals of, say, 30°, and tlien find x, the circular measure of this angle, in radians, and y from the Table of Ait. 4. Angle in X Angle in X degrees radians y degrees radians y 30 .52 ..50 - 30 J.-2 - .50 60 1.04 .86 - 60 -1.C4 - .86 90 1.56 1.00 - 90 -1.56 -1.00 120 2.08 .86 -120 -2.08 - .86 150 2.60 .50 - 150 - 2.60 - .50 180 3.14 -180 - 3.14 TRANSCENDENTAL CURVES AND EQUATIONS 107 Thus, for 30°, y = sin 30° = .50. For 150°, y = sin 150°= sin (180°- 30°) = sin 30° = .50 (30, p. 3). To plot, ciioose a convenient unit of lengtli on XX' to represent 1 radian, and use the same unit of length for ordinates. The divisions laid off on t!ie X-axis in the figure are 1 radian, 2 radians, etc. Plotting the points (x, y) of the table, the curve APOQB is the result. The course of the curve beyond B is easily determined from the relation sin(2 7r + x) = sinx. Hence ?/ = sinx = sin(2 7r + x); that is, the curve is unchanged if x + 2Tr be substituted for x. This means, however, that every point is moved a distance 2 tt to the right. Hence the arc APO may be moved parallel to XX' until A falls on B, that is, into the position BRC, and it will also be a part of the curve in its new position. This property is expressed by the statement : The curve y = sinx is a periodic curve with a period equal to 2 7r. Also, the arc OQB may be displaced parallel to A"X' until falls upon C. In this way it is seen that the entire locus consists of an indefinite number of congruent arcs, alternately above and below XX'. General discussion. 1. The curve passes through the origin, since (0, 0) satisfies the equation. 2. In (3), if X = 0, y = sin = = intercept on the axis of y. Solving (3) for x, (4) X = arc sin y. In (4), if 2/ = 0, X = arc sin = nir, n being any integer. Hence the curve cuts the axis of x an indefinite number of times botli on the right and left of O, these points being at a distance of tt from one another. 3. Since sin(— x) =— sinx, changing signs in (3), — y = — sin X, or — y = sin(— x). Hence the locus is unchanged if (x, y) is replaced by (— x, — y), and the curve is symmetrical with respect to the origin (Theorem II, p. 43). 4. In (3), X may have any value, since any number is the circular measure of an angle. In (4), y may have values from — 1 to + 1 inclusive, since the sine of an angle has values only from — 1 to + 1 inclusive. 5. The curve extends out indefinitely along XX' in both directions, but is contained entirely between the lines ?/ = + 1, y =— 1. 108 np:w analytic geometry The locus is called the wave curve, from its shape, or the sine curve, from its equation (3). The maximum value of y is called the £implitude. Again, let us construct the locus of irx (5) y = 2sin — . Solution. We now choose for x the values 0, J, 1, 1|, etc., radians, and arrange the vs'ork of calculation as in the table. X radians iTTX radians degrees . -KX sin — 3 y i l-^ 30 .50 1.00 1 iTT 60 .86 1.72 H 4 7r 90 1.00 2.00 2 JTT 120 .86 1.72 2^ i^ 150 .50 1.00 3 ■n 180 The figure represents a sine curve of period 6 and amplitude 2. For the curve crosses the x-axis ot intervals of 3, and the maximum value of j/ equals 2. v 3X E(iuatiou (5) is of the form 1/ = a sin Icji'. '1 he iuuplitude in this case equals a. To find the period, set Jcx = 'Zir. Solvmg for x, x = -— = period. TRANSCENDENTAL CURVES AND EQUATIONS 109 As it is important to sketch sine curves quickly, the follow- ing directions are useful : 1. Find the aurvplitude and the jpercod. 2. Choose the same scales on hath axes. 3. L - e- « ft 11 — 2 9. y = e* — cos4x. x = ^tt. 10. y = sin x + sin 2 x. x = 0.8. TRANSCENDENTAL CURVES AND EQUATIONS 113 . TTX TTX 11. U = Sin h t-os X 4 3 12. J/ = sin ax 4- cos ax. x = -^^ a. 13. ?/ = 2sinx + 5cosx. x = 0.5. 14. y = 2 sin 2 x + 3 cos | x. x = 2. 15. ?/ = sin ax + sin hx. 16. y = VO — X- + sin 27rx. x = 2^. 17. 2/ = e-^ + 4x-. X =— 2.4. • 2 TTX 18. y = logipX + sm ^- . x = 2 /- 1 /tTX • TT 19. w = 2 Vx 4- - cos — + - 2 \ 2 3 20. y = -{e" + e a). 2\a. The locus in Problem 20 is called the catenary (see figure). The shape of the curve is that as.sumed by a heavy flexible- cord freely sus- pended from its extremities. The student may have observed from the preceding exam- ples the truth of the following Theorem. The curve obtained by adding corresj^onding ordi- nates of sine curves with the same period is also a sine curve with eqtial period. For example, consider the equation (5) /2 7rt \ /2'irt \ y = asml-^ -faj+/>sin(— +/31, 2'7rt in which a, /8, and P are constants. The period of both sine curves equals P. Expand the right-hand member by the rule (33, p. 3) for sin {x + y) and collect the terms in sin 2 irt and cos Then equation (5) assumes the form . 2'Trt 2 irt Cy (6) // = A sm — — + B cos -— - ? where A and B are constants, independent of t. Let us now introduce the angle y of the right triangle whose legs are A and B. Let the hypotenuse V.4^ + B^ = C. Then B — C sin y, 114 NEW ANALYTIC GEOMETRY A =C cos y. Substituting these values in (6) gives y- -^ COS y + cos — (() ?/ = C[sin —^ cosy + cos— ^ smy 1 = C sinl — -- + y )• This is a sine curve with period P and amplitude C = '\/ A^-\-E^. Q.E.D. The curve resulting from the addition of ordinates of sine curves with unequal periods is, however, not a sine curve. 43. Boundary curves. In plotting the locus of an equation of the form (1) y = jjroduct of two factors one of which is a sine or cosine, as, for example, TTt y = e' sm x, or s = t' cos — ? much aid is obtained by the following considerations : For example, consider the locus of (2) y = e^ sm— -• We now make the following observations : 1. Since the numerical value of the sine never exceeds unity, the values of ij in (2) will not exceed in numerical value the value of the first factor e~^^. Moreover, the extreme values of sin -^ TTX are + 1 J^i^d — 1 respectively. Hence y has the extreme vaiues e * and — e * . Consequently, if the curves (3) y = e~^^ and y=-e-*" are drawn, the locus of (2) will lie entirely between these curves. They are accordingly called boundary curves. Draw these curves. The second is obviously symmetrical to the first with respect to the a:-axis. To plot, find three points on the first curve, as in the table. (Use the Table, p. 104.) X y 2 4 1 e- ^ = .61 e-i = .37 TRANSCENDENTAL CURVES AND EQUATIONS 115 2. When sin ^ttx =0, then in (2) y = 0, since the first fac- tor is always Jinlte. Hence the locus of (2) vieets the x-axls In the same points as the auxiliary sine curve (4) y = sin ^ ttx. 3. The required curve touches * the boundary curves when the second factor, sin i ttx, is -|- 1 or — 1 ; that is, when the ordinates of the auxiliary curve (4) have a maximum or minimum value. Hence draw the sine curve (4). The period is 4 and the amplitude is 1. This curve is the dotted line of the figure. The discussion shows these facts : The locus of (2) crosses XX' at x = 0, ± 2, ± 4, + 6, etc., and touches the boundary curves (3) at x = ± 1, ±3, ±,5, etc. We may then readily sketch the curve, as in the figure ; that is, the winding curve between the boundary curves (3). 4. For a check remember that the ordinate of (2) is the product of the ordinates of the boundary curve y = e~^^ and the sine curve (4). In the figure, for example, the required curve lies above A'A'' between x = and x — 2, for the ordinates of 7j = e~^^ and of the sine curve are now all positive. But between x = 2 and .x = 4 the required curve lies below A'A'', for the ordinates of y = e~^'^ and the sine curve now have unlike signs. *The discussion shows merely that the curve (2) reaches the boundary <5urves. Tangency is shown by calculus. 116 NEW ANALYTIC (JEO.METRY PROBLEMS Draw the following loci and calculatt- y accurately for the given values of x : ^ ■ n , ,, sinx/ ] . \ 1. y = - smx. x = 2; lir. 11. i/ = (=-.siiix|. x=0.1. 4 " X \ X J r2 , „ sin 2x „ , , 2. i/ = — cos2x. j; = l;7r. 12.?/=:— j- = 0.1;l. cos X ^.ir.'^^ . _ Q . 1 13. // = — . X = 1 ; TT. 3. 7/ = - sin X = S ; I. ■'3 3 A X- TTX o oi 4. « = — cos X = 3 ; 2 ,', . 10 5 X U.y = ^^. . = 0.2; r x^ . TTX 5. y = e-^sinx. x =^Tr: ^ir. sin -^ 15. y = X =0.1 ; 2. B. y = e-^cos 2x. x = J tt ; 2. 7. v/ = e - Sin x = — 2 ; .•. , . .1 „ , 4 16. // = sm - X cos 2 x . x = ^ tt 8. w = e^i'^cos!!^ 3._Q._i / ]\ 1 3- ^-^' ^-- 17. ,/ = (x + -)sin-a;. .—It /tX 7r\ o 1 ■$ -, cos I - cos-x COS -a;. 4 2 4 2 - u-x /2 TTX \ ,„ _/ . , _!/ . 7rf 10. y = ae cos ( !-«)• 19- '/ = c suiirt + e ^ sin — ' P / 2 20. Draw the two loci obtained (1) l)y adding and (2) by multiplying the ordiiiates in the following pairs of curves : C x2 f _?^ I ?/ = 2 H , y = cos (c ^ i> -^ ' (e) < U = sinx. [2/ = sin7r./'. | ^.^^os — I '3 r 16 - X2 y — — 8 (t) ' TTX •II — COS I 2 2/ = 3 + —. (b) ^ ?/ = e^ (d) j J^ 1^.(7 = cos7r.f. I y = sin- - l 2 44. Transcendental equations. Graphical solution. The solution of certain equations of frequent occurrence may be simplified by using graphical methods. TRANSCENDENTAL CURVES AND EQUATIONS 117 Consider the equation (1) cot X = X, or cot X — X := 0. To find values of x {in radians) for which this equation holds. To aid in determining the roots, let us draw the curves (2) y = cot X and y = x. Now the abscissa of each point of intersection is a root of equa- tion (1), for, obviously, at each point of intersection of the curves (2) we must have cot x = x; that is, equation (1) is satisfied. In plotting it is well to lay off carefully both scales (degrees and radians) on OX. y =cot X Degrees X radians y CO 10 .174 .5.67 20 .342 2.75 30 .524 1.73 40 .698 1.19 45 .785 1.000 50 .873 .839 60 1.047 .577 70 1.222 .364 80 1.396 .176 90 1.571 .0 1§0 190 200°Jf Number of solutions. The curve y = cot x consists of an infinite number of branches congruent to AQJJ of the figure. 118 NEW ANALYTIC GEOMETRY The line y = x will obviously cross each branch. Hence the equation (1) has an infinite number of solutions. Smallest solution. From the figure this solution lies between 45° and 50°, or, in radians, between x = .785 and x = .873. Hence the first significant figure of the smallest root is 0.8. Inter];)olation is necessary to determine subsequent figures. For this purpose arrange the work thus, using the preceding table. X (radians) cotx cotx — x .873 .839 -.034 .785 1.000 +.215 difference + .088 - .249 We wish to know what change in x above .785 will produce a decrease in cot x — x equal to .215 ; that is, make cot x — x equal to zero. Call this change z. Then, by proportion, :o88 = 3:249' ■■' = -^'^ Hence x = .785 + .076 = .86 (to two decimal places). PROBLEMS Determine graphically the number of solutions in each of tlie following, and find the smallest root (different from zero). Ans. One solution; x = 0.74. Ans. Three solutions; x = 0.95 Ans. Infinite number. Ans. Three. Ans. Two. Ans. Two. 13. 3sinx = 2cos4x. 19. e^ = tanx. , 14. 2 sin - = cos 2 X. ^0. sin x = log^,, x. 2 21. cosx = logjgX. 15. sinSx = cos2x. 16. e- = x. 22. tanx = log,ox. 17. e^ = sinx. 23. e-' = log«x. 18. e-* = cosx. 24. e-^' = x^. 1. cos X = X. 2. sin2x = X. 3. tan X = X. 4. sinx = |x. 5. sinx = x^. 6. cosx = x^. 7. tanx = x^. 8. cotx = x^. 9. X cosx= -. 3 10. tan X = 1 — X. 11. cosx = 1 — X. 12. 3sinx = cosx CHAPTER VII POLAR COORDINATES 0\ . 45. Polar coordinates. In this chapter we shall consider a second method of determining points of the plane by pairs of real numbers. We suppose given a p fixed point 0, called the pole, and a fixed line OA, passing through 0, called the polar axis. Then any point P determines a length OP = p (Greek letter " rho ") and an angle AOP = 0. \ The numbers p and 6 are called the \ polar coordinates of 1\ p is called the radius vector and the vectorial angle. The vectorial angle 6 is jjositive or negative as in trigonometry. The radius vector is positive if P lies on the terminal line of 0, and , \ , ,,, negative if P lies on that line produced through the pole O. Thus in the figure the radius vector of P is positive, and that of P' is negative. It is evident that every pair of real nnm- hers (p, 6) determines a single point, which may be plotted by the "Rule for plotting ajjoint whose jjolar coordinates (p, 6^ are given. 119 120 NEW ANALYTIC GEOMETRY First step. Construct the terminal line of the vectorial angle 0, as in trigonortietry. Second step. If the 7'adius vector is positive, lay off a length OP = p on the terminal line of 6 ; if negative, prodtice the termi- nal line through the pole and lay off OP equal to the numerical value of p. Then P is the required p)oint. In the tigure on page 119 are plotted the points whose polar coordinates are (6, 60°), (s, ^) , (- 3, 225°), (6, 180°), and Every point determines an inf- nitenumber of pairs of numbers (p, &). Thus, if OB = p, the coordinates of B may be written in any one of the forms (p, 6), (- p, 180° + 0), (p, 360° + e), (~p,6- 180°), etc. Unless the contrary is stated, we shall always suppose that $ is 2>ositive, or zero, and less than 360°; that is, 0^6<360°. . PROBLEMS <5, ,r). Plot the points (4, 45°), (6, 120°), i- 2,— V U, ^V (- 4, - 240°), 2. Plot the points U, ^ -\ (-2, ± ^Y {3,7r), (- 4,7r), (6, 0), (- 6, 0). 3. Show that the points (p, 6) and (p, — d) are symmetrical with respect to the polar axis. 4. Show that the points (p, 6), {— p, ff) are symmetrical with respect to the pole. 5. Show that the points (— p, 180° — 0) and (p, 0) are symmetrical with respect to the polar axis. 46. Locus of an equation. If we are given an equation in the variables p and 0, then the locus of the equation is a curve such that 1. Every point whose coordinates (p, 6) satisfy the equation lies on the curve. POLAR COORDINATES 121 2. The coordinates of every point on tlie curve satisfy the eqx:ation. The curve may be plotted by solving the equation for p and finding the values of p for particular values of $ until the coordinates of enough points are obtained to determine the form «f the curve. The plotting is facilitated by the use of polar coordinate paper, which enables us to plot values of 6 by lines drawn through the pole and values of p by circles having the pole as center. The tables on page 6 are to be used in constructing tables of values of p and 0. EXAMPLES 1. Plot tlie locus of the equation (1) /o = 10 cos (9. Solution. The calculation is made by assuming values for 6, as in the table, and calculating p, making use of the natural values of the cosine given in Art. 4. For example, if e = 105°, /3 = 10 cos 105° = 10 cos (180°- 75°) = - 10 cos 75° = - 2.6. 90° p = lOcos0 e P e P 10 105° - 2.G 15° 9.7 120° - 5 30° 8.7 135° — 7 45° 7 150° - 8.7 G0° 5 1G5° - 0.7 75° 2.6 180° -10 90° 1 The complete locus is found in this example without going beyond 180° for 9. The curve is a circle (Art. 50). Since cos(— ^) = cos^ (29, p. 3), equation (1) may be written p = 10 cos (— 6) , that is, for every point (p, 6) on tho locus there is also 122 NEW ANALYTIC GEOMETRY a second point {p, — 6) on the locus. Since tliese points are symmetrical with respect to the polar axis, we have the result : The locus of (1) is symmetrical with respect to the polar axis. 2. Draw the locus of (2) p^ = a2 cos 2 0. Solution. Before plotting, we make the following observations : 1. Since the maximum value of cos 2 ^ is 1, the maximum value of p is a, and the curve must be closed. 2. When cos 2 ^ is negative, p will be imaginary. Now cos 2 ^ is nega- tive when 2 ^ is an angle in the second or third quadrant. That is, when 90° < 2 6* < 270°, that is, 45° < ^ < 135°, p is imaginary. There is no part of the curve between the 45° and 135° lines. 3. We may change ^ to — ^ in (2) without affecting the equation, and hence the locus is symmetrical with respect to the polar axis. The complete curve is obtained if is given values from 0° to 45°, as in the table. p2 =a2cos2e e 20 cos 2 6 P 1 ±rt 15° 30° .866 ± .93 a 30° 60° .500 ± .7 a 45° 90° The complete curve results by plotting these points and the points symmetrical to them with respect to the polar axis. The curve is called a. lemniscate. In the figure a is taken equal to 9.5. 3. Discuss and plot the locus of the equation iS) p = a sec2 ^ ft. POLAR COORDINATES 12S For convenience we change the form of the equation. Using (26), p. 3, _ ^ Then by (41), p. 4, cos^l^ = A + ^ cos^. Hence the result : P = 2a 1 + COS( P = 2 - (1 + cos 9) e cos e 1 + cos e P e cos 1 + cos e P 1 2 1 105° -.259 .741 2.7 15^ .966 1.966 1.02 120° -.500 .500 4 30° .866 1.866 1.07 135° -.707 .293 6.7 45° .707 1.707 1.2 150° -.866 .134 14 60° .500 1.500 1.3 165° -.906 .034 50 75° .259 1.259 1.6 180° -1 CO 90° 1 2 Solution. Before plotting, we note 1. The curve is symmetrical with respect to the polar axis, since & may be replaced by — 0. 2. p becomes infinite when 1 + cos^ = 0, or cos = —I, and hence 6 = 180°. The curve re- cedes to infinity in the direction = 180°. 3. p is never imaginary. On account of 1 the table of values is com- puted only to =180°, and the rest of the curve is ob- tained from the .symmetry witli re.spect to the polar axis. Take a =1. The locus is a parabola. Before plotting polar equations, the student should establish such simple facts as result froni a discussion, as illustrated above. 124 NEW AXALYTIC (GEOMETRY PROBLEMS Plot the loci of the following equations : 1. p = 10. * 2. = 4.5^ -3. p = 16 cos^. 4. p cos 6 — Q. 6. p sin ^ = 4. 6. 4 1 - cos (9 7. 8 ^ 2-cos(9 8. 8 ^ 1 - 2 cos ^ 9. p = a sin ^. 10. 10 ^ 1 + tan 11. p^ sin 2^ = 16. 12. p" cos 2 ^ = a^. 13. p cos = a sin^ ^. 15. p = a{l— cos6). CARDIOID 14. p = a sec ± b. b < a. CONCHOID OK NICOMEDES 16. p- = a2 sin 2 0. TWO-LEAVEU ROSE LEMNISCATK F Q 17. p = b — a cos^. 6 < (X. LIMAgON POLAR COORDINATES 125 18. Plot the conchoid (Problem 14) for h = a; b > a. 19. Plot the lima^on (Problem 17) for b > a. 47. The student should acquire skill in plotting polar equa- tions rapidly when a rough diagram will serve. For example, to draw the locus of (1) p = a sin 3^, we proceed as follows : Let 6 increase from 0°. Follow the variation of p from (1) as 3 tescribes the successive quadrants. When 3 e varies from 0° to 90° 90° to 180° 180° to 270° 270° to 360° 3G0° to 450° 450° to 540° then varies from 0° to 30° 30° to 60° 60° to 90 90° to 120° 120° to 150° 150° to 180° anil p varies from to n a toO to - rt - rt to to a a to For example, when 3 & varies from 270° to 360°, that is, is an angle in ihe fourth quadrant, then p is negative and increases from — a to 0. Now draw the radial lines corresponding to the inter- vals of d; that is, 0°, 30°, 60°, 90°, 120°,, 150°, 180°. Noting the variation of p, we sketch the curve as follows : The curve starts from the pole in the direction 0°, crosses the 30° line perpen- dicularly at p = a, returns to and passes through the pole on the 60° line, crosses the 90° line produced at p = — a, returns to and passes through the pole on the 120° line (produced), crosses the 150° line at /o = a, and returns to the pole on the 180° line. This gives the complete locus. The pencil point has moved continu- ously without abrupt change in direction, and has returned to the original position and direction. The curve is called the three-leaved rose. 126 NEW ANALYTIC GEOMETRY PROBLEMS Draw rapidly the locus of each of the following equations ; 1. p = acos3^. 3. p = acos2^. 3_ y THREE-LEAVED ROSE FOUR-LEAVED ROSE 4. /o = a sin4( FOUR-LEAVED ROSE 10. p = acos{0 + 45°). 11. p = a sin I ^ + ^|. 12. p = a sin 1 0. 13. p = a cos-- EIGHT-LEAVED ROSE 5. p = a cos 4^. 6. p = a sin 5 0. 1, p = a cos 5^. 8. p =a{l + sin^) 9. p = a(l-^ cos (9) 14. p = a sin 6 15. p — asin^^f 16. p = acos2^( 17. p =: a sin^l ( 18. p = a cos^ \ { POLAR COORDINATES 127 48. Points of intersection. By a method analogous to that used in rectangular coordinates we find the coordinates of the points of intersection of two polar curves by solving their equations simultaneously. This is best done by eliminating p, which will give rise in general to a transcendental equation in 6 which can be solved either by inspection or by the graphical method employed in Art. 44. The following example will illustrate the method. EXAMPLE rind the points of intersection of (1) /D = l+COS(9, (2) P ^ 2(1- cos ^) Solution. Eliminating /3, 1 + cos (9 1 — C0S2 COS 6 = ± 2(1— COS (9) h V2 2 .-. e = ± 45°, ± 135° Substituting these values in either equation, we obtain the following four points, 1 + -:^. ±45°), (1--^, ±135°). 2 / \ 2 The result checks in the figure. The locus of (1) is a cardioid ; of (2), a parabola. PROBLEMS Find the points of intersection of the following pairs of curves and check by drawing the figure : r4/3cos6' = 3, • t2p = 3. 4 p cos ^ = 3, p = 3 cos 6. '2p = S, /) = 3 sin 0. 4. ^ fp =V3, [p = 2 sin (9. r /3 = cos 0, ' \4/3 = 3sec(9. p = 1 + cos 0, 2p = 3. '^ 2 p = 2. r 3 /o = 4 cos 0, 1 2/3COS2- = l. 1128 NEW ANALYTIC GEOMETRY 10. i = sin 6/, = cos 2 0. Am. (J, 30°), {i, 150P). p = 1 + cos 6, /)(! + cos 6') = 1. A}is. (1, ± 90°). rp = 2{l-sin^), ■ \p(H-sin^) = L Ans. (2 tV2, ± 45°), (2tV2, ± 135°). fp = 4(1 + cosi9), |p(l-cos^) = 3. Ans. (6, ± 60°), (2, ± 120°). 13. i p = 5— 2sinff, 6 L l + sin^ C /3 = 3 — 2 cos ( 14. ^ [' 3 + 2 cos ^ 15. ■> ( p- = cos : IP ^' G cos ( 16. f p2 = sin 2 e, \p = V2smd 17. -i f p = cos 3 ( 12 cos^. 18. -; {2p (p=:0, IP = cos 8 Let ^ 49. Transformation from rectangular to polar coordinates. OX and OF be the axes of a rectangular system of coordinates, and let O be the pole and OX the i)olar axis of a system of r polar coordinates. Let (.r, ?/) and (p, 6) be respectively the rec- tangular and polar coordinates of any point P. It is necessary to distinguish two cases according as p is positive or negative. When p is positloe (Fig. 1) we have, by definition, cos p = - , sui Q = - 1 P P whatever quadrant P is in. Hence (1) X = p cos 6, y = p sin 6. POLAR coordinatp:s 129 When p is negative (Fig. 2) we consider the point /*' sym- metrical to /' with res])ect to 0, whose rectangular and polar coordinates are respectively (— .r, — y) and (— p, 6). The radius vector of ]'', — p, is positive, since p is negative, and we can therefore use equations (1). Hence for /-"' — X = — p cos 9, — y = — p sin ; and hence for P X = p cos $, y = p sin ^, as before. Hence we have the Theorem. If the pole coincides icith the origin and the polar axis witli the positive x-axis, then \y — p sin B, ivhere (.r, ?/) are tlie rcctangaiar coordinates and (^p, 6) the polar coordinates of any point. Equations (I) are called the equations of transformation from rectangular to polar coordinates. They express the rectangular coordinates of any point in terms of the polar coordinates of that point and enable us to find the equation of a curve in polar coordinates when its equation in rectangular coordinates is known, and vice versa. From the figures we also have (2) ' X ^ y « ^ s\nO = 1 cos^ I ± V^T? ± ^^ + y' These equations express the polar coordinates of any point in terms of the rectangular co(')rdinates. They are not as con- venient for use as (T), although the first one is at times very convenient. 130 NEW ANALYTIC GEOMETRY EXAMPLES 1. Find the equation of the circle x'^ + y" = 25 in polar coordinates. Solution. From the first equation of (2), we have at once p^ = 25 ; lience p = ± 5, which is the required equation. It expresses the fact that the point (p, 0) Is live units from the origin. 2. Find the equation of the lemniscate (Ex. 2, p. 122) p- = a^ cos 2 in rectangular coordinates. Solution. By 39, p. 4, since cos 2^ = cos"^^ — sin^^, p- — a'^{cos"d— sin-^). Substituting from (2), X- + y- = a-( ^ ■ .•. (x- + y-)- = a"-^(x- — ?/2). Ans. 50. Applications. Straight line and circle. Theorem. The general equation of the straight line in jJolar coordinates is (II) p(AeosO -{- B sin 6) + C = 0, where A, B, and C are arbitrary constants. Proof. The general equation of the line in rectangular coordi- iiates is ^4 3. ^ By -\- C = 0. By substitution from (I) we obtain (II). q.e.d. Special cases of (II) are pcos^=«, psin^=6, which I'esult respectively when JS = 0, or ^t = ; that is, when the line is parallel to OY or OX. In like manner we obtain from (II), p. 93, the Theorem. The general equation of the circle In polar coordinates Is (III) p-+ p(D eosO + E sin &) + F=0, where D, E, and F are arhltrarij constants. We may easily show further that if the pole is on the cir- cumference and the polar axis is a diameter, the equation of the circle is p = 2rcos^, where r is the radius of the circle. POLAR COORDIXATES 131 For if the center lies on the polar axis, or a:"-axis, E = 0, and if the circle passes through the pole, or origin, F= 0. The abscissa of the center equals the radius, and hence — ■;j = >", or 1) = — 2 r. Substi- tuting these values of D, E, and /-' in (III) gives p — 2r cos ^ = 0. This result is easily seen also directly from the figure on page 130. Similarly, if the circle touches the polar axis at the pole, the equation is p = 2 r sin 6. Theorem. Tlte h'mjth I of the Ime Joining tiro jjolnts l\{p., 0) and /'.,(po! ^o) '*' ffu-en hy (IV) P = pi +p^-2 p,p, cos (6, - e,) . Froof. Let the rectangular coordinates of P^ and P^ be re- spectively (x^, 7/j) and (a-.,, y.,). Then by (I), p. 129, x^ = p^ cos 6^, :i\^ = p,^ cos ^.„ y^ = p^ sin 0^, ?/., = p., sin ^.,. But /•^ = (.,.^_^j^ + (_y^_y./, and hence /- = (p^ cos 6^ — p., cos ^.,)- + (p^ sin 6^ — p.^ sin O.^y. Removing parentheses and using 28 and 36, p. 3, Ave o\>- tain (IV). O-E.D. Formula (IV) may also be derived directly from a figure by using the law of cosines (44, p. 4). rdinates of the points (s, 7 ) (—2, — j PROBLEMS 1. Find the polar coordinates of the points (3,4), (—4, 3), (5, —12), (4, 5). 2. Find the rectangular coon (3, TT). 3. Transform the following equations into polar coordinates and plot their loci: (a) X — Sy = 0. Ans. = tau-^i. (b) y^ + 5x = 0. Ans. p = — 5 cot cosec 9. 132 NEW ANALY'lIC (iKOAIKTRY (c) x^ + 2/^ = 1(). Alls, p = ± 4. ^ (d) x^ + y'^ — 0,\ LITUUS SriKAI. OF Al!( Ul.MKDES HYFKHHOLIC OH KKl Sl'lliAI. l.tWAKITIlMIC OK E«}IIIAN- GII.AR gPIKAL POLAR COORDINATES 133 (b) its rarlius vector is inversely proportional to its vectorial angle. Ans. The hyperbolic or reciprocal spiral, pd = a. (c) the square of its radius vector is inversely proportional to its vectorial aniile. Ans. The lituus, p^O = a^. (d) the l(»i;aritlnn of its radius vector is proportional to its vectorial angle. Ans. The logarithmic spiral, log p = a^. Theorem on the logarithmic spiral. When two points, Pj and P,, have been plotted on a logarithmic spiral, points between them on the locus may be constructed geometrically by the following theorem : If the angle P^OP-^ is bisected, and if on this bisector OP3 is laid off equal to a mean proportional between OP.^ and 0P„, then Pg is on the locus. Proof. By hypothesis, since Pj and P^ are on the curve log p = aO, (1) log /)j = a^j and log/?., = aO^. Adding and dividing by 2, h + ^2^ (2) log 2 l0gpi+ I logP2 /^1 + (14 and 17, p. 1). If Pg is (pj, ^3), then, by construction. = (^3 - ^1, or (93 + : , and p^ = V, P\P-2 Hence, by (2), logpg = ad^, and P^ is also 011 the locus. R.{po,9o) PiiPudi) Q.E.D. PROBLEMS FOR mDIVIDUAL STUDY Plot carefully the following loci : 1. p = a sin ^ + bsecO. 2. ip--) = a2cos2^. 3. p = a (cos 2 (9 + sin 2^). i. p = a cos 2 -\- - sec 0. 5. p = a sill 2ff+- sec 0. 2 6. p = a cos 20 + b cos 0. 7. p = a sin 20 + b cos 0. 8. p - a cos2 + b{ii\n0 + 1). 9. p = a cos 3^ — 6 cos 0. 10. p = cos 3 ^ + cos + 1 11. p = cos 3 ^ + cos 2 0. 12. p = cos 3 6* — sin 2^. 13. . .0 p = asin3-. 14. p = a cos^ - . '^ 2 15. p'^cos0 = a2sin3^. 16. 2cos^ ^ ~cos2^ ■ 17. 2 cos 2 (9 P^ = ^-^ + 1- cos (9 + 2 CHAPTER VIII FUNCTIONS AND GRAPHS 51. Functions. In many practical problems two variables are involved in such a manner that the value of one depends upon the value of the other. For example, given a large num- ber of letters, the postage and the weight are variables, and the amount of the postage depends upon the weight. Again, the premium of a life-insurance policy depends upon the age of the applicant. Many other examples will occur to the student. This relation between two variables is made precise by the definition : A variable is said to he a function of a second variable ichen its value depends upon the value of the latter and is determined when a definite value is assnvied for the second variable. Thus the jyostage is determined when a definite weight is as- sumed ; the premium is determined when a definite age is assumed. Consider another example : Draw a circle of diameter 5 in. An indefinite number of rectangles may be inscribed within this circle. But the student will notice that the entire rectawjle is determined as soon as a side is drawn. Hence the area of the rectangle is ^function of its side. Let us now find the equation ex- pressing the relation between a side and the area of the rectangle. Draw any one of the rectangles and denote the length of its base by x in. Then by drawing a diagonal (which is, of 134 FUNCTIONS AND GRAPHS 135 x'f. course, a diameter of the circle), the altitude is found to be equal to (25 — x^y. Hence if A denotes the area in square inches, we have (1) A = x(25 This equation gives the functional relation between the func- tion .4 and the variable x. From it we are enabled to calculate the value of the function A corresponding to any value of the variable x. For example : if x = l in., .1 = (24)^ = 4.9 sq. in. ; if ic = 3in., yl=12sq. in. ; if x = A in., .4=12 sq. in. ; etc. To obtain a representation of the equation (1) for all vraues of X, we draw a graph of the equation. This we do by draw- ing rectangular axes and plotting the. values of the variable (x) as abscissas, the values of the function (.4) as ordinates. Any functional relation may be graphed in this way. We must, however, first discuss the equation (1). The values of x and A are positive from the nature of the problem. The values of x range from zero to 5, inclusive. The student should now choose a suitable scale on each axis and draw the graph. In this case, unit length on the axis of abscissas represents 1 in., and unit length on the axis of ordinates represents 1 sq. in. These two unit lengths need not be the same. What do we learn from the graph ? 1. If carefully drawn, we may measure from the graph the area of the inscribed rectangle corresponding to any side we choose to assume. 136 NEW ANALYTIC GEOMETRY ' 2. There is one horizontal tangent. The ordinate at its point of contact is greater than any other ordinate. Hence this discov^ery : One of the inscribed rectangles is greater in area than any of the otliers\ that is, there is a maximum rectangle. In other words, the function defined by equation (1) has a maximum value. Careful measurement will give for the base of the maximum rectangle, x = 3.5, and for the area, A = 12.5. These results, as may be shown by the methods of the differential calculus, are, in fact, correct to one place of decimals. The maximum rectangle is a square ; that is, of all rectangles inscribed in a given circle, the square has the greatest area. The fact that a maximum rectangle exists can be seen in advance by reasoning thus : Let the base x increase from zero to 5 in. The area A will then begin with the value zero and return to zero. Since .4 is always positive, the graph must have a " highest point." Hence there is a maximum value of A, and therefore a maximum rectangle. Take one more example : A wooden box, open at the top, is to be built to contain 108 cu. ft. The base must be square. This is the only condition. It is evi- dent that under this condition any number of such boxes may be built, and that the number of square feet of lumber used will vary accordingly. If, however, we choose any length for a side of the square base, only one box with this dimension can be built, and the material used is determined. Hence the material used is a function of a side of the square base. Let us now find the functional relation between the number of square feet of lumber necessary and the length of one side of the square base measured in feet. Consider any one box. A / ) FUNCTIONS AND GRAPHS 137 Let and Then and Hence M = amount of lumber in square feet, X = length of side of the square base in feet, ]i = height of the box in feet, area of Vjase = x^ sq. ft., area of sides = 4 hx sq. ft. M= x^ -\- A Jix. But a relation exists between A and a-, for the value of M must depend upon the value of x alone. In fact, the volume equals 108 cu. ft. Hence hx"^ Therefore 432 x (2) 108, and It = — —• M= x^ + This equation enables us to calculate the number of square feet of lumber in any box with a given square base which has a capacity of 108 cu. ft. The calculation is given in the table : X 1 1 2 4 5 G 7 8 20 etc. feet M 00 i 433 1 220 153 124 111 108 111 118 421 etc. sq. ft. Thus, if X = 1 ft., M - 433 sq. ft. ; if a; = 4 ft., .1/ = 124 sq. ft. ; if a; = 8 ft., .1/ = 118 sq. ft. ; etc. The student should now graph equation (2), choosing units thus : unit length on the axis of abscissas represents 1 ft. ; unit length on the axis of ordinates represents 1 sq. ft. We must, however, choose a very small unit ordinate, since the values of .1/ are large. A preliminary discussion of (2) shows that x may have any value (ix)sitive). 138 NEW ANALYTIC GEOMETRY M 1 2 3 4 5 6 Feet 7 8 9 lOx What do we learn from the graph? 1. If carefully drawn, we may measure from the graph the number of square feet of lumber in any box which contains 108 cu. ft. and has a square base. 2. There is one horizontal tangent. The ordinate at its point of contact is less than any other ordinate. Hence this discovery: One of the boxes takes less himher than any other ; that is, M has a minimum value. This point on the graph can be deter- mined exactly by calculus, but fiareful measurement will in this case give the correct values, namely, a? = 6, M = 108. That is, the construction will take the least lumber (108 sq. ft.) if the base is 6 ft. square. The fact that a least value of M must exist is seen thus. Let the base increase from a very small square to a very large one. In the former case the height must be very great, and hence the amount of lumber will be large. In the latter case, while the height is small, the base will take a great deal of lumber. Hence M varies from a large value to another large value, and the graph must have a " lowest point." In the following problems the student will work out the functional relation, draw the graph, and state any conclusions to be drawn from the figure. Care should be exercised in the selection of suitable scales on the axes, especially in the scale adopted for plotting values of the function (compare p. 137). The graph should be neither very flat nor very steep. To avoid the latter w^e may select a large miit of length for the variable. The plot should be accurate and the maximum and minimum values of the function should be measured and calcu- lated, additional values of the variable being used, if necessary. FUNCTIONS AND GRAPHS 139 PROBLEMS 1. Rectangles are inscribed in a circle of radius 2 in. Plot the perimeter P of the rectangles as a function of the breadth x. Ans. P = 2x + 2{l6-x^)i. 2. Right triangles are constructed on a line of length 5 in. as hypote- nuse. Plot (a) the area A and (b) the perimeter P as a function of the length X of one leg. Ans. (a) A=ix{2P>- x^)^ ; (b) P = x + 5 + (25 - x'^)^. 3. Right cylinders* are inscribed in a sphere of radius r. Plot as func- tions of the altitude x of the cylinder, (a) the volume V of the cylinder, (b) the curved surface S. Ans. (a) V= — (4 r2x - x^) ; (b) ,S = irx (4 r^ - x'^)^. 4. Right cones* are inscribed in a sphere of radius r. Plot as func- tions of the altitude x of the cone, (a) the volume V of the cone, (b) the curved surface S. _ , Ans. (a) F = - (2 rx^ - x^) ; (b) S = tt (4 r^-x^ - 2 rx^)^. o 5. Right cylinders are inscribed in a given right cone. If the height of the cone is h and the radius of the base r, plot (a) the volume V of the cylinder, (b) the curved surface S, (c) the entire surface T, as functions of the altitude x of the cylinder. {h-x); (a) F = 7rr2x {h- -X)2; (b) S = 2Trrx h (c) T = 2Trr {h- - x) [rh + {h - -r)x] 6. Right cones are circumscribed about a sphere of radius r. Plot as a function of the altitude x of the cylinder, the volume V of the cone. T^x'^ Ans. F = i TT ^ X - 2 r 7. Right cones are constructed with a given slant height L. Plot as functions of the altitude x of the cone, (a) the volume V of the cone, (b) the curved surface S, (c) the entire surface T. Ans. (a) F = i TT (L2x - x^) ; (b) S = tt Z {U - x^)*. 8. A conical tent is to be constructed of given volume F. Plot the amount A of canvas required as a function of the radius x of the base. , (7r2x6 + 9 F2)^ J.TIS. A=- —■ X *Use formulas .")-9, p. 1. 140 NEW ANALYTIC GEOMP:TPvY 9. A cylindrical tin can is to be constructed of given volume V. Plot the amount A of tin required as a function of the radius x of the can. . o o 2 V Ann. A = 2 wx- -\ X 10. An open box is to be made from a sheet of pasteboard 12 in. square by cutting equal squares from the four corners and bending up the sides. Plot the volume V as a function of the side x of the square cut out. Ans. T" = X (12 — 2 x)"^. 11. The strength of a rectangular beam is proportional to the product of the cross section by the square of the depth. Plot the strength S as a function of the depth x for beams which are cut from a log 12 in. in diameter. Ans. S = kx3{lU - x^)i. 12. A rectangular stockade is to be built to contain an area of 1000 sq. yd. A stone wall already constructed is available for one of the sides. Plot the length L of the wall to be built as a function of the length x of the side of the rectangle parallel to the wall. ^ „ t 2000 ° ^ Ans. L = 1- x. x 13. A tower is 100 ft. high. Plot the angle y subtended by the tower at a point on the ground as a function of the distance x from the foot of X 14. A tower .5.5 ft. high is surmounted by a statue 10 ft. high. If an observer's eyes are 5 ft. above the ground, plot the angle y subtended by the statue as a function of the observer's distance x from the tower. ,60 ,50 Ans. //=tan-i tan-i — X X 15. A line is drawn through a fixed point (a, b). Plot as a function of the intercept on XX' (= x) of the line, the area A of the triangle formed with the coordinate axes. ^„„ a bx^ Ans. A = 2{x — «) 16. A ship is 41 mi. due north of a second ship. The first sails south at the rate of 8 mi. an hour, the second east at the rate of 10 mi. an hour. Plot their distance d apart as a function of the time t which has elapsed since they were in the position given. ^^^^ ^ ^ ^^jy^a _ gsei + I681)i 17. Plot the distance e from the point (4, 0) to the points (x, y) on the parabola y- = 4x. Ans. e = (x^ — 4x + 16)^. 18. A gutter is to be constructed whose cross section is a broken line made up of three pieces, each 4 in. long, the middle piece being horizon- tal, and the two sides being equally inclined, (a) Plot the area A of FUNCTIONS AND GRAPHS 141 a cross section of the gutter as a function of the width x of the gutter across the top. (b) Plot the area J. as a function of the angle of incli- nation of the sides to the horizontal. Ans. (a) A =\{x+ 4) (48 + 8x - x2)i; (b)^ = 8(sin 2^ + 2 .sin^). 19. A Norman window consists of a rectangle surmounted by a semi- circle. Given the perimeter P, plot the area J. as a function of the width x. Ans. A = -£P x2 x2. 2 2 8 20. A person in a boat 9 mi. from the nearest point of the beach wishes to reach a place 15 mi. from that point along the shore. He can row at the rate of 4 mi. an hour and walk at the rate of 5 mi. an hour. The time it takes him to reach his destination depends on the place at which he lands. Plot the time as a function of the distance x of his landing place from the nearest point on the beach. /— ° ^ . r^. V 81 -f- x^ 15 — X Ans. Time = 1 4 5 21. The illumination of a plane surface by a luminous point varies directly as the cosine of the angle of incidence, and inversely as the square of the distance from the surface. Plot the illumination Z at a point on the floor 10 ft. from the wall as a function of the height x of a gas burner on the wall. j^^g j _ ^^ (100 + ^2)2 22. A Gothic window has the shape of an equilateral triangle mounted on a rectangle. The base of the triangle is a chord of the window. The total length of the frame of the window is constant. Express, plot, and discuss the area of the window as a function of the width. 23. A printed page is to contain 24 sq. in. of printed matter. The top and bottom margins are each 1^ in., the side margins 1 in. each. Express, plot, and discuss the area of the page as a function of the width. 24. A manufacturer has 96 sq. ft. of lumber with which to make a box with a square base and a top. Express, plot, and discuss the contents of the box as a function of the side of the base. 25. (a) Isosceles triangles of the same perimeter, 12 in., are cut out of rubber. Express, plot, and discuss the area as a function of the base. (b) Isosceles triangles of the same area, 10 sq. in., are cut out of rubber. Express, plot, and discuss the perimeter as a function of the base. 26. Small cylindrical boxes are made each with a cover whose breadth and height are equal. The cover slips on tight. Each box is to hold TT cu. in. Express, plot, and di.scuss the amount of material used as a function of the length of the box. 142 NEW ANALYTIC GEOMETRY 27. A circular filter paper has a diameter of 11 in. It is folded into a conical shape. Express the volume of the cone as a function of the angle of the sector folded over. Plot and discuss this function. 28. Two sources of heat are at the points A and B. Remembering that the intensity of heat at a point varies inversely as the square of the distance from the source, express the intensity of heat at any point betvpeen A and B as a function of its distance from A. Plot and discuss this function. 29. A submarine telegraph cable consists of a central circular part, called the core, surrounded by a ring. If x denotes the ratio of the radius of the core to the thickness of the ring, it is known that the speed of signaling varies as x^ log-. Plot and discuss this function. X 30. A wall 10 ft. high surrounds a square house which is 1.5 ft. from the wall. Express the length of a ladder placed without the wall, resting upon it and just reaching the house, as a function either of the distance of the foot of the ladder from the wall, or of the inclination of the ladder to the horizontal. Plot and discuss this function. 31. The volume of a right prism having an equilateral triangular base is 2. Express its total surface as a function of the edge of the base. Plot and discuss. 32. A letter Y .stands a ft. high and measures b ft. across the top. Express the total length of the leg and two arms as a function of the length of the leg. Plot and discuss. 33. The sum of the perimeters of a square and a circle is constant. Express their combined areas as a function of the radius of the circle. Plot and discuss. 34. A water tank is to be constructed with a square base and open top, and is to hold 64 cu. yd. The cost of the sides is |1 a square yard, and of the bottom |2 a square yard. Plot and discuss the cost. 35. A rectangular tract of land is to be bought for the purpose of lay- ing out a quarter-mile track with straightaway sides and semicircular ends. In addition a strip 3-5 yd. wide along each straightaway is to be bought for grand stands, training quarters, etc. If the land costs |200 an acre, plot and discuss the cost of the land required. 36. A cylindrical steam boiler is to be constructed having a capacity of 1000 cu. ft. The material for the side costs |2 a square foot, and for the ends |.3 a square foot. Plot and discuss the cost. 37. In the corner of a field bounded by two perpendicular roads a spring is situated 6 rd. from one road and 8 rd. from the other. How FUNCTIONS AND GRAPHS 143 -should a straight road be run by this spring and across the corner so as to cut off as little of the field as ijossible ? Ans. 12 and 16 rd. from the corner. 38. When the resistance of air is taken into account, the inclination of a pendulum to the vertical is given by the formula = ae-^' cos (nt + e). Plot ^ as a function of the time t. 52. Notation of functions. The symbol f (x) is used to de- note a function of x, and is read "/ of x." In order to distin- guish between different functions, the prefixed letter is changed, as F(x), (x) (read "^7u* of x "),f(x), etc. During any investigation the same functional symbol always indicates the same law of dependence of the function upon the variable. In the simpler cases this law takes the form of a series of analytical operations upon that variable. Hence, in such a case, the same functional symbol will indicate the same operations or series of operations, even though applied to different quantities. Thus, if f{x)=x^-9x + U, then /(y) =2/-9 }/-\- 14. Also f(a)=a''-9a + 14:, f{b ^l)=(b + iy-9{b + l) + 14: = b^-7b + 6, /(0)=02-90 + 14 = 14, /(-l) = (-lf-9(-l) + 14=24, /(7) = 72 - 9 ■ 7 + 14 = 0, etc. PROBLEMS 1. Given 0(x) = log^a;. Find ^(2), ^(1), 0(5), 0(a-l), 4>{b'^), {x+ 1), (Vx). 2. Given 0(x) = e^^. Find 0(0), 0(1), 0(- 1), 0(2?/), >/ coordinates are y '•\ Y \ • y \0'(h. kj .V" y /^ y X (111) ( X = :^ COS. 9 — y^ &\n9 -Jf- h, ly = X? sin 6 -\- y' cos, 6 -\- k. Proof. To tiunslate the axes to O'X" and o'Y" we have, Ijy (I), :r = .•■" + ]>, y = //" + /.•, where (.r", //") are the coordinates of any point V referred to ^^'A" and C'F". To rotate the axes we set, by (II), . ./•" = ,'r' cos Q — //' sin ^, //" ;= a-' sin 6 + f cos 6. Substituting these values of ,'•" and //", we obtain (HI). Q.e.d. 57. Classification of loci. The loci of algebraic equations are classified according to the degree of the ecjuations. This classifiration is justified by the following theorem, which shows that the degree of the ef]uation of a locus is the same, no matter how the axes are chosen. TRANSFORMATION OF COORDINATES 149 Theorem. The degree of the equation of a locus is unchanged by a transformation of coordinates. Proof Since equations (III) are of the first degree in x' and y', the degree of an equation cannot be raised when the values of X and y given by (III) are substituted. Neither can the degree be lowered; for then the degree must be raised if we transform back to the old axes, and we have seen that it cannot be raised by changing the axes.* As the degree can neither be raised nor lowered by a trans- formation of coordinates, it must remain unchanged. o.e.d. 58. Simplification of equations by transformation of coordinates. The principal use made of transformation of coordinates is to simplify a given equation by choosing suitable new axes. The method of doing this is illustrated in the following examples. EXAMPLES 1. Simplify the equation ?/2_8x-l-62/+17 = Oby translating the axes. Solution. Set x = x' -{■ h and y = y' + k. This gives {?/' + A-)^ _ 8 (x' + /;) + 6 (;/' + A:) + 17 = 0, or (1) y'2_8x' + 2^- + 0. if + A;2 -8A + Qk + 17 If, now, we choose for h and k such numbers that the coefficient of y' shall be zero, that is, (2) 2A;+6=:0, and also the constant term shall be zero, that is, (3) A:2-8A + 6fc + 17 = 0, the transformed equation is simply (4) ?/'2_8x' = 0. * This also follows from the fact that when equations (III) are solved for x' and ?/', the results are of the first degree in x and ?/. t These vertical bars play the part of parentheses. Thus 2 A + 6 is the coefti- cient of y' and A;^ - 8 /i + 6 A: + 17 is the constant term. Their use enables us to collect like powers of x' and 7/ at the same time that we remove the parentheses in the preceding equation. 150 NEW ANALYTIC GEOMETRY From (2) and (3) we obtain A = 1, fc = — 3, and these are the coordinates of the new origin. The locus may be readily plotted by draw- ing the new axes and then plotting (4) on these axes. A second method often used is the fol- lowing : Rewrite the given equation, collecting the terms in ?/, (5) 0/ + 6 2/) = 8x-17. Complete the square in the left-hand member, (6) (y^ + Qy +9) = 8x-n + 9 = 8x-8. Writing this equation in the form (7) {y + 3)2 = 8{x- 1), it is obvious by inspection that if we substitute in this equation (8) X = x' + 1, y = y'- 3, the transformed equation is y"- = 8 x'. But equations (8) translate the axes to the new origin (1, — 3), as before. 2. Simplify x'^ + 4 2/'^ — 2 x — 16 ;/ + 1 = by translating the axes. Solution. Set x = x' ■{■ h and y = y' + k. This gives y r /* y^ y /' / / / X / o a. -3; X' V \ \ \ \ s N (9) x'2 + 4y'^ + 2h -2 x'+ 8A; 2/'+ h^ -16 + 4fc2 -2/i -16k + 1 1 = 0. Let us choose the new origin so that in (9) the coefficients of x' and y' shall be zero ; that is, so that (10) 2 /i - 2 = and 8 fc - 16 = 0. From (10), h = 1, k — 2, and these values substituted in (9) give the transformed equation (11) x'2 -I- 4 ?/'2 = 16. The locus of the given equation is now readily drawn by constructing parallel axes through (1, 2) and plotting equation (11) on these axes. A second method is the following : Collect corresponding terms in the given equation thus : (12) (x2-2x) + 4(y2_42/) =-1. TRANSFORMATION OF COORDINATES 151 Complete the squares within the parentheses, adding the corresponding numbers to the right-hand member, (13) (a;2-2x + l) + 4(2/2-42/ + 4) = -1 + 1 + 16 = 16. Writing (13) in the form (a:_l)2 4.4(^_2)2 = 16, it is obvious by inspection that by substituting (14) x = x'+l, j/ = y' + 2, the simple new equation x"^ -if Ay"^ = 16 results. But equations (14) trans- late the axes to the new origin (1, 2), the same as in the first method. 3. Kemove the x?/-term from x^ + 4 jy + ?/- = 4 by rotating the axes. Solution. Set x = x' cos — y' sin 6 and y = x' sin ^ + if cos 0, whence r|r| ^ r '—"I ■^ / \ V 0' ri, I) / x' >^ U^ X + 4 sin 6 cos 6 + sin'- d x"- — 2 sin 9 cos 6 + 4(cos2^-sin26') + 2 sin cos 9 x'y' + sin2 — 4 sin 9 cos 9 + cos2 9 V 4, or, since 2 sin 9 cos ^ = sin 2 ^ and cos^ 9 — sin2 9 = cos 2 9^ (15) (1 + 2 sin 2 ^) x'2 + 4 cos2 (9 • xV + (1 - 2 sin 2 (9)2/'2 = 4. The new equation is to contain no x'y'-term. Hence, setting the coefficient of x'y' equal to zero, cos 2^ = 0. .-. 2^ = - and ^ = -. 2 4 Substituting in (15), since sin — = 1, the trans- 2 formed equation is 3x'2_y'2^4. The locus of this equation is the hyperbola plotted on the new axes in the figure. These examples show that it is often wise not to plot the locus of an equation as it stands, but ratlier to endeavor first to simplify by transformation to new axes. .N . 1" /x ^ \ \ / \ \ / \ \ / \ V / \ \ / ~ \ / \ti ^ \ \ u Si Y. X / \ \ ^ . / \ / k / \ \ / \ \ / \ \ / \ k Ans. X'' + 4 2/' = 0. Ans. x'2 = 4 y'. Ans. X'2 + 2/'2 : = 16. Ans. y'2 = 6x' Ans. X'2 - 2/'2 : = 2. Ans. X'2 + 4 ij"- 2 = 16, Ans. 8x'+?/3 = 0. 152 NEW ANALYTIC GEOMETRY PROBLEMS 1. Simplify tlie following equations by translating the axes. Plot both pairs of axes and the curve. (a) x2 + 6x + 4?/ + 8 = 0. (b) x2-42/ + 8 = 0. (c) x2 + 2/2 + 4x - 6 y - 3 = 0. (d) 2/2 _ 6x - lOy + 19 = 0. (e) x2 - 2/2 + 8x - 14 ?/ - 35 = 0. (f) x2 + 42/2 -16x + 24 2/ + 84 = 0. (g) 2/3 + 8x- 40 = 0. 2. Remove the xy-term from the following equations by rotating the axes. Plot both pairs of axes and the curve. (a) x2 — 2x2/ + 2/^ = 12. Ans. %j'^ = Q. (b) x2 — 2 X2/ + 2/^ + 8 X + 8 2/ = 0. Ans. V2 y'" + 8 x' = 0. (c) xy = 18. Ans. x'2 - ?/2 = 36. (d) 25 x2 + 14 X2/ + 25 ?/ = 288. Ans. 16 x'2 + 9 2/'2 = 144. (e) 3 x2 — 10x2/ + 3 2/2 = 0. ^ns. x'2 _ 4 2/'2 = 0. 3. Translate the axes so that each of the following equations is trans- formed into a new equation without any terms of the first degree in the new coordinates. Draw the locus. (a) x2 — 4 X2/ + 6 2/ = 0. Ans. /i = f , k — \. (b) 2/2- 2x2/ + 3x = 0. (e) 3x2 — x?/ — 2/2 + 4x = 0. (c) x2 + X2/ + 2/2 + 6x = 0. (f) 2x2/ + 6x-8 2/ = 0. (d) x2 - x?/ + 2 2/2 + 62/ = 0. (g) 3xj/ + 4 2/ - 2 = 0. CHAPTER X PARABOLA, ELLIPSE, AND HYPERBOLA 59. The parabola. Consider the following locus problem. A point moves so that its distances from a fixed line and a fixed point are equal. Determine the nature of the locus. Solution. Let DD' be the fixed line and F the fixed point. Draw the a;-axis through F perpendicular to DD'. Take the origin midway between F and DD'. Let (1) distance from F to DD' = p. Then, if P (x, y) is any point on the locus, (2) FP = MP. But FP = V(.T — ^py-hV% MP = MN -{- Nl' = ^p +x. Substituting in (2), ^{x-\py + if=\p + x. Squaring and reducing, (3) if = 2px. The locus is called a, parabola. The fixed line DD' is called the directrix, the fixed point F, the focus. From (3), it is clear that the .X-axis is an axis of symmetry. For this reason, the x-axis is called the axis of the parabola. Furthermore, the origin is on the curve. This point, midway between focus and directrix, is called the vertex. 1.58 1) Y. N ^^ M y D Y V ^(^,0 ) X FP=MP 154 NEW ANALYTIC GEOMETRY Theorem. If the origin is the vertex and the x-axis the axis of a parabola, then its equation is (I) y^ = 2px. and the equation of the directrix ^,0 The focus is the point P IS X ^ 2 A discussion of (I) gives us the following properties of the parabola in addition to those already obtained. 1. Values of x having the sign opposite to that of 2^ are to be excluded. Hence the curve lies to the right of YY^ when p is positive and to the left when ^; is negative. 2. No values of y are to be excluded ; hence the curve extends indefinitely up and down. The chord drawn through the focus parallel to the directrix is called the latus rectum. To find its length, put X = \p in (I). Then ?/ =±p, and the length of the latus rectum = 2^? ; that is, equals tlie coefficient ofx in (I). It will be noted that equation (I) contains two terms only ; namely, the sqicare of one coordinate and tJoe first power of the other. Obviously, the locus of x^ = 2py is also a parabola, and thus we have the Theorem. If tlie origin is the vertex and the y-axis the axis of a parabola^ then its equation is (II) x' = 2py. The fociis is the point [ 0, - )> and the equation of the directrix is y =— -• -^ 2 Equations (I) and (II) are called the typical fo)'ms of the equation of the parabola. PARABOLA, ELLIPSE, AND HYPERBOLA 155 Equations of the forms Ax^ + Ei/ = and Cif — Dx = Q, where A, E, C, and D are different from zero, may, by ti'ansposition and division, be written in one of the forms (I) or (II), To plot a parabola quickly from its typical equation, its position (above or below A' A'', to the right or left of YY') is best determined by discussion of the equation. The value of 2p is found by comparison with (I) or (II), and the focus and directrix are then plotted. EXAMPLES 1. Plot the locus of x^ + 4y = and plot the focus and directrix. Solution. The given equation may be written x^ = — 4 y. The t/-axis is an axis of symmetry ; positive values of y must be ex- cluded. Hence the parabola lies below the a;-axis. The table gives a few points on the curve. X y ±2 ±4 - 1 -4 5M > D ,-[ D X' ^ N X / -'•(O.-ll \ / \ / \ t ' I 1 \ y Comparing with (II), p = — 2. The focus is therefore the point (0,-1) and the directrix the line ?/ = l. The length of the latus rectum is 4. Every point on the locus is equidistant from (0, — 1) and the line y = '[. 156 NEW ANALYTIC GEOMETRY 2. Find the equation of the parabola whose focus is (4, — 2) and ■directrix the line x = l. Solution. In the figure, by definition, .(1) FP = FM. ^ But FP and V{x - if + {y + 2)2, PJlf = x-l. Substituting in (1) and reducing, (2) 2/2 — 6x + 42/ + 19 = 0. Ans. If tlie axes are translated to the vertex (|, — 2) as a new origin, that is, if we substitute in (2) x = x' + | and y = y' — 2, the equation reduces to the typical form ?/'2 — 6 x' = 0. A second and useful method is the following : Draw the axis VX' of the parabola and the tangent VY' at the vertex. Referred to these lines as temporary axes, the equation must have the typical form (3) 2/2 = 6 X, since p = 3. Now translate the temporary axes so that they will coincide with the given axes. The coordinates of referred to the temporary axes are ( — f , 2). Sub- stituting in (3) x=x'— I, y = y'-\- 2, and reducing, we obtain the equation (2). PROBLEMS 1. Plot the locus of the following equations. Draw the focus and directrix in each case and find the length of the latus rectum. (a) 2/2 = 4x. (d) y^-Qx- 0. (b)2/2 4-4x = 0. (e) x2 + 102/ = 0. (c)x2-82/ = 0. (f)2/^ + x = 0. 2. Find the equations of the following parabolas : (a) directrix x = 0, vertex (3, 4). Ans. (y — 4)2 = 12 (x — 3). (b) focus (0, - 3), vertex (2, - 3). Ans. y^ + 8x + 6y -7 = 0. PARABOLA, ELLIPSE, AND HrPERBOLA 157 (c) axis X = 0, vertex (0, — 4), passes through (6, 0). Ans. x" = 9y + 86. (d) axis y = 0, vertex (6, 0), passes through (0, 4). Ans. 3y" = 48- 8x. (e) directrix x + 2 y — 1 = 0, focus (0, 0). Ans. (2 X — ?/)2 + 2 X + 4 ?/ - 1 = 0. 3. Transform each of tlie following equations to one of the typical forms (I) or (II) by translation of the axes. Draw the figure in each case. (a) ?/2 + 4x + 4y-2 = 0. (b) x2+ 6x + ?/-2 = 0. (c) x2+3x + 4?/-l=0. (d) y2+5x + 8y = 0. (e) 2x-+ 5y + 4 = 0. (f) 2/2+ 6x- 9 = 0. (g) 7x2 + 8y + 10 = 0. (h) x2 + 4 y/ + 4 = 0. Ans. y'" + 4x' = 0. Ans. x'2 -f- 2/' = 0. (i) 2?/ + 3x — 8 = 0. (j) 5x2 + 102/ + 12 = 0. (k) 3x2-6?/ + 8 = 0. (1) 2x2- Ox + 2/ = 0. 4. Show that abscissas of points on the parabola (I) are proportional to the squares of the ordinates. 5. Find the equation in polar coordinates of a parabola if the focus is the pole, and if the axis of the parabola is the polar axis. A P Ans. p= i l-cos^l 60. Construction of the parabola. A parabola Avhose focus and directrix are given is readily constructed by rule and com- passes as follows : Draw the axis MX. Con- struct the vertex V, the middle point of MF. Through any point A to the right of V draw a line AB parallel to the direc- trix. From F as a center with a radius equal to MA strike arcs to intersect AB at P and Q. Then P and Q are points on the parabola. For FP = MA, by construction, and hence. P is equidistant from focus and diivctrix. 158 NEW ANALYTIC GEOMETRY C a' I) n VI I 1 1 1 1 ' ^^^^ 1 [ 1 ""' 1 1 1 1 1 ' 1 r^^^ t 1 A a \}_ c -^-^pan — > B a, h, c, I, m, n. By changing the position of A we may construct as many points on the curve as desired. Parabolic arch. When the span AB and height OH of a para- bolic arch are given, points on the arch may be constructed as follows : Draw the rectangle A BCD. Divide ^//and AC into the same number of equal parts. Starting from A, let the suc- cessive points of division be on AH, on A C, Now draw the perpendicular aa' to AB, and draw 01. Mark the intersection. Do likewise for the points b and vi, c and n. The intersections are points on the parabola required. Proof. Take axes OX and OY, as in the figure. Let (1) OM' = X, M'P = ij, AB = 2a, OH = h. Y By construction, NC and MH are equal parts of ^C and AH respectively. .^ . NV_MH NC _x ^"^ ' ■ 'aC ~ Ih' ^^ h ~ a From the similar triangles OM'P and OCN, y _NC _NC c il/'p a^- N \\ X B(c A(r a,h)M H \ I' — — a ^> (3) OC Substituting the value of NC from (2) into (3), and reducing, (4) 2 * PARABOLA, ELLIPSE, AND HYPERBOLA 159 This is the typical form (II), and the locus passes through 0, A (— a, h) and B(a, Ji), as required. Solving (4) for y, we get (5) ^ a} ' X la ia fa a y -h h ih T%/' h showing that y varies as the square of x, and giving a simple formula for computing y, as in the table. 61. Equations (I) and (II) are extraordinarily simple types of equations of the second degree. The question, To derive a test for determining if the locus of a given equa- tion of the second degree is a parabola, will be answered in Art. 70 ; but at this point, if the previous results are borne in mind, we may state the Theorem. The locus of an equation of the second degree is a parabola if the only tervi of the second degree * is the square of one coordinate, and if also the first power of the other coordi- nate is present in the equation. For illustration, see Problem 3, p. 157. 62. The ellipse. Let us solve the following locus problem : Given two fixed points F and F'. A point P moves so that the sum of its distances from F and F' remains constant. Determine the nature of the locus. Solution. Draw the ar-axis through F and /•", and take for origin the middle point of F'F. By definition, (1) PF + PF' = a constant. PF+ PF'=2a *The student should not forget tiiat the product zy is a term of the second degree. 160 NEW ANALYTIC GEOMETRY Let us denote this constant by 2 a. Then (1) becomes (2) PF + PF' = 2 a. Let FF' = 2c. Then PF = ■\/{x — (■)'- + 2/^ PF' = V(cc + cf + y*, since the coordinates of F are (c, 0), and of F', (— c, 0). Hence (2) becomes (3) V(x- - c)^ + f + V(at + cy + if = 2 a. Transposing one of the radicals, squaring and reducing, the result is (4) {a' - c^ X- + ahf = d\a? - FF', or 2 a > 2 c ; that is, a > c, and q2 _ (.2 jg 2i positive number. PARABOLA, ELLIPSE, AND HYPERBOLA 161 AA ' through O is called the major axis ; the shortest chord BB', the minor axis. Obviously, (7) major axis = 2 a, minor axis = 2h. , Dividing (6) through by a^"^, and summarizing, gives the Theorem. The equation of an ellipse whose center is the orirjin and whose foci are on the x-axis is (III) where 2 a is the major axis and 2 b the minor axis. Ifc^ ^a^ — 6^, then the foci are (± c, 0). If the foci are on the ?/-axis, and if we keep the above nota- tion, the equation of the ellipse is obviously (8) a^x^ + by = a%% or Equations (6), (8), and (III) are typical equations of the ellipse, and are of the form (9) Ax"^ -\-Bf=C, where ^4, 2?, and C agree in sign. In the figure 'BF'^ = h'^ + c\ Substituting the value of c^ from (5), then BF = a\ Hence the property : The distance from either focus to the end of the minor axis equals the sem imajor axis. The chord drawn through either focus perpendicular to the major axis is called the latus rectum. Its length is deter- mined by setting a; = c in (III), and solving for y. This gives 1/ = - Va** — c^ = — • Hence F; , ^^ B ^-^c,#) > A h \ i\ / C \[ \ ^'\ o \F JU-^ X (10) length of latus rectum = • 162 NEW ANALYTIC GEOMETRY Eccentricity. When the foci are very near together the ellipse differs but little from a circle. The value of the ratio OF : OA may, in fact, be said to determine the divergence of the ellipse from a circle. The value of this ratio is called the eccentricity of the ellipse, and is denoted by e. Hence OF c The value of e varies from to 1. If the major axis A A' remains of fixed length, then the " flatness " of the ellipse in- creases as e increases from to 1, the limiting forms being a circle of diameter ^A' and the line segment ^1.4'. From (11) and (5), (12) b^ = a'-c' = a\l-e^). To draw an ellipse quickly when its equation is in the typical form, proceed thus : 1. Find the intercepts, mark them off on the coordinate axes, and set the larger one equal to a, the smaller equal to h. Letter the major axis A A' and the minor axis BB'. 2. Find c from c^ = a^ — V^. Mark the foci F and F' on the major axis. 3. Calculate directly one or more sets of values of the coordi- nates, and sketch in the curve. EXAMPLE Draw the ellipse ix^ + 2/^ = 16. Solution. The intercepts are, on XX', ± 2 ; on YY', ±4. Hence the major axis fallson YY', and a = 4, 6 = 2, c = V12 = 2 VS = 3.4. The foci are on the 2/-axis. The length of the latus rectum equals = 2. The eccentricity e — ~ =\ v 3. ^ The points found in the table ai-e the ends of the latus rectum. If P is any point on the ellipse, then PF + PF' = 2 a = 8. X y ±1 ±3.4 PARABOLA, ELLIPSE, AN^D HYPERBOLA 163 PROBLEMS * 1. Plot each of the following equations. Letter the axes and mark the foci. Find the eccentricity, the length of the latus rectum, and draw the latus rectum. ' (a) a;2 + 92/2 =: 9. -(e) 9x^ + iy^ = 36. ' (b) 9x2 + 16 2/2 = 144^ (f j 2 x2 + 2/2 = 25. (c) 2 x2 + 2/2 = 4. (g) 4 x2 + 8 2/2 = 32. (d) 4x2 + Qy2 ^ 36. (h) 7^2 + 3j/2^ 21. • 2. Transform each of the following equations by translation of the axes so that the transformed equation shall lack terms of the first degree in the new coordinates. Draw the figure. . (a) x2 + 4 2/2 + 6 X — 8 2/ = 0. Ans. x'"- + 4 y'- = 13. .(b) 9x2 + 4y2 + 30j. _ 4y + 1=0. (c) x2 + 5 2/2 + 102/ = 20. (d) 5x2 + 2/2 + lOx + 4 2/ = 6. (e) 3 x2 + 2/2 + 6 X — 4 ?/ = 2. . (f) 4x2 + 52/2 + 4x + 20 2/ = 20. 3. Find the equation of each of the following ellipses: (a) major axis = 8, foci (5, 2) and (— 1, 2). Ans. 7 (x- 2)2 + 16 (2/ -2)2 = 112. (b) major axis = 10, foci (0, 0) and (0, 6). Ans. 25x2 + 16(2/- 3)2 = 400. (c) minor axis = 8, foci (—1,0) and (4, 0). (d) minor axis = 4, foci (0, — 2) and (0, 4). 63. Construction of the ellipse. The definition (2) of the pre- ceding section affords a simple method of drawing an ellipse. Place two tacks in the drawing board at the foci F and F' and wind a string about them as indicated. If now a pencil be placed in the loop FPF' and be moved so as to keep the string taut, then J^F -f PF' is constant and P describes an ellipse. If the major axis is to be 2 a, then the length of the loop FPF' must be 2 a + 2 c. 164 NEW ANALYTIC GEOMETRY A useful construction of an ellipse by rule and compasses is the following: Draw circles on the axes A A' and BB' as diameters. From the center draw any radius intersecting these circles in M and N respectively. From M draw a line MB parallel to the minor axis, and from N a line JVS paral- lel to the major axis. These lines will intersect in a point P on the ellipse. Proof. Take the coordinate axes as in the fiq-ure below. Let OA = x, AP = y = OD. Clearly, OB OC ZMOX = cf>. semimajor axis = a, semiminor axis = b. Then in the right triangle OAB, (1) OA X Similarly, in the right triangle ODC, AOCD = ZCOA = ^, and (2) «in* = ^^f■ But cos^ + siu^- ^ = 1. Hence, 2 2 from (1) and (2), -^ + '^ = 1, and P(x, y) lies on the ellipse whose semiaxes are a and b. Q.e.d. The angle ^ is called the eccen- tric angle of P. The construction circles used in this problem are called, respectively, the major and minor auxiliary circles. PARABOLA, ELLIPSE, AND HYPERBOLA 165 64. Equations (6) and (8) of Art. 62 are simple equations of the second degree. We may ask the question, What is the test that the locus of a given equation of the second degree shall be an ellipse ? Reserving for a later section the answer to this question, we have, however, some light on it now. For we have observed in Problem 2, p. 163, that the locus was in each case an ellipse. These equations agree in the respect that there is no xy-term, and the squares of x and y have unequal positive coefficients. Consider such an equation, for example, (1) x'+^y'+^x-^y + N=0, where N is some nun^ber. If we translate the axes to the new origin (— 2, 1), the transformed equation is (2) a;''+4?/'^=8-iV. If N is less than 8, the locus is an ellipse. If N = 8, the locus is the single point (0, 0), often called a point-ellipse. If N is greater than 8, there is no locus. This discussion is general, and may be summarized in the Theorem. If an equation of the second degree contains no xy-term, oMd if x? and y^ occur with coefficients having like signs, the locus is necessarily an elVqyse or point-ellipse. The case when x^ and y^ have equal coefficients has been dis- cussed in Art. 38. The circle and point-circle may, of course, be regarded as special cases of the ellipse and point-ellipse. 65. The hyperbola. Let us next turn our attention to a third locus problem. Given two fixed points F and F'. A point 7' moves so that the difference of its distance from F and F' remains constant. Determine the nature of the locus. 166 NEW ANALYTIC GEOMETRY Solution. Draw the x-axis through the fixed points, and take for origin tlie middle point of F'F. By definition (1) PF' — PF = a constant. Let us denote tliis constant by 2 a. Then (1) becomes (2) PF' - PF=2a. Let FF' =2c. Tlien PF = ^{x — cf-{-if, and PF = V(^ + cf + if, since the coordinates of F are (c, 0), and of F' , (- c, 0). Substituting in (2), (3) V(.r + if + if - V(a; - cf + f = 2 a. Transposing either radical, squaring and reducing, the result is (4) (a^ — c^) X- + a^y' = a^ (a^ — c^). For added simplicit}',* set (5) a^ — c^ = — Jy-, or 1. The relation of the value of e to the shape of the curve will be made clear later. From (5) and (11), (12) b'^ = c" -a^ = cr {tP - 1). To draw a hyperbola quickly when its equation is in the typical form (9), proceed thus : 1. Find the intercepts and mark them off on the proper axis. Set a equal to the real intercept and b equal to the coefficient of V— 1 in the imaginary intercept. Lay off the conjugate axis ; letter it BB' and the transverse axis A A '. 2. Find c from c^ = a^ + b\ Mark the foci F and F* en the transverse axis. 3. Calculate directly one or more sets of values of the coor- dinates, and sketch the curve. PARABOLA, ELLIPSP:, AND HYPERBOLA 169 EXAMPLE Draw the hyperbola 4x2- 52/2 + 20 = 0. Solution. Th e intercept s ar e, on XX\ ± V - 5 = ± VS V^ ; on YY\ ±2. Hence 6 = Vs, a = 2, c = V a2 -|- 62 = 3, and the transverse axis and the foci are on YY' . The ec- centricity is f . The length of the latus 2 62 rectum is 5. If P is any point on the hyperbola, then -p-E" - PF = 'i. X y ±2 ±3 PROBLEMS 1. Plot each of the following equations, letter the axes, and mark the foci. Find the eccentricity, the length of the latus rectum, and draw the latus rectum. (a) 5x2-4 2/2 = 20. (b) x2 - 8 2/2 + 8 = 0. (c) 9 x2 - 2/2 _ 9. (d) 3x2-2/2 = 12. (e) x2 -32/2 + 3 = 0. (f) 7x2-9t/ = G3. (g) 2x2-7 2/2 = 18. (h) 7x2 -2 2/2 =-8. 2. Transform each of the following equations by translation of the axes so that the transformed equation shall lack terms of the first degree in the new coordinates. Draw the figure. (a) 4x2-2/2 + 8x-2 2/-l = 0. (d) 4x2 - 2/2 - 6x - 4y = 0. (b) 9x2-i/2 + l8x-42/+ 14 = 0. (e) x2 - 52/2 + 6x - lOy = 0. (c) 3x2-2/2 + 12x + 22/ + 14 = 0. (f) x^ - 2y^ + 10y = 0. 3. Find the equations of the following hyperbolas : (a) transverse axis = 6, foci (— 2, 0) and (6, 0). Ans. 7 (x- 2)2 -9 2/3 = 63. (b) conjugate axis = 6, foci (0, 2) and (0, — 8). (c) conjugate axis = 4, foci (1, 2) and (— 4, 2). (d) transverse axis = 2, foci (0, 0) and (— 4, 0). 170 NEW ANALYTIC GEOMETRY 66. Conjugate hyperbolas and asymptotes. Two hyperbolas are called conjugate hyperbolas if the transverse and conjugate axes of one are respectively the conjugate and transverse axes of the other. If the equation of a hyperbola is given in typical form, then the eqiiatlon of the conjugate hyperbola Isfouyid hij cJuinging the signs of the coefficients ofx^ and if' in the gloen ef^uation. Thus the loci of the equations (1) 1Q> x" - 7f =10, and -1Q> x" + if =1Q are conjugate hyperbolas. They may be written 1-1«=1 =">* -1 + 16=1- The foci of the first are on the ic-axis, those of the second ■on the ?/-axis. The transverse axis of the first and the conju- gate axis of the second are equal to 2, while the conjugate axis ■of the first and the transverse axis of the second are equal to 8. The foci of two conjugate hyperbolas are equally distant from the origin. For c^ equals the sum of the squares of the semitransverse and semiconjugate axes, and that sum is the same for two conjugate hyperbolas. Thus in the first of the hyperbolas above c^ =1 + 16, while in the second c"^ = 16 + 1. If in one of the typical forms of the equation of a hyper- bola we replace the constant term by zero, then the locus of the new equation is a pair of lines (Theorem, p. 40) which are called the asymptotes of the hyperbola. Thus the asymptotes of the hyperbola (2) 6V - ahf = a%'' are the lines (3) ^»v-ay = o, or (4) hdi- -\- ai/ = and bx — ay = 0. PARABOLA, ELLIPSE, AND HYPERBOLA 171 These may be written (5) y ^ — — X and 7/ = - a;. They pass through the origin and their slopes are respectively h , I and -• a a The property of these lines which they have in common with. the vertical or horizontal asymptotes of Art. 22 is expressed in the Theorem. The branches of the hijpevhola ajjproach indefinitehj near its asymptotes as they recede to infinity. Proof. Let P-^(x^, y^ be a point on either branch of (2) near the asymptote ^^ _ ^^ ^ 0. The perpendicular distance from this line to P^ is hx^ — ay^ _ Yk (6) d = — -\llr + a^ We may find a value for the numerator as follows : Since P^ lies on (2), Ip-xl — i^yl = a^lP-. Factoring and dividing, a^h"- bx^ — ay. = ; 1 "^1 bx^ + ay^ Substituting in (6), d . — — V^'-^ + a^ (bx^ + ay^ As Pj recedes to infinity in the first quadrant, x^ and y^ be- come infinite and d approaches zero. Hence the curve approaches closer and closer to its asymp- totes. Q.E. D. Two conjugate hyperbolas have the same asymptotes. Thus the asymptotes of the conjugate hyperbolas (1) are respectively the loci of 16x2-^2 = and -16x2 + 2/2 = 0, which are the same. 172 NEW ANALYTIC GEOMETRY A hyperbola may be drawn witb fair accuracy by the fol- lowing Construction. Lay off OA = OA' = a on the axis on which the foci lie, and 05 = 05'= 6 on the other axis. Draw lines through A, A', B, B', parallel to the axes, forming a rectangle. Draw the diagonals of the rectangle. Then the length of each diago- nal is obviously 2 c (since cv^ + 1/ = c"). Moreover, the diagonals produced are the asymptotes. For the equations of the diagonals ate readily seen to be Itx — ai/ ^= and hx -\- ay = 0, and these are the same as (4). Construct the circle which circumscribes the rectangle. Draw the branches of the hyperbola tangent to the sides of the rectangle at A and A' and approaching nearer and nearer to the diagonals. The conjugate hyperbola may be drawn tangent to the sides of the rectangle at B and B' and approaching the diagonals. The foci of both are the points in which the circle cuts the axes. From this construction the influence of the value of the eccentricity upon the shape of the hyperbola can be easily dis- cussed. In the figure, let yl.4' be fixed. Now from (12), Art. 65, When e diminishes towards unity, b decreases, the altitude BB' of the rectangle diminishes, the asymptotes turn towards the a;-axis, and the hyperbola flattens. When e increases, the asymptotes turn from the a;-axis, and the hyperbola broadens. PARABOLA, ELLIPSE, AND HYPERBOLA 173 67. Equilateral or rectangular hyperbola. When the axes of a hyperbola are equal (a = Ji), the hyperbola is said to be equilateral. If we set a = b\\\ equation (IV), we obtain (1) x^ — if= a^, which is accordingly the equation of an equilateral hyperbola whose trans- verse axis lies on A'A''. Its asymptotes are the lines X — y = and x + ?/ = 0. These lines are perpendicular, and hence they may be used as coordinate axes. The designation " rectangular " hyj)erbola arises from this fact. Theorem. The equation of an equilateral hyperbola referred to its asymptotes is (V) %xy = d'. Proof The axes must be rotated through — 45° to coincide with the asymptotes. Hence we substitute (Art. 55) x' 4- ?/' _ — x' + y' V2 m (1). This gives y V2 (- x^ + y'f 2 Reducing and dropping primes we have (V). Q.e.d. It is important to observe that (V) has the simple form (2) cfy = a constant. 68. Construction of the" hyperbola. A mechanical construction, depending upon the definition (1) of Art. G5, is the following : Fasten thumb tacks at the foci. Pass over F' and around F a string whose ends are held together (Fig. 1, p. 174). If a pencil be tied to the string at P, and both strings be pulled in or let out together, then PF' — PF will be constant 174 NEW ANALYTIC GEOMETRY and P will describe a hyperbola. If the transverse axis is to be 2 a, the strings must be adjusted at the start so that the difference between PF' and PF equals 2 a. A construction often used for an equilateral hyperbola when the asymp- totes and one point A are given, is as follows (Fig. 2) : Let OX and Y be the asymptotes and A the given point. Draw any line through A to meet OX at M and O F at iV. Lay off MP =AN. Then P is a point on the required hyperbola. Proof. Choose the asymptotes as axes. Let the coordinates of A be {a, h) and of P, (x, y). Then OS = x, SP = y, OB = 1), BA = a. By construction, AN = MP. ,". triangle PSM = triangle NBA, and BN =SP = y, SM = AB = a. Since the triangles 03IN and ABN are similar, BN _ UN ' ' Ib ~~om Substituting, y_ ^J + y a OB + BN OS -}- sm' or xy = ab. My Mo Fig. 3 a + ■*■ Comparing with (V), we see that P («, y) lies upon an equilat- eral hyperbola which has OX and OY for its asymptotes and which passes through (a,h). Q. e.d. By drawing different lines through A, and laying off -V^P^ = AN^, M^P^ = AN,^, etc., we determine as many points J\, P.,, etc., as we wish on the hyperbola (Fig. 3). PARABOLA, ELLIPSE, AND HYPERBOLA 175 PROBLEMS 1. Find the equations of the asymptotes and of the hyperbolas conju- gate to the following hyperbolas, and plot : (a) 4 X- - y-^ = 36. (c) 16 x^ - y^ + M = 0. (b) 9x2-251/2 = 100. (d) 8a;2-16?/2+ 25 = 0. 2. The distance from an asymptote of a hyperbola to either focus is numerically equal to 5. >j 3. The distance from the center to a line drawn through a focus of a hyperbola perpendicular to an asymptote is numerically equal to a. J 4. The product of the distances from the asymptotes to any point on the hyperbola is constant. ,•" 5. The focal radius of a point P^ (x^, y^) on the parabola 2/2 — 2px is 2+^1- ' 6. The ordinates of points on an ellipse and the major auxiliary circle which have the same abscissas are in the ratio of 6 : a. f 7. The area of an ellipse is irab. Hint. Divide the major axis into equal parts. With these as bases inscribe rectangles in the ellipse and major auxiliary circle (p. 164). Apply Problem 6 and increase the number of rectangles indefinitely. 69. The examples of Problem 2, p. 1G9, illustrated the fact that any equation of the second degree lacking an ajy-term, but containing a-'-^ and ?/'^ with coefficients of unlike signs, can by translation of the axes be transformed into the form (9) Ax^ + Btj'^ = a, in which .4 and B differ in sign. From the preceding it is clear that the locus of this equation is a hyperbola if C is not zero, and a pair of intersecting lines if C is zero. Hence the Theorem. If an equation of the second degree contains no xy-tevni, and If x^ and if occur ivlth coefficients differing In sign, the locus is either a hyperbola or a pair of intersecting lines. 176 NEW ANALYTIC GEOMETRY 70. Locus of any equation of the second degree. The locus problems of this chapter have led to the equations of the sec- ond degree, (1) y = 2^;»cc and x^ = '^py^ (2) iP'x^ + aSf = d-J}^ and V'x^ — a?y- = a^V^. These are simple types, of course. The question is, however, this : Given an equation of the second degree, can the equation he transfoi'med by translating and rotating the axes so that the transformed equation will reduce to one of these simple types? To answer this question, take the general equation of the sec- ond degree, namely, (3) Ax^ + Bxy -}- Cf + Dx + Ey + F = 0. This equation contains every term that can appear in an equation of the second degree. We begin by rotating the axes through an angle 6. To do this, set in (3), ic = cc' cos d — y' sin 0, and y = ^' sin & + y' cos 6. This gives, after squaring, multiplying, and collecting, the transformed equation (4) A cos"-' d x''^-2 A sin ^ cos ^ x'y' + As\n^d + 5sin(9cos^ +5(cos-^— sin-6') — ^sin^cos^ y'^ + C sin" ^ + 2 C sin ^ cos 6 + C cos- + Dcos6 x' — DsinO y' + F = 0. + Esm6 -\-Ecose The angle $ is, as yet, any angle at all. But let us now, if possible, choose this angle so that the equation (4) shall not contain the x'y'-teTva.. To do this, we must set the coefficient of x'y' equal to zero ; that is, (5) -2 A sin ecos6 + B (cos^^ - sin^^) + 2 C sin ^ cos ^ = 0. But 2 sin 6 cos 6 — sin 2 0, cos'^O — sin'-^ = cos 2 6. PARABOLA, ELLIPSE, AND HYPERBOLA 177 Hence (5) becomes (6) (C - ^) sin 2 ^ + i^ cos 2 6 = 0. Dividing through by cos 2 6, and transposing, (7) tan2^ = j4-^- Since any number may be the tangent of an angle, it is always possible to find a value for 6 from this equation. If, then, the axes are rotated through the angle determined by (7), equation (3) reduces to (8) A 'a;'2 4- C'lf^ + D'x' + E'y' + F = 0, where from (4), (9) A' = A cos'B + B sin ^ cos ^ + C sin'^^, (10) C'= A &\n^d — B sin^ cos^ + C cos^B. The discussion gives the Theorem. The term in xy may alivays he removed from an equation of the second degree, Ao-} + Bxy -\- Cy"^ -[- Dx -{- Ey -[- F = 0, by rotating the axes through an angle 6 such that (VI) ^"2^ = 73^- Now equation (8) is of a form which we have met frequently in this chapter, and we have learned to simplify it by transla- tion of the axes. We saw in Art. 61 that if only one square (yl'= 0, or C' = 0) and the first power of the other coordinate were present, the equation could be transformed into one of the typical forms (1) of the parabola. Suppose, however, that the first power of the other coordi- nate does not appear. For example, suppose in (8) that /I ' = and /)' = 0. Then the equation is 178 NEW ANALYTIC GEOMETRY This is an ordinary quadratic in y. If tlie roots are real, the locus will be two lines parallel to the cc'-axis. These lines will coincide if the roots are equal. There will be no locus if the roots are imaginary. If neither A^ nor C is zero, we may, by translation to the new origin { — 7-— > — -^-—^^ \ transform the equation into \ Z A Z C I (11) A'x'"" + C'y"'^ + F' = 0. The locus of this equation has been discussed in Arts. 64 and 69. The result we have established is expressed in the Theorem. The locus of an equation of the second degree is either a parabola, an ellipse, a hyperbola, two straight lines (which may coincide), or a point. The following conclusion also may be drawn : The presence of the xy-term indicates that the axes of the curve are not paral- lel to the axes of coordinates. We seek now a test to apply to an equation containing an icy-term in order to decide in advance the nature of the locus. To do this we eliminate the angle from equations (9) and (10), making use of (6). The result is the simple equation, (VII) -4:A'C'=B^-^AC. The steps in the ehmination process are as follows : Adding and subtracting (9) and (10), (12) A'+ C' = A + C (since sin2^ + cos2^ = 1). (13) A'- C'={A- C)cos20 + Bsm2e. Squaring (13), (14) ( j^'_ C')2 = {A- Cf cos2 2 6" + 2 B (J- - C) sin 2 ^i cos 2 (9 + 52 sin2 2 0. Squaring (6), (16) = (4- C)2sin22(9+ 2 B(C - J.)sin2(9cos2^ + -K^cos22^. Adding (14) and (15), (16) {A' - C'f = {A- C)2 + B2. PARABOLA, ELLIPSE, AND HYPERBOLA 179 Squaring (12), (17) {A' + cy = {A + Cf. Subtracting (16) and (17), we obtain (VII). If the locus of (8) is a parabola, .4 ' = or C = 0. Hence from (VII), £'-4^C = 0. If the locus of (8) is an ellipse, A' and C agree in sign. Hence ^'C is positive, and from (VII), ii^— 4: AC is negative. If the locus of (8) is a hyperbola, A' and C" differ in sign. Hence A'C is negative, and from (VII), E'— ^.AC is a posi- tive number. Collecting all the results in tabular form, we have the Theorem. Given any equation of the second degree, Ax^ + Bxy + Cif -\- Dx + El/ -i- F = 0. The possible loci may he classified thus: Test General case Exceptional cases "■ zero parabola two parallel lines one line B2_4^C negative ellipse point-ellipse m-AAc positive hyperbola two intersecting lines A point-ellipse is often called a " degenerate ellipse," two intersecting lines a " degenerate hyperbola," and two parallel lines a " degenerate parabola." Note that B'^ — A: AC is the discriminant of the terms of the second degree in the equation. *For tests to distinguisli the exceptional cases, see Smith and Gale's " Ele- ments of Analytic Geometry," p. 277. t This case is recognizable by inspection, for the terms of the second degree, Ax^ + Bxy + Cy'^, now will form a perfect square. 180 XEW AXALYTIC GEOMETRY The exceptional cases are recognizable by the condition that the equation is then factorable into two factors of the first degree in x and y. A number of problems of this kind were given on page 41. When the equation is not readily factored by trial, it may appear by the first method of the following section (Art. 71) that factors do nevertheless exist. Moreover, under the two lirst cases in the table (parabola and ellipse) there may be no locus. This fact will also readily appear by the first method of Art. 71. 71. Plotting the locus of an equation of the second degree. In this section we discuss methods of plotting second-degree equations which contain a:-?/-terms. First Method. Bij direct jjlotting. Test by the theorem at the end of the preceding section, and then plot the equation directly. EXAMPLES 1. Plot the locus of (1) x^ — 2xy + -iy^ — Ax = 0. Solution. Here ^=1, ii=-2, C = 4. .•. ii- — 4^(7 = 4 — 10 = — 12 = a negative number. Hence the locus is an ellipse. Y,- X y 1 J -f 0,4 3 ± Vfj 1 4 2 ^ -1 (2) Solve the equation for x as follows : [Collecting terms in x and completing the square.] ^3) .-. X = 2/ + 2 ±V(2-2/){2 + 3?/). PARABOLA, ELLIPSE, AND HYPERBOLA 181 Solving also for y, (4) y = \x± iVa:(16-3x). From the radicals in (3) and (4) we see that (see p. 49) y may have values from — | to 2 inclusive ; X may have values from to -*/ inclusive. Hence the ellipse lies within the rectangle y=-h y = ^^ x = o, X = if-. Points on the locus may be found from (3) as in the table. 2. Determine the locus of 5 x- + 4 xy — ?/■- + 24 X — 6 ?/ — 5 = 0. Solution. ^ = 5, 5 = 4, G'=-l. .-. F^ - 4^C=: 16 + 20 = 36. Hence, from the table of Art. 70, we may expect a hyperbola or a pair of intersecting lines. Solve the equation for y as follows : y^ -(4x - 6)2/ + (2x - 3)" = 5x- + 24x - 5 + (2x - 3)^ = 9x2 + 12x + 4 =(3x + 2)-2. [Collecting terms in y and completing the square.] ... 2/_(2x-.3) = ±(3x + 2). Hence the locus is the intersecting lines y = bx — 1 and y =— x— b. PROBLEMS 1. Test and plot the following equations : (a) x^ — 2xy + 2/^ — 5x = 0. (c) 4,xy + 4?/^ + 4?/ + 4 = 0. (b) 4x?/ + 4y2_2x + 3 = 0. (d) 2x2 + 4x2/ + 4?/'-2 + 2x - 3 = 0. (e (f (g {^ (i (J (k (1 (m; (o x2 + 2x2/ + 2^2^2x + 22/-l = 0. 3x2-12x2/ + 92/2 + 8X-12 2/ + 5 = 0, 5x2 — 12x^ + 92/2 + 8x-12 2/ + 3 = 0. x2 + X2/ + 2/2 + 3 2/ = 0. x2 + 2xy + 42/2 + 6^ = 0. 4x2 + 4x2/ + ?/ + 6x - 9 = 0. 3x2 — 2x2/ + 2/'^-4x-6 = 0. x2 — 2 xy + 5 2/2 - 8 2/ = 0. x2 — 4x2/ + 42/2 + 4x + 22/ = 0. 3x2 + 4x2/ + y2 — 2x — 1 = 0. 3x2 + 8x2/+4 2/2 + 2x + 42/ = 0. 182 NEW ANALYTIC GEOMETRY Second Method. By transformation. If the icy-terra is lacking, we have seen that the equation may be simplified by translating the axes. The transformed equation is then readily plotted on the new axes. When the xy-texrci is present, rotate the axes through the angle d given by (VI), (5) tan2^ = j4^. The term in xy will then disappear and further simplification is accomplished by translation. ' To rotate, we substitute (6) X = x' cos — ?/'sin $, y = x' sin -{- y' cos 0. We find sin 6 and cos 6 as follows. First compute cos 26 from (7) cos 2 ^ = ± —=!=■ (26 and 28, p. 3) ^ ^ Vl + tan-2^ ^ From (5), 2 ^ must lie in the first or second quadrant, so the sign in (7) must be the same as in (5). 6 will then be acute ; and from 40, p. 4, we have /I -cos 2^ . /I + cos 2 ^ (8) sine=+yj , cos^ = + ^^^^ EXAMPLES 1. Construct and discuss the locus of (9) x2 + 4x?/ + 4?/2 + 12x— 6?/ = 0. Solution. Here ^ = 1, B = i, C = 4. .-. B'^ — i AC = 0, and the locus is a parabola. "Write the equation (9) in the form (10) (x + 2?/)2 + 12x-6?/ = 0. We rotate the axes through an angle 0, such that 4 4 tan2^ = = 1-4 3 Then by (7), and by (8), (11) PARABOLA, ELLIPSE, AND HYPERBOLA cos 2 ^ = - I, 183 sin^ --= and cos ( V5 1 The equations for rotating tlie axes are therefore y = 2x' + y' Substituting in tlie equation (10), we obtain 6 V5 0. r| fx' \, l,D \, / \ s. "^ \ / ^ ■v. \ h "^ ^ > J / ^ u r __^ ^ f- \ X /j \ \ k. / ^& Hence the locus is a parabola 3 for which 'p — , and whose V5 focus is on the ?/'-axis. The figure shows both sets of axes, the parabola, its focus and directrix. The axis OX' lias the slope tan = = 2, from (11). Hence to draw 0X\ simply draw a line through the origin whose slope equals 2. In the new coordinates the focus is the point / 0, — ^ ) and the ,...,,,., 3 \ 2V5/ directrix is the line ?/ = =• 2V5 2. Construct the locus of 5x2 _|. 6xy + 5?/2 + 22x — 6?/ + 21 =0. Solution. Here .4 = 5, B = 6, C = 5. .-. 52 _ 4 j^(7 = 36 _ IQO = _ 64 = a negative number. Hence the locus is an ellipse. We rotate the axes through the angle ^, given by tan 2 (9 5-5 .-. 2^ = 90°, (9* = 46°. Hence the equations of the transformation are __ x' —v' x' + y' V2 y V2 *■' n 4 = C, the angle d always equals 45°. 184 NEW ANALYTIC GEOMETRY 1^' ^'t ^'' N / -i- \ \ / \ ^ / / \ / I \ \ X Nj \ A\ \ < A \\ \ / y V \ ^ / / X \) X / \ Substituting in the given equation and reducing, 4x'2 + ?/'2 + 4 V2x'- 7 Vi/ + -2/ = 0. Translating to the new origin (— |\/2, |V2), the final equation is 4x"-^ + ?/"2 = 16. Hence the locus is an ellipse whose major axis is 8, whose minor axis is 4, -j^ and whose foci are on the F"-axis. The figure shows the three sets of axes and the ellipse. The coordinates of the new origin 0'(— |V2, fVi) refer to the axes OX' and 0Y\ and this must be remembered in plotting. The equation (12) Bxij + Dx-^Ey + F =^, in which a;^ and if' are lacking, offers an exception to the above process, for, by translation, the equation may be reduced to (13) Bx'y' + F'=0; and the locus of (13) is, by (V), Art. 67, an equilateral hyper- bola referred to its asymptotes as axes. Hence to plot (12), translate so that the terms of the first degree disappear and then plot the new equation. To show that (12) may be transformed into (13) by translation, proceed thus : Substitute x = x' + /i, ?/ = ?/' + i, in (12), multiply out and collect the terms. We obtain (14) Choose the new origin (A, k) so that the coefficient of x' vanishes ; that is, Bk + D = 0, coefficient of y' vanishes ; that is, Bh + E = 0. E D — , k= , and (14) reduces to the B B Bx'y' + Bk x' + Bh y' + Bhk + B + E + Dh + Ek + F Solving these equations, h = — form (13). PARABOLA, ELLIPSE, AND HYPERBOLA 185 PROBLEMS 1. Simplify the following equations and construct the loci. Check the figure by finding the intercepts on the original axes. V, (a) x^ + xy + 2j" = 3. (b) X2 + 3XT/ + ?/ + 4y = o. (c) x'' + 2xy + 7/' + Sx - Sy = 0. (d) 3x2 -4xv/ + 8x-l = 0. (^ (e) 4x2 + 4x1/ + 2/'^ + 8x — 16?/ = 0. (f) 3xy + 4x + 6y + l = 0. (g) 17x2-12x?/ + 8?/2-68x + 24?/-12 Ans. 3x'2 + ?/'2:=6. Ans, 25x"2-52/"2 + 32^0, Ans. 2x'2-3V2?/ = 0. Ans. x"2_4 2/"2 + l = 0. >. Ans. 5x'"-8Vly' = 0. Ans. 3xy- 7 = 0. !/-12 = 0- Ans. x"2 + 4?/'2-l6 = 0, Ans. ?/'2 + 6x'= 0. Ans. 6xy + 11 = 0. Ans. 4x"2-92/"2 = 36. 12 = 0. Ans. 52 ?/'2 _ 49 = 0. E/ + 43 = 0. (h) 2/2 + 6x- 6?/ + 21 = 0. (i) 6xi/ + 4X-12 2/ + 3 = 0. ( j ) 12x2/ - -5'/ + 48 ?/ - 30 = 0. (k) 4x2-12x?/ + 9?/2 + 2x-3?/ (1) 12x2 + 8x2/ + 18 2/2 + 48x + 16t/ + 43 Ans. 4x2 + 2 2/2 = 1. (m) 7 x2 + 50 X2/ + 7 2/2 = 50. Ans. 16 x'2 - 9 ?/2 = 25. (n) x2 + 3x2/ -32/2 + 6x = 0. ^ns. 21 x"2 _ 49 2/"2 = 72. I (o) 16x2-24x2/ + 9?/2-60x- 802/ + 400 = 0. J.ns. 2/"^ — 4x" = 0. 2. Show that the general equation ^x2 + Bxy + Cy^ + Dx + Ey + F=0 may be simplified by translation only, so that the new equation contains no terms of the first degree in x and y, if the coordinates of the new origin {h, k) satisfy the equations 2Ah + Bk + D = 0, Bh + 2 Ck + E = 0. Ilence show that the new origin {h, k) is the center of the locus, unless B- — 4AC = 0. In the latter case the transformation fails. 72. Conic sections. Historically, the parabola, ellipse, and hyperbola were discovered as plane sections of a right circular cone. Hence the generic term used for them, — conic sections or conies. A definition often used, which will include all conic sections, is the following : When a point P moves so that its distances 186 NEW ANALYTIC GEOMETRY from a given fixed point and a given fixed line ewe in a constant ratio, the locus is a conic. The given fixed line is called the directrix, the fixed point the focus, and the number representing the ratio of the dis- tances of P from the focus and directrix is called the eccen- tricity. In Problem 3, p. 51, we found the equation for any conic to be (1) (1 - e") x^ ^if-^-px^f^ 0, if e is the eccentricity, YY^ is the directrix, and (p, 0) is the focus. Now (1) has no .T//-term. Hence we see at once by ■comparison with our previous results that a conic is a parabola tvhen e = 1, an ellipse when e <. 1, a hyperbola when e > 1. Clearly, when e = 1 the definition of the conic agrees with that already given for the parabola. The ellipse and hyperbola, each having a center, are called central conies. Focus and eccentricity, as used in this section, agree with these terms as already introduced. This fact is left to the student to prove in the following problems. The equation of a conic in polar co- ordinates is readily found. We may show that if the pole is the focus and the polar axis the principal axis of a conic section, then the polar equation of the conic is _ ep D E T(p,e) y 9/ /' /\e II p' F M J I) (2) e cos 6 ' where e is the eccentricity and p is the distance from the •directrix to the focus. PARABOLA, ELLIPSE, AND HYPERBOLA 187 For let P be any point on the conic. Then, by definition, FP Ef = '- From the figure, FP = p, and EP = HM ^p -}- pcosO. Substituting these values of FP and EP, we have — ^ = .; . J) + p COS or, solving for p, p = z ;;• Q.E. D. '^ 1 — ecos^ PROBLEMS 1. Simplify (1), p. 186, by translation of the axes when e ^zi 1. 22 Ans. (1 - e2) x2 + 2/2 - ^^ l-e2 2. Show that in a central conic the focus coincides with the focus already adopted. Hence show that a central conic has two directrices, one associated by the above definition with each focus. 3. Prove that e in Problem 1 agrees with e as defined in Arts. 62 and 65. 4. Prove that the focal radii of appoint {x, y) on the ellipse (HI), p. 161, are u + ex and a — ex. 5. Prove that the focal radii of a point on the hyperbola (IV), p. 167, are ex— a and ex + a. LOCUS PROBLEMS It is expected that the locus in each problem will be constructed and discussed after its equation is found. • 1. The base of a triangle is fixed in length and position. Find the locus of the opposite vertex if (a) the sum of the other sides is constant. Ans. An ellipse. (b) the difference of the other sides is constant. Ans. A hyperbola. (c) one base angle is double the other. Ans. A hyperbola. (d) the sum of the base angles is constant. Ans. A circle. (e) the difference of the base angles is constant. Ans. A conic. ( f ) the product of the tangents of the base angles is constant. Ans. A conic. 188 NEW ANALYTIC GEOMETRY (g) the product of the other sides is equal to the square of half the base. ^"^- A lemniscate (Ex. 2, p. 122). (h) the median to one of the other sides is constant. Am. A circle. 2. Find the locus of a point the sum of the squares of whose distances from (a) the sides of a square, (b) the vertices of a square, is constant. Ans. A circle in each case. 3. Find the locus of a point such that the ratio of its distance from a lixed point Pj (Xj, y^) to its distance from a given line Ax + By + C = is equal to a constant k. Ans. {A-^ + £2 _ k^A'~) x^-2 k^ABxy + {A' + B^- k^B^) y^ - 2 {AH^ + B^x^ + k^A C)x-2 {A'^y^ + B^y^ + k^BC) y + (xf + y{) {A^ + J32) _ k^C' = 0. 4. Find the locus of a point such that the ratio of the square of its distance from a fixed line to its distance from a fixed point equals a constant k. Ans. x^ — k^{x — p)^ — k^y^ = if the ^/-axis is the fixed line and the X-axis passes through the fixed point, p being the distance from the line to the point. Systems of conies. When an equation of the second degree contains one arbitrary constant, the locus is a system of conies. PARABOLA, ELLIPSE, AND HYPERBOLA 189 EXAMPLE Discuss the system represented by ■ -\ = 1. ^o — tc y — fc Solution. When k<9 the locus is an ellipse whose foci are (± c, 0), where c- = (25 — A;) — (9 — A;) = 16. When 9 <.A: < 2.5 the locus is an hyperbola whose foci are (± c, 0), where c^ = (25 — k) — (9 — fc) = 16. When k > 25 there is no locus. Since the ellipses and hyperbolas have the same foci (± 4, 0), they are called confocal. In the figure the locus is plotted for fc =— 56, — 24, 0, 7, 9, 11, 16, 21, 24, 25. As k increases and approaches 9, the ellipses flatten out and finally degenerate into the x-axis, and as fc decreases and approaches 9, the hyper- bolas flatten out and degenerate into the x-axis. As fc increases and approaches 25, the two branches of the hyperbolas lie closer to the 2/-axis, and in the limit they coincide with the y-nxis. PROBLEMS 1. Plot the following systems of conies and show that the conies of each system belong to the same type. Draw enough conies so that the degenerate conies of the system appear as limiting cases. (a)^ + '^' = fc. (c)^-^' = fc. ^ ^ 16 9 ^ ' 16 9 (b) 2/2 = 2 fcx. (d) x^ = 2 fcy - 6. 2. Plot the following systems of conies and show that all of the conies of each system are confocal. Discuss degenerate cases and show that two conies of each system pass through every point in the plane. , , x^ y- , , . x^ "2 (a) TTT— 7 + l^TT-^ = 1- (0 TT. 7 + 16 -fc .36 -fc ^ ' 64-fc 16 -fc (b) y^ = 2kx + fc2. (d) x^=2ky + fc2. 3. Plot and discuss the systems : (a) 16 (X - fc)2 + 9 y2 = 144. (c) {y - fc)2 = 4 x. (b) xy = fc. (d) 4 (X - fc)2 - 9 {y - kf = 36. 4. Plot the following systems and discuss the locus as fc approaches zero and infinity : (X-fc)2 ^2^ (^_fc)2_^^ fc-2 36 ^ ' fc2 36 CHAPTER XI TANGENTS -^ 73. Equation of the tangent. A tangent to a curve at a point P^ is obtained as follows. Take a second point P^ on the curve near P^. Draw the secant through Pj and P^. Now let P^ move along the curve toward P^. The secant will turn around P^. The limiting position of the secant when P^ reaches P^ is called the tangent at P^. We wish to calculate the slope of the tangent at a point on a curve. Let the coordinates of P^ be (x^, y^ and of P^ (a-^ + li, y^ + k). Then ^ ^ . „ _ yt slo2Je of secant P^P^ h To find the slope of the tangent, we begin by finding a value for - > the slope of the secant, as in the following example. EXAMPLE Find the slope of the tangent to the curve C : Sy = x^a,t any point P.^(oc^, 7/j) on C (see figure on "page 191). Solution. Let P^(x^, y^ and Pjx.^+h, y^+k) be two points on C. Then since these coordinates must satisfy the equation of C, (2) 8 2/x = ^T, and 8(y^ + A-)=(a-^ + /0'; or (3) Sij^ + 8k = xl + 3 x^h + 3 xjv" + h\ 190 TANGENTS 191 Subtracting (2) from (3), we obtain Sk = h(3x^ + 3a;jA + A'"); k _Sx^ + 3 xji + /«-" h~ 8 Factoring, and hence = slope of secant P^P^. Now as P^ approaches P^, h and k approach zero, and when the secant becomes a tangent to the curve, h and k are both equal to zero. Hence the slope m of the tangent at P^ will be obtained from the above value of the slope of the secant, namely, S-Tj- + 3 xji + Ji- ■ ~8 ' and also k = 0, if k appeared in the expres- 3.rf m = —z — A ns. by setting h sion. Hence The method employed in this example is general and may be formulated in the following Rule to determine the slojye of the tangent to a curve C at a point P^ on C. First step. Let P^(x^, y^ and Pjx^ -\- h, y^ + k) he two points on C. Substitute their coordinates in the eqtiation of C and subtract. Second step. Find a value for — j the -^ h slope of the secant through P^ and P^. Third step. Find the limiting value of the result of the second step lohen h and k approach zero. This value is the required slope. Having found the slope of the tangent at P^, its equation is found at once by the point-slope formula. The point /'^ is called the point of contact. 192 ^EW ANALYTIC GEOMETRY EXAMPLE Find the equation of the tangect to the circle x2 + ?/2 = r^ at the point of contact (x^, y^). Solution. Let Pj (Xj, y^) and P„ {x-^ + h, y^ + k) be two points on the circle C. Then these coordinates must satisfy the x equation of the circle. Therefore (1) x'l + 2/2 = r2, and (x^ + hy- + (?/i + k)'^ = r^ ; or (2) cc 2 + 2 x^/i + /i2 + 2/2 + 2 z/^A: + fc'^ = f^. Subtracting (1) from (2), we have 2x.^h + Jfi + 2y^k + k^ = 0. Transposing and factoring, this becomes k{2y^ + k)=-h{2x^ + h). k 2x, + h Whence - = — — -* h 2y^ + k = slope of the secant through Pj and P^. Letting P„ approach P^, h and k approach zero, so that m, the slope of the tangent at Pj, is ^1 m = -• Vi The equation of the tangent at P^ is then 2/ - 2/i = - — (-c - a^i), or XjX + 2/,?/ = xf + 2/2. This equation may be simplified. For by (1), so that the required equation is x^x + y{y = r^. Q- E- 1*- TANGENTS 193 Theorem. The equation of the tangent to the circle C : X- -{- If — r^ at the point of contact P.^(x^, y^ is (I) XiX + y^y = r^. The point to be observed in this proof is this : Always simplifj the equation of the tangent by making use of the equation obtained when x^ and y^ are substituted for x and y in the equation of the given curve. In the equation (I) the point of contact is {x^, y^, while (ic, y) is any point on the tangent. In like manner we may prove the following Theorem. The equation of the tangent at the point of contact Pi (•«!» Vi) io the elliqjse iV + a^'if = aHP- is b^x^x + a^y^^y = a^t^ ; hr/perbola V^x^ — ahf = crlr' is b^x^x — a'^y^^y = a^ti^ ; parahola 'if-=2px is y^y = p(x-\-x^. PROBLEMS 1. Find the equations of the tangent to each of the following curves at the point of contact (x^, y-^ -. (a)x2 = 2/)y. Ans. x^x-p{y + y^). 1/ (b) x^ + y^ — 2rx. Ans. x-^x + y{y = r(x + x^). ]/(c) ?/2 _ 4a; ^. 3, Ans. y^y = 2x + 2x^ + S. (d) xy = a^. Ans. x-^y + y-^x = 2 a~. (e) x^ + xy = 4. Ans. 2XjX + x^y + y^x = 8. (i) x^ + y^ + Dx + Ey + F=0. ^^ ^ Ans. XjX + y^y + -{x + x^) + -{y + y^) + F = 0. ■' {g)y = ^^- -^'^- 3xf X — 2/ + 2?/i = 0. (h) y^^x^. (i) y = -4x2 + Bx+ C. (m) xy^ + a = 0. ( j ) Ax^ + By^ + Cx = 0. (n) x^y + b = 0. (k) Ax^ + By^ = 0. (o) xy^ + a^x - a% = 0. l\) Axy + Bx + Cy = 0. (p) y-{2a- x) = x^. 194 NEW ANALYTIC GEOMETRY 74. Taking next any equation of the second degree, we may prove the Theorem. The equation of the tangent to the locus of Ax^ + Bxij + Cif + Dx + Ey + F = at the point of contact P^(x^, y^ is Proof. Let P^ (j-^, y^ and P^ (a;^ + 7i, y^ + k) be two points on the conic. Then (1 ) ^xf + Bx^y^ + Cy^ + Dx^ + %i + F = and A{x^ + hf ■\-B (x^ + h) (y^ + ^-) + C {y^ + A:)2 + Z» (x^ + A) + £(l/i + A:) + ^=0. Clearing of parentheses, (2) Ax^ + 2^x^A + 4/i2 ^ J5xi?/^ + Bx^fc + By^h + BM + C?/i- + 2 Cj/^fc + Cfc2 + i)Xj + Z»A + £'?/^ 4- £'4 + F = 0. Subtracting (1) from (2), (3) 2 ^Xjft + Ah- + Bx^fc + By^h + B/ifc + 2 Cy^k + CA:2 + DA + ^-fc = 0. Transposing all the terms containing A, and factoring, (3) becomes k{Bx^ + 2Cy^-\- Ck+ E)=-h(2Ax^ + Ah + By^ + Bfc + D) ; k 2 Ax, + By, + D + Ah + Bk whence -= h Bx^ + 2Cy^ + E+ Ck This is the slope of the secant P^Po- Letting P., approach Pj, h and k will approach zero and the slope of the tangent is 2Ax, + By, + B m= Bx^ + 2 C2/1 + £ The equation of the tangent line is then 2 Ax, + By, + D, y-y, = ^-^ — ^^LZ — (x — x,). •"' Bx^ + 2Cy^ + E^ '' To reduce this equation to the required form we first clear of fractions and transpose. This gives {2AXj^ + By^ + B)x+ {Bx^ + 2Cy^ + E) y - (2 Axl + 2 Bx^y^ + 2 Cy^ + Dx^+ Ey^) = 0. TANGENTS 195 But from (1) the last parenthesis in this equation equals -{Dx^-\-Ey^ + 2F). Substituting, the equation of the tangent line is {2Ax^ + By^ + D)x + (Bx^ + 2 Cy^ + E)y + {Dx^ + Ey^ + 2F) = 0. Removing the parentheses, collecting the coefficients of ^, B, C, D, E, and F, and dividing by (2), we obtain the eqi;ation of the theorem. Q. E. D. The above result enables us to write down the equation of the tangent to the locus of any equation of the second degree. For by comparing the equation of the ciu've and the equation of the tangent we obtain the following Rule to irrlte the equation of the tangent at the point of con- tact P^(x^, ?/j) to the locus of an equation of the second degree. ?/ IT ~\~ K* II Sid)stitute x^x and y^j for x^ and y'^, — — - — — for xy, and — — and — TT'^ Jo>' •"' «^« y i^n ine given equation. For example, the equation of the tangent at the point of contact (x^, ^,) to the conic x- + 3 x?/ — 4 2/ + 5 = is x^x 4- \ (x^2/ + 2/iX) _!(?/ + 2/^) + 5 = ; or, also, (2 x j + 3 y,) x + (3 Xj - 4) ?/ - 4 j/j + 10 = 0. 75. Equation of the normal. The normal to a curve at a point P^ is the line drawn through P^ perpendicular to the tangent at P . When the equation of the tangent has been found, we may find at once the e(iuatiou of the normal in the manner of Chapi- ter IV. Thus, using the equations of. the tangents given on page 193, we find the Theorem. The equation of the normal at P^(x^,y^ to the ellipse b'^x^ + «"y = a%'^ is c^y^x — H^x^y — (d^ — b"^) x^y^ ; hyperbola Ip-x^ — a'^if = a-lr is d^y^x -f- b^x^y =(0^ -\-b^) x^^y^ ; parabola y^ = 2 px is y^x -\- py — x^y^ -}- py^. 196 NEW ANALYTIC GEOMETRY For example, for the ellipse : The slope of the tangent A IS m = B b% Hence the equation of the normal is 2/ - 2/1 = ~ (-c - Xi), b-x. and this reduces to the equation in the tli^rem. In numerical examples the student should use the Rule given to write down the equation of the tangent, find the normal as a perpendicular line, and not use the special formulas. 76. Subtangent and subnormal. If the tangent and normal at P^ intersect the .r-axis in T and X respectively, then we define P^ T = length of tangent at P^, Y' (1) P,N = leny-th of normal at P, The projections on A'A'' of P^Tand P^N are called respectively the sub- tangent and s\(hnorvial at 1^^. That is, in the figure, ]\I^T = suhtangmt at P^, M^N ^ subnonnal at P,. (2) The subtangent and subnormal are readily found when the equations of the tangent and normal are known. For, from the figure. (3) and M^T = 0T OM,. M^N = ON — OM^ OM, = x„ while OT and ON are respectively the intercepts on A'A' of the tangent and normal at 7*^^. Since the subtangent and sub- normal are measured in opposite directions from the foot of the ordinate M^P^, they will have opposite signs. TANGENTS 197 EXAMPLE Find the equations of tangent and normal, and the lengths of subtan- gent and subnormal at the point on the parabola x^ = 4 y whose abscissa equals 3 Solution. The point of contact (x,, y.) is The formula for the tagigent at (Xj, 2/i) is, by the Rule, p. 195, XjX = 2{y + 2/j). Substituting the values of a:^ and ?/^, 3x = 2(i/ + I) or 6x — 4?/- 9 = 0. This is the required equation of the tangent. The slope of this line is |. Hence the equation of normal at (3, |) is 2/- |=- t(-c-3), or 8X + 122/-51 = 0, TJie intercept on A' A"' of the tangent Is f ; of the normal \i. Also x^ = 3. .-. subtangent = | — 3 = — |, and subnormal = y — 3 = --f-. The lengths of the tangents and normals may be found by geometry, for the lengths of the legs of the triangles P^M^ T and P^M^N are now known. PROBLEMS 1. Find the equations of the tangent and normal at the point indicated to each of the following. Find also the lengths of subtangent and subnor- mal. Draw a figure in each case. (a) 2x2 + 3 2/2 = 35^ x^ = 2, y^ positive.* Ans. Tangent, 4 x + 9 ?/ = 35 ; normal, 9 x — 4 y = 6. Subtangent = \'- ; subnormal =— |. (b) x2 — 4 ?/2 + 15 = 0, x^ =: 1, y^ negative, (c) 2/2 = 4x + 3, yy-2. (d) xy = 4, Xj = 2. (e) x2-f-2/2-4x-3 = 0, Xi = 3. (f) x2 + 4?/2 + 5x = 0,2/1 = 1. (g) 4x2 + 3y2 = 1 ; positive extremity of latus rectum. * Substituting a; = 2 in the given equation, we find y = ± 3. Hence ?/, =+3. 198 NEW ANALYTIC GEOMETRY (h ) x2 + x?/ + 4 = 0, Xj = 2. (i) 2/2 + 2a;2/-3 = 0, 2/j=-l. ( j ) x2 — 3 x?/ — 4 1/2 + 9 = 0, Xj positive, y^ = 2. (k) x2 + xy + ?/2 = 4, x^ = 0, j/j^ negative, ( 1 ) x2 + 42/2 + 4x - 82/ = 0, x^ = 0. (m) 4 2/ = x^, x^ = 2. (n) 4^/2 = x^, x^ = 2. 2. Show that the subtangent in the parabola y^ = 2j3x is bisected at the vertex, and that the subnormal is constant and equals p. ■/ 77. Tangent whose slope is given. Let it be required to find the equation of a tangent to the ellipse (1) 5x^ + 2/^ = 5 whose slope equals 2. Solution. Draw the system of lines whose slope equals 2 (Art. 36). We observe that some of the lines intersect the ellipse in two points, and also that some of them do not inter- sect the ellipse at all. Furthermore, two of them are tangent. We wish to find the equations of these two tangents. The equation of the system of lines whose slope equals 2 is (2) y = 2x + k, where k is an arbitrary parameter. Let us now start to solve for the points of intersection. Substituting from (2) into (1), (3) 5 x' + {2x + kf = 5. Squaring and collecting terms, (4) 9 .T^ -H 4 kx + 7r -5 = 0. If the line (2) is the tangent AB of the figure, by solving equation (4) we shall obtain the abscissa of the point of con- tact. But (4) is a quadratic and has two roots. Hence these roots must be equal. TANGENTS 199 We learn in algebra that the roots of the quadratic (5) Ax- -{-Bx+C = are equal when (6) B''-4:AC = 0. Comparing (4) with (5), A = 9, B = 4:k, C = k''~5. Substituting in (6), (7) 16F-36(P_5^^^0, or A; = ± 3. Hence the equations of the required tangents are AB:y=2x-^3 and CD:y = 2x-Z. Check. AVriting A- = 3 in (4), it becomes 9 ic- + 12 a: + 4 = 0, or (3 a; + 2f = 0. The equation is now a 'perfect square, and this fact consti- tutes the check desired. Hence the equal roots have the com- mon value ic = — §. This is the abscissa of the point of contact /'. The ordinate is found from y = 2 a; + 3 to be ^ = |. Hence J' is (- f , I). Similarly, putting A- = — 3 in (4), we find Q to be (§, — §). The method followed in the preceding may be thus outlined. To find the equation of the tangent to a conic when the slope of the tangent is given. 1. Write down the equation of the system of lines with the given slope (y = mx -f- A:). This equation contains a parameter (7j) whose value must be found. 2. Eliminate x ox y from the equations of the line and conic and arrange the result in the form of a quadratic (8) A^f + By-l^ C = 0, or Ax^ + Bx + C = 0. 200 NEW ANALYTIC GEOMETRY 3. The roots of this quadratic must be equal. Hence set (9) B'^-4.AC = 0, and solve this for the parameter k. 4. Substitute the values of the parameter k in the equation of the system of lines. 5. Check. When each value of the parameter satisfying (9) is substituted in (8), the quadratic becomes a perfect square. PROBLEMS 1. Find the equations of tlie tangents to tlie following conies which satisfy the condition indicated, check, and find the points of contact. Verify by constructing the figure. (a) 7/2 = 4 X, slope = |. Ans. x — 2y + 4: = 0. (b) x2 + ?/2 = 16, slope = - |. Ans. ix + Sy ±20 = 0. (c) 9 x2 + 16 2/2 = 144, slope = - i. Ans. x + 4 ?/ ± 4 VlO = 0. (d) X- — 4 ?/2 = 36, perpendicular to6x — 4?/ + 9 = 0. Ans. 2x + 3y ±3\^ = 0. _„(e) x2 + 2y2_x + ?/ = 0, slope = — 1. Ans. x + ?/ = l, 2x + 2?/ + 1 = 0. (f ) xy + y'^ — 4 X + 8 y = 0, parallel to2x — 4y = 7. Ans. x = 2y,x — 2y + 48 = 0. (g) x^ + 2xy + y^ + 8x — Gy = 0, slope =: |. Ans. ix — Sy = 0. (h) x2 -f 2x?/ — 4x + 2?/ = 0, slope = 2. Ans. y = 2x,2x — y + 10=0. (i ) 2 x2 + 3 2/2 = 35, slope = |. ( 1 ) 2/2 + 4 x - 9 = 0, slope = - 1. ( j ) x2 + 2/^ = 2.5, slope = — |. (m) x^ — y^ = 16, slope = f . (k) x^ + 4 2/ — 8 = 0, slope = 2. ( n ) xy — 4 = 0, slope = — |. 78. Formulas for tangents when the slope is given. For later reference we collect in this section formulas giving the equa- tions of tangents to the conies in terms of the slope m of the tangent. The student should derive these formulas, following the method of the preceding section. Theorem. The equation of a tangent in terms of Its .flops m to the / circle x^ -\- y^ = r^ is y = mx±r Vl + m'' / ellipse ly^x^ + ahj'^ = cc^lp- is y=mx± ^a^m^ -\- b"^ ; hyperbola IP-x^ — ahf — a%^ is y = mxd: ^a^m^ — b^ ; P parabola y = 2px is y = mx -\- TANGENTS 201 79. Properties of tangents and normals to conies. If we draw the tangent AB and the normal CD at any point P^ on the ellipse, and if we draw also the focal radii P^F and P^F', we may prove the property : The tanjent (ind normal to an cUlpse bisect respectively the exter- nal and internal angles formed hi/ the focal radii of the point of contact. Proof In the figure we wish to l)rove 6= cf). To do this we find tan (f> and tan by (VI), Art. 35. The slopes of the lines joining P^ (x^, y^ on the ellipse to the foci F' {c, 0) and F (- c, 0) are slope of F'P^ = -'^ ; slope of FP^ = x^-\- c The equation of the tangent AB is (Theorem, Art. 73) b'^x^x + a'^y^y = a'b^. Now tan = b^x slope of AB — ~ • J where m^ = slope of AB, m,^ = slope ofP^F'. 1 + ^1^2 Substituting the above values of the slopes, tan 6 bh'^ JLl «7/i X I ~ c = - b^x f + Wcx^ - « V 1 U% '/l - ^''^^Ji ahj i(-^i — ') {"'iff + b-'x^) - - b^cx^ aVi - - (a' - U )^-iZ/, 202 NEW ANALYTIC GEOMETRY But since I\ lies on the ellipse, a '■yl -\- U'Xi = a%'^ and also a^ — V^ = c^. a%^ — J?cx. Iria^ — cx^ V' Hence tan 6 = —„ ;; = , ., ~ = — • a'cij^ — G-x^y^ cy^itr—cx^ cy^ In like manner, _ — Irx^ — Jrcxy — ci^yf _ (li^x^ + ft^yf ) + b'^cx-^ — «"^',3/i — «%i + ^'^iV^ «%i + («^ — ^^) x^y^ ahy^ + c\yi cy^ Hence tan = tan <^ ; and since and are both less than -TT, = . That is, AB bisects the external angle of FP^ and F'P^, and hence, also, CD bisects the internal angle. Q. E. D. An obvious application of this theorem is to the problem : To draw a tangent and normal at a g'loen point P^ on an ellipse. This is accomplished hy connecting P^ with the foci and bisecting the internal and external angles formed by these lines. The phenomenon observed in " whispering galleries " depends upon this property; namely, let the elliptic arc A' PA be a vertical section of such a gallery. The waves of sound from a person's voice at the focus F will, after meeting the ceiling of the gallery, be reflected in the direction F'. For if PN is the normal at P, angle NPF = angle NPF', and the law of reflection of sound waves is precisely that the angles of incidence (= Z. NPF) and reflection (= Z. XPF') are equal. Hence sound waves emanating from F in all directions will converge at F'. A whisper at F, which would not carry over the distance FF', might consequently, through reflection, be audible at F'. TANGENTS 203 In like manner we prove the following properties : The tangent and normal to a hyperbola bisect respectively the internal and external angles formed by the focal radii of the point of contact. The tangent and normal to a parabola bisect respectively the internal and external angles formed by the focal radius of the point of contact and the line through that point parallel to the cuxis. These theorems give rules for constructing the tangent and normal to these conies by means of ruler and compasses. Construction. To construct the tangent and normal to a hyper- bola at any point, join that point to the foci and bisect the angles formed by these lines. To construct the tangent and normal to a parabola at any point, draw lines through it to the focus and parallel to the axis, and bisect the angles formed by these lines. The principle of parabolic reflectors depends upon the prop- erty of tangent and normal just enunciated ; namely, the reflect- ing surface of such a reflector is obtained by revolving a para- bolic arc about its axis. If, now, a light be placed at the focus, the rays of light which meet the surface of the reflector will all be reflected in the direction of the axis of the parabola ; for a ray meeting the surface at P^ in the figure will be reflected in a direction making with the normal PD an angle equal to the angle FP^D. But this direction is, by the above property, parallel to the axis OX of the parabola. 204 NEW ANALYTIC GEOMETRY PROBLEMS 1. Tangents to an ellipse and its major auxiliary circle (p. 164) at points with the same abscissa intersect on the x-axis. 2. The point of contact of a tangent to a hyperbola is midway be- tween the points in which the tangent meets the asymptotes. 3. The foot of the perpendicular from the focus of a parabola to a tangent lies on the tangent at the vertex. 4. The foot of the perpendicular from a focus of an ellipse to a tangent lies on the major auxiliary circle (p. 164). 5. Tangents to a parabola from a point on the directrix are perpen- dicular to each other. , 6. Tangents to a parabola at the extremities of a chord which passes through the focus are perpendicular to each other. 7. The ordinate of the point of intersection of the directrix of a parab- ola and the line through the focus perpendicular to a tangent is the same as that of the point of contact. 8. How may Problem 7 be used to draw a tangent to a parabola ? 9. The line drawn perpendicular to a tangent to a central conic from a focus, and the line passing through the center and the point of contact intersect on the corresponding directrix (Art. 72). 10. The angle which one tangent to a parabola makes with a second is half the angle which the focal radius drawn to the point of contact of the first makes with that drawn to the point of contact of the second. 11. The product of the distances from a tangent to a central conic to the foci is constant. 12. Tangents to any conic at the ends of the latus rectum pass through the intersection of the directrix and principal axis. 13. Tangents to a parabola at the extremities of the latus rectum are perpendicular. 14. The equation of the parabola referred to the tangents in Problem 13 IS x2 - 2 xi/ + 2/2 - 2 y/2p {x + y) + 2p2 - o. Show that this equation has the form x- + y- = vpV2. 15. The area of the triangle formed by a tangent to a hyperbola and the asymptotes is constant. 16. An ellipse and a hyperbola which are confocal intersect at right angles. CHAPTER XII PARAMETRIC EQUATIONS AND LOCI 80. If X and y are rectangular coordinates, and if each is ex- pressed as a function of a variable parameter, as, for example, (1) x = \t\ y = \t\ in whiQh Hs a variable, then these equations are called the jjara- metr'ic equations of the curve, — the locus of (x, y). To plot the curve, give values to t and compute values of x and y, arranging the work in a table. When the computation is finished, plot the points (.r, y) and draw a smooth curve through them. EXAMPLES 1. Plot the curve whose parametric equations are (2) x = \r\ y^itfi- _jYA t X y 1 .5 .25 2 2 2 3 4.5 0.75 etc. etc. etc. - 1 .5 - .25 -2 2 -2 -8 4.5 - (S.75 etc. etc. etc. Solution. The table is easily made. For example, if i = 2, then x = 2, y = 2, etc. The curve is called a semicubical parabola. 205 206 NEW ANALYTIC GEOMETRY 2. Draw the locus of the equa- tions (3) a; = 2rcos^ + rcos2^, ?/ = 2 ?• sin 9 — r sin 2 6, where ^ is a variable parameter. Solution. Take r = 5. Arrange the computation as below : The three-pointed curve thus obtained is called a hypocycloid of three cusps. j:= 10 cos 6-1- 5 cos 2 0, i/= 10 sin 9 — 5 sin 2 6 cosO 20 cos 2 X sin0 sin 2 U» 1 1 15 1 C 30° .86 60° .50 11.1 .50 .86 0.7 60° .50 120° -.50 2.5 .86 .86 4.3 90° 180° -1 -6 1 ^^ 120° -.50 240° -.50 — 7.5 .86 -.86 12.9 150° -.86 300° .50 -6.1 .50 -.86 9.3 180° -1 360° 1 -5 "4 *'? 210° - .86 420° .50 -6.1 -.50 .86 -9.3 240° -.50 480° -.50 -7.5 -.86 .86 -12.9 270° 540° -1 -5 -1 -10 300° .50 600° -.50 2.5 -.86 -.86 ^'^ 330° .86 660° .50 11.1 -.50 - .86 * ^'0.7 360° 1 720° 1 15 To obtain the rectangular equation from the parametric equa- tions, the parameter must be eliminated. The method used depends upon the example. PARAMETRIC EQUATIONS AND LOCI 207 EXAMPLES 1. Find the rectangular equation of the curve whose parametric equa- tions are (4) x = 2t + 3, y = ir--4. Solution. The first equation may be solved readily for t. We find f = |(x — 3), andsubstitutingin the second equation gives J/ = i(x — 3)^—4; or, expanding and simplifying, a;^ — 6x — 8?/ — 23 = 0, a isai'abola. 2. Find the rectangular equation of the curve whose parametric equa- tions are (5) X =: 3 -h 4 cos ^, ?/ = 3 sin 0. Solution. Remembering that sin^ ^+ cos- ^=: 1, we solve the first equa- tion for cos^, the second for sin 0. This gives (6) cos^ = ^(x-3), sin0 = ^y. Hence the rectangular equation is ^'' 16 ^9 ' an ellipse. PROBLEMS 1. Plot the following parametric equations, t and being variable parameters. Find the rectangular equation in each case : "^ (a) X = i - 1, 2/ = 4 — (2. ( i ) X = cos (9, 2/ = cos 2 ^. (b) X = 2 ^2 _ 2, 7/ = i - 3. ( j ) X = 1 sin 0, y = sin 2 0. (c) X = 3 cos 0, y = sin 0. ( k ) x = 1 — cos 0, y = \ sin I 0. (d) X = 3 tan 0, y = sec0. {\ ) x = St^, y = St - t^. ■ (e) x = 2t v = -- (m) x = 2sini9 + 3cos(9, ^/^siiK?. ^' t' (n) x = 2cosi9 + l,|/=sin^ + 4cosi9. (f ) X = 2 + sin 0,y =^2cos0. (o) x = t- t'\ y = t + t^. {g)x = lt^,y = lt. (p)x = 3-2i,z/ = l + l (h) X= «2_2i, ,y = l_<2. ^^ ' ' ^ ^ t \l 2. Plot the following parametric equations : (a) X = 2 r cos — r cos 2 0, ^ y = 2r sin — r sin 2 0. (b) X = 3 r cos + r cos S0, y = 3rs\n0 — r sin 3 0. (c) X = 3 r cos — r cos S0, y = Sr sin ^ — r sin 3 0. (d) X — r cos — r cos 2 0, y = rsin0 — r sin 2 0. (e) X = 2 r cos + ^r cos 2 0, y = 2r sin — \r sin 2 6. 208 NEW ANALYTIC GEOMETRY (f) X = a{6 ■ sm 6'), cos^). . , (x-= a{Q -^ smO), ^^' \y = a{\-cosd). Y ^ a .^c a \ A \/ w^ ^^ a \/ r^^ V A' X D, CUSP AT ORIGIN CYCLOID, VERTEX AT ORIGIN (h) x = aO — ^asin^, y = a — \a cos 6. { i ) x = ae -2a sin 6, y — a — 'ia cos 6. ( j ) X = r cos 9 ■\- r sin 6^ y = r sin 6 — rO cos ^. ( k ) X = 4 r COS ^ — r cos 4 ^, y — ir sin ^ — r sin 4 ff. (1) X = alogi, ?/ = |a/i + -|. (ni) X = f + sini, y = 1 + cos^. { n ) X = 2 cos t + t, y = 3 cos i + sin 2 ^. ( o ) X = b cos'^ 0^ y = a tan ^. V 81. Various parametric equations for the same curve. When the rectangular equation of a curve is given, any number of para- metric equations may be obtained for the curve. For example, given the ellipse (1) 4.T^ + v/-^ = 16. Let a; = 2 cos 6, where ^ is a variable parameter. Substitut- ing in (1), 16 cos^^ + y- = 16, or 2/^=16(l-cos"'^)=16sin-^. Hence the equations (2) a; = 2cos^, ?/ = 4sin^, are parametric equations of the ellipse (1). Again, substitute in (1), y = fx + 4, where t is a. variable parameter. This gives (3) 4 a;2 -K !^V -f 8 At + 16 = 16, or (4 -f- i;')^-'-!- 8 ^a; = 0. (4) ■'^ = -TTf' PARAMETRIC EQUATIONS AXD LOCI 209 Substituting this value in y = fee + 4 and reducing, (5) y 16 - 4 j'^ 4. + f' Hence the equations (4) and (5) are also parametric equa^- tions of the ellipse. The point is : We obtain parametric equations by setting one of the coordinates equal to a function of a 'parameter, substitut- ing in the given rectangular equation and solcing for the other coordinate in terms of the parameter. To obtain simple parametric equations we must, of course, assume the right function for one coordinate. No general rule applicable to all cases can be given for this purpose, but the study of the problems below will aid the student. Many rectangular equations difficult to plot are treated by deriving parametric equations and plotting the latter. EXAMPLES ' 1. Draw the locus of the equation (6) x3 + ?/3 _ 3 ctxy = 0. Solution. Set y = te, where I is the parameter. Then, from (6), (7) a:3 + <3j.3 _ 3 a^a;2 = 0. Dividina: out the x^, solving for x, and remembering that y = te, we obtain the desired parametric equations 3 at 3 ai^ (^) 1 + i^ y = 1 + The locus is the curve of the figure, called the folium of Descartes. The line drawn in the figure is an oblique asymptote. Its equation is X 4- y + a = 0. The parameter t in (7) is obviously the slope of the line y is, of the line joining a point on the curve and the origin. tx ; that A 210 NEW ANALYTIC GEOMETRY The reason for assuming the relation y = tx in the preceding example is that x'- divides out in (7), leaving an equation of the first degree to solve for x. Problems 1 (a), (d), (e), (f),and (j) below are worked on the same principle. In many cases trigo- nometric functions are employed with advantage, as in (b) and (c). PROBLEMS 1. Find parametric equations for each of the following curves by making the substitution indicated in the given equation. The parameter is t or 0, as the case may be. Plot the locus. \\ (a) 2/2 = 4 a;2 _ x3, ?/ = tx. Ans. x = 4: — t^,y=:4:t — t^. (b) x^y^ = b-x'^ + a^ y'^, x = a sec 6. Ans. y = b esc d. I (c) x^y^ = d-y- — b-x-, x — asm 6. Ans. y = 6tan^. y (d) y^ -2 ax-— x^, y — tx. ^ " X (h) x^ + y^ = a^,x — a co^O. PARABOLA \ (e) 2/2 (2 a — x) = x^, y — tx. CISSOID OF DIOCLES (f) 2/2 = x2|+-5,2/ = te. 2 — X 2 <2 _ 2 2 i3 _ 2 « Ans. X — 1 y = • (g) x2 + xy + 22/2 + 2x + 1=0, X =■ tV — 1. ' i22 , 2 + ?- 1 - ( i ) x*+ yi = a^,x = a sin^ 6. Ans. X = — , y = — • ^ ' t1 J^ t + 2 t- -{■ t -ir 2 HYPOCYCLOID OF FOUR CUSPS„ PARAMETRIC EQUATIONS AND LOCI 211 ( j ) a;* + 2 ax^y — ay^ = , y = tx. (k) (a;2 + y2 + 4ay- a-) {x^ - a^) + 4 a"y^ = 0, x^ = a^ - i-y-. (1) x'' = y{y-2f,y-2 = tx. (in) ( X2 - I 62)2 ^ y2 (a.2 _ ^2) = 0, X2 = 1 ^2 ^. ty. 82. Locus problems solved by parametric equations. Parametric equations are important because it is sometimes easy in locus problems to express the coordinates of a point on the locus in terms of a parameter, when it is otherwise difficult to obtain the equation of the locus. The following examples illustrate this statement : EXAMPLES 1. ABP is a rigid line. The points A and B move along two perpen- dicular intersecting lines. What is the locus of the point P on ^iJ ? In the figure, A moves on XX', B moves on YY' ; required the locus of the point P(x, y). Solution. Take the coordinate axes as indi- cated, and consider the line in any one of its positions. Choose for parameter the angle XAB = e. Let AP -a, PB = b. Now OM = X, MP = y. In the right triangle MPA, . n MP y sni 6 = = - • PA a In the right triangle BSP, Z. PBS = 6. .-. cos PBS = cos — = - • BP b From (1) and (2), (3) X = b cos 0, y = a sin 0. These are the parametric equations of the locus. Squaring (1) and (2) and adding, (1) (2) 62 '^ a^ = 1. Hence the point P moves on an ellipse whose axes 2 a and 2 b lie along the given perpendicular lines. A method commonly employed for drawing ellipses depends upon this result. The instrument consists of two grooved perpendicular bars X'X 212 NEW ANALYTIC GEOMETRY and YY' and a crossbar ABP. At A and B are screw nuts fitting the grooves and adjustable along ABP. If the crossbar is moved, a pencil at P will describe an ellipse whose semiaxes are PA and PB. © 2. The cycloid. Find the parametric equations of the locus of a point P on a circle which rolls along the axis of x. Solution. Take for origin a point at which the moving point P touched the axis of x. Let the circle drawn be any position of the rolling circle. Let a be the radius of the circle and take for the variable param- eter 6 the variable angle CBP. Then PC = a sin 0, CB = a cos 0. By definition, OA — arc AP = ad. [For an arc of a circle equals its radius times the subtended angle, from the definition of a radian.] Hence from the figure, if (x, y) are the coordinates of P, X = OD = OA - PC = ad - a sin 6, y = BP = AB - CB = a - a cos (9. (4) (X a{9 — sin 6), \^y = a{l — cosO). These are the parametric equations of the cycloid. The cycloid extends indefinitely to the right and left and consists of arcs equal to OMN. 3/ Construction of the cy- cloid. The definition of the cycloid suggests the follow- ing simple construction : Lay off 0N=2Tra ^ cir- cumference of the generat- ing circle. Draw the latter touching at C, the middle point of ON. Divide OC into any number of equal parts, and the semicircle C3f into PARAMETRIC EQUATIONS AND LOCI " 213 the same number of equal arcs. Letter as in the figure. Through M\, M„, etc., draw lines parallel to ON. Lay off M^Dj^ = CC^, M^D^ = CCg, M^D^ = CC^, etc. Then D^, D„, Dg, etc. are points on the cycloid. For, let the generating circle roll to the left, the point M tracing the curve. When the circle touches ON at C^, M will lie on a level with jVj, and at a distance to the left of ilf^ equal to CC-^^. Similarly for D^., Dg, etc. The arc MN of the cycloid may be constructed by using CM as an axis of symmetry. n\ 3. The hypocycloid of four cusps. Find the parametric equations of the locus of a point P on a circle which rolls on the inside of a fixed circle of four times the radius. Y Solution. Take the center of the fixed circle for the origin and let the X-axis pass through a point A where the tracing point P touched the large circle. Then OA = 4 C2>, by hypothesis. _,„ OA a ^ CB = = - . Draw 4 4 the rolling circle in any of its positions. Take for the variable parameter the Z A OB. Then Z BCP = 4 ^. 214 NEW ANALYTIC GEOMETRY [For, by hypothesis, arc PB — arc A B ; and, from the definition of a radian, arc P5 = - Z-BCP, arc ^5= a^. .-. - Z BCP=aO,orZBCP=4d.] But ZOCE+ ZECP + ZPCB = -jr. .: --0 + ZECP + 4e = 7r. 2 Whence Z ECP = --S0. Now OF = x, FP =: y. From the figure, (5) 0F= OE + DP, FP = EC - CD. Finding the lengths of the segments in the right-hand members, OE = OCcos0 = — cos 0, EC = OC sin = — sin ff. 4 4 DC= CP cos (^^-3(9') = - sin 3 (9, . (by 31, p. 3) J)P= CP sin/- -3 (9") = -cos 3^. (by 31, p. 3) Substituting in (5), f X = f a cos + ia cos 3 d, ly = iashi — ianinS 6. These are parametric equations for the hypocycloid of four cusps. Anotlier form of (6) from wliich tlie rectangular equation may easily be derived is obtained by expressing cos 3 and sin 3 in terms of cos and sin respectively. Thus, cos 3 (9 = cos (2 ^ + ^) = cos 2 (9 cos ^ - sin 2 (9 sin (by 35, p. 3) = (2 cos2 ^ — 1) cos ^ — 2 sin2 cos = 2 cos3 ^ _ cos^ - 2 (1 - cos'^ 0) cos = 4 cos^^— 3 cos^. sin 30 = sin {2 + 0) = sin 2 6' cos ^ + cos 2 ^ sin (by 33, p. 3) = 2 sin cos2(9 + (1 — 2 sin2(9) sin = 2 sin (9 (1 - sin^ ^) + sin ^ - 2 sin3 = 3sin(9-4sin3^. Substituting in (6) and reducing, the result is (7) X = aGOs^0, y = asm^0. From these, x^ = aJ cos^^, ?/3" — a^ sin-^. Adding, (8) x^ + y^ = a*, which is the rectangular equation of the hypocycloid of four CMsps PARAMETRIC EQUATIONS AND LOCI 215 PROBLEMS In the following problems express x and y in terms of the parameter and the lengths of the given lines of the figure. Sketch the locus. 1. Find the parametric equations of the ellipse, using as parameter the eccentric angle 0, that is, the angle between the major axis and the radius of the point B on the major auxiliary circle (p. 164) which has the same abscissa as the point P (x, y) on the ellipse. (See figure.) Ans. X = a cos ^,y = b sin ^. Y "^^""^--.^ 1 1 \± 11 ^ ;mm^x 2, In the figure, ABP is a rigid equilateral triangle. A moves on YY', B moves on XX'. Find the locus of the vertex P. Ans. x = acos(9+ a cos (120°- ^), ?/ =: a sin(120°- (?). Ellipse, x^ — V3 xy ■\-y'^—\a^. 3. Two vertices^ and B of a rigid isosceles right triangle ABP move on perpendicular lines. Find the locus of the vertex P. Ans. x = acosd+ a sin 0, y = a cos 0. Ellipse, X'—2xy + 2y^= a^. i. AB is a, fixed line and R a fixed point. Draw RQ to any point Q in AB and erect the perpendicular QP, making QP -i- QR equal to a con- stant e. What is the locus of P ? ^2 ^,2 Ans. x-pcotO,y = epcscff. Hyperbola,^ -^ = — 1. 216 NEW ANALYTIC GEOMETRY 5. AB is a fixed line and a fixed point. Througli draw OX parallel to AB and ON perpendicular to AB. Draw a line from through any point Q in AB. MSrk on this line a point P such that MP - NQ, MP being J. to OX. What is the locus of P? Ans. X — a cot^^, y = a cot^. Parabola, y^ = ax. li?(a,6) Ans. 'i, ' 6. Through the fixed point U (a, 6) lines are drawn meeting the coordinate axes in A and B. What is the locus of the middle point ot AB? Ans. 2x = a , 2y = h — at, where t = slope of AB. Equilateral hyperbola, (2 x 7. Find the locus of a point Q on the radius BP (Fig., Ex. 2, p. 212) itBQ = b. fx = aO — b sin 9, [^y = a — b cos^. The locus is called a prolate or cur- tate cycloid according as fl|y^ greater or less than bk Describe a •construction for the curve analogous to that given for the cycloid in Art. 82. 8. Given a string wrapped around a circle ; find the locus of the end of the string as it is unwound. Hint. Take the center of the cir- cle for origin and let the .T-axis pass through the point A at which tlie end of the string rests. If the string is im- wound to a point B, let ZAOIJ:=d. (See figure.) _.,.,., fx = rcos^ -t- r^sio^, Ans. The involute of a circle -{ . ^ „ ^ (_ij = r sin ff — rff cos 0, o PARAMETRIC EQUATIONS AND LOCI 217 9. A circle of radius r rolls on the inside of. a circle whose radius is r'. Find the locus of a point on the rolling circle. Ans. The hypocycloid "V '^'^'^^M ^• r Y ('•' r) cos 6 -\- r cos 6, r t' — T y — {r' — r) sin 6 — r sin 6. The curve is closed when r and r' are commensurable. The hypocycloid of four cusps, p. 213, is a special case. Describe a construction for the curve analogous to that given for the cycloid in Art. 82. rr> 10. A circle of radius r rolls on the outside of a circle whose radius- is r'. Find the locus of a point on the rolling circle. Ans. The epicycloid r' 4- r Kc = (r' + r) cos 9 — r cos I y = (r' + r) sin ^ — r sin r •/ + r e. The curve is closed when r and r' are commensurable. Describe a construction for the curve analogous to that given for the cycloid in Art. 82. 11. Given a fixed point on a fixed circle and a .fixied line AB, Draw the X-axis through perpendicular to AB and the ?/-axis through O parallel to AB. Draw any line through to meet AB m L and the fixed circle in S. Draw LP II to OX to meet SM drawn II to OY. Re- quired the locus of P. Ans. X = 6 cos^ ^, ?/ = a tan ^. Cubic, xy^+ a^x — a% = 0.' Give a full discussion of the equation. Show that the y-axis is an asymptote. What modifications, if any, are necessary in the equations when yl J5 is a tangent ? when AB does not intersect the circle ? 218 NEW ANALYTIC GEOMETRY J 12. OB is the crank of an engine and AB the connecting rod. B moves on the crank circle whose center is 0, and A moves on the fixed line OX. What is the locus of any point P on AB ? A ns. x = bcos0 + Vr2 — (a + Oy^ sin'-^6', y = a sin 0. Ellipse, when r = a + b; other- wise an egg-shaped curve. 13. OB is an engine crank re- volving about O, and A Bis the con- necting rod, the point A moving on OX. Draw AP ± to OX to meet OB produced at P.* What is the loc us of P ? Ans. X = r cos ^-|- Vc'-^ — r'-'sin'"^, y = r sin + tan OVc- — r^ sin"-^ When c = r, the locus is the circle a;2 -f 7/2- 4^2 83. Loci derived by a construction from a given curve. Many important loci are defined as the locus of a point obtained by a given construction from a given curve. The method of treatment of such loci is illustrated in the follow- ing examples. EXAMPLES 1. Find the locus of the middle points of the chords of the circle X- + y" = 25 which pass through P2(3,4). Solution. Let P^(Xi, ?/j) be any point on the circle. (1) ••• x2 -1- 2/f = 25. Then a point P (x, y) on the locus is obtained by bisecting PjP,. By (IV), Art. 13, .-. Xi = 2a:-3, v^^2y-4. * P \s the " instantaneous center " of the motion of the conuet;ting rod. rA k ^ •^ ;^ ^ ^!-p,J ; / / / / ' y \ / 1 / V 'h •y, A' (; — ^ X y 1 \ xr, n) \ ^ I ^ ^ ^ ^ ~1 — — — y — — — — — , — PARAMETRIC EQUATIONS AND LOCI 219 Substituting in (1). ■ (2x- 3)2 + (22/ -4)2 = 25, or x^ + y^ — S X — 4 y =^ 0. Ayis. The locus is a circle constructed upon OP^ as a diameter. 2. The witch. Find the equation of the locus of a point P constructed as follows : Let OA be a diameter of the circle X' + ij'^ — 2 ay = 0, and let any line OB be drawn through to meet the circle at P^ anil the tangent at A at B. Draw PjP ± to OA and BP II to OA. Required tlie locus of P. Solution. Let (x, y) be the coordinates of P and (Xj, y^) of Pj. Then the coordinates of P^ (a;^, ?/^) must satisfy the equation a;2 j^ yi — 2ay — 0. (2) .-. x2 + 2/2- 2 02/1 = 0. From the figure, (3) y, = y. From the similar triangles OCP^ and 0MB we have OC GP, X, = -1 or — OM MB X [For OC = Xj, OJVf : A B yf^ ">> A yT " ^^ -- 1 ^[ V "^nlT" --^K' h (4) 2a CP^ = y^, MB = 2a.li Solving (3) and (4) for Xj and t/j, we obtain (5) "^^'S' ^»"^" Substituting from (5) in (2), 2ay = 0, ^ y n 4a^ or (6) ?/(x2 + 4a2) = 8a3. The locus of this equation is known as the witch of Agnesi. The method followed in Examples 1 and 2 may evidently be described as follows : Rule fo7' finding the equation of a locus derived by a construc- tion from, a given curve. 220 NEW ANALYTIC GEOMETRY First step. The construction will give rise to a figure from which we viay find expressions for the coordinates of any point P^(x^, y^ on the given curve in terms of a point P(x, y) on the required curve. Second step. Substitute the results of the first step for the coor- dinates x^ and y^ in the equation of the given curve and sivipdify. The. result is the required equation. PROBLEMS 1. Find tlie locus of a point whose ordinate is half the ordinate of a point on the circle x^ + y- — 64. Arts. The ellipse x^ + 4 2/2 = 64. 2. Find the locus of a point which cuts off a part of an ordinate of the circle x^ -\- y- = a^ whose ratio to the whole ordinate is b : a. Ans. The ellipse b^x'^ + a^y"^ = a^b^. 3. Find the locus of the middle points of the chords of (a) an ellipse, (b) a parabola, (c) a hyperbola which pass through a fixed point P^i^zi ^2) on the curve. Ans. A conic of the same type for which the values of a and 6 or of p are half the values of those constants for the given conic. 4. Lines are drawn from the point (0, 4) to the hyperbola x'^ —4y^ — 16. Find the locus of the points which divide the.se lines in the ratio 1:2. Ans. 3x2-12?/2 + 64 2/-90f = 0. 5. A chord OP^ of the circle x^ + y^-2ax = meets the line x = 2a at a point A. Find the locus of a point P on the line OP^ such that OP = P^A. Ans. The cissoid of Diodes y'^{2a — x) = x^ (see figure). 6. DIX is the directrix and F the focus of a given conic (Art. 72). Q is any point on the conic. Through Q draw QN ± to the axis of the conic and construct P on NQ so that NP = FQ. What is the locus of P ? Ans. A straight line. 84. Loci using polar coordinates. When the required locus is described by the end-point of a line of variable length whose other extremity is fixed, polar coordinates may be employed to advantage. PARAMETRIC EQUATIONS AND LOCI 221 EXAMPLE The conchoid. Find the locus of a point P constructed as follows : Through a fixed point 0, a line is drawn cutting a fixed line ^J5 at Pj. On this line a point P is taken so that P^P — ±^, where 6 is a constant. Solution. The required locus is the locus of the end-point P of the line OP, and is fixed. Hence we use polar coor- dinates, taking for the pole and the perpendicular OM to AB for the polar axis. Then (1) OP = p, ZMOP = d. By construction, (2) p= 0P= 0P^± b. But in the right triangle OMP^, (3) OP^ = OM sec ZMOP^ = a sec 0. Substituting from (3) in (2), (4) p = a sec ±b. The locus of this equation is called the conchoid of Nicomedes. It has three distinct forms according as a is greater, equal to, or less than 6. PROBLEMS 1. OA is a diameter of a fixed circle, and OB is any chord drawn from the fixed point 0. In the figure below, BP = AB. Find the locus of P. Ans. The circle p = a (sin + cos^). 2. The chord OB of a fixed circle drawn from is produced to P, making BP = diameter = a. What is the locus of P ? Ans. The cardioid /o = «(1 -I- cos^).. 222 NEW ANALYTIC GEOMETRY 3. In problem 2, if BP = any length = b, the locus of P is the limapon of Pascal, p = b+ a cos^. The limagon has three distinct forms accord- ing as b = a. In the figure on p. 124, b < a. The rectangular equation is (x2 + 2/2 + aa;)2 = 62 (3.2 + ^z^). 4. F is the focus and DD' the directrix of a conic (figure below). Q is any point on the conic. On the focal radius FQ lay off FP = QM, where QM is II to DD'. Find the locus of P (see Art. 72). Ans. p = ep sin ff 1 — e cos ( ^■"^^^'Cb.^ L C ^^^zKi'y^ X M 5. Lines are drawn from the fixed point on a fixed circle to meet a fixed line LM which is ± to the diameter through O. On any such line OG lay off OP = BC. What is the locus of P? Ans. p = bsecO — acos0. Draw the locus for b>a,bI2 b X -\- a ij = (lb K^liich bisects all chords witli the slope m is b^x-\-a^my = Q. In like manner (see the figures on p. 227) we may prove the Theorem. TJie diameter wliich bisects all chords with the slope VI of the hijperbola lrx~ — a'^ij^ = crlr is IP'X — cp-my = ; jiarabola y'^lpxis myz=.p. Every line through the center of an ellipse or hyperbola is a diameter, while in a f»arabola every line parallel to the axis is a diameter. PROBLEMS 1. Find the equation of the diameter of each of the following conies which bisects the chords with the given slope m. (a) X- — 4 y2 _ 16^ ,^ _ 2. Alls, x — 8 ?/ = 0. . A, (b) y- = 4x, 7/1 = — |. ^ns. y + 4 = 0. /,^(c) xy = 6, m = 3. Ans. 2/ + 3x = 0. (d) x2 + X2/ — 8 = 0, m=— 3. Ans. x — tj - 0. (e) X'^ — 42/2 + 4X — 16 = 0, m = - 1. Ans. x + 4 y + 2 = 0. (f)xy + 2if-4x — 2y + 6 = 0,m=^. Ans. 2x + 1] ?/ — IG = 0. 2. Find the equation of that diameter of (a) 4x2 + 9?/2 = 36 passing through (3, 2). Ans. 2x — 3y = 0. (b) ?/2 = 4x passing through (2, 1). Ans. y = 1. (c) xy = 8 passing through (— 2, 3). Ans. 3x + 2^ — 0. (d) x2 - 4 ?/ + 6 = passing through (3, - 4). Ans. x = 3. (e) x?/— (/2 + 2x — 4 = Opassing through (5. 2). Ans. 4x — 9?/— 2 = 0. PARAMETRIC EQUATIONS AND LOCI 229 3. Find the equation of the chord of the locus of A' (a) x^ + y'^ = 25 which is bisected at the point (2, 1). Ans. 2x + 2/ — 5 = 0. (b) 4x-^ — (/"^ = 9 which is bisected at the point (4, 2). Ans. 8x — 2/ — 30 = 0. (c) x;/ = 4 which is bisected at the point (5, 3). Ayis. 3x + 5?/ — 30=0. (d) X- — xy — 8 = which is bisected at the point (4, 0). » Ans. 2x — 2/ — 8 = 0. 4. Show that if two lines thi'ough the center of the ellipse 6-x2 + ay- = a%" have slopes m and ?n' such that mm' = -, then each line bisects all chords parallel to the other. " Draw two such lines. They are called conjugate diameters. 5. Through the point (Xq, t/q) on the ellipse b-x^ + a'^y- = aW a diam- eter is drawn ; prove that the coordinates of the extremities of its conjugate diameter are x = ± — - , y = :f — L' . b a 6. If a' and b' are the lengths of two conjugate semidiameters of the ellipse, prove that a''^ + b"- = a' + 6- (use Example 5). 7. Prove that the tangent at any point of the ellipse is parallel to the diameter which is conjugate to the diameter through the given point; and hence that the tangents at the extremities of two conjugate diameters form a parallelogram. 8. Pi'ove that the area of the parallelogram formed by the tangents at the extremities of two conjugate diameters of an ellipse is constant and is equal to 4 ah. Hint. The area in question is eight times the area of the triangle whose vertices are (0, 0), (a:o. ?/o). and [^ ' - ~-^) (see Example 5). 9. Two tangents with the slopes m^ and m., are drawn from a point P to an ellipse 6'-x- + a-y- = a-h"^. Find the locus of P (a) when m^ + m.^ = 0. Ans. x = and y = 0. \y' '(b) when m^ + m._, = 1. Ans. x^ — 2xy — a- = 0. (c) when m^m.^ = 1. Ans. x^ — y^ — a^ — b^. CHAPTER XIII CARTESIAN COORDINATES IN SPACE 86. Cartesian coordinates. The foundation of plane analytic geometry depends upon the possibility of determining a point in the plane by a pair of real numbers (x, y). The study of solid analytic geometry is based on the determination of a point in space by a set of three real numbers x, y, and z. This determination is accomplished as follows : Let there be given three mutually perpendicular planes intersecting in the lines A'A'', YY', and ZZ', which will also be mutually perpendicular. These three planes are called the coordinate planes and may be distinguished as the A'F-plane, the yZ-plane, and the ZA'-plane. Their lines of intersection are called the axes of coordinates, and the positive directions on them are indicated by the arrowheads.* The point of inter- section of the coordinate planes is called the origin. Let P be any point in space and let three planes be drawn through P parallel to the coordinate planes and cutting the axes at A, B, and C. These three planes together with the * A'A^and ZZ' are supposed to be in the plane of the paper, the positive direction on XX' being to tlie right, tliat on ZZ' being upioard. YY' is sup- posed to be perpendicular to the plane of the paper, the positive direction be- ing in front of the paper, that is, from the plane of the paper toward the reader 230 CARTESIAN COORDINATES IN SPACE 231 coordinate planes form a rectangular parallelepiped, of which P and the origin O are opposite vertices, as in the figure. The three edges OA = x, OB = y, and OC = z are called the rectangular coordinates of P. Any point P in space determines three numbers, the coordi- nates of P. Conversely, given any three real numbers x, y, and z, a point P in space may always be constructed whose coordinates are x, y, and z. For if we lay off OA = x, OB = y, and OC = z, and draw planes through A, B, and C parallel to the coordinate planes, they will intersect in a point P. Hence Every 2^0 int determines three real numbers, and conversely, three real numbers determine a point. The coordinates of P are written (x, y, z), and the symbol P (x, y, z) is to be read, " The point P whose coordinates are X, y, and ^." From the figure we have the relations AP ^ OS = ■V(0By -\-(ocy', BP=OR^ ■V{OC)--h(OAy ; CP = OQ = V((9J)- + (0£/; OP = V(a4 y + (pBf + {ocf. Hence, letP (cc, y, z) be any point in space; then its distance from the A' F-plane is z, from the FZ-plane is x, from the Z^J^-plane is y, from the A-axis is Vy^ + z^, from the F-axis is V,?- + x^, from the Z-axis is Vx-'^ -|- //-, from the origin is Va;^ + y"^ + «^- 232 NEW ANALYTIC GEOMETRY The coordinate planes divide all space into eight parts called octants, designated by 0-XYZ, 0-X'YZ, etc. The signs of the coordinates of a point in any octant may be determined by the Rule for signs. X is positive or negative accord- ing as P lies to the right or left of the YZ-plane. y is positive or negative accord- ing as P lies in front or in back of the ZX-pIane. z is positive or negative accord- ing as P lies above or belotv the X Y-plane. Points in space may be con- veniently plotted by marking the same scale on A'A'' and ZZ' and a somewhat smaller scale on YY'. Then to plot any point, for example (7, 6, 10), we lay off OA = 7 on OX, draw AQ jjarallel to Y and equal to 6 units on O Y, and QP parallel to OZ and equal to 10 units on OZ. PROBLEMS 1. What are the coordinates of the origin ? 2. Plot the following sets of points : (a) (8, 0,2), (-3, 4, 7), (0,0, 5). (b) (4, - 3, 6), (- 4, 6, 0), (0, 8, 0). (c) (10, 3, - 4), (- 4, 0, 0), (0, 8, 4). (cl) (3, - 4, - 8), (- 5, - 6, 4). (8, 6, 0). (e) (-4, -8, -0), (3, 0,7), (6, -4,2). (f) (-6,4, - 4), (0, - 4, 6), (9, 7, -2). 3. Calculate the distances of each of the following points to each of the coordinate planes and axes and to the origin : (a) (2, - 2, 1), (b) (3, - 4, - .3), (c) /-i, - 1, ^ 4. Show that the following points lie on a sphere whose center is the origin and whose radius is 3 : (VS, - 2, V2), (2 a/2. 0,-1), (-2, 2, 1), (- Vs, Vs, 1). CARTESIAN COORDINATES IN SPACE 233 5. Show that the following points He on a circular cylinder of radius 5 whose, axis is the F-axis : (3, - 8, 4), (2 VI, 6, V5), (- 4, 0, - 3), (1, J, 2 V6). 6. Where can a point move ifx = 0? ii y = 0? it z = 0? 7. Where can a point move if a; = and y = 0? ii y = and 2 = 0? if z = and x = ? 8. Show that the points (x, y, z) and (— x, ?/, z) are symmetrical with respect to the FZ-plane ; (x, y, z) and (x, — y, z) with respect to the ZX- plane ; (x, y, z) and (x, y, — z) with respect to the A''Y'-plane. 9. Show that the points (x, y, z) and (— x, — y, z) are symmetrical with respect to ZZ'; (x, y, z) and (x, — y, — z) with respect to XX'; (x, y, z) and (— X, ?/, — z) with respect to YY'; (x, ?/, z) and (— x, — ?/, — z) with respect to the origin. 10. What is the value of z if P (x, y, z) is in the XY-plane ? of x if P is in the FZ-plane ? oi y ii P is in the ZX-plane ? 11. What are the values of y and z if P (x, ?/, z) is on the X-axis ? of z and X if P is on the F-axis ? of x and y ii P is on the Z-axis ? 12. A rectangular parallelepiped lies in the octant 0-XYZ with three faces in the coordinate planes. If its dimensions are a, 6, and c, what are the coordinates of its vertices ? 87. Orthogonal projections. To extend the first theorem of projection, Art. 31, we define the angle between two directed lines in space which do not intersect to be the angle between two intersecting directed lines drawn parallel to the given lines and having their positive directions agreeing with those of the given lines. The definitions of the orthogonal projection of a point upon a line and of a directed length AB upon a directed line hold when the points and lines lie in space instead of in the plane. It is evident that the projection of a point upon a line may also be regarded as the point of intersection of the line and the plane passed through the point perpendicular to the line. As two parallel planes are equidistant, then the projections of a directed length A B upon two parallel lines whose positive direc- tions agree are equal. 234 NEW ANALYTIC GEOMETRY A J/ /. i> / ^'1 / y c' A^^ \-'1f^ 1 ^p' c 1 ,D X / Q / First Theorem of Projection. If A and B are points ■upon a directed line making an angle of y with a directed line CD, then the (I) projection of the length AB upon CD = AB cos y. P7vof. Draw CD' through A parallel to CD. Then, by defi- Bition, the angle between AB and CD' equals y. Since CD' and AB intersect we may apply the first theorem of projection in the plane, and hence the projection of the length AB upon C'D' = AB cos y. Since the projection of siB on CD equals the projection of AB upon CD' we get (I). q.e.d. Second Theorem of Projection. If each segment of a broken line in space he given the direction determined vn pcbssing continuously from one extremXtij to the other, then the algebraic sum, of the projections of the segments iipjon any directed line equals the projection of the closing line. The proof given on page 69 holds whether the broken line lies in the plane or in space. Corollary I. The projections of the line joining the origin to any point P on the axes of coordinates are respect ioely the coordi- nates of P. For the projection of OP (Fig., p. 230) upon OX equals OA, since A is the })rojection of P on OX. Similarly for the pro- jections on OY and OZ. Corollary II. Given any two points P^ (x^, y^, z^ and P^ (^2' y-v -2)' ^^^en jTj — JTi = projection of P^P^ upon XX', y^— y^= projection of P^Pz upon YY\ z^— z^= projection of P^P^ ^po'^ ^^' • CARTESIAN COORDINATES IN SPACE 235 For if we project P^OP^ and P^P.^ upon A' A'', we have the projection of P^O-'r projection of 0P,^= projection of P^P^. But by Corollary I, projection of P^O — — x^, projection of OP^ = x^. .'. x^ — x^= projection of P-^P.^ upon A A'. In like manner the other formulas are proved. Corollary III. If the sides of a jjolygon he given the direction established by jjassing contimiously around tlie jierimeter, the •sum of the projections of the sides upon any directed line is zero. PROBLEMS 1. Find the projections upon each of the axes of the sides of the tri- angles whose vertices are the following points, and verify the results by Corollary III. ^^^ ^_3^ ^^ _ g^^ ^.^ _ ^^ ^^^ ^g^ ^^ q^_ (b) (- 4, - 8, - G), (3, 0, 7), (6, 4, - 2). (c) (10, 3, -4), (-4, 0,2), (0,8, 4). (d) (- 0, 4, - 4), (0, - 4, G), (9, 7, - 2). 2. If the projections of PjPg ^" ^^^ 2clQ'S, are respectively 3, — 2, and 7,. and if the coordinates of P^ are (— 4, 3, 2), find the coordinates of P„. Ans. (- 1, 1, 9). 3. A broken line joins continuously the points (6, 0, 0), (0, 4, 3), (— 4, 0, 0), and (0, 0, 8). Find the sum of the projections of the segments and the projection of the closing line on (a) the AT-axis, (b) the F-axis,^ (c) the Z-axis, and verify the results. Construct the figure. 4. A broken line joins continuously the points (6, 8, — 3), (0, 0, — 3), (0, 0, 6), (— 8, 0, 2), and (— 8, 4, 0). Find the sum of the projections of the segments and the projection of the closing line on (a) the A'-axis, (b) the Y-axis, (c) the Z-axis, and verify the results. Construct the figure. 5. Find the projections on the axes of the line joining the origin to each of the points in Problem 1. 6. Find the angle between each axis and the line drawn from the origin to ^ g ^ (a) the point (8, 6, 0). Aw^. cos-i - , cos-i - , - • o o 2 (b) the point (2, — 1, — 2). Ans. cos-i-,cos-M ),cos-M )• 236 NEW ANALYTIC GEOMETRY 7. Find two expressions for the projections upon tlie axes of the line drawn from the origin to the point P {x, y, z), if the length of the line is p and the angles between the line and the axes are a, /3, and 7. 8. Find the projections of the coordinates of P (x, ?/, z) upon the line drawn from the origin to P if the angles between that line and the axes are a, /3, and y. Ans. x cos a, y cos/3, 2 cos 7. 88. Direction cosines of a line. The angles a, /3, and y between a directed line and the axes of coordinates are called the direc- tion angles of the line. If the line does not intersect the axes, then a, /3, and y are the angles between the axes and a line drawn through the ori- gin parallel to the given line and agreeing with it in direction. The cosines of the direction angles of a line are called the direction cosines of the line. Reversing the direction of a line changes the signs of the direction cosines of the line. For reversing the direction of a line changes a, /8, and y into TT — a, TT — /?, and tt — y respectively, and (30, p. 3) cos (tt — x) = — cos X. Theorem. If a, (3, and y are the direction angles of a line, then (II) cos'^a + cos'^yff + cos^y = 1. That is, the sum of the squares of the direction cosines of a line is unity. Proof. Let AB he 2i line whose direction angles are a, f3, and y. Through draw OP parallel to A B and let OP = p. By definition Z XOP = a, Z YOP = (3, ZZOP = y. Projecting OP on the axes, (1) X = p cos a, y — p cos /3, z = p cos y. CARTESIAN COORDINATES IN SPACE 237 Projecting OP and OCQP on OP, (2) p = X cos a -\- y cos ^ + z cos y. Substituting from (1) in (2) and dividing by p, we obtain (II). Q.E.D. „ „ ^- COS a cosfl cosy ^, Corollary. If = — — ^ = '- > fAew (III) cos a = — » cos p ± Va^ + &2 _^ ^2 ± Va2 + 62 _|_ c2 c cos y = =r • ± Va^ + &2 + c2 That is, if the direction cosines of a line are proportional to three numbers, they are respectively equal to these numbers each divided by the square root of the sum of their squctres. For if r denotes the common value of tlie given ratios, then (3) cos a = ar, cos /3 = br, cos y = cr. Squaring, adding, and applying (II), l=7^(«2+Z,--^-}-c^). 1 ± Va'^ + b'^ + c^ Substituting in (3), we get the values of cos a, cos ^8, and cos y to be derived. The important conclusion just derived may be thus stated : Any three numbers a, b, and c determine the direction of a line in space. This direction is the same as that of the line joining the origin and the point (a, b, c). If a line cuts the XF-plane, it will be directed upward or downward according as cos y is positive or negative. If a line is parallel to the XF-plane, cos 7 = 0, and it will be directed in front or in back of the ZX-plane according as cos /3 is posjiiye or negative. If a line is parallel to the A'-axis, cos /S = cos 7=0, and its positive direction will agree or disagree with that of the X-axis according as cos a = 1 or — 1. 238 NEW ANALYTIC GEOMETRY These considerations enable us to choose the sign of the radical in the Corollary so that the positive direction on the line shall be that given in advance. 89. Lengths. Theorem. The length I of the line joining two points i\ {^v Vv ^i) ^'^^ ^2 (^2' y-v ^2) ^ ^"^^^ ^y (IV) / = V(ar, - x^f + (y, - y^Y + (z, - z^\ Proof. Let the direction angles of the line P^P,^ ^® ^' A ^-'^^ y- Projecting PjP^ on the axes, we get, by the first theorem of projection and Corollary II, p. 234, (1) I cos a = a-., — x^, I cos ^ = y_^ — y^, I cos y = z,^ — z^ Squaring and adding, ^^(cos^ a + cos- ^ + cos- y) = (.T, - a^j)2 + (_y.^ — 7/^)2 + (,-<^ _ z^^ = {^x - ^■^' + O/i - Vo)' + ('^'i - -2)'- Applying (II), and taking the square root, we have (IV). Q. E.D. Corollary. The direction cosines of the line drawn from P^ to P^ are proportional to the projections of P^P^ on the axes. For, from (1), cos a _ cos fi cos y ^2 - ^1 2/2 - yi -■! since each ratio equals Also .l'"" X the denominators are the pro- jections of PJ^,^ on the axes. If we construct a rectangular parallelepiped by passing planes through P^ and P^ parallel to the coordinate planes, its edges will be parallel to the axes and equal numerically to the projections of P^^ upon the axes. P^^ will be a diagonal of this parallelepiped, and hence V^ will equal the sum of the squares of its three dimensions. We have thus a second method of deriving (IV). CARTESIAN COORDINATES IN SPACE 239 PROBLEMS 1. Find the length and the direction cosines of the line drawn from (a) P, (4, 3, - 2) to P2 (-2, 1, - 5). Ans. 7, - f , - |, - f. (b) P, (4, 7, - 2) to P„ (3, 5, - 4). Ans. 3, - 1, - f, - f. (c) P^ (3, - 8, 6) to Pg (6, - 4, 6). Ans. 5, f, |, 0. 2. Find the direction cosines of a line directed upward if they are proportional to (a) 3, 6, and 2 ; (b) 2, 1, and — 4 ; (c) 1, — 2, and 3. /x362^,^2 1 4 1-2 3 Ans. (a) -.-,-; (b) — , —, — ; (c) — = , -— ^ , —^ • 7 7 7 _ V2I - V21 + V21 V14 Vl4 Vl4 3. Find the lengths and direction cosines of the sides of the triangles whose vertices are the following points ; then find the projections of the sides upon the axes by the first theorem of projection and verify by Corollary III, p. 236. (a) (0,0,3), (4,0,0), (8,0,0). (b) (3, 2, 0), (- 2, 5, 7), (1, - 3, - 5). (c) (-4,0,6), (8,2, -1), (2, 4,6). (d) (3, - 3, - 3), (4, 2, 7), (- 1, - 2, - 5). 4. In what octant {0-XYZ, 0-X'YZ, etc.) will the positive part of a line through lie if (a) cosar>0, cos/3 >0, cos7>0? (e) cos ar<0, cosj8>0, cos7>0 ? (b) cosa>0, cos/3>0, cos7<0? (f) cos a-<0, cosi3<0, cos7>0 ? (c) cosa>0, cos/3<0, cos7<0? (g) cos a<0, cos/3<0, cos7<0 ? (d) cos ct > 0, cos /S < 0, cos 7 > ? (h) cos a < 0, cos /S > 0, cos 7 < ? 5. What is the direction of a line if cos a = 0? cos /3 = 0? cos 7 = 0? cos a = cos p — 0? cos /3 = cos 7 = 0? cos 7 = cos or = ? 6. Find the projection of the line drawn from the origin to P^ (5, — 7, 6) upon a line whose direction cosines are f, — f, and |. Ans. 9. Hint. The projection of OPj on any line equals the projection of a broken line whose segments equal the coordinates of Pj. 7. Find the projection of the line drawn from the origin to P^ (Xj, ?/^, Zj) upon a line whose direction angles are a, /3, and 7. Ans. a;j cos a + ?/j cos/3 4- Zj COS7. 8. Show that the points (- 3, 2, - 7), (2, 2, - 3), and (- 3, 6, - 2) are the vertices of an isosceles triangle. 9. Show that the points (4, 3, - 4), (- 2, 9, - 4), and (- 2, 3, 2) are the vertices of an equilateral triangle. 240 NEW ANALYTIC GECMETRY 10. Show that the points (-4,0,2), (- 1, 3V3, 2), (2,0,2), and (— 1, Vi, 2 + 2 Ve) are the vertices of a regular tetrahedron. 11. What does formula (IV) become if P^ and P^ lie in the XF-plane ? in a plane parallel to the ATF-plane ? 12. Show that the direction cosines of the lines joining each of the points (4, — 8, 6) and (— 2, 4, — 3) to the point (12, — 24, 18) are the same. How are the three points situated ? 13. Show by means of direction cosines that the three points (3, — 2, 7), (6, 4, — 2), and (5, 2, 1) lie on a straight line. 14. What are the direction cosines of a line parallel to the X-axis ? to the F-axis ? to the Z-axis ? 15. What is the value of one of the direction cosines of a line parallel to the XF-plane '? the FZ-plane ? the ZX-plane ? What relation exists between the other two ? 16. Show that the point (— 1, — 2, — 1) is on the line joining the points (4^ _ 7^ 3) and (— 6, 3, — 5) and is equally distant from them. TT TT 17. If two of the direction angles of a line are — and — , what is the third? ^ ^ TT 2 7r A ns. — or — • 3 3 18. Find the direction angles of a line which is equally inclined to the three coordinate axes. Aiis. a = ^ = y = cos-i ^Vs. 19. Find the length of a line whose projections on the axes are respectively (a) 6, — 3, and 2. Ans. 7. (b) 12, 4, and - 3. Ans. 13. (c) —2,-1, and 2. Ans. 3. 90. Angle between two directed lines. Theorem, /fa, (3, y and a', /3', y' are the direction angles of two directed lines, then the angle 6 between them is given bij (V) cos 9 =1 cos a cos a' + cos p cos ^' + cos y cos y'. Proof. Draw OP and OP' (figure, p. 241) parallel to the given lines and let OP = p. Then, by definition, Z POP' = 0. CARTESIAN COORDINATES IN SPACE 241 Now, if the coordinates of P are (x, y, z), then, in the figure, OA—x, AB = y, BP = z. Project OP and OABP on OP'. Then (1) p cos 6 = x cos a' -\- y cos /3' + z cos y'. Projecting OP on the axes, (2) X = p cos a, y = p cos /?, Substituting in (1) from (2) and dividing by p, we ob- tain (V). Q.E.D. Theorem. If a, (3, y and a', /3', y' are the direction angles of two lines, then the lines are (a) parallel and in the same direction* when and only when a = a\ (3 = P', y = y'; (b) perpendicular^ when and only when cos a cos a' -\- cos jB cos /3' -f- cos y cos y' = 0. That is, tivo lines are parallel and in the same direction when and only when their direction angles are equal, and perpen- dicular when and only when the sum of the products of their direction cosines is zero. Proof The condition for parallelism follows from the fact that both lines will be parallel to and agree in direction with the same line through the origin when and only when their direction angles are equal. The condition for perpendicularity follows from (V), for if TT ^ = — , then cos 5 = 0, and conversely q.e.d. Li * They will be parallel and have opposite directions when and only when the direction angles are supplementary. t Two lines in space are said to be perpendicular when the angle between them is -, but the lines do not necessarily intersect. 242 NEW ANALYTIC GEOMETRY In the applications we usually have given not the direction cosines, but three numbers to which they are proportional. Hence the importance of the following Corollary. If the direction cosines of two lines are proportional to a, h, G and a\ h\ c', then^the conditions for pa^'allelism and pteipendicularity are respectively — = — = -;5 aa' + bb' + cc' = 0. a' b c' 91. Point of division. Theorem. The coordinates (x, y, z) of the point of division P on the line joining P.^{x^, y^, z^ and P^{x^, y.^, z,^ such that the ratio of the segvients is p p are given by the formulas (VI) x= ^/ \ y= \ ,% z= ' ' l+A 1+A 1+A This is proved as in Art. 13. Corollary. The coordinates (x, y, z) of the middle point P of the line Joining P^(x^, y^, z^ and P^(x,^, y.„ s;.,) are ^ = |(-^l + J^2). y = 1(^1 +1/2), ^ = 1(^1+^2)- PROBLEMS 1. Find the angle between two lines whose direction cosines are respectively (a) 7, f, - f and f, - f, f. Avs. ^. (b) h-hi and - J-, j% if. Ai,^. cm-^^. (c) t, - I, \ and f, f, f . Am. cos-i(- ^\). 2. Show that the lines whose direction cosines are f, f, f ; — f , f , — 7 ; and — 7, I , f are mutually perpendicular. 3. Show that the lines joining the following pairs of points are either parallel or perpendicular. (a) (3, 2, 7), (1, 4, 6) and (7, - 5, 9), (5, - 3, 8). (b) (13, 4, 9), (1, 7, 13) and (7, 16, - 6), (3, 4, - 9). (c) (- 6, 4, - 3), (1, 2, 7) and (8, - 5, 10), (15, - 7, 20). CARTESIAN COORDINATES IN SPACE 243 4. Find the coordinates of the point dividing the line joining the fol- lowing points in the ratio given. (a) (3, 4, 2), (7, -6,4), \=l. Ans. (V, |, |). (b) (- 1, 4, - 6), (2, 3, - 7), X = - 3. Ans. {|, §, - V^). (c) (8, 4, 2), (3, 9, 6), X = - i. Ans. {\}, i, 0). (d) (7, 3, 9), (2, 1, 2), X = 4. Am. (3, |, V)- 5. Show that the points (7, 3, 4), (1, 0, 6), and (4, 5, — 2) are the ver- tices of a right triangle. 6. Show that the points (- 6, 3, 2), (3, - 2, 4), (5, 7, 3), and (— 13, 17, — 1) are the vertices of a trapezoid. 7. Show that the points (3, 7, 2), (4, 3, 1), (1, 6, 3), and (2, 2, 2) are the vertices of a parallelogram. 8. Show that the points (6, 7, 3), (3, 11, 1), (0, 3, 4), and (- 3, 7, 2) are the vertices of a rectangle. 9. Show that the points (6, — 6, 0), (3, — 4, 4), (2, - 9, 2), and ^— 1, — 7, 6) are the vertices of a rhombus. 10. Sliowthatthepoints(7, 2,4), (4, — 4,2), (9, - 1,10), and (G, -7,8) are the vertices of a square. 11. Show that each of the following sets of points lies on a .straight line, and find the ratio of the segments in which the third divides the line joining the first to the second. (a) (4, 13, 3), (3, 6, 4), and (2, - 1, 5). Ans. - 2. (b) (4, - 5, - 12), (- 2, 4, 6), and (2, - 2, - 6). Ans. \. (c) {- 3, 4, 2), (7, - 2, 6), and (2, 1, 4). Ans. 1. 12. Find the lengths of the medians of the triangle whose vertices are the points (3, 4, - 2), (7, 0, 8), and {- 5, 4, 6). Ans. VllS, V89, 2 V29. 13. Show that the lines joining the middle points of the opposite sides of the quadrilaterals whose vertices are the following points bisect each other. (a) (8, 4, 2), (0, 2, 5), (- 3, 2, 4), and (8, 0, - 6). (b) (0, 0, 9), (2, G, 8), (- 8, 0, 4), and (0, - 8, G). (C) Pj(Xi, V/j, Zj), P.i{X^, 7/2, Zo), P3(X3, l/g, Z.^, P^{.r^, IJ^, Z^). 14. Show that the lines joining successively the middle points of the sides of any quadrilateral form a parallelogram. 16. Find the projection of the line drawn from P^ (3, 2, — 6) to P^ (—3, 6, —4) upon a line directed upward whose direction cosines are proportional to 2. 1, and — 2. Ans. 4i. 244 NEW ANALYTIC GEOMETRY 16. Find the projection of the line drawn from P^ (6, 3, 2) to P^ifi, 2, 0) upon the line drawn. from P^il, — 0, 0) to P^i— 5, — 2, 3). Ans. if. 17. Find the coordinates of the point of intersection of the medians of the triangle whose vertices are (3, 0, — 2), (7, — 4, 3), and (— 1, 4, — 7). Ans. (3, 2, - 2). 18. Find the coordinates of the point of intersection of the medians of the triangle whose vertices are any three points P^, P^, and Pg. Ans. [} {x^ + x^ + Jg), i (j/i + 2/„ + 2/3), 1 {Zy + Z2 + 23)]- 19. The three lines joining the middle points of the o^jposite edges of a tetrahedron pass through the same point and are bisected at that point. 20. The four lines drawn from the vertices of any tetrahedron to the point of intersection of the medians of the opposite face meet in a point which is three fourths of the distance from each vertex to the opposite face (the center of gravity of the tetrahedron). CHAPTER XIV SURFACES, CURVES, AND EQUATIONS 92. Loci in space. In solid geometry it is necessary to con- sider two kinds of loci : 1. The locus of a point in space which satisfies one given con- dition is, in general, a surface. Thus the locus of a point at a given distance from a fixed point is a sphere, and the locus of a point equidistant from two fixed points is the plane which is perpendicular to the line join- ing the given points at its middle point. 2. The locus of a point in space which satisfies two conditions * is, in general, a curve. For the locus of a point which satisfies either condition is a surface, and hence the points which satisfy both conditions lie on two surfaces, that is, on their curve of intersection. Thus the locus of a point which is at a given distance r from a fixed point 7*^ and is equally distant from two fixed points P^ and Pg is the circle in which the sphere whose center is Pj and whose radius is r intersects the plane which is perpendicular to PJ^z at its middle point. These two kinds of loci must be carefully distinguished. 93. Equation of a surface. First fundamental problem. If any point P which lies on a given surface be given the coordinates (ic, y, z), then the condition which defines the surface as a locus will lead to an equation involving the variables x, y, and ,*;. * The number of conditions must be counted carefully. Thus if a point is to be equidistant from three fixed points Pj, P^, and P^, it satisfies two coiuli- tions, namely, of being equidistant from P^ and P2 and from Pj ^^^ ^^a- 245 246 NEW ANALYTIC GEOMETRY The equation of a surface is an equation in the variables x, y, and z representing coordinates such that : 1. The coordinates of every point on the surface will satisfy the equation. 2. Every point whose coordinates satisfy the equation will lie upon the surface. If the surface is defined as the locus of a point satisfying one condition, its equation may be found in many cases by a Rule analogous to that in Art. 17. EXAMPLE Find the equation of the locus of a point whose distance from Pj (3, 0, - 2) is 4. Solution. Let P {x, ?/, z) be any point on the locus. The given con- dition may be written p p _ 4 By (IV), P^P = V(x - 3)2+2/2 + (2 + 2)2. ... V(a; - 3)2 + 2/2 + (. + 2)2 = 4. Simplifying, we obtain as the requii-ed equation x2+2/2+z2-6x+4z-3 = 0. That this is indeed the equation of the locus should be verified as on page 3L PROBLEMS 1. Find the equation of the locus of a point which is (a) 3 units above the A'F-plane. (b) 4 units to the right of the FZ-plane. (c) 5 units below the AT'-plane. (d) 10 units back of the ZA'-plane. (e) 7 units to the left of the FZ-plane. (f) 2 units in front of the ZA'-plane. 2. Find the equation of the plane which is parallel to (a) the A'F-pIane and 4 units above it. (b) the A'l"-plane and 5 units below it. (c) the ZA'-plane and 3 units in front of it. (d) the i'Z-plane and 7 units to the left of it. (e) the ZA'-plane and 2 units back of it. (f ) the FZ-plane and 4 units to the right of It. SURFACES, CURVES, AND EQUATIONS 247 3. What are the equations of the coordinate planes? 4. What is the form of the equation of a plane which is parallel to the AT-plane ? the FZ-plane ? the ZX-plane ? 5. What are the equations of the faces of the rectangular parallele- piped which has one vertex at the origin, three edges lying along the coordinate axes, and one vertex at the point (3, 5, 7) ? 6. Find the equation of the locus of a point whose distance from the point (a) (2, - 2, 1) is 3. (d) (- 2, |, 0) is Vs. (b) (0, 1, - 2) is |. (e) (a, 6, c) is d. (c) (-1,3,1) is v'i. (f) {a, 13, y) is r. 7. Find the efjuation of the sphere whose center is the point (a) (3, 0, 4) and whose radius is 5. Ans. x- + y- + z'^ — Q X — 8 z = 0. (b) (— 3, 2, 1) and whose radius is 4. Ans. x^ + 1/^ + z^ + 6 X — iy — 2 z — 2 = 0. ■ (c) (6, 4, 0) and whose radius is 7. (d) (a, p, y) and whose radius is r. Ans. x'^+ y^+ z^—2a£ — 2^y — 2yz + a^+ l3- + y"—r^ = 0. 8. Find the equation of a sphere (a) having the line joining (3, 0, 7) and (1, —2,-1) for a diameter. (b) of radius 2, which is tangent to all three coordinate planes in the first octant. (c) of radius 3, which is tangent to all three coordinate planes in the third octant. (d) whose center is the point (3, 1, — 2) and which is tangent to the XF-plane. (e) whose center is (6, 2, 3) and which passes through the origin. (f) passing through the four points(-2,0,0),(0,-4,0),(0, 0,4), (8, 0,0). 9. Find the equation of the locus of a point which is equally distant from the points (a) (3, 2, -1) and (4, -3, 0). Ans. 2x- 10?/ + 22 - 11 = 0. (b) (4, - .3, 6) and (2, - 4, 2). Ans. 4x-l-2?/ + 8z-37=0. (c) (1, 3, 2) and (4, -1, 1). Ans. Sx- Ay - z - 2 = 0. (d) (4, - 6, - 8) and (-2, 7, 9). Ans. 6x- 13?/- 17^ + 9 = 0. 10. Find the equation of a plane perpendicular at the niiddle point to the line ioining * (a) (1, - 2, 1) and (2, - 1, 0). (b) (- 3, I, 0) and (0, 0, i). (c) (- 2, 1, i) and (i, 0, 0). 248 NEW ANALYTIC GEOMETRY' 11. Find the equations of the six planes drawn through the middle points of the edges of the tetrahedron whose vertices are the points (5, 4, 0), (2, — 5, — 4), (1, 7, — 5), and (— 4, 3, 4), which are perpendicular to the respective edges, and show that they all pass through the point (-1,1,-2). 12. Find the equation of the locus of a point which is three times as far from the point (2, 6, 8) as from (4, — 2, 4), and determine the nature of the locus by comparison with the answer to Problem 7 (d). 13. Find the equation of the locus of a point the sum of the squares of whose distances from (1, 3, — 2) and (6, — 4, 2) is 50, and determine the nature of the locus by comparison with the answer to Problem 7 (d). 14. Find the equation of the locus of a point whose distance (a) from the A''-axis is 3. (b) from the F-axis is ^. _ (c) from the Z-axis is Vs. 15. i ind the equation of a circular cylinder (a) whose axis is the F-axis and whose radius is 2. _ (b) whose axis is the Z-axis and whose radius is Vs. (c) whose axis is the A'-axis and whose diameter is V7. 16. A point moves so that the sum of its distances to the two fixed points (V3, 0, 0) and (— V3, 0, 0) is always equal to 4. Find the equa- tion of its locus. Ans. x^ + i z'^ + i y- — 4 = 0. 17. Find the equation of the locus of a point (a) whose distance from the point (1, 0, 0) equals its distance from the rZ-plane. Ans. y^ + z-— 2x + 1 = 0. (b) wdiose distance from the point (1, 0, 0) equals its distance from the Z-axis. Ans. z^— 2x + 1 = 0. (c) whose distance from the A'^-axis is one half of its distance from the FZ-plane. Arts. 4 y"^ + i z" — x^ = 0. (d) whose distance from the Z-axis is twice its distance from the F-axis. (e) whose distance from the origin equals the sum of its distances from the A"Z-plane and the FZ-plane. Ans. z"^ — 2 xy = 0. (f ) the sum of whose distances from the three coordinate planes is constant. (g) whose distance from the origin equals the sum of its distances from the three coordinate planes. Ans. xy + yz + zx = 0. (h) whose distance from the X-axis is half the difference of its dis- tances from the A'F-plane and the A'Z-plane. SURFACES, CURVES, AND EQUATIONS 249 (i) whose distance from the point (0, 0, 1) equals its distance from the A'F-plane increased by 1. (j) whose distance from the Z-axis equals its distance from the point (1, 1, 0). 18. Find the equation of the locus of a point the sum of whose dis- tances from the A'-axis and the Y"-axis is unity. 19. Find the equation of the locus of a point the sum of whose dis- tances from the three coordinate axes is unity. 94. Planes parallel to the coordinate planes. We may easily prove the Theorem. The equation of a plane which is ' parallel to the XY-plane has the form z = constant; parallel to the YZ-plane has the form x = constant; parallel to the ZX-plane has the form y = constant. 95. Equations of a curve. First fundamental problem. If any point P wliich lies on a given curve be given the coordinates (.T, y, «), then the two conditions which define the curve as a locus will lead to two equations involving the variables x, y, and z. The equations of a curve are two equations in the variables X, y, and z representing coordinates such that : 1. The coordinates of every point on the curve will satisfy both equations. 2. Every point whose coordinates satisfy both equations will lie on the curve. If the curve is defined as the locus of a point satisfying two conditions, the equations of the surfaces defined by each condi- tion separately may be found in many cases by a Rule anal- ogous to that of Art. 17. These equations will be the equations of the curve. It will appear later that the equations of the same curve may have an endless variety of forms. 250 NEW ANALYTIC GEOMETRY EXAMPLES 1. Find the equations of the locus of a point whose distance from the origin is 4 and which is equally distant from the points Pj (8, 0, 0) and Po{0, 8, 0). Solution. Let V (x, ?/, z) be any point on the locus. The given conditions are (1) F0 = 4, rr^ = pp„. By (IV), PO = Vx2 + 2/-i + z\ PP. = V(X - 8)- + 2/2 + 2-2, Substituting in (1), we get Vx'- + y~ + z^ = 4, V(X - 8)2 + 2/2 + 22 = Vx2 + (2/ - 8)2 + Z^. Squaring and reducing, we have the required equations, namely, x2 + 2/2 + -2 = 16, X — 2/ = 0. These equations should be verified as in Art. 16. 2. Find the equations of the circle lying in the A'T-plane whose center Is the origin and whose radius is 5. Solution. In plane analytic geometiy the equation of the circle is (2) x2 + 2/2 = 25. Regarded as a problem in solid analytic geometry we must have two equations which the coordinates of any point P (x, y, z) which lies on the circle must satisfy. Since P lies in the AT-plane, (3) z = 0. Hence equations (2) and (3) together express that the point P lies in the A'F-plane and on the given circle. The equations of the circle are therefore x2 + 2/2 = 25, z = 0. The reasoning in Ex. 2 is generaL Hence If the equation of a curve in the XY-pUnie is known, then the equations of that curve regarded as a curve in sjjace are the given equation and s = 0. SURFACES, CURVES, AND EQUATIONS 251 An analogous statement evidently applies to the equations of a curve lying in one of the other coordinate planes. From Art. 94 we have at once the Theorem. The equations of a line which is parallel to the X-axis have the form y = constant, z = constant; the Y-axis have the forni z = constant, x = constant; the Z-axis have the form x = constant, y = constant. PROBLEMS 1. Find the equations of the locus of a point which is (a) 3 units above the JTF-plane and 4 units to the riglit of tlie FZ-plane, (b) 5 units to the left of the FZ-plane and 2 units in front of the ZX- plane. (c) 4 units back of the ZX-plane and 7 units to the left of the FZ-plane. (d) 9 units below the A^i'-plane and 4 units to the right of the FZ-plane. 2. Find the equations of the straight line which is (a) 5 units above the A'F-plane and 2 units in front of the ZA'-plane (b) 2 units to the left of the FZ-plane and 8 units below the ATy-plane. (c) 3 units to the right of the FZ-plane and 5 units from the Z-axis. (d) 13 units from the X-axis and 5 units back of the ZA'-plane. (e) parallel to the F-axis and passing through (3, 7, — 5). (f) parallel to the Z-axis and passing through (— 4, 7, 6). 3. What are the equations of the axes of coordinates ? 4. What are the equations of the edges of a rectangular parallelepiped whose dimensions are a, 6, and c, if three of its faces coincide with the coordinate planes and one vertex lies in 0-XYZ ? in 0-XY'Z ? in O-A'T'Z ? 5. Find the equations of the locus of a point which is (a) 5 units from the origin and 3 units above the XF-plane. (b) 5 units from the origin and 3 units from the X-axis. (c) 6 units from the F-axis and 3 units behind the XZ-plane. (d) 7 units from the Z-axis and 2 units below the XZ-plane. 6. Find the ecpiations of a circle defined as follows : (a) center on the Z-axis, radius 4, and lying in the XF-plane. (b) center on the X-axis, radius 7, and lying in a plane parallel to the yZ-plane and 3 units to the right of it. (c) center on the Y-axis, radius 2, and lying in a plane 2 units behind the XZ-plane. 252 NEW ANALYTIC GEOMETRY (d) center at the point (1, 0, 1), parallel to the JTY-plane, and cutting the Z-axis. 7. The following equations are the equations of curves lying in one of the coordinate planes. What are the equations of the same curves regarded as curves in space ? (a) y^ = 4x. (e) a;^ + 42 + 6x = 0. (b) x2 + 2=2 = 16. (f ) 2/2 - z2 _ 4 2/ = 0. (c) 8x2 _ 2/2 = 64. (g) yz2 ^ z^-6y = 0. (d) 4^2 + 92/2 = 36. (h) z^-ix^ + 8z = 0. 8. Find the equations of the locus of a point which is (a) 5 units above the Xy-plane and 3 units from (3, 7, 1). Ans. 2 = 5, x2 + 2/2 + z2 — 6 X — 14 ?/ — 2 z + 50 = 0. (b) 2 units from (3, 7, 6) and 4 units from (2, 5, 4). Ans. x2 + 2/2 + z^ - 6x - 14 2/ - 122 + 90 = 0, x2 + 2/2 + z2 - 4 X- 10 2/ — 8 2 + 29 = 0. (c) 5 units from the origin and equidistant from (3, 7, 2) and ( _ 3, _ 7, _ 2) . Ans. x2 + 2/2 + ^2 — 25 = 0, 3 x + lij + 2z =0. (d) equidistant from (3, 5, — 4) and (— 7, 1, 6), and also from (4, - 6, 3) and (- 2, 8, 5). Ans. 5x + 2// — 2 + 9 = 0, 3x— 72/ — 2 + 8 = 0. (e) equidistant from (2, 3, 7), (3, — 4, 6), and (4, 3, - 2). Ans. 2 X - 14 2/ - 2 2 + 1 = 0, X + 7 2/ - 8 2 + 16 = 0. 9. Find the equations of the locus of a point which is equally distant from the points (6, 4, 3) and (6, 4, 9), and also from (— 5, S, 3) and (— 5, 0, 3), and determine the nature of the locus. Ans. z = 6, y = 4. 10. Find the equations of the locus of a point which is equally distant from the points (3, 7, — 4), (— 5, 7, — 4), and (— 5, 1, — 4), and deter- mine the nature of the locus. Ans. x =—1, y = 4. 11. Determine the nature of each of the following loci after finding their equations. The moving point is equidistant from (a) the three coordinate planes. (b) the three coordinate axes. . (c) the three points (1, 0, 0), (0, 1, 0), and (0, 0, 1). (d) the A'F-plane, the Z-axis, and the point (0, 0, 1). (e) the A'F-plane, the A'-axis, and the point (0, 0, 1). (f ) the points (1, 0, 0), (0, 1, 0), and the Z-axis. (g) the X-axis, the F-axis, and the point (1, 0, 0). (h) the Z-axis, the XF-plane, and the FZ-plane. SURFACES, curvp:s, and p:quations 253 96. Locus of one equation. Second fundamental problem. The locus of one equation in three variables (one or two may be lacking) representing coordinates in space is the surface passing through all points whose coordinates satisfy that equation and through such points only. The coordinates of points on the surface niay be obtained as follows : Solve the equation for one of the variables, say z, assume pairs of values of x and y, and compute the corresponding values of z. A rough model of the surface might then be constructed by taking a thin board for the AT-plane, sticking needles into it at the assumed points (.r, y) whose lengths are the computed values of z, and stretching a sheet of rubber over their extremities. 97. Locus of two equations. Second fundamental problem. The locus of two equations in three variables representing coordinates in space is the curve passing through all points whose coordi- nates satisfy both equations and through such points only. That is, the ' locus is the curve of intersection of the surfaces defined by the two given equations. The coordinates of points on the curve may be obtained as follows : Solve the equations for two of the variables, say x and y, in terms of the third, z, assume values for z, and compute the corresponding values of x and y. 98. Discussion of the equations of a curve. Third fundamental problem. The discussion of curves in elementary analytic geom- etry is largely confined to curves which lie entirely in a plane which is usually parallel to one of the coordinate planes. Such a curve is defined as the intersection of a given surface with a plane parallel to one of the coordinate planes. The method of determining its nature is illustrated as follows : 254 NEW ANALYTIC GEOMETRY EXAMPLE Determine the nature of the curve in which the plane z = 4 intersects the surface whose equation is ?/2 + z2 _ 4 x. Solution. The equations of the curve are, by definition, (1) y" + z-^ = 4x, z = 4. Eliminate z by substituting from the second equation in the first. This gives <2) y"--4x + 16 = 0, 2 = 4. Equations (2) are also the equations of the curve. For every set of values of {x, y, z) which satisfy both of equations (1) will evidently satisfy both of equations (2), and conversely. If we take as axes in the plane z = 4 the lines (XX' and O'Y' in which the plane cuts the ZA'-plane and the I'Z-plane, then the equation of the curve when referred to these axes is the first of equations (2), namely, <3) 2/2 -4x + 16 = 0. The locus of (.3) is a pa- rabola. The vertex, in the plane z = 4, is the point ^ (4, 0); alsop = 2. In plotting the locus of (3) in the plane A'O'y' the values of x and y must be laid off parallel to O'A'' and O'Y' respectively, as in plotting oblique coordinates (Art. 9). From the preceding example we may state the Rule to determine the nature of the curve in which a plane parallel to one of the coordinate planes cuts a given surface. ■Eliminate the variable occurring in the equation of the plane frovi the equations of the plane and surface. The result is the equation of the curve referred to the lines in ivhich the given plane cuts the other two coordinate planes as axes. Discuss this curve by the methods of plane analytic geometry. SURFACES, CURVES, AND EQUATIONS 255 PROBLEMS 1. Determine the nature of the following curves and construct their loci : (a) x2 _ 4y2 ^ 8z, z = 8. (e) x^ + 4y^ + 9z2 = 36, y = l. (b) x^ + 9y- = 9z2, z = 2. (f) x^ - 4y^ + z^ = 25, x =- 3. (c) x2 — 4 2/2 = 4 z, ?/ = - 2. (g) x2 — 2/2 — 4 z2 + 6 X = 0, X = 2. (d) x2 + 2/2 + z2 = 25, x = 3. (h) 2/2 + z2-4x + 8 = 0, 2/ = 4. 2. Construct the curves in which each of the following surfaces inter- sects the coordinate planes : (a) x2 + 42/2 + 16z2 = 64. (d) x2 + 9^2 ^ joz. (b) x2 + 42/2-16z2 = 64. (e) x^-9y'^ = 10z. (c) x2 - 42/2 - 16z2 = 64. (f) x2 + 42/2 - 16z2 = 0. 3. Show that the curves of intersection of each of the surfaces in Problem 2 with a system of planes parallel to one of the coordinate planes are conies of the same species (see Art. 70). 4. Determine the nature of the intersection of the surface x2 + y'^ + 4 z2 = 64 with the plane z — k. How does the curve change as k increases from to 4 ? from — 4 to ? What idea of the appearance of the surface is thus obtained ? 5. Determine the nature of the intersection of the surface 4x — 2 2/ = 4 with the plane y = k; with the plane z = k'. How does the intersection change as fc or Ar' changes ? What idea of the form of the surface is obtained ? 6. In each of the following lind the equations of the locus, determine its nature, and construct it : (a) A point is 5 units from the origin and 3 units from the Z-axis. (b) A point is 3 units from both the X-axis and the Z-axis. (c) The distance of a point from the Z-axis is equal to twice its distance from the XF-plane and its distance from the origin is 2. (d) A point is 5 units from the A'-axis and 4 units from the AZ-plane. (e) A point is equidistant from the TZ-plane and the XZ-plane and its distance from the X-axis is 7. Ans. An ellipse. (f ) A point is equidistant from the Z-axis, the FZ-plane, and the point (2, 0, 0). Ans. A parabola. 256 NEW ANALYTIC GEOMETRY 7. The ratio of the distances of a point to the Z-axis and the I'-axis respectively is |. Determine the nature of its locus if it is also (a) one unit above the XF-plane. (b) one unit in front of the XZ-plane. (c) one unit to the left of the FZ-plane. (d) in the A'Z-plane. (e) equidistant from the A'Z-plane and the Y"Z-plane. (f ) in the plane 4x — 3z — 12 = 0. 8. Find the equations of the locus of a point whose distance from the point (2, 0, 0) is always equal to three times its distance from the Z-axis, and whose distance from the FZ-plane is always unity. Name and draw the locus. 9. Find the equations of the locus of a point which is equidistant from the point (1, — 2, 0) and the Z-axis, and which is 3^ units behind the ATZ-plane. Name and draw the locus. 10. Find the equations of the locus of a point which is equidistant from the y-axis and the A'Z-plane and equidistant from the origin and the point (0, 0, — 4). Name and draw the locus. 99. Discussion of the equation of a surface. Third fundamental problem. Theorem. The locus of an algebraic equation jj asses through the origin if there is no constant term in the equation. The proof is analogous to that on page 47. Theorem. If the locus of an equation is unaffected by chang- ing the sign of one variable throughout its equation, then the locus is symmetidcal with respect to the coordinate plane from which that variable is measured. If the locus is unaffected by changing the signs of two variables throughout its equation, it is symmetrical with respect to the axis along which the third variable is measured. If the locus is unaffected by changing the signs of all three variables throughout its equation, it is symmetrical with respect to the origin. The proof is analogous to that on page 42. SURFACES, CURVES, AND EQUATIONS 257 Rule to find the- intercepts of a surface on the axes of coordinates. Set each pair of variables equal to zero and solve for real values of the third. The curves in which a surface intersects the coordinate planes are called its traces on the coordinate planes. From the Rule, p. 254, it is seen that The equations of the traces of a surface are obtained by succes- sively setting a:; = 0, ?/ = 0, and z=:0 in the equation of the surface. By these means we can determine some properties of the sur- face. The general appearance of a surface is determined by con- sidering the curves in which it is cut by a system of planes parallel to each of the coordinate planes. This also enables us to determine whether the surface is closed or recedes to infinity. EXAMPLE Discuss the locus of the equation ^/^ + z^ = 4x. Solution. 1. The surface passes through the origin since there is no constant term in its equation. 2. The surface is symmetrical with respect to the A'y-plane, the ZX- plane, and the JT-axis. For the locus of the given equation is unaffected by changing the sign of z, of y, or of both together. 3. It cuts the axes at the origin only. 4. Its traces are respec- tively the point-circle y'^ 4- z" = a and the parabolas z^ = 4iX and y"^ = 4iX. 5. It intersects the plane X = k in the curve 2/2 + z2 ^ 4 k. This curve is a circle whose center is the origin, that is, is on the A'-axis, and whose radius is2 v^ if fc>0, but there is no locus if A;<0. Hence the surface lies entirely to the right of the FZ-plane. 258 NEW ANALYTIC GEOMETRY If A; increases from zero to infinity, the radius of the circle increases from zero to infinity while the plane x = k recedes from the FZ-plane. The intersection with a plane z — k or y = kf, parallel to the XY- or the ZA'-plane, is seen to be a parabola whose equation is y"^ = 4x — k^ or z^ = ix — k"^. These parabolas have the same value of p, namely p = 2, and their ver- tices recede from the YZ- or the ZA'-plane as k ork' increases numerically. PROBLEMS 1. Discuss and draw the loci of the following equations: (a)x2 + z2 = 4x. (k) x^ + y"- - z"" = 0. (b) x2 + ?/2 4- 4 z2 = 16. ( 1 ) a;2 - y^ -z^ = 9. (c) x2 + 2/2 - 4 z2 = 16. (m) x'^ + y^ - z'^ + 2xy = 0. (d) 6x + iy + Sz = 12. (w) x + y - Gz = 6. (e) 3x + 2y + z = 12. (o) y'^ + z^ = 25. (f) X + 23 - 4 = 0. (p) x2 + 2/2 - 22 _ 1 ^ 0, (g) x2 + 2/2 - 2 z = 0. (q) x2 + 7/2 - -2 + 1 ^ 0. (h) x2 + 2/2 - 2x = 0. ( r) 4x2 - 2/2 - 22 = 0. (i) x2 + 2/2 — 4 = 0. (r) z^ - X - y = 0. {{) y^ + z'^-x-4 = 0. ( t ) x2 + 2/2 - 2 2X = 0. 2. Show that the locus of Ax + By + Cz + D = is a plane by con- sidering its traces on the coordinate planes and the sections made by planes parallel to one of the coordinate planes. 3. In each of the following find the equation of the locus of the point and draw and discuss it : (a) The sum of the distances of a point from the A'Z-plane and the FZ-plane equals twice its distance from the A'F-plane increased by 4. (b) The square of its distance from the Z-axis is equal to four times its distance from the A'F-plane. (c) Its distance from the Z-axis is double its distance from the XY- plane. (d) Its distance from the F-axis is twice the square root of its distance from the YZ-plane. (e) It is equally distant from the point (2, 0, 0) and the FZ-plane. Ans. y^ + z^ — 4x + 4 -0. (f) It is equally distant from the point (0, 2, 0) and the AT-axis. (g) Its distance from the Z-axis is equal to its distance from the FZ-plane increased by 2. SURFACES, CURVES, AND EQUATIONS 259 (h) Its distance from the point (0, 0, — 2) is equal to double its distance from the JTZ-plane increased by unity. (i) Its distance from the point (i, 0, 0) is equal to half its distance from the yZ-plane diminished by one. Ans. 3x- + 4 y- + 4 z^ — 3 = 0. (j) The product of the sum and the difference of its distances from the A'Z-plane and the FZ-plane respectively is equal to twice its distance from the A'l'-plane. 4. Find the equation of the locus of a point whose distance from the point (0, 0, 3) is twice its distance from the A'F-plane, and discuss the locus. Ans. a;2 + 2/"^ — 3 z2 _ 6 z 4- 9 = 0. 5. Find the equation of the locus of a point whose distance from the point (0, 4, 0) is three fifths its distance from the ZX-plane, and discuss the locus. A ns. 25 x^ + 16y' + 25 z^ — 200y + 400 - 0. CHAPTER XV THE PLANE AND THE GENERAL EQUATION OF THE FIRST DEGREE IN THREE VARIABLES 100. The normal form of the equation of the plane. Let ABC be any plane, and let ON be drawn from the origin perpen- dicular to ABC at D. Let the jjositive direction on ON he from toward N, that is, from the origin to- ward the plane, and denote the directed length OD by j) and the direction angles of ON by a, j3, and y. Then the jjosltlon of any plane is deter- mined hy given posi- tive values of p, a, P, and y. If p = 0, the positive direction on OiV, as just defined, becomes mean- ingless. Ifp — 0, we shall suppose that ON is directed upward, and hence TT cos 7 >0 since y < - ■ If the plane passes through OZ, then ON lies in the Xy-plane and cos 7 = 0; in this case vje shall suppose ON so directed that TT P<- and hence cos ^>0. Finally, if the plane coincides with the yZ-plane, the positive direction on ON shall be that on OX. Let us now solve the problem : Given the perpendicular distance p from the origin to a plane and the direction angles a, y8, y of this perpendicular, to find the equation of the plane. 260 THE PLANE 261 Solution. Let P(x, y, z) be any point on the given plane ABC. Draw the coordinates OE = x, EF = y, FP = z of P. Project OEFP and OP on the line ON. By the second theorem of projection, projection of OE + projection of EF + projection of FP = projection of OP. Then by the first theorem of projection and by the defini- tion of p, a: cos a 4- y cos /? + « cos y = j9. Transposing, we obtain the Theorem. Normal form. The eqiiatlon of a plane is (I) jrcosa + y cosyff + ^cosy — /> = 0, where p is the perpendicular distance from the origin to theplane, and a, /3, and y are the direction cosines of tltat 2i^^p>endlcular. Corollary. The equation of any plane is of the first degree in X, y, and z. 101. The general equation of the first degree, Ax -\- By -\- Cz -\-D = Q. The question now arises: Given an equation of the first degree in the coordinates x,y,z\ what is the locus ? This question is answered by the Theorem. The locus of any equation of the first degree In X, y, and z, (II) Ax-\-By-\-Cz-\-D = Q, is a plane. Proof. We shall prove the theorem by showing that (TI) may be reduced to the normal form (I) by multiplying by a proper constant. To determine this constant, multiply (II) by /c, which gives (1) kAx + IcBy + kCz + kD = 0. 262 NEW ANALYTIC GEOMETRY Equating corresponding coefficients of (1) and (I), (2) kA = cos a, kB = cos /3. kC = eosy, kD = — p. Squaring the first three of equations (2) and adding, k'^{A^-\- Br-{- C^) = cos-rt + cos'^yS + cos'y = 1. 1 (3) .-. k ±^A^+ E"+ C'^ From the last of equations (2) we see that the sign of the radical must be opposite to that of D in order that p shall be positive. Substituting from (3) in (2), we get cos/5 (4) ± V.-l''^+5-+ C^ ± V. 1- + i?2 4- c^ C —D eOSy = . =r; p = - , =• ±^A--\-B-+ C ±^A-^+B''+C^ We have thus determined values of a, fS, y, and p such that (I) and (II) have the same locus. Hence the locus of (II) is a plane. q.e.d If D = 0, then p = ; and from the third of equations (2) the sign of the radical must be the same as that of C, since when p = 0, cos7>0. If D = and C = 0, then p = and cos 7 = 0; and from the second of equations (2) the sign of the radical must be the same as that of B, since when p = Q and cos 7 = 0, cos j8 > 0. Equation (II) is called the general equation of the first degree in X, y, and z. The discussion gives the Rule to reduce the equation of a plane to the normal form. Divide the equation by ± V /1^+ B--\- C', chooslmj the sign of the radical opposite to that of D. When Z) = 0, the sign of the radical must be the same as that of C, the same as that of 5 if C := i* = 0, or the same as that of A\iB=C = D = Q. THE PLANE 263 From (4) we have the important Theorem. The coefficients of x, y, and z in the equation of a plane are proportional to the direction cosines of any line per- pendicular to the plane. From this theorem and Art. 90 we easily prove the following : Corollary I. Two planes whose equations are Ax + By-\-Cz-\-D=^Q, A'x + B'y + C'z -\- D' = are parallel when and only when the coefficients ofx, y, and z are proportional, that is, ABC T'^B'^'C'' Corollary II. Two jylanesaj'eperpendlcnlarwhen and only when AA'+BB'+ CC'= 0. Corollary III. A plane whose equation has the form Ax -\- By -\- D — is perpendicular to the XY-pjlane ; By -\- Cz -\- D := () Is pjerpendlcular to the YZplane ; Ax -\- Cz -{- D = Is jyerjjendlcitlar to the ZX-jylane. That is, if one variable is lacking, the pilane is pjerjjendlcular to the coordinate plane correspjondlng to the two variables which occur in the equation. For these planes are respectively perpendicular to the planes s: = 0, aj = 0, and ?/ = by Corollary II. Corollary IV. A plane whose equation has the form Ax 4- 7) = is perpendicular to the axis ofx; By -f Z) = is p)erpendlcular to the axis of y ; Cz -\- D = is pe7pendlcular to the axis of z. That is, if two variables are lacking, the pdane Is perpendicular to the axis coiTespondlng to the variable which occurs in the equation. For two of the direction cosines of a perpendicular to the plane are now zero, and hence this line is parallel to one of the axes and the plane is therefore peri)endicular to that axis. 264 NEW ANALYTIC GEOMETRY PROBLEMS 1. Find the intercepts on the axes and the traces on the coordinate planes of each of the following planes and construct the figures : (a) 2x + 3y + 4z- 24 = 0. (e) 5x-72/-35 = 0. (h) 7x-Sy + z-21 = 0. (f) 4x + 3z + 36 = 0. (c) 9x- 7?/-9z + 63 = 0. (g) 5?/ - 8z - 40 = 0. (d) 6x + 4?/-z + 12 = 0. (h) 3x + 5z + 45 = 0. 2. What are the intercepts and the equations of the traces on the coor- dinate planes of the plane Ax + By+Cz + B = 0? 3. Find the equations of the planes and construct them by drawing their traces, for which (a) ct = -, /3 = -, 7 = -, p = 6. Ans. V2x + y + z — 12 = 0. ,, s 2 TT „ 37r "" _ . /— -rt ^ (b) a = ^,l3 = — ,7 = ^,p = 8. Ans. x + ■\2y — z + 16 - 0. o 4 3 , ^ cos a COSjS cos 7 . < /. « , o no r. (c) = — , p = 4. Ans. 6x — 2w + 3z — 28 = 0. ^ ' 6 - 2 3 ,., cosor cos/3 cos 7 (d) —^ = -—J = —^, p = 2. Ans. 2x + y + 2z + 6 = 0. 4. Find the equation of the plane such that the foot of the perpen- dicular from the origin to the plane is the point (a)(-3, 2, 6). Ans. 3x - 2y - 6z + 49 = 0. (b) (4, 3, - 12). Ans. 4x -F 3y - 12z - 169 = 0. (c) (2,2,-1). Ans. 2x + 2y-z-9-0. 6. Reduce the following equations to the normal form and find a, /3, 7, and p : (a) 6x — 3y + 2z-7 = 0. Ans. cos-i f, cos-i (— f), cos-i f, 1. . 2 7r TT 2 7r . (b) x- V27/ + z + 8 = 0. Ans. —,-,—,4. (e) 2x — 2y — z + 12 = 0. Ans. cos-i (— |), cos-i |, cos-i ^, 4. {d)y-z+10 = 0. ^ns. -,—,-, 5 V2. 2 4 4 {e) Sx + 2y-6z = 0. Ans. cos-i (- f), cos-i(- f), cos-i ^, 0. 6. Find the distance from the origin to the plane 12x — 42/4-3z— 39 = 0. THE PLANE 265 7. Find the area of the triangle which the three coordinate planes cut from each of the following planes : (a) 2 X + 2 ?/ + z - 12 = 0. Am. 54. (b) 6a: -2?/ -32 + 21 = 0. (c) 12x- 3?/ + 42-13 = 0. r- (d) X + 5?/ + 72 - 3 = 0. Anfi. — ^ . (e) X - 2 2/ + 3 2 - 6 = 0. ^'* (f) dx + 2y-z + 18 = 0. Hint. Find the volume of the tetrahedron formed by the four planes by find- ing the intercepts. Set this equal to the product of the required area by one third the distance of the given plane from the origin, and solve. 8. Find the distance between the parallel planes 6x + 2?/ — 32 — G3 = and 6x + 2?/ — 32 + 49 = 0. Ans. 16. 9. Find the equation of a plane parallel to the plane 2x + 2y + z — 15 = and two units nearer to the origin. 10. Show that the following pairs of planes are either parallel or per- pendicular : r2x + 57/-62 + 8 = 0, J6x-3y + 2z-7 = 0, ^'' \6x + 15?/-182-5 = 0. ^^' \3x + 2?/-62 + 28 = 0. r3x- 52/- 42+ 7 = 0, Ji4x- 7?/ -212-50 = 0, {6x + 2y + 2z-7 = 0. ^ ' \2x - y - Sz + 12 = 0. 11. What may be said of the position of the plane (I), Art. 100, if (a) cosa=:0? (c) cos7 = 0? (e) cos/3 = cosy = ? . (b) cos /3 = ? (d) cos a = cos /3 = ? (f ) cos y = cos a = ? 12. For what values of ex, ft 7, and p will the locus of (I), Art. 100, be parallel to the A'F-plane ? the YZ-plane ? the ZA'-plane ? coincide with one of these planes '? 13. For what values of a, /3, 7, and p will the locus of (I), Art. 100, pass through the AT-axis ? the Y-axis ? the Z-axis ? 14. Find the coordinates of the point of intersection of the planes x + 2?/ + z = 0, x-2?/-8 = 0, x + 2/ + 2-3 = 0. Ans. (2, - 3, 4). 15. Show that the. plane x + 2y— 2z — 9 = passes through the point of intersection o\ the planes x + y + z — 1 = 0, x — y — z — l = 0, and 2x + 3?/-8 = 0. 16. Show that the four planes x + 2/ + 2z — 2 = 0, x + y— 2z + 2 = 0, X — ?/ + 8 = 0, and Sx — y — 2z + 18 = pass through the same point. 17. Show that the planes 2x — y + z + S = 0, x — y + iz = 0, 3x + 2/-22 + 8 = 0, 4x-2?/ + 22-5 = 0, 9x + 3?/- 6z- 7 = 0, and 7x- 7(/ + 282 — 6 = bound a parallelepiped. 266 NEW ANALYTIC GEOMETRY 18. Show that the planes 6x — 3y + 2z = 4,3x + 2y — 6z=10,2x + 6y + 3z = 9, Sx + 2y-Gz = 0, \2x + 36 ?/ + I82 - 11 ^ 0, and 12x- 6y + 4z— 17 = bound a rectangular parallelepiped. 19. Show that the planes a; + 2?/ — z = 0, ?/4-7z — 2 = 0, x — 2y — z — 4 = 0, 2x + ?/ — 8 = 0, and 3x + 3?/ — s — 8 = bound a quadrangu- lar pyramid. 20. Derive the conditions for parallelism of two planes from the fact that two planes are parallel if all their traces are parallel lines. 102. Planes determined by three conditions. The equation (1) Ax+B!j^rz + D = represents, as we know, all planes. The statement of a problem, to find the equation of a certain plane, may be such that we are able to write down three homogeneous equations in the coeffi- cients A, B, (', J), which we can then solve for three coefficients in terms of the fourth. When these values are substituted in (1), the fourth coefficient will divide out, giving the required equation. EXAMPLES 1. Find the equation of the plane which passes through the point p^ (2, — 7, f ) and is parallel to the plane 21 x — 12 ?/ + 28 2 — 84 = 0. Solution. Let the equa- tion of the required plane be (2) Ax+ By + Cz + D = 0. Since P-y lies on (2), we may substitute a; = 2, ^ = — 7, z-l, giving (.3) 2A-1 B+ I C + X> = 0. Since (2) is parallel to the given plane (Corollary I, p. 263), ^ ' 21 -12 28 Equations (3) and (4) are three homogeneous equations in ^, B, C, D. Solving (3) and (4) for A, B, and D in terms of C, A = IC, B=- I C, D = - 6C. THE PLANE 267 Substituting in (2), ^ Cx — I Cy + Cz — 6 C = 0. Clearing of fractions and dividing by C, 21 X — 12 y + 28 2-168 = 0. Ans. The answer should be checked by testing whether the coordinates of Pj satisfy the answer. 2. To find the equation of a plane passing through three points, sub- stitute for X, y, and z in (1) the coordinates of each of the three points. Then three equations involving -4, B, C, and D will be obtained, which may be solved for three o? these coefficients in terms of the fourth. It is convenient to write down the equation of a plane passing through three given points (x^, ?/j, 2j), (x,, y^, z^), {x^, y^, Zg) in the form of a deter- minant. This is (5) X y z 1 X, y^ ^1 1 ^2 2/2 ^2 1 Xg Vz Z3 1 = 0. In fact, when (5) is expanded in terms of the elements of the first row, an equation of the first degree in x, y, and z results. Hence (5) is the equation of a plane. Further, (5) is satisfied when the coordinates of any one of the three given points are substituted for x, y, and 2, since then two rows become identical. Hence the plane (5) passes through the given points. The equation (5) may be used also to determine whether four given points lie in a plane. If we write (5), when expanded, in the form Ax + By -\- Cz + D- 0, then the coefficients are the determinants of the third order. A = Vx H 1 X, z, 1 X, y, 1 Vi z^ 1 , B=- X2 2, 1 , C = X., 2/2 1 Vz 23 1 •*3 ~3 ^ ■^3 Vz 1 D=- Vi Vi Vz PROBLEMS Check the answer in each of the following : 1. Find the equation of the plane which passes through the points (2, 3, 0), {- 2, - 3, 4), and (0, 6, 0). Am. 3x-f22/-F62-r2 = 0. 2. Find the equation of the plane which passes through the points (1, 1, - 1), (- 2, - 2, 2), and (1, - 1, 2). Ans. x - 3 // - 2 2 = 0. 3. Find the equation of the plane which passes through the point (3, — 3, 2) and is parallel to the plane 3x — ?/-l-2 — 6 = 0. Ans. 3x — ?/ + 2 — 14 = 0. 268 NEW ANALYTIC GEOMETRY 4. Find the equation of the plane which passes through the points (0, 3, 0) and (4, 0, 0) and is perpendicular to the plane 4x — 6y — z=12. Ans. 3a; + 4?/-12z-12 = 0. 6. Find the equation of the plane which passes through the point (0, 0, 4) and is perpendicular to each of the planes 2x — Sy = b and x-42 = 3. Ans. 12x + 8?/ + 3z-12 = 0. 6. Find the equation of the plane whose intercepts on the axes are 3, 5, and 4. Aiis. 20 x + 12 2/ + 15 z - 60 = 0. 7. Find the equation of the plane whicl* passes through the point (2, — 1, 6) and is parallel to the plane x — 2y — 3z + 4 = 0. Ans. x — 2y-Sz + U = 0. 8. Find the equation of- the plane which passes through the points (2, — 1, 6) and (1, — 2, 4) and is perpendicular to the plane x — 2y — 22 + 9 = 0. Ans. 2x + 4?/-32 + 18 = 0. 9. Find the equation of the plane whose intercepts are — 1,-1, and 4. Ans. 4x + 4?/ — 2 + 4 = 0. 10. Find the equation of the plane which passes through the point (4, — 2, 0) and is perpendicular to the planes x + y — z — and 2 x — 4 ?/ + 2 = 5. Ans. X + y + 2 z — 2 = 0. 11. Show that the four points (2, - 3, 4), (1, 0, 2), (2, - 1, 2), and (1, — 1, 3) lie in a plane. 12. Show that the four points (1, 0, - 1), (3, 4, - 3), (8, - 2, G), and (2, 2, — 2) lie in a plane. 13. Find the equation of the plane which is perpendicular to the line joining (3, 4, — 1) and (5, 2, 7) at its middle point. Ans. X — ?/ + 42 — 13 = 0. 14. Find the equations of the faces of the tetrahedron whose vertices are the points (0, 3, 1), (2, - 7, 1), (0, 5, - 4), and (2, 0, 1). A7W. 25x + 5?/ + 22 = 17, 5x-22 = 8, 2 = 1, 15x + 10?/ + 4z = 34. 15. The equations of three faces of a parallelepiped are x — 4 ?/ = 3, 2x — ?/ + 2 = 3, and Sx + y — 2z — 0, and one vertex is the point (3, 7, — 2). What are the equations of the other three faces ? Ans. X — 4?/ + 25 = 0, 2x — ?/ + 2 + 3 = 0, 3x + ?/ — 22 = 20. 16. Find the equation of the plane whose intercepts are o, b, c. Ans. ? + ^ + ? = l. a b c 17. What are the equations of the traces of the plane in Problem 16? How might these equations have been anticipated from plane analytic geometry ? THE PLANE 269 18. Find the equation of the plane which passes through the point P\ (Xp 2/1, Zj) and is parallel to the plane A^x -}- B^y + C^z + i*^ = 0. Aiis. A^ (X - x^) + L', {y - y^) + C\ {z — z^) = 0. 19. Find the equation of the plane which passes through the origin and Pj(Xj, ^j, z■^) and is perpendicular to the plane J^jX + B^y + C-^z + Z>j = 0. Ylns. {B^z■^— C^y^)x+ {C-^x^ — A-^z^)y -{■ {A^y^— B^x^)z = 0. 103. The equation of a plane in terms of its intercepts. Theorem. If a, b, and c are respectively the intercejjfs of a plane on the axes of X, Y, and Z, then the equation of the jJ lane is (III) ^ + | + _ = 1. ^ -^ a b c Proof Let the equation of the required plane be (1) Ax + Bij -\-Cz-\-D= 0. • Then we know three points in the plane, namely (./, 0, 0), (0,^^,0), (0,0, f). These coordinates must satisfy (1). Hence Aa + ]) = 0, Bb-\-D = 0, Cc + D = 0. Whence A = Substituting in (1), dividing by — D, and transposing, we obtain (III). Q. e. d. 104. The perpendicular distance from a plane to a point. The positive direction on any line perpendicular to a plane is assumed to agree with that on the line drawn through the ori- gin perpendicular to the plane (Art. 100). Hence the distance fro7n a plane to the point P^ is positive or negative according as Pj and the origin are on opposite sides of the plane or not. If the plane passes through the origin, the sign of tiie distance from the plane to P, must be determined by the conventions for the special cases in Art. 100. D D D — 5 B=— -, C = a b c 270 NEW ANALYTIC GEOMETRY We now solve the problem : Given the equation of a plane and a point, to find the perpendicular distance from the plane to the point. Solution. Let the point be P^ (x^, y^, z^ and assume that the equation of the given plane is in the normal form (1) ajcosar+z/cos^+^cosy— ^ = 0. Let d equal the required distance. Dravsr OP^. Projecting OP^ on ON, we evidently get j'j + d. Projecting OE, EF, and FP^ on ON, we get respectively x^ cos a, y^ cos /?, and z^ cos y. Then, by the second theorem of projection, Y>( p -\- d = x^ cos a -\- ij^ cos fi -\- z^ cos y. .•. d = x^ cos a + ^/j cos ji -\- z^ cos y — p. Hence the perpendicular distance d is the number obtained by substituting the coordinates of the given point for x, y, and z in the left-hand member of (1). Whence the Rule to find the perpendicular distance d from a given plane to a given pjoint. Reduce the equation of the plane to the normal form. Place d equal to the left-hand memher of this equation. Substitute the coordinates of the given point for x, y, and z. The result is the required distance. For example : To find the perpendicular distance from the plane 2a- + ?/ — 2z4-8 = to the point (— 1, 2, 3). Dividing the equation by — 3, we have d = _ 2x + J/-22 + 8 _ 2(-l)+ 2-2(3) + 8 :— t- Ans. -3 -3 Hence the given point is on the same side of the plane as the origin. THE PLANE 271 The rule gives for the perpendicular distance d from the plane Ax + By + Cz -{- D = ^ to the point (a^^ y^, z^ the result (2) ^_^., + %,+ C.,+£ ^ ^ ± V^2 ^_ ij2 _f_ (^,2 the sign of the radical being determined as above (Art. 101). 105. The angle between two planes. The plane angle of one pair of dihedral angles formed by two intersecting planes is evi- dently equal to the angle between the positive directions of the perpendiculars to the planes. That angle is called the angle between the planes. Theorem. The angle 6 hetiveen the two ■planes A^x + J5j// + C\z + Z)j = and A^x + B,^j + C^z + D^= Q is given by (IV) cos^= A,A, + B,B, + C,C, ±^Al + Bi + C,2 X ± V>1 1 + Bi + Ci the signs of the radicals being chosen as in Art. 101. Proof. By definition the angle Q between the planes is the angle between their normals. The direction cosines of the normals to the planes are cos CTi = ^1 ± V.42 + BI + ci COS/81 = B, ± VI7 + B^ + c\^ cos V, c\ COS a.2 = A. ± ^Ji + BI + CI cos 13.2 = B, ± -Jai -\-B.I + c| cos v„ c. ± V.4 -^ + Bf + C^ ' ± V^l + Bl + C| By (V), Art. 90, we have cos Q — cos a^ cos a^ + cos ^j cos ^^ -\- cos y^ cos y„. Substituting the values of the direction cosines of the normals, we obtain (IV). q.e.d. 272 NEW ANALYTIC GEOMETRY PROBLEMS 1. Find the distance from the plane (a) 6x — 3?/ + 2z - 10 = to the point (4, 2, 10). Ans. 4t. (b) X + 2?/ — 2z — 12 = to the point(l, - 2, 3). Ans. —7. (c) 4x + 3?/ + 122 + = to tlie point(9, -1, 0) Ans. — 3.__ (d) 2x- 5?/ + 3z - 4 = to the point(.- 2, 1, 7). . Ans. i'VV38. 2. Do the origin and tlie point (3, 5, — 2) lie on the same side of the plane 7x — y — 3z + Q = 0? Ans. Yes. 3. Does the point (1, 6, 0) lie on the same side of the plane X + 2y — 3z = 6 as the origin ? 4. Find the length of the altitude which is drawn from the first vertex of the tetrahedron whose vertices are (0, 3, 1), (2, — 7, 1), (0, 5, — 4), and (2, 0, 1). Ans. i^ V29. 5. Find the volume of the tetrahedron formed by the point (1, 2, 1) and the points where the plane 3x+4?/ + 22 — 12 = intersects the coordinate axes. 6. Find the volumes of the tetrahedrons having the following vertices : (a) (3, 4, 0), (4, - 1, 0), (1, 2, 0), (6, - 1, 4). Ans. 8. (b) (0, 0, 4), (3, 0, 0), (0, 2, 0), (7, 7, 3). (c) (4, 0, 0), (0, 4, 0), (0, 0, 4), (7, 3, 2). (d) (3, 0, 0), (0, - 2, 0), (0, 0, - 1), (3, - 1, - 1). Ans. |. (e) (1, 0, 0), (0, 1, 0), (0, 0, - 2), (4, - 1, .3). (f ) (3, 0, 0), (0, 5, 0), (0, 0, - 1), (3, - 4, 0). 7. Find the angles between the following pairs of planes : (a) 2x + ij — •2z — 9 = 0,x — 2y + 2z = 0. Ans. cos-i(— |). (b) X + 2/ - 4z = 0, 3 ?/ - 3 2 + 7 = 0. (c) 4x + 2?/ + 4z-7 = 0, 3x- 4?/ = 0. (d) 2x-2/ + z = 7, x + ?/ + 2z = 11. (e) 3X-22/ + 6z = 0, x + 2?/-2z + 5 = 0. (f) x + 5?/-3z + 8 = 0, 2x-3i/ + z-5 = 0. 8. Show that the angle given by (V) is that angle formed by the planes •which does not contain the origin. 9. Find the vertex and the dihedral angles of that trihedral angle formed by the planes x + 7/ + z = 2, x — y — 2z = 4, and 2x+?/ — z = 2 in which the origin lies. . ,, , „, , 1 /- 27r Ans. cos-i(- Ans. cos-i 1 Ans. cos-i(- Ans. TT Ans. (4, - 4, 2), cos-i - V2, — , cos-i (- - V2'\ 3 3 \ 3 / the' plane 273 10. Find the equation of the plane which passes through the points 27r (0, — 1, 0) and (0, 0, — 1) and which makes an angle of — with the plane Ans. ± V6x + ?/ -f z + 1 = 0. 11. Find the locus of points which are equally distant from the jjlanes 2x — 2/ — 22-3 = and 6x — 32/ + 2z + 4 = 0. Am. 32x-162/-8z- 9 = 0. 12. Find the locus of a point which is three times as far from the plane 3 J _ 6 ?/ — 2 z = as from the plane 2x— ?/ + 2z = 9. Ans. 17x-13y + 122- 63 = 0. 13. Find the equation of the locus of a point whose distance from the plane x + 2/+z — l=Ois equal to its distance from the origin. 14. Find the equation of the locus of a point whose distance from the plane x + y = 1 equals its distance from the Z-axis. Ans. (X - j^)2 + 2 (X + ?/) - 1 = 0. 15. Find the equation of the locus of a point, the sum of the squares of whose distances from the planes x + y — z— 1 = and x + ?/ + z + l = is equal to unity. Ans. 2(x + ?/)2 + 2 z(z + 2) — 1 = 0. 106. Systems of planes. The equation of a plane which sat- isfies two conditions will, in general, contain an arbitrary con- stant, for it takes three conditions to determine a plane. Such an equation therefore represents a systevi of planes. Systems of planes are used to find the equation of a plane satisfying three conditions in the same manner that systems of lines are used to find the equation of a line (Art. 36). Three important systems of planes are the following : The system of planes parallel to a rj'iv en plane Ax + ny-^ Cr. -f X> = is represented by (V) Ax + By-irCz + k = Q, where k is an arbitrary constant. The plane (V) is obviously parallel to the given plane (Corol- lary I, Art. 101). 274 NEW ANALYTIC GEOMETRY The system of planes passing through the line of intersection of two given planes A^x + B^j + C^z + i), = 0, A^x + i3,3/ + ( './ + il. = is represented by (VI) A^x + B^y + Qz + Z?i + ft {A^x + B^y + Qz + ZJ^) = 0, where k is a>i arbitrary constant. Clearly, the coordinates of any point on the lifte of intersec- tion will satisfy the equations of both of the given planes, and hence will satisfy (VI) also. The equation of a system of planes which satisfy a single condition must contain two arbitrary constants. One of the most important systems of this sort is the following : The system of planes passing throngli n gir en point I\(x-^, y^, z^ is represented by (VII) ^(x-jrO + 5(i/-i/0+C(z-zO = 0. Equation (VII) is the equation of a plane which passes through P^, for the coordinates of P^ obviously satisfy it. Again, if any plane whose equation is Ax + 7>'// + Cz + /) = passes through 7^^, then Subtracting, we get (VII). Hence (VII) represents all planes passing through P^. Equation (VII) contains two arbitrary- constants, namely, the ratio of any two coefficients to the third. In the following problems write down the equation of the appropriate system of planes and then determine the unknown parameters from the remaining data. THE PLANE 275 PROBLEMS 1. Determine the value of k such that the plane x + fcy — 2z — 9 = shall (a) pass through the point (5, — 4, — 6). Ans. 2. (b) be parallel to the plane &x — 2y — Viz = 1. Aim. — \. (c) be perpendicular to the plane 2j; — 4?/ + 2 = 3. Ans. 0. (d) be 3 units from the origin. Ans. ± 2. IT (e) make an angle of — with the plane 2a; — 2t/ + z = 0. ^ Ans. - 3 V35. 2. Find the equation of the plane which passes through the point (3, 2,-1) and is parallel to the plane 7 x — ?/ + z = 14. Ans. 7x — y + z — 18 = 0. 3. Find the equation of the plane which passes through the inter- section of the planes 2x-|-?/— 4 = and j/ + 2 2 = 0, and which (a) passes through the point (2, — 1, 1); (b) is perpendicular to the plane 3x-l-2?/-3z = 6. Ans. {a)x + 2/ + z-2 = 0; (b)2x + 3?/ + 4z-4 = 0. 4. Find the ecjuations of the planes which bisect the angles formed by the planes (a) 2 X — ?/ + 2 z ^ and x + 2 y — 2 z = 6. Ans. 3x + ?/-6 = 0, X-3 2/ + 4Z+6 =0. (b) 6x — 22/-3z = and 4 x + 3 ?/ - 13 z = 10. 5. Find the equations of the planes passing through the line of inter- section of the planes 2x-f-i/ — z = 4 and x— ?/-i-2z = which are per- pendicular to the coordinate planes. Ans. 5x4-2/ = 8, 3x4-2 = 4, 3y — 5z = 4. 6. Find the equation of a plane parallel to the plane 6x— 32/ + 2z-f- 21 = and tangent to a sphere of unit radius whose center is the origin. 7. Find the equationof aplane parallel to theplane6x— 2?/— 3z-|-35=0 and such that the point (0, —2,-1) lies midway between the two planes. 8. Find the equation of a plane through the point (2, — 3, 0), and having the aame trace on the ^Z-plane as the plane x — 3?/ + 7z — 2 = 0. 9. Find the equationof aplane parallel to the plane 2x-l-2/-|-2z + 5 = 0, and forming a tetrahedron of unit volume with the three coordinate planes. 10. Find the equation of aplane parallel to the plane 5x-l-3 j/-|- z — 7=0 if the sum of its intercepts is 23. 270 NEW ANALYTIC GEOMETRY 11. Find the equation of a plane parallel to the plane 2x + 6y + 8^ — 8 = 0, upon which the area intercepted by the coordinate planes in the first octant is |. Ans. 2x + 6?/ + 3z — 3 = 0. 12. Find the equation of a plane parallel to the plane 2x + y + 2z — 5 = and such that the entire surface of the tetrahedron which it forms with the coordinate planes is unity. Ans. 2x + y + 2z±l=0. 13. Find the equation of a plane having the trace x + Sy — 2 = and forming a tetrahedron of volume | with the coordinate planes. Ans. 3x + 9y + z — 6 -0. 14. Find the equation of a plane passing through the intersection of the two planes 6x-\-2y+Sz — G = and x + y + z — 1 — and forming a tetrahedron of unit volume with the coordinate planes. Ans. 12x-8y -3z-12 = 0. 15. A point moves so that the volume of the tetrahedron which it forms with the three points (2, 0, 0), (0, 6, 0), and (0, 0, 4) is always equal to 2. Find the equation of its locus. 16. A point moves so that the sum of its distances from the three coordinate planes is unity. Determine the equation of the locus of a second point which bisects the line joining the first with the origin. 17. Find the equation of the plane passing through the intersection of the planes A^x + B^y + C^z + 1)^-0 and A.,x + B„y + C^z + D^ = which passes through the origin. Ans. (A^D^ - A^D^)x +{B^D, - B.-,D^) y + {C^D.^- C.^D^)z = 0. 18. Find the equations of the planes which bisect the angles formed by the planes A^x + B^y + C^z + 2>j = and A„x + B^y + C^z + Dg = 0. . A,x + B^y + C^z + Di , A^x + B-^y + C^z + B^ Ans. — i — ^ ' — - = + — - — ^ — -• 19. Find the equations of the planes passing through the intersection of the planes A^x + B^y + CjZ + D, = and A^x + B^y + CgZ + ^2 = *^ which are perpendicular to the coordinate planes. Ans. {A^B^ - A„B^)y- {C^A^ - C,A^)z + A^D^ - A^D^ = 0, (^^^2 - ^2^i)a; -{B,C2 - B.^C^)z- {B^D^ - B^l),) = 0, (C'1^2 - ^2^1)0; - (B1C2 - BiC^)y + C^D^ - C^I)^ = 0. CHAPTER XVI THE STRAIGHT LINE IN SPACE 107. General equations of the straight line. A straight line may be regarded as the intersection of any two planes which pass through it. The equations of the planes regarded as simultaneous are the equations of the line of intersection, and hence the . Theorem. The equations of a straight line are of the first degree in x, ?/, and z. Conversely, the locus of two equations of the first degree is a straight line unless the planes which are the loci of the separate equations are parallel. Hence w^e have the Theorem. The locus of two equations of the first degree, is « straight line unless the coefficients of x, y, and z are proportional. To plot a straight line we need to know only the coordinates of two points on the line. The easiest points to obtain are usually those lying in the coordinate planes, which we get by setting one of the variables equal to zero and solving for the other two, as in the following example. The direction of a line is known when its direction cosines are known. The method of obtaining these will now be illustrated. 277 278 NEW ANALYTIC GEOMETRY EXAMPLES 1. Find the direction cosines of the line whose equations are (1) Sx + 2y-z-l = 0, 2x-2/ + 2z-3 = 0. Solution. Let us find the point where the line pierces the A"F-plane. To do this, let z = in both eijuations. Then solving the resulting equations 3x + 2y — 1 = and 2x — ?/ — 3 — for x and ?/, we find the requii-ed point is (1, — 1, 0). Similarly, putting y = 0, the point on the line in the ZJT-plane is (f, 0, |). Hence A (1, — 1, 0) and B (|, 0, ^) are two points on the line. Let the required direction cosines of AB be cos a, cos/3, and cos 7. Then, by the corollary of Art. 89, cos a _ cos /3 _ cos 7 _ or, reducing (multiplying the denominators by 8), cos a _ cos /3 _ cos 7 ^ ' 3 ~ -8 ^ - 7 ■ The direction cosines may now be found as usual (Art. 88). A second method is the following : , ,^ , cos a cosS cos 7 (4) Assume = = ^ ' a b c The coefficients 3, 2, and — 1 in the first plane of (1) are proportional to the direction cosines of a perpendicular to that plane. The required line lies in this plane. Hence (corollary. Art. 90) (5) 3a + 26-c = 0. For the same reason, using the second plane in (1), (6) 2a-b + 2c = 0. Solving (5) and (6) for the ratios of a, 6, and c, the result is 8a=-36, 7a = -3c. (7) ...^ = -^ = ^. ^ ' 3-8-7 Combining (7) and (4), we have the previous result (3). 2. Find the direction cosines of the line (I). Solution. The direction cosines cos a, cos ;3, cos 7 must satisfy Aj^cof^a + B^ cos ^ + C^ cos 7 = 0, A^ cos a + B^cos^ + C^ cos 7 = 0, reasoning as in Ex. 1. THE STRAIGHT LINE IN SPACE 279 Solving these equations for the ratios, we have the Theorem. If a, /3, and 7 are the direction angles of the line (J), then cos a _ cos P _ cos -y B1C2—B2C1 ~ C1A2—C2A1 ~ A1B2—A2B1 ' The denominators are readily remembered as the three determinants of the second order B, C, B„ C„ C, A^\ \A, B, Co A^\ \ At, Bg '2 ^2 formed from the coefficients of x, ?/, and z in (I). PROBLEMS 1. Find the points in which the following lines pierce the coordinate planes, and construct the lines : (a) 2x + 2/-z = 2, X-2/ + 22-4. (c) a; + 2?/ = 8, 2x- 4?/ = 7. (b) 4x + 3y-6z = 12, 4x-32/ = 2. (d) ?/ + z = 4, X-2/ + 22 == 10. 2. Find the direction cosines of the following lines : (a) 2X-2/ + 2z = 0, x + 22/-2Z = 4. _ Ans. i^^sVeS, Tg'VV^, Tt\V65. (b) x + 2/ + z = 5, x-?/ + z = 3. An&. ±\ V2, 0, T^ V2. (c) 3x + 22/-z = 4,x-2y-2z = 5. ^TW. i^^Vs, TiVs, ±5*5 VB. (d) x + ^-3z = 6,2x-?/ + 3z = 3. ^ns._0, ±t%V10, ±i:^-\^. (e) x + ?/ = 6, 2x-3z = 5. Ans. ±^35V22, T3j\V22, ±JjV22. (f) 2/4- 3z = 4,3 2/- 5z = l. ^«s. ±1,_0, 0. (g) 2x-3y + 2 = 0, 2x-32/-2z = 6. Ans. ±^3 Vl3, i^^g Vl3, 0. (h) 5x-142-7 = 0, 2x + 7z = 19. Ans. 0, ±1, 0. 3. Show that the lines of the following pairs are parallel and construct the lines : (a) 2 2/ + z = 0, 3 2/ - 4 z =: 7 ; and 5 ?/ - 2 z = 8, 4 2/ + H z = 44. (b) x + 22/-z=7,2/+z — 2x=:6;and3x + 6^— 3z = 8,2x-2/-2 = 0. (c) 3x + z = 4, 2/ + 2z = 9; and 6x — 2/ = 7, 3 2/ + 6z = l. 4. Show that the lines of the following pairs meet in a point and are perpendicular : (a) x + 22/ = l, 2?/ — z = l; and x — 2/ = l, x — 2z = 3. (b) 4x + 2/ — 32 + 24 = 0, z = 5; and x + ?/4-3 = 0, x + 2 = 0. (c) 3x + 2/-z = l,2x— 2 = 2; and2x — ?/ + 22 = 4, x — 2/+22 = 3. 280 NEW ANALYTIC GEOMETRY 5. Find the angles between the following lines, assuming that they are directed upward, or in front of the ZX-plane : 77" (a) x + y — z~0, y+z = 0; and x — y = 1, x — 3 y + z = 0. Ans. -. o (b) x + 2y + 2z = l,x — 2z=zl; and 4,x + 3//-z + l=0, 2x + 32/=0. Ans. cos-i^^. (c) X — 2y+z = 2,2y— z = l; and x — 2y+z = 2, x — 2y + 2z = 4. Ans. cos-i^. 6. Find the equations of the planes through the line X + 2/ — z = 0, 2 X — ?/ + 3 2 = 5, which are perpendicular to the coordinate planes. Ans. 3x + 2z = 5, 3r/— 5z + 5 = 0, 5x + 22/ = 5. 7. Show analytically that the intersections of the planes x — 2y — z = 3 and 2x — 4?/ — 2z = 5 with the plane x + y — 3z = are parallel lines. 8. Verify analytically that the intersections of any two parallel planes with a third plane are parallel lines. 108. The projecting planes of a line. The three planes pass- ing through a given line and perpendicular to the coordinate planes are called the project- ing planes of the line. If the line is perpendicular to one of the coordinate planes, any plane containing the line is per- pendicular to that plane. In thi.-^ case we speak of but two project- ing planes, namely, those drawn through the line perpendicular to the other coordinate planes. If the line is i^arallel to one of the coordinate planes, two of the projecting planes coincide. By (VI), Art. 106, the equation of any plane through the line (1) 3.'K-f22/-^-l = 0, 2x -y + 2z-S = has the form 3 .^ + 2 ?/ - « - 1 -f /o (2 .T - y + 2 ,- - 3) = 0. THE STRAIGHT LINE IN SPACE 281 Multiplying out and collecting terms, (2) (3 + 2k)x+(2-k)7/+{-l-\-2k)z-l-Sk = 0. This plane will be perpendicular to the XF-plane when the coefficient of z equals zero, that is, if A; = ^. Writing this value of k in (2) and reducing, (3) 4a- + |y-f = 0, 0TSx + 3y -5 = 0. This is therefore the equation of the projecting plane of the line (1) on XY, that is, of the plane ABA^D^ of the iigure. Now equation (3) is simply the result obtained by cUmlnotlng zfrom the equations (1); namely, we multiply the lirst of equar tions (1) by 2 and add it to the second. Hence the result : To find the ejnatlons of the prqjectlng planes of (i line., elim- inate X, y, and z In turn from the gioen equations. Thus, to finish the example begun, eliminating // from (1), we find 7 ic + 3 ;s — 7 = for the projecting plane on XZ. Eliminating r, we get 7?/ — 8«4-7 = for the equation of the projecting plane on YZ. Special forms of the projecting planes will indicate special positions of the line relative to the coordinate planes. These cases should be noted in the following problems. PROBLEMS 1. Find the equations of the projecting planes of the following lines; (a) 2 X + ?/ — z = 0, x — ?/ + 2 z = 3. Arts. 6a; + 2/ = 8, 3x + z = 8, 3(/-5z + (5 = 0, (b) a; + 7/ + z = 0, z — ?/ — 2z = 2. Ans. 3x + 2/ = 14, 2a; — z = 8, 2?/ + 3z = 4. (c) 2x + y — z = 1, X — 2/ + z = 2. Ans. Line parallel to YZ. x = l, ?/ — z + l = 0. (d) x + y-4z = l,2x + 2?/ + z = 0. Ans. Line parallel to XY. 9x + Oy = 1, 9z + 2 - 0. (e) 2y + 3z = G, 2?/-3z = 18. Ans. Line parallel to OX. ?/ = 0, z =— 2. (f) 2x-?/+ z = 0, 4x + 3;/ + 2z = 0. Ans. 6?/ = G, lOx + 5z = 0. (g) X + z = 1, X — z = 3. Ans. x = 2. z = — 1. 282 NEW ANALYTIC GEOMETRY 2. Reduce the equations of the following lines to the given answers and construct the lines : (a)x + 2/ — 22 = 0, X — y + 2 = 4. Ans. x = lz + 2, y =^^z — 2. {h) x + 2y — z = 2, 2x + 4y + 2z = 5. Ans. z = \, y =- Ix + I. (c) X — 2y + z = 'i, x + 2y — z = Q. Ans. x = 5, y = iz + l. (d) X + 3z = 6, 2x + 6z = 8. Ans. z = 4, x =- 6. (e) X + 22/ — 2z = 2, 2x + ?/ — 4z = 1. Ans. x = 2z,y = l. (i) x — y + z = S,3x-3y + 2z = 6. Ans. z = 3, y = x. 3. Find the equations of the line passing through the points {— 2, 2, 1) and (- 8, 5, - 2). Ans. x = 2z-4, ?/=-z + 3. 4. Find the equations of the projection of the line x = z4-2, y = 2z — i upon the plane x + y — z = 0. Ans. x = ^ z + \^, y = iz— y . 5. Find the equations of the projection of the line z = 2, y = x — 2 upon the plane x — 2y — 3z = i. Ans. x=— 5z + 4, y =— 4z. 6. Show that the equations of a line may be written in one of the forms ( y = nix + a^ j x — a, Jx \z = nx +b, iz = my + b, \y according as it pierces the FZ-plane, is parallel to the FZ-plane, or is pai-allel to the Z-axis. 7. Show that the condition that the line x = mz + a, y = nz + b should , . . , , / ,, . a — a' h — h' intersect the line x = m z + a ., y = n z ■\- b is m n — n 109. Various forms of the equations of a straight line. Theorem. Parametric form. The coordinates of any point P (x, I/, z) on the line through a given jJolnt Pi(x^, y^, z^) whose direction angles are a, jS, and y are given by (II) x = Xj^-\- pcosa, y = y^-^ p cos p, z = Zj^-\- p cosy, where p denotes the variable directed length P^P- Proof. The projections of P^P on the axes are respectively ^ - ^v y - Vv •' - ^r But, by the first theorem of projection, these are also equal to p cos a, p cos p, p cos y. Hence X — x^ — p cos if ?/, = j/g and 2^ =z,, ? if z, = z^ and x^ = Xg =* 4. Do the following sets of points lie on sti-aight lines ? (a) (3, 2, - 4), (5, 4, - 6), and (9, 8, - 10). Ans. Yes. (b) (3, 0, 1), (0, - 3, 2), and (G, 3, 0). Ans. Yes. (c) (2, 5, 7), (- 3, 8, 1), and (0, 0, 3). Ans. No. 5. Show that the conditions that the three points P^(Xj, y■^, Zj), ^2 (•''2' 2/21 ^2)' ^'^^^ ^s (•''S' Vzi %) should lie on a straight line are X2 - Xj y.-, - 2/j Zo — z, 6. Find the equations of the line passing through the point (2, — 1,-3) whose direction cosines are proportional to 3, 2, and 7, and reduce them to the given answer. Ans. x = ^z+^f',y = fZ— }. 7. Find the equations of the line passing through the point (0, — 3, 2) which is parallel to the line joining the points (3, 4, 7) and (2, 7, 5). . X y+3 z-2 Ans. - = = 1 -3 2 «o, ., .., 1- X — 2 y + 2 z ,x + l ?/ — 5 z + 3 8. Show that the lines = ^ = - and — ■ — = = -- — are parallel. "" ~ ~ X — I u — 6 2 — 3 * From (V), - — - = ^ — - = - — - • The value of the last ratio is infinite unless 2-3=0. If 2 - 3 = 0, then the last ratio may have any value and may be equal X — 1 V — 6 to the first two. Hence the equations of the line become — — = ' — — '2=3. Geometrically it is evident that the two points lie in the plane 2 = 3, and hence the line joining them also lies in that plane. THE STKAIGHT LINE IN SPACE 285 9. Find the equations of the line through the point (—2, 4, 0) which is parallel to the line - = ^ = , and reduce them to the answer. ^ ^ -'^ Am. x = -4z-2,y = -3z + -i. ,«.^, , , ,. x + 2 w — 3 2-1 ,x-S y z + 3 10. Show that the lines — ^— = = and = - = — ^ ,. , 6-3 2 2 6 3 are perpendicular. «• Q ?/4-l z 3 11. Find the angle between the lines = = and ° 9 1 1 X + 2 V 7z 27r = = -, if both are directed upward. Ans. 12 1 3 12. Find the parametric equations of the line passing through the point (2, — 3, 4) whose direction cosines are proportional to 1, — 2, and 2. Ans. X = 2 + 1/3, y=-3-|p, z = 4 + |p. 13. Construct the lines whose parametric equations are (a)x = 2 + |p, 2/ = 4-ip, z = 6 + |p. (b)x=-3-f/), ^ = 6-fp, z = 4 + fp. 14. Find the distance, measured along the line x = 2 — y\ p, ?/ = 4 + i| p, z = — 3 + y*3 p, from the point (2, 4, — 3) to the intersection of the line with the plane 4x — i/ — 2z = 6. Ans. 1|. 15. Sliow that the symmetric equations of the straight line become i = i, z = z,, if COS 7 = 0. What do they become if cos a =0 ? cos a cos /3 if cos /3 = ? 16. Show that the symmetric equations of the straight line become z = z^, X = Xj, if cos 7 = cos o" = 0. What do they become if cosa = cos /3 = ? if cos /3 = cos 7 = ? 17. Reduce the equations of the following lines to the symmetric form (IV). (a) X - 2 y + z = 8, 2 X - 3 y = 13. Ans. ^ = - = ^ • 3 :;^ 1 Solution. Find the equations of two projecting planes. The second plane is already the projecting plane on XY. Eliminating x, we get y — 2z =— S. Now in the two projecting planes thus found, (1) 2x-Sy-lS and y-2z=-3, solving each for y and equating results, ,„, 2x-13_7/_22-3 (2) ^^______. Multiplying the numerators through by I, we have the answer.. 28G NEAV AXALYTIC GEOMETRY Comparison with (IV) gives x^ = Y, j/j = 0, z^ = |, a = 3, 6 = 2, c = 1. Hence the line passes through (i/, 0, |) and its direction cosines are proportional to 3, 2, 1. A remark here is important. In (IV), Xj, ?/j, and Zj are the coordinates of any fixed point on the line. Hence for a given line the numerators in (IV) may be quite different. For example, putting z = in (1), we find X — 2 y + 3 z X = 2, y =— S. Hence the equations = = - represent the given line also. Notice that in equations (IV) the coefficients of x, y, and z must be unity. This explains the step, after deriving (1), of removing the 2 from the 2x and the 2z. (b) 4x-5?/ + 3z = 3, 4x-5;/ +z + 9 = 0. Ans. - = 1^^1,2 = 6. 5 4 (c) 2x + z + 5 = 0, x + 3z-5 = 0. Ans. z = 3, x = — 4. (d) x + 2y +6z = 5, 3x-2j/-10z = 7. Ans. ?-^ = '/^ = - . (e) 3x-2/-2z = 0, 6x-32/-4z + 9 = 0. Ans. ^-^-=?,?/ = 9. (f) 3x — 4j/ = 7, X + 3?/ = 11. Ans. x = 5, ?/ = 2. y 3 2 (g) 2x + ?/ + 22 = 7, X + 31/ + 6z = 11. Ans. = , x = 2. (h) 2x — 3?/ + z = 4, 4x— 62/-Z = 5. Ans. -^^-JL-, z = l. (i) 3z + y = l,4z-Sy = 10. Ans. y=- 2, z = 1. X — (X y f) (]) X = rnz + a., y = nz + b. Ans. = 1 18. Find the equations of the line passing through the point (2, 0,-2), J. Q y Z -1- 1 which is perpendicular to each of the lines = - = and '^: = ^±I = "-±!. , %-2 i/% + 2 q 1 9 Ans. = - = 19. Find the equations of the line passing through the point (3, — 1, 2) which is perpendicular to each of the lines x = 2z — 1, y = z + 3, and x_y_z . _ X — 3_y + l_ g — 2 2""3~4' "^' 1 ~ -6 ~~i 20. Find the equations of the line through P^^ (x^, y^, Zj) parallel to a b c a b c (b) X — 7nz + a, y = nz + b. Ans. a; — x^ _ y — Vi _ 2 m 71 1 THE STRAIGHT LINE IN SPACE 287 (0) z = a, y = mx + b. Ans. ^ = i, z = z.. 1 m (d) A^x + B^y + C^z + D^ = 0, A^x + B^y + C.,z + Z>2 = 0. Ans. x-x^ _ y-y^ B^C^ - B^C, C,A^ - A^C^ A^B., - A^B^ 21. Find the equations of the line passing through P^ (a;^, j/^, z^) which is perpendicular to each of the lines X-X2 ^ y-y<2 . ^ Z-Z.2 ^^^ x-Xg ^ y-y^ , ^ z-Zg _ 02 62 ^2 ' «3 ^3 Cg ^oCg — 63C2 fottg — CgOg Og^g — ttg^g 110. Relative positions of a line and plane. If the equations of the line have the form (IV), and if Ave substitute the values of two of the variables given by (IV) in the equation of the plane, then if the result is true for all values of the third vari- able, the line lies in the plane. We next easily prove the Theorem. A line tvlwse direction angles are a, /8, and y and the jilane Ax -\- By -\- Cz -\- D = are (a) parallel when and only when A cos a -f 5 cos y? + C cos y = ; (b) perpendicular when and only when A. _ B _ C cos a cos yff cos y Proof. The direction cosines of a perpendicular L^ to the plane are proportional to A, B, and C. The line and plane are parallel when and only when the line is perpendicular to the line 7.., ; that is, when and only when A cos a ■\- B cos ^-\- C cos y = 0. The line and plane are perpendicular when and only when the line is parallel to L^ ; that is, when and only when cos a cos )8 cos y 288 NEW ANALYTIC GEOMETRY PROBLEMS 1. Show that the line = = - is parallel to the plane 4x + 2 2/ + 2z = 9. ^ ~^ ^ 2. Show that the line - = - = - is perpendicular to the plane 3 x + 22/ + 7z = 8. 3 2 7 3. Show that the line x = z — 4, y = 2z — Z lies in the plane 2x — 4. Find the equations of the line passing through (1, — 6, 2 ) and per- pendicular to the plane 2x — w + 6z = 0. . x — \ 2/ + 6 z — 2 Ans. = = . 2-16 5. Show that the lines x = 2z + l, 2/ = 32 + 2, and 2x = z + 2, 3 2/ = 6 — z intersect, and find the equation of the plane determined by them. Ans. 20x — 9y — 13z — 2 = 0. J. 2 w4-2 z 3 6. Show that the line = = lies in the plane 2 x + 2i/-z + 3 = 0. ^ ~^ "* 7. Find the equations of the line passing through the point (3, 2, — 6) which is perpendicular to the plane 4x — 2/ + 3z = 5. x-3 y -2 z+6 Ans. -1 3 8. Find the equations of the line passing through the point (4, — 6, 2) which is perpendicular to the plane x4-2y — 32 = 8. x-4 2/ + 6 z-2 Ans. = = 12-3 9. Find the equations of the line passing through the point (— 2, 3, 2) which is parallel to each of the planes 3x — ?/ + z = and x — z = 0. . x + 2 y-d, z-2 Ans. = = 1 4 1 10. Find the equation of the plane passing through the point (1, 3, — 2) which is perpendicular to the line = = ^ o * — 1 Ans. 2 X + 5 1/ — z = 19. 11, Find the equation of the plane passing through the point (2, — 2, 0) which is perpendicular to the line z = 3, y = 2x — 4. Ans. x + 2y ■\- 2 — 0. THE STRAIGHT LINE IN SPACE 289 12. Find the equation of the plane passing through the line x + 2 z = 4, J* Q 7/ _i_ 4 z 7 y — z = 8 which is parallel to the line = = 2 3 4 Ans. a; + lOy- 8z- 84 = 0. 13. Find the equation of the plane passing through the point (3, 6, — 12) which is parallel to each of the lines = = and z+2 3-132 = ——'2/ = 3. Ans. 2x + Sy — z=zS6. 14. Find the equations of the line passing through the point (3, 1, — 2) "which is perpendicular to the plane 2x — y — 5z = 6. Ans. 2 = — I z -f y-, ?/ = ^ z + |. ,, „, , , ,. X — 2 ^ + 1 z ,x — 2 y + l z 16. Show that the lines = = and = = - 3 4-2 -18 2 intersect, and find the equation of the plane determined by them. Ans. 14x — 4y + 13z = 32. /g 2 7/4-3 16. Find the equation of the plane determined by the line = ~ z — 1 " = and the point (0, 3, — 4). Ans. x + 2y + 2z + 2 = 0. 17. Find the equation of the plane determined by the parallel lines x+1 V —2 z ,x— 3 w+4 z— 1 t^^ = ^ = Iand ^!LLZ = ^ — i. Ans. 8x + y- 26z + 6 = 0. 3 2 13 2 1 -T-y -r 18. Find the equations of a line lying in the plane x + Sy — 2z + 4^ = ^ 4 ?/ + 2 z 2 and perpendicular to the line = = at the point where it meets the plane. 19. Find the equations of a line tangent to the sphere x^ + y"^ + z^ = 9 at the point (2, — 1, — 2), and parallel to the plane x + 3j/— 5z — 1 = 0. 20. Find the equations of a line tangent to the sphere x^ + y'^ + z"^ = 9 at the point (2, 2, — 1), and perpendicular to the line = = - . o 1 O 21. Find the equations of the line passing through Pi(Xj, y^, Zj) which is perpendicular to the plane Ax + By + Cz + D = 0. A B C 22. Find the equation of the plane passing through tlic point X — x, _ ?/ — 2/2 _ z — z^ a b c Ans. a(x — Xj) + 6(y — ?/i) + c(z — Zj) = 0. Pi(Xj, 2/j, z,) which is perpendicular to the line a b c 290 NEW ANALYTIC GEOMETRY 23. Find the angle 6 between the line ^- = ^ = and the plane ^x+Bi,+ C.+ Z- = 0. « ^„ %, ^ ^^ ^ns. sin = V^=2 4- ir-i + C2 Va2 + 62 + c2 i/i/zi. The angle between a line and a plane is the acute angle between the line and its projection on the plane. This angle equals — increased or decreased by the angle between the line and the normal to the plane. 24. Find the equation of the plane pa.ssing through F^ (x.,, 2/3, Zg) which is parallel to each of the lines ?Z1^ = yjzll = ^ - ~\ ^^^^^ -''■-' 2 ^ V-Vi a, 61 Ci a„ 62 ^2 Ans. {b^c^ - Vi) {x - x^) + {c^a^ - a^c^) {y - 2/3) + {ajb^ - aj>{) (2 - Zg) = 0. 25. Find the condition that the plane A^x + B^y + C^z + D^ = should be parallel to the line A„x + B^y + G^z + D„ = 0, A^x + B^y + C^z + D3 = 0. Ans. A^{B^C^ - bIc^) + B, (C^g - ~C,A,) + C, {A„B^ - A,B„) = 0. 26. Find the equation of the plane determined by the point P, (x^, ?/^, z,) and the line J-^x + B-^y + C^z + D^ = 0, A^x + B^y + CgZ + I)., = 0. ^jts. (^2^1 + So2/i + <^22i + -D2) (^ia; + Bi2/ + C^z + X>,) = {A^x^ 4- -Ki2/i + C^Zj + Di) (^„x + i^22/ + C^z + X»2). 27. Find the equation of the plane determined by the intersecting lines «! 6j Cj tto ft., c., ^ns. (ft^Co - 62C1) (X - Xj) + (Ci«2 - C2ai) (2/ - ?/i) + {a A " «2'^i) (- - ^i) = 0. 28. Find the equation of the plane determined by the parallel lines x-^i ^ y-Vx ^ ^-^1 and ^~^'^ = '^ ~ ''- = ^~~2 . a b c a b c Ans. [(2/1 - 2/2) c - (z, - Z2) b]x+ [(Zi - Za) « - (^1 " ^2) c] ^ + [(-^1 -^2)^- iVi - Vi) «] 2 + (2/1^2 - 2/2^1) « + (z,X2 - ZoX^) b + (X12/2 - a^oJ/i) c = 0. 29. Find the conditions that the line x = mz -\- a,y = nz + b should lie in the plane Ax + By + Cz + D = 0. Ans. Aa + Bb + D = 0, Am + Bn + C = 0. 30. Find the equation of the plane passing through the line y-y-i z — z, ^. , . „ w ., T a; — X, y — y^ z — z^ — - — -^ = 1 which is parallel to the line = — ; — - = -■ bi Ci «2 ^2 ^2 Ans. (61C2 - 62C1) {x-x^) + (Cia2 - f otti) {y - 2/0 + (ai&2 - «2^) (^ " ^i) = 0- CHAPTER XVII SPECIAL SURFACES ^ 111. In this chapter we shall consider spheres, cylinders, and cones* (surfaces considered in elementary geometry), and surfaces which may be generated by revolving a curve about one of the coordinate axes, or by moving a straight line. 112. The sphere. We begin Avith the Theorem. Tlie equation of the sphere tvhose center is the point (a, (3, y) and whose radius is r is (I) {X - ay + (y - py + (z - yy = r\ Proof. Ijct P (x, y, z) be any point on the sphere, and denote the center of the sphere by C. Then, by dehnition, PC = r. Sul)stituting the value of PC given by the length formula, and squaring, we obtain (I). q. e. d. When (I) is multiplied out, it is that is, it is in the form ^■'^ -\- y'^ -V z^ -\- Gx -\- liy + Iz + A' = 0. The question now is. When is the locus of tills equation a sphere ? To answer this, collect the terms thus : (a:^ + Gx) + (//2 4- ///y) + («' + /-) = - K. *In analytic geometry the terms "sphere," "cylinder," and "cone " are usually used to denote the spherical surface, cylindrical surface, and conical surface of elementary geometry, and not the solids bounded wholly or in part by such surfaces. 291 292 NEW ANALYTIC GEOMETRY Completing the squares within the parentheses, we obtain (x + 1 Oy +(!/ + h_ Hf +{z + \ If = \ {G-' + 7/2 _^ /2 _ 4 A'). Comparing with (I), we have at once the Theorem. The loms of an equation of tJie foiin (II) x^ + z/2 4- z2 + Gat + fii/ 4- Iz +K = Is determined as foUoivs : (a) IVhen 6''^+ 7/^-1- Z^— 4 K > 0, the locus is a sj^here^whose center is I — -^j ~ "o ' ~" o ) ^^^ whose radius is r = ^ VG'- + 7/- + 7'-4A:. (b) When 6'^+ //'+ J^— 4 7v = 0, the locus is the point-sphere* ~Y ~ 2' ^2y (c) When G'- + 77" + 7^ — 4 7v < 0, there is no locus. In numerical examples it is recommended that the theorem be iiot used, but that the squares be completed as in the proof, and the center and radius be found by comparison with (I). EXAMPLE What is the locus of the equation x2+2/2 4.22_2x + 32/ + l = 0? Solution. Collecting terms, (x2-2x) + (2/' + 32/) + z2 = -l. Completing the squares, (x2-2x + l) + (2/2+32/ + |) + 2^ = -l + 1+1, or (X-l)2+(y+|)-2+z2 = | This equation is in the form (I); r=|, cr = 1, /3 =— |, 7 = 0. That is, tlie locus is a sphere of radius | and center (1, — |, 0). * That is, a poiut or sphere of radius zero. SPECIAL SURFACES 293 PROBLEMS 1. Find the equation of the sphere whose center is the point (a) (a, 0, 0) and whose radius is a. Ans. x^ + y^ + z^ — 2 ax = 0. (b) (0, /3, 0) and whose radius is (3. Ans. x^ + y^ + z'^— 2/3?/ = 0. (c) (0, 0, 7) and whose radius is 7. Ans. x^ + y^ + z^ — 2 yz = 0. 2. Determine the nature of the loci of the following equations and find the center and radius if the locus is a sphere, or the coordinates of the point-sphere if the locus is a point-sphere. (ay'j24.2/2+z2_6x + 4z = 0. (c) x^+y^+ z^+ 4x — z+ 7 = 0. (b) 'x^ + y^ + z^+ 2x - 4y - 5 = 0. (d) x^+ y^+ z^-12x + 6y + 4z=0. 3. Where will the center of (II) lie if (a)G = 0? (c) 7 = 0? (e) 11=1 = 0? (b)fl" = 0? {d)G = n = Of {i)I=G = 0? 4. Prove that each of the following loci is a sphere, and find its radius and the coordinates of its center. (a) The distance of a point from the origin is proportional to the square root of the sum of its distances from the three coordinate planes. (b) The sum of the squares of the distances of a point from two fixed points (2, 4, — 8) and (- 4, 0, 2) is equal to 104. _ Ans. ct = — 1, /3 = 2, 7 = - 3, r = Vl4. (c) The distance of a point from the origin is half its distance from the point (3, — 6, 9). (d) The distance of a point from the point (7, 1, — 3) is twice its dis- tance from the point (— |, — 2, |). /— — - Ans. a=— 4, /3=— 3, 7 = 1, r = • (e) The sum of the squares of the distances of a point from the three planes -x + 2y + 2z — l = 0,2x — y+2z — l = 0,2x + 2y—z — l=0 is unity. 5. Show that a sphere is determined by four conditions and formulate a rule by which to find its equation. 6. Find the equation of a sphere passing through the three points in any one of the following columns and through a fourth point selected from the other two. ^(-1,-1,1). -D (0,0,1), G(0, -4, 5), B{-1,- 3, 1), E(3, 0, 2), H{2, - 4, 5), C(-l, -4, 4); i^(2, 0,1); 7(3,-1,5). Ans. x^ + y^ + z^ — 2x + 4y — 6z + b = 0. 294 NEW ANALYTIC GEOMETRY 7. Find the equation of a sphere which (a) has the center (3, 0, — 2) and passes through (1, 6, — 6). Ans. x^+ y'^ + z^— Qx + 4 z — 36 = 0. (b) passes through the points (0, 0, 0), (0, 2, 0), (4, 0, 0), and (0, 0, — 6). Ans. x^+ y^ + z^—4x — 2y + 6z = 0. ( c ) is concentric with the sphere x^ + y^ + z^ — 6x + 4-;; = and passes througli the point (3, 1, 0). (d) has the line joining (4, — 6, 5) and (2, 0, 2) as a diameter. ( e ) lias the center (2, 2, — 2) and is tangent to the plane 2x + y — 3z + 2 = 0. ( f ) has a unit radius and is tangent to each of the coordinate planes in the first octant. ( g ) passes through the three points (1, 0,2), (1,3, 1), and (—3,0,0) and has the center in the A'Z-plane. Ans. x^ + y"^ + z^ — 2x + 6z — 15 = 0. (h) passes through the three points (1, — 3, 4), (1, — 5, 2), and (1, — 3, 0) and has its center in the plane x + ?/ + z = 0. Ans. x- + y^ + z" — 2 x + 6y — 4 z + 10 — 0. ( i ) has its center on the I'-axis and passes through the points (0, 2, 2) and (4, 0, 0). Ans. x" + y" + z^ + iy -16 ^0. ( j ) passes through the points (1, 5, — 3) and (— 3, 0, 0), and whose center lies on the line of intersection of the planes 3x + y + z = 0, x + 2y — l=0. Ans. x"^ + y~ + z- - 2x + 6z — 15 = 0. (k) is tangent to the three coordinate planes and to the plane 6a; + 22/ + 3z — 4 = 0. Ans. x^ + y- + z- - 2x - 2y -2z — 2 = 0. ( 1 ) has its center at (3, 1, 1) and is tangent to the sphere x^ + y'^ + 22_2x-4y + 2z + 2 = 0. Ans. x^ + y^ + z- -6x — 2y - 2z +10 = 0; x^ + y^ + z- — 6x — 2y — 2z — U = 0. (m) passes through the points (1, 1, 0), (0, 1, 1), and (1, 0, 1) and whose radius is 11. -4ns. x^ + 2/'^ + z- — 14x — 14?/ — 14z + 26 = 0. ( n) is tangent to the plane x+y — z + l = 0?it the point (3, — 2, 2) and has its center in the XF-plane. (o) passes through the three points (2, 0, 1), (2, — 1, 0), and (1, — 1, 1) and is tangent to the plane 2x + 2y — z + 2 = 0. Ans. x"^ + y'^ + z^ — 4 x + 2 y — 2 z + 5 = 0. ( p ) passes through the intersection of the two spheres x^ + y^ + z^ — 6x = 0, x^ + y^ + z^ + 9y — 5z — 7 = 0, and through the point (0, 1, 1). 8. Find the equations of the tangent plane and the normal line to the sphere x^ + y^ + z^ — 14 = a.t the point (3, - 2, 1). 9. Find the equations of the tangent plane and normal line to the sphere x^ + j/2 + j;2 _ 2x + 4?/ — 6z + 5 = at the point (3, — 4, 2). SPECIAL SURFACES 295 10. Find the equations of tlie planes tangent to tlie sphere x^ + y^ -\-z'^ — lOx + 5y—2z — 24 = at the points where it intersects the coordinate axes. 11. Find tlie equation of a sphere inscribed in the tetrahedron formed by any four of the following planes : 14j;+ 52/-2Z-168 = 0, lOx + Uy + 2z+ 88 == 0, 14x— 5y + 2z+28 = 0, 2x- y-2z+ 12 = 0, 10a;-ll2/4- 2z+ 33 = 0, 2x- y + 2z+ 8 = 0. 12. Find the equation of the smallest sphere tangent to the two spheres x^ + y^ + z^ -2x-6y + 1 = 0, x"^ + y- + z^ + 6x + 2y- 4:Z+ 5 = 0. Ans. x^ + i/^ + z- + 2 x — 2 y — 2 z + S = 0. 113. Cylinders. A surface which is generated by a straight line which moves parallel to itself and intersects a given fixed (iurve is called a crjlinde): The fixed curve is called the directrix. We now consider equations whose loci are cylinders. EXAMPLES 1. Determine the nature of the locus of y"^ = 4nX. Solution. The intersection of the surface with a plane x = k, parallel to the FZ-plane, is the pair of lines (1) x = k, y = ±2\^, which are parallel to the Z-axis. If A: > 0, the locus of equations (1) is a pair of lines ; if A; = 0, it is a single line (the Z-axis) ; and if fc < 0, equations (1) have no locus. Similarly, the intersection with a plane y = k, parallel to the Z J-plane, is a straight line whose equations are x = \k'', y = k, and which is therefore par- allel to the Z-axis. The intersection with a plane z = k parallel to the A'F-plaue is the parabola z = k, 2/2 = 4 X. For different values of k these parabolas are equal and placed one above another. The surface is therefore a cylinder whose elements are parallel to the Z-axis and intersect the parabola y^ = ix, z = 0. 296 NEW ANALYTIC GEOMETRY It is evident from Ex. 1 that the locus of any equation which contains but two of the variables x, y, and z will intersect planes parallel to two of the coordinate i)lanes in one or more straight lines parallel to one of the axes, and planes parallel to the third coordinate plane in congruent curves. Such a sur- face is evidently a cylinder. Hence the Theorem. The locus of an equation in which one \iariahle is lacking is a cylinder wliose elements are j)arallel to the axis along which that variable is tneasured. The student should not infer from this statement that the equations of all cylinders have one variable lacking. In case the elements are inclined, all three variables will appear in the equation. This is illustrated by the following example : 2. Determine the nature of the locus of X- + 2 .cz + z^ = 1 — y"^. The intersection of this locus by the i^lane y = k is y = k, X + z = zb Vl — ^-, a pair of parallel lines whose direction is independent of k. In fact, the direction cosines of these lines are proportional to — 1, 0, 1 ; that is, they are parallel to the line joining the point (—1, 0, 1) to the origin. "We conclude then that the surface is a cylinder. To construct the surface, draw its traces and pass lines through them hav- ing the above direction. The trace in the FZ-plane is the circle y" + z^ = l; in the XY-plane, the circle x2 + 2/2 = 1. It is evident from Ex. 2 that in order to prove that a surface is cylindrical it is only necessary to find a system of planes which cut from it a system of parallel lines. SPECIAL SURFACES 297 PROBLEMS 1. Determine the nature of the following loci, and discuss and con- struct them : (a) x2 + t/2 = 36. (f ) z^ + x^ = f\ (b) x2 + ?/ = 3x. (g) x^ + 6y = 0. (c) x^-z^ = 16. , (h) yz-4 = 0. (d) 2/2 + 4 z2 = 0. (i) 2/2 + 2-4 = 0. (e) x2+22/-4 = 0. (j) 2/2_a.3^0. 2. Find the equations of the cylinders whose directrices are the folJow- ing curves and whose elements are parallel to one of the axes : (a) 2/2 + 22 _ 4 ?/ =0, X = 0. (c) b-^x^ - a^y"- = a?b^, 2 = 0. (b) 22 + 2 X = 8, 2/ = 0. (d) ?/2 + 2p2 = 0, X = 0. 3. Prove that the following loci are cylinders. Discuss and construct them, (a) X + ?/ - 22 = 0. (d) x2 - 4 (2 + ^) + 8 = 0. (b) X2 + ?/2 — 1 = 0. (e) x2 + 2x2/ + Ip- = 2- (c) y2 =^ 3x + 2. (f ) x2 - 2x2/ + ?/2 = 1 - z^- 4. A point moves so that its distance from a fixed point is always equal to its distance from a fixed line. Prove that the locus is a parabolic cylinder. 6. A point moves so that the difference of the squares of its distances from two intersecting perpendicular lines is constant. Prove that the locus is a hyperbolic cylinder. 6. A point moves so that the sum of its distances from two planes is equal to the square of its distance from a third plane. The three planes are mutually perpendicular. Prove that the locus is a parabolic cylinder. 7. A point moves so that the sum of its distances from two planes is equal to the square root of its distance from a third plane. Prove that the locus is a parabolic cylinder when the three planes are mutually perpendicular. 114. The projecting cylinders of a curve. The cylinders whose elements intei'sect a given curve and are parallel to one of the coordinate axes are called the projecting cylinders of the curve. The equations may be found by eliminating in turn each of the variables a-, y, and z from the equations of the curve, l^'or if we eliminate z, for example, the result, by the preceding section, is 298 NEW ANALYTIC GEOMETRY the equation of a cylinder which passes through the curve, since values of a-, y, and ;;; which satisfy each of two equations satisfy an equation obtained from them by eliminating one variable. The equations of two of the projecting cylinders may be conveniently used as the equations of the curve.* Hence the problem of constructing the original curve reduces to that of constructing the curve of intersection of two cylinders whose elements are parallel to the coordinate axes. The method is illustrated in the following examples. EXAMPLES 1. Construct the curve of intersection of the two cylinders x"- + 1/2 - 2 ?/ = 0, ?/ + z2 _ 4 = 0. Solution. Draw the trace of each cylinder on the coordinate plane to v^hich its elements are pei'pendicular. Then consider a plane perpendicu- F \ () / A i h '/ ^' ' A <- c 1 1 E ^ G lar to the coordinate axis to which the elements of neither cylinder are parallel. In this case such a plane is ?/ = fc. Let this plane intersect the * In general, the equations of a curve may be replaced by any two inde- pendent equations to which they are equivalent ; that is, by two independent equations which are derived by combining the given equations. SPECIAL SURFACES 29^ axis at the point K. It will intersect the traces at the points J^, B, C, and I). Through each of these points will pass an element of the corresponding cylinder, all four elements lying in this plane. The points of intersection E, F, G, and H of these elements are points on the curve of intersection of the two cylinders. By taking several positions of the plane y = k, we obtain a sufficient number of points to construct the entire curve as shown in the second figure on page 298. 2. Construct the curve whose equations are 2y^ + z^ + 4x = 4z, y"^ + Sz^ - 8x = 12z. Eliminating x, y, and z in turn, we obtain the equations of the project- ing cylinders 2/2 + z2 _ 4 2^ z'^ — 4 X = 4 r, y'^ -\- Ax — 0. The figure shows the first and third of these cylinders, intersecting in the original curve constructed by the method explained in the previous example. It is usually wise to deduce the equations of all three of the projecting cylinders, for it may be that two of them are distinguished for simplicity and hence are most convenient to construct. If the curve lies in a plane parallel to one of the coordinate planes, then two of its projecting cylinders coincide with the plane of the curve, or part of it. For a straight line the projecting cylinders are the j)roject- ing planes. 3P0 NEW ANALYTIC GEOMETRY PROBLEMS 1. Construct the curve in which the following, in each case a plane and a cylinder, intersect : 2. Construct the curve in which the following pairs of cylinders intersect : rx2-42/ = 0, rx2+2/2 = 25, ^'\y-^+4z = 0. ^ ' \5z + y^ + 10y = 0. ri/2 + 4 z = 0, r?/- + z' - 36 = 0, ^ ' |x'2 + 2/2 - 4 = 0. ^°^ \x2 + ^2_7y^0. rx2- 9z + 36 = 0, r2/2 + z2_ 36 = 0, ^^^ |x2 + 2/2 _ 36 = 0. ^ ^ |x2 + 2/2 - 5 y = 0. r2/-^ + 4. = 0, r2/2+x2_36 = 0, ^ ^ \x2+2/2-42/ = 0. ^ ^ \z2+2/2-62/ = 0. rx2+z2_25=0, rz2/ = 12, (^) I2/2 - z = 0. ^^^ |x2 + 2/2- 72/ + 6 = 0. 3. Find the equations of the projecting cylinders of the following curves and construct the curve as the intersection of two of these cylinders : (a) x2 + 2/2 + z2 = 25, x2 + 4 y'^ - z^ = 0. (b) a;2 + 4 2/2 - z2 = 16, ix^ + y^ + z^ = 16. (c) x2+ 2/2= 4z, x'-— 2/2 = 8 z. (d) x2 + 22/2+ 4z2= 32, x2 + 42/2 = 4z. f 2/2- x2+ 2^2+ 72/- 72 = 0, (e) 2/2 + z,c = 0, 2/- + 2x + 2/ - 2 = 0. ^ ^^ |x2 _ z2 _ 7 ^^ ^ 36 = 0. rx2_ 102/ -5z- 25 = 0, r2x2+2/2-9z = 0, ^ ^ V-+ 2 2/2+ 5z+102/-25 = 0. ^^' l?/^ + 9z-72 = 0. rx2+ 22/2 + 4z- 4 = 0, f2x2 + 22/2 + Z2/- 142/ = 0, ^°^ |2x2+ 52/2+ 12z- 8 = 0. ^-"^ |x2 + 2/2 + 2z2/-72/-18 = 0. 4. A point is two units from the Z-axis and the sum of its distances from the A'F-plane and the FZ-plane is equal to its distance from the ZX-plane increased by 2. Construct its locus. 5. A point is equidistant from the Z-axis and the A'F-plane, and its distance from the origin is equal to its distance from the FZ-plane increased by 2. Construct the locus. 115. Parametric equations of curves in space. If the coordi- nates X, j/, and 2; of a point P in space are functions of a variable parameter, then the locus of P is a curve (compare Art. 80). SPECIAL SURFACES 301 For example, if (1) x = ii^ 2/ = l-2i, 2 = 3^3+2, where tisa variable parameter, then the locus of {x, y, z) is a curve in space. This curve may be drawn by assuming values for t, computing x, y, and z, plotting the points, and then joining these points in order by a continuous curve. Equations (1) are called the parametric equations of the curve. The equations of the projecting cylinders of the curve, the locus of (1), result when the parameter t is eliminated from each pair of the equations. Thus, taking the first two, (2) x = lt^ y = l-2t, we find from the second, t = l(^ — y), and substituting in the first, (3) 4a; = i(l-?/)2, or (?/ - 1)2- I6x = 0, and the locus lies on this parabolic cylinder. Similarly, eliminating t from the first and third equations of (1), x = \t-, 2 = 3(3+2, we obtain the cubic cylinder (4) (2 -2)2 =.5 76x3. Hence the curve (l) is the curve of intersection of the cylinders (3) and (4). In some cases it is convenient to find the equations of a curve in space by using a parameter. EXAMPLE Equations of the helix. A point moves on a right cylinder in such a man- ner that the distance it moves parallel to the axis varies directly as the angle it turns through around the axis. Find the equations of the locus. Solution. Choose the axes of coordi- nates so that the ecjuation of the cylin- der is (5) X- + y- = a", as in the figure. Let Pq on OX be one position of the moving point, and P any other position. Then, by definition, the distance NP (= z) varies as the angle XON {=6); that is, z = bd, where 6 is a constant. Furthermore, from the figure, X = OM = ON cosd = a cos 0, y = MN = ON sin = o sin 0. 302 NEW ANALYTIC GEOMETRY Hence the equations of the helix are : (6) X = a cos 0, y = asind, z = bd, where ^ is a variable parameter. Ans. Eliminating from the first two of equations (6), we obtain (5), as we should. Given the equations of the projecting cylinders, to find para- metric equations for the curve. It was shown in Art. 81 that an indefinite number of parametric equations could be obtained for the same plane curve. The same statement holds for space curves, as illustrated in the following example. EXAMPLE Find parametric equations for the curve of intersection of the sui'faces (see Example 2, Art. 114), 2^/2 +.2- + 4x = 42, ?/- + 322- 8a; = 12z. Solution. The projecting cylinders are (7) 2/2 + 2^ = 42, z^ — \x = \z, 2/2+4a; = 0. If we assume y = 2t, then the last equation will give x =— t^. From either of the other two cylinders we find z = 2±2 Vl - i2. Hence the given curve is the locus of (8) X =-f\ y = 2t, z = 2±2 Vl - t-. Other parametric equations result when we set one of the coordinates in (7) equal to some other function of a parameter. The aim is, of course, to find simple parametric equations. The method adopted must depend upon the given problem. PROBLEM Find simple parametric equations for the curves of Problems 2 and 3, p. 300. Ans. For Problem 2. (a) x = 2t, y = t^, 2 = — | <*. (b) x = 2cos^, y = 2smd, z=—s\n^O. (c) X = 6 cos 0, y = 6 sin 0, z = 4(1 + cos^ 0). 116. Cones. The surface generated by a straight line turning around one of its points and intersecting a fixed curve is called a.6'07ie. SPECIAL SURFACES 303 EXAMPLE Determine the nature of the locus of the equation 16 x^ + y^ — z^ = 0. Solution. Let Pj (x^, y^, z^) be a point on a curve C on the surface in which the locus intersects a plane, for example z = k. Then (1) 16x{ + y^ -z^ = 0, Zi = k. Now the origin O lies on the surface. "We shall show that the line OPj lies entirely on the surface. The direction cosines of OP.^ are -^, -^, and -^ , where p? = x? + yr + z? Pi Pi Pi A-i 1 1 1 = OP^. Hence the coordinates of any point on OP^ are, by (II), Art. 109, (2) X = ~p, Px Vi z = ^p. Pi Substituting these values of x, ?/, and 2 in the left-hand member of the given equation, we obtain P'l ' (3) ^Q^iP'^y

r, a =■ r, and ar the surface is called an anchor ring or torus. 6. Find the equations of the cylinders of revolution whose axes are the coordinate axes and whose radii equal r. Ans. y- + z'^ = r^ ; z^ + x'^ = r"^ ; x^ + y"^ = r'^. SPECIAL SURFACES 307 7. Find the equations of the cones of revolution whose axes are the coordinate axes and wliose elements make an angle of with the axis of revolution. Ans. y'^+ z'^ = xHan^(p; z'^+ x'^=y'^tan^. 8. Show that the following loci are surfaces of revolution : (a) 2/2 + z2 = 4x. (f) (x2+22)2/:^4a2{2a-?/). (b) x2-4?/2+22= 0. (g) x^ + y^+zx^+zy^-z + 3 = 0. (c) 4x2 + 4 2/2 _ ^2 ^ 16. (h) X* -y* + z*+2 x'^z"- = 1. (d) x2-4?/2+22_3y = 0. (i) X2+ i/2+z3_2y + 1 =0. (e) X22 + X2/2 — 3. 9. A point moves so that its distance from a fixed plane is in a con- stant ratio to its distance from a fixed point. Show that the locus is a surface of revolution. 10. A point moves so that Its distance from a fixed line is in a constant ratio to its distance from a fixed point on that line. Prove analytically that the locus is a cone of revolution. What values of the ratio are excluded ? 118. Ruled surfaces. A surface generated by a moving straight line is called a ruled surface. If the equations of a straight line involve an arbitrary constant, then the equations represent a system of lines which form a ruled surface. If we eliminate the parameter from the equations of the line, the result will be the equation of the ruled surface. For if {pc^, 7/j, z^ satisfy the given equations for some value of the parameter, they will satisfy the equation obtained by eliminating the parameter; that is, the coordinates of every point on every line of the system satisfy that equation. Cylinders and cones are the simplest ruled surfaces. EXAMPLES 1. Find the equation of the surface generated by the line whose equations are , 1 x + y — kz,x — y=-z. k Solution. We may eliminate k from these equations of the line by multiplying them. This gives (1) x2-?/2=:22_ This is the equation of a cone (Art. 116) whose vertex is the origin. As the sections made by the planes x = k are circles, it is a cone of revolu- tion whose axis is the A'-axis. 308 NEW ANALYTIC GEOMETRY We may verify that the given line lies on the surface (1) for all values of /c as follows : Solving the equations of the line for x and y in terms of z, we get x = -{k + -\z, y = -[k )z. 2\ kj 2\ k) Substituting in (1), ^(-r^'4('-*/ an equation which is true for all values of k and z, as is seen by removing the parentheses. Hence every point on any line of the system lies on (1), since its coordinates satisfy (1). 2. Determine the nature of the surface z^ — 3zx 4- 8?/ = 0. Solution. The intersec- tion of the surface with the plane z = k is the straight line A-3-3A:a; + 8?/ = 0, z = k. Hence the surface is the ruled surface generated by this line as k varies. To construct the surface con- sider the intersections with the planes x == and x = 8. Their equations are respec- ti^^ly X = 0, 8 (/ + ^3 = ; and X = 8, By- 24 z + 23 = 0. Joining the points on these curves which have the same value of z gives the lines generating the surface. The method used in Ex. 2 is adapted to the determination and construction of ruled surfaces. An examination of the equation of such a surface will suggest a system of planes whose intersections with the surface are a system of lines, as illustrated in Problem 2 on the following page. SPECIAL SURFACES 309 PROBLEMS 1. Show that the following loci are ruled surfaces whose generators are parallel to one of the coordinate planes. Construct and discuss the loci : (a) z — xy = 0. (f ) j/2 _ x^z. (b) x^y - z2 = 0. (g) y = xz{2- z)^. (c) z'^ - zx + y = 0. (h) 2/2 _ a.2(22 + 1^ (d) xhj + xz = y. (i) 2/2 = x2{z2-l). {e)y-xz"' = 0. (j) 2/2^x2(1 -22). Remark. The surfaces may be easily constructed from string and cardboard. 2. Show that the following loci are ruled surfaces : (a) (x + 2/)z+ (x + ?/)2-l = 0. (b) j;2 _ 2 a-2 - ?/2 + 22 = 3. (c) 2/2 + 4 ^2 ^ jy _ 4 ^2 _ 2 xz + 3 = 0. (d) X3 + 3 ?/x2 - X22 _ 3 2/22 _ .^2 ^ ^2 _ Q. (e) x2 - 2/2 = 2. (f) X2 - 2/2 =22-1. /fi/i^ Find a system of planes which cut the surface in a system of straight lines. 3. Find the equations of the ruled surfaces whose generators are the following systems of lines, and discuss the surfaces : (a) X + y = k, k {x — y) = a?. Ans. x^ — y^ = a^. (h) 'ix — 2y = kz,k{4:X + 2y) =z. Ans. 16x^ — 4y^ = z-. (c) X — 2 y — 4 kz, k {x — 2 y) = 4. Ans. x^ — 4 2/2 = 16 2. (d) X + ky + 4 z = 4 k, kx — y — 4 kz = 4. Ans. x2 + 2/2 — 16 22 = 16. (e) X — y — kz = 0, X — z — ky =: 0. (f ) Sx- z-k = 0, ky -z = 0. 4. Given two planes, one with a variable intercept on the A'-axis, the other with a variable intercept on the Y-axis. The remaining intercepts being unity, find the equation of the ruled surface generated by the line of intersection of these planes (a) when their variable intercepts are in the ratio 1 : 2. (b) when their distances from the origin are in the ratio 1 : 3. Ans. [y (z + 7/)f - [3x (2 + x)]^ = {4xy)^. (c) when the sum of their distances from the origin is unity. CHAPTER XVIII TRANSFORMATION OF COORDINATES. DIFFERENT SYSTEMS OF COORDINATES 119. Translation of the axes. Formulas applicable to space, entirely analogous to those established in Chapter IX for the plane, are derived as ex- Z'< z\- (h,k,/ ±, /X plained below. Theorem. 17ie equations for translating the axes to a neiv origin O' (h, k, V) are (I) x = x^ + h, y = y' + k, z = z' -{-1. Proof. Let the coordi- nates of any point before and after the translation of the axes be (x, y, z) and {x\ y\ «') respectively. Projecting OP and OO^P on each of the axes, we get equations (I). q.e.d. 120. Rotation of the axes. Simple formulas for rotation arise if two of the axes are rotated about the third. For example, when the axes OX and OY are turned through an angle about the Z-axis, the ^.'-coordinate of any point P does not change, and the new x- and ?/-eoordinates are given by formulas (II), Art. 55. Hence the Theorem. The equations for rotating the axes about the Z-axis through an angle 6 are (II) jr= jr'costf — y'sin^, y = x^ s\nd -\- y' co&d , z = z\ 310 TRANSFORMATION OF COORDINATES 311 Similar formulas result when the axes are rotated about OY or OX. If the axes are rotated about the origin into the new- position 0-X'Y'Z', and if the coordinates of any point P before and after the rotation are respectively (x, y,z) and (x\ y\ z'), we have the Theorem. Jf a^, (3^, y^; a^, fS,^, y^- and a^, /S^, y.^, are resj^ectiveli/ the direction angles of the three mtitu- ally perpendicular lines OX', OY', and OZ', then the equations for rotating the axes to the position O-X'Y'Z' are (III) X = x' cosa^ + z/'cosa2 + z' cosa^. i y = x'cos/3j^ -{-y'cosp^ + z'cos^3, }^z = jr'cosy^ + i/'cosy2 + ^'cosy3.* Proof Projecting OP and OA'B'P on each of the axes OX, OY, and OZ, we obtain immediately equations (III). q. e.d. Theorem. The degree of an equation is unchanged by a trans- forination of coordinates. This may be shown by reasoning as in Art. 57. PROBLEMS 1. Transform the equation x'^-\-y'^ — Ax-\-2y — Az + l^Qhj trans- lating the origin to the point (2, — 1, — 1). Ans. x^ + 2/^ — 4 2 = 0. 2. Derive the equations for rotating the axes through an angle 6 about (a) tlie X-axis ; (b) the Y-axis. * The direction cosines of OX', OY', and OZ' obviously satisfy the six equations cos- at + cos2 /3i + cos2 71 = 1, cos a\ cos a.i + cos ^i cos ^2 + cos 7^ cos 72 = 0, cos2 ai + cos^ ^2 + cos2 72 == 1, cos a^ cos a^ + cos /Sg cos ^^ + cos 72 cos 73 = 0, cos^ u'g + cos2 /33 + cos2 73 = 1 , cos a^ cos a I + cos /Sg cos /3i + cos 73 cos 7i = 0. Hence only three of the nine constants in (III) are independent. 312 NEW AXALYTIC GEOxMETRY 3. Show that the following equations may be transformed into the given answers by translating the axes, or by rotating them about one of the coordinate axes (see Art. 71) : (a) X- + y-^ - Z' - 6x - 8y + lOz = 0. Ans. x- + y- - z"^ = 0. (b) 3x2 — 8x2/ + 32/2— 522+ 5 = 0. -^ns. x''- - 1 y^ + bz^ = b. (c) 2/2 + 4z2 - 16x - 6y + 16z + 9 = 0. Ans. y^ + 4z2 = \Qx. (d) 2x2-52/2- 522- 62/3 = 0. Ans. x^-4y^-z- = 0. (e) 9x2 _ 252/2 + 16z2 - 242X - 80x - GOz = 0. Ans. x^-y^ = Az. 4. Show that Ax + By + Cz + D = may be reduced to the form x = by a transformation of coordinates. Hint. Remove the constant term by translating the axes, then remove the 2-teim by rotating the axes about the F-axis, and finally remove the y-term by rotating about the Z-axis. 5. Transform the equation 5 x2 -|- 8 2/'^ + 522 — 4 2/z + 8 2X + 4 x?/ — 4 x + 2?/ + 4z = by rotating the axes to a position in which their direction cosines are respectively f, f, i ; i, — I, f ; J, — i, — I- Ans. 3 x2 4- 8 2/2 = 2 2. 6. Show that the xy-term may always be removed from the equation .4x2 + 2)2/2 + C22 + Fxy + ii = by a rotation about the Z-axis. 7. Show that the 2/2-term may always be removed from the equation ^x2 + By^ + Cz^ + Dyz + ^ = by rotating about the X-axis. 8. What are the direction cosines of OX, OF, and OZ (Fig., p. 311) referred to 0X\ OY', and OZ' ? What six equations do they satisfy ? 9. Show that the six equations obtained in Problem 8 are equivalent to the six equations in the footnote, p. 311. 10. If (x, y, 2) and (x', y', z') are respectively the coordinates of a point before and after a rotation of the axes, show that X' -\- y" + Z^ = X'2 + ?/'2 -I- 2'2_ 11. The possibilities of simplifying an equation by rotation of the axes will be stated here without proof. Consider the equation of the second degree Ax" + By- + Cz' 4- Byz + -E'2X ^- Fxy -^^ Gx + Hy ^ Iz ^ K = 0. It is shown in more advanced treatises that it is always possible to de- termine at least one new system of rectangular axes OA'', 01"', and OZ', having the same origin as the old axes, and .such that the tran.sformed equation with reference to these new axes will lack terms in xy, yz, and zx. This simplification of the equation is accomplished by a transformation of the form (III), that is, a rotation of the axes about the origin. TRANSFORMATION OF COORDINATES 313 121. Polar coordinates. The line OP drawn from the origin to any point P is called the radius vector of P. Any point P determines four numbers, its radius vector p, and the direction angles of OP, namely a, p, and y, which are called the polar coordi- nates of P. These numbers are not all independ- ent, since a, /3, and 7 satir.fy (II), Art. 88. If two are known, the third may then be found, but all three are retained for the sake of symmetry. / Conversely, any set of values of p, a, /?, and y which satisfy (II), Art. 88, determine a point whose polar coordinates are p, a, (3, and y. Projecting OP on each of the axes, we get the Theorem. The equations of transformation from rectangiilar to polar coordinates are (IV) x = pcosa, y = pcosp, z=y9cosy. Obviously (1) p' = x' + y-' + z\ which expresses the radius vector in terms of x, y, and z. 122. Spherical coordinates. Any point P determines three numbers, namely, its radius vector p, the angle 6 be- tween the radius vector and the Z-axis, and the angle <^ between the projection of its radius vector on the A'F-plane and the A-axis. These numbers are (tailed the spherical coordinates of P. 6 is called the colatitude and the longitude. Conversely, given values of p, 6, and <^ determine a j)oint P whose spherical coordinates are (p, 6, 4>)- 314 NEW ANALYTIC GEOMETRY Projecting OP on OA , OM = p sin 6, and projecting OP and OMP on. eacli of the axes, we prove the Theorem. The equations of transformation from rectangular to spherical coordinates are (V) x = ps,\ndcos, y = p sin sin ^, z z= p cos 6. The equations of transformation from splierical to rectangular coordinates may be obtained by solving (V) for p, 6, and . 123. Cylindrical coordinates. Any point P (x, ?/, z) determines three numbers, its distance ?: from the A'F-plane and the polar coordinates (r, ) of its projection (x,u,0) on the A'F-plane. These three numbers are called the cylindrical coordinates of P. Conversely, given values of r, , and z de- ^.^ termine a point whose cylindrical coordi- nates are (r, ^, z). Then we have at once the Theorem. Tlie equations of transforma- tion from rectanrjular to cylindrical coordi- nates are (VI) j:=rcos^, z/=rsin^, z = z. The equations of transformation from cylindrical to rectangu- lar coordinates may be obtained by solving (VI) for ?•, «^, and z. PROBLEMS 1. AVhat is meant by the "locus of an equation" in the polar coordi- nates /D, cr, /3, and 7? in the spherical coordinates p, (9, and 0? in the cylindrical coordinates r, 0, and z ? 2. How may the intercepts of a surface on the rectangular axes be found if its equation in polar coordinates is given? if its eciuation in spherical coordinates is given? if its equation in cylindrical coordinates is given? 3. Transform the following equations into polar coordinates : (a) z" 4- v" + z^ = 25. Ans. p = 5. (b) x^ + y- — z~ = 0. Ans. 7 = - • (c) 2 x2 — !/■- — 22 = 0. -4ns. a = cos-i^V3. TRANSFORMATION OF COORDINATES SIS 4. Transform the following equations into spherical coordinates : (a.) x^ + y^ + z" = 16. Ans. p = i. (h) 2x + Sy = 0. Ans. = tan- 1 (- |). (c) Sx^+Sy- = 7z\ Ans. ^ = tan-iiV21. 6. Transform the following equations into cylindrical coordinates: (a) 5x — y = 0. Ans. (^ = tan-i 5. (b) a;2 + y2. - 4, ^^s_ ,. _ 2. 6. Find the equation in polar coordinates of (a) a sphere whose center is the pole. (b) a cone of revolution whose axis is one of the coordinate axes. Ans. (a) p = constant; (b) a = constant, /3 = constant, or 7 = constant. 7. Find the equation in spherical coordinates of (a) a sphere whose center is the origin. (b) a plane through the Z-axis. (c) a cone of revolution whose axis is the Z-axis. Ans. (a) p = constant ; (b) <;& = constant ; (c) — constant. 8. Find the equation in cylindrical coordinates of (a) a plane parallel to the A'F-plane. (b) a plane through the Z-axis. (c) a cylinder of revolution whose axis is the Z-axis. Ans. (a) z = constant; (b) = con!?tant; (c) ?■ = constant. 9. In rectangular coordinates a point is determined as the intersec- tion of three mutually perpendicular planes. Show that (a) in polar coordinates a point is regarded as the intersection of a sphere and three cones of revolution which have an element in common. (b) in spherical coordinates a point is regarded as the intersection of a sphere, a plane, and a cone of revolution which are mutually orthogonal. (c) in cylindrical coordinates a point is regarded as the intersection of two planes and a cylinder of revolution which are mutually orthogonal. 10. Show that the square of the distance r between two points whose polar coordinates are (pj, a^, /3^, y^) and (p.,, tr.,, /Sj, 7") is j'2 = p~ 4. p I _ 2 p, P2 (cos = 0. 12. Find the general equation of a sphere in polar coordinates. Ans. p^ + p{G cos a + H cos ^ + 1 cos 7) 4- K = 0. CHAPTER XIX QUADRIC SURFACES AND EQUATIONS OF THE SECOND DEGREE IN THREE VARIABLES 124. Quadric surfaces. The locus of an equation of the sec- ond degree in x, y, and z, of which the most general form is (1) ^ x^-\- Bif+ Cz'-^ Dijz + Ezx + Fxy -\-Gx+Hy-\- Iz + A' = 0, is called a quadric surface or conicoid. We may learn something of the nature of such a surface by taking cross seetioas. We first obtain Theorem I. The intersection of a quadric with any plane is a conic or a degenerate conic. Proof. By a transformation of coordinates any plane may be made the A'F-plane, z = 0. Referred to any axes the equation of a quadric has the form (1) (Theorem, p. 311). Hence the equation of the curve of intersection referred to axes in its own plane 5; = is A x^ + Fxy + B7f + Gx -\-Hy-{-K = 0, and the locus is therefore a conic or a degenerate conic, by Art. 70. Q.E.D. As already pointed out in Art. 71, the parabola, ellipse, and hyperbola were originally studied as conic sections, — plane sec- tions of a conical surface. From the preceding theorem and by intuition, the truth of the following statement is manifest. Corollary. The curve of intersection of a cone of revolution lo'ith a plane is an ellipse, hyperhola, or parabola, accordlncj as the 2:)lane cuts all of the elements, is parallel to two elements 316 QUADRIC SURFACES 317 (cutting the other elements — some on one side of the vertex and some on the other'), or is paixdlel to one element (cutting all the others on the same side of the vertex). For sections of a quadric by a set of parallel planes, the following result is important : Theorem II. The sections of a quadric with a system of paral- lel planes are conies of the same species. The truth of this statement is established in the following sections. Tiie meaning of the theorem is this : A set of parallel sections will all be ellipses, or all hyperbolas, or all parabolas, the exceptional cases (Art. 70) under each species being included. 125. Simplification of the general equation of the second degree in three variables. If equation (1) be transformed by rotating the axes, it can be shown that the new axes may be so chosen that the terms in yz, zx, and xy will drop out (Problem 11, p. 312). Hence (1) reduces to the form A 'x"^ + i3y + C'z'^ + G'x + iVy + I'z + A'' = 0. Transforming this equation by translating the axes, it is easy to show that the axes may be so chosen that the transformed equation will have one of the two forms (1) A "x' + B'Y + C"z- + K" = 0, (2) A"x^ -\- B'Y -\- ^"^ = 0. Note the difference in (1) and (2). In (1) all the squares and no first powers are represented, in (2) only two squares and the hrst power of the other variable. If all of the coefficients in (1) and (2) are different from zero, they may, with a change in notation, be respectively writ- ten in the forms (3) ±rU|-!±5' = -'- (4) !-.±7i = 2... a^^ b^ I 318 NEW ANALYTIC GEOMETllY The purpose of the following sections is to discuss the loci of these equations, which are called central and noncentral quadrics respectively. If one or more of the coefficients in (1) or (2) are zero, the locus is called a degenerate quadric. Certain cases are readily disposed of by means of former results. If /v" = 0, the locus of (1) is a cone (Theorem, Art. IIG) unless the signs of A", B", and C" are the same, in which case the locus is z,XJomt, namely the origin. If one of the coefficients A", B", and C" is zero, the locus is a eijUnder whose elements are parallel to one of the axes and whose directrix is a conic of the elliptic or hyperbolic type. If also A'" = 0, the locus will be a pair of intersecting jAanes. If two of the coefficients A ", B", and C" are zero, the locus is a pair of parallel jolanes (coincident if 7v" = 0), or there is no locus. If one of the coefficients in (2) is zero, the locus is a cylinder whose directrix is a parabola, or di, pair of intersecting p)lanes. If tioo of the coefficients are zero, the locus is ajmir of coinci- dent planes. (A" and B" cannot be zero simultaneously, as the equation would cease to be of the second degree.) PROB) EMS 1. Construct and discuss the loci of the following equations: (a) 9x^-362/2+ 422 = 0. (e) 4y^-25 = 0. (b) 16x2- 4 2/2 _ 22 = 0. (f) 3 2/2 + 702 = 0. (c) 4x2 + z2_ 10 = 0. (g) 8?/2+25z = 0. (d) 1/2 - 9z2 + 36 = 0. (h) z2 + 16 = o. 2. Show by transformation of coordinates that the following quadrics are degenerate : (a) X2 - ?/ + 22 _ 2 + 9 = 0. (b) x2 + 4 2/2 _ z2 _ 2 x + 8 2/ + 5 = 0. (c) x2 + 2/2 + z2 + 2 X - 2 2/ + 4 z + 6 = 0. (d) x2 + 2/2 • - 2 22 + 2 2/ + 4 2 - 1 =^ 0. (e) x2 + 2/z = 0. QUADRIC SURFACES 319 A" x^ ^ ^ 126. The ellipsoid 1 1 =1. If all of the coefficients in c or k^— r, there is no locus. Hence the ellipsoid lies entirely between the planes z =±c. 320 NEW ANALYTIC GEOMETRY 111 like manner tlie sections parallel to the FZ-plane and the ZA'-plane are ellipses whose axis decrease as the planes recede. Hence the ellipsoid lies entirely between the planes x = ±_a and y ^ -^h. The ellipsoid is therefore a closed surface. If a = b, the section (1) is a circle for values of k such that — c < J: < c, and hence the ellipsoid is now an ellipsoid of revolution whose axis is the -Z-axis. If ^> = c or c = a, it is an ellipsoid of revolution whose axis is the A'- or I'-axis. If a. = i = c, the ellipsoid is a sphere, for its equation may be written in the form x^ + y^ + ^" = o'^- x^ ip' z^ 127. The hyperboloid of one sheet 1 = 1. If two of q2 ^ ^ the coefficients in (3), Art. 125, are positive and one is negative, the locus is called a hyperboloid of one sheet. Consider first the equation (1) a- h~ c^ ^2 ~2 A discussion of this equation gives us the following properties : 1. The hyperboloid is symmetri- cal with respect to each of the coordi- nate planes and axes and the origin. 2. Its intercepts on the X-axis and the F-axis are respectively •'■ = ± 2'72~' 2^^-^J ^ ^ a- b- (•'■ cr b^ r are the equations of hyperboloids of one sheet which lie along the I'-axis and the ^-axis respectively. li a = b, the hyperboloid (1) is a surface of revolution whose axis is the Z-axis, because the section (2) becomes a circle. The hyperboloids (3) will be hyperboloids of revolution if a = c and b = c respectively. X^ y2 2^2 128. The hyperboloid of two sheets = 1. If only a^ b^ (^ one; of the coefficients in (3), Art. 125, is positive, the locus is called a hyperboloid of two sheets. Consider first the equation iC ?/ Z' ^ ■' a^ V c^ 322 NEAV ANALYTIC GEOMETRY 1. The hyperboloid is symmetrical with respect to each of the coordinate planes and axes and the origin. 2. Its intercepts on the A'-axis are x = ± a, but it does not cut the F-axis and the Z-axis. 3. Its traces on the A'F-plane and the AZ-plane are respec- tively the hyperbolas -- = 1 ..2 -^' ^ ~ i^ " ' a^ r which have the same transverse axis AA' = 2a, but it does not cut the FZ-plane. 4. The equation of 'the curve in which a plane parallel to the FZ-plane, x = k, intersects the hyperboloid (1) is Z.-2 >fi -y^ V'^ z- k' _ — -\ — = 1 or a ■ + r.2 = 1. «■") -2(^'-«') This equation has no locus if — a < kr Qz"^ = ZQ. (b) 4x2 + 92/2- 16z2 = 144. (f) z- - Ax^ - ^y"- = U. ' (c) 4x2_9^2_i6z2 = i44. (g) 16x2 + ?/2 + l6z2 = 64. (d) x2 + 16 2/2 + z2 = 64. (h) x2 + 2/- - z^ = 25. 2. Reduce, by translation of the axes, each of the following to a standard form and determine the type of central quadric it represents: ^.(a) x2 + 2 2/2 + 2-2 _2x + 42/ -82 + 10 = 0. , (b) x2 - y2 ^_ 2 z2 _ 6 X + 2 2/ + 4 2 + 9 = 0. ^ (c) 2/2 - x2 _ 9 v2 ^ 6x - 2 2/ - 42 + G = 0. (d) x2 - 2 2/2 - 4 22 - 2x - 8 2/ - 8 = 0. (e) 4x2 -2/2 -22 -8x -22/ + = 0. (f ) 4 x2 — 2/2 - 22 — 8 X — 2 2/ + 4 = 0. (g) 3x2 + 4 2/2 - 8 2/ - z2 = 0. ( ^ 3. Find the equations of the planes whose intersections with the ellip- soid 9x2 + 25 2/2 ^ 169 z2 = 1 are circles. Ans. 4x = ± 12z + A:. 4. The square of the distance of a point from a line is equal to the square of its distance from a perpendicular plane (a) increased by a con- stant ; (b) diminished by a constant. How do the two loci differ ? What property have they in common ? 5. A point moves so that its distances from a fixed point and a fixed line are in constant ratio n. Determine and name the locus (a) when/;i1. (d) when the point is on the line. /i^ 324 NEW ANALYTIC GEOMETRY 6. A point moves so that its distances from a fixed point and a fixed plane are in a constant ratio. Prove that the locus is an ellipsoid of revo- lution when the ratio is less than unity, and a hyperboloid of revolution when greater than unity. 7. A point moves so that the sum of the squares of its distances from two intersecting perpendicular lines in .space is constant. Prove that the locus is an ellipsoid of revolution. 129. The elliptic paraboloid ^ — = 2cz. If the coefi&cient of if \\\ (4), Art. 125, is positive, the locus is called an elliptic paraboloid. A discussion of its equation gives us the following properties : 1. The elliptic paraboloid is symmetrical with respect to the }'Z-plane and the ZA'-plane and the Z-axis. 2. It passes through the origin, but does not intersect the axes elsewhere. 3. Its traces on the coordi- nate planes are respectively the conies o o of which the first is a point-ellipse and the other two are parabolas. 4. The equation of the curve in which a plane parallel to the A'F-plane, z = k, cuts the paraboloid is | + '^=2o., + r 2d-ck 2bhk The curve is an ellipse if c and k have the same sign, but there is no locus if c and k have opposite signs. Hence, if c is positive, the surface lies entirely above the A'F-plane. If A- increases from to oo, the plane recedes from the A'F-plane QUADRIC SURFACES 325 and the axes of the ellipse increase indefinitely. Hence the surface recedes indefinitely from the AT-plane and from the Z-axis. In like manner the sections parallel to the FZ-plane and the ZA-plane are parabolas whose vertices recede from the X F-plane as their planes recede from the coordinate planes. The paraboloid is said to " lie along the Z-axis." The loci of the equations (1) fj + | = 2«a., ^;+|'=26y, are elliptic paraboloids which lie along the A'-axis and the F-axis respectively. It a = b, the first surface considered is a paraboloid of revolu- tion whose axis is the Z-axis ; and if b = c and a = c, the parab- oloids (1) are surfaces of revolution whose axes are respectively the A'-axis and the F-axis. An elliptic paraboloid lies along the axis corresponding to the term of the first degree in its equation, and in the positive or negative direction of the axis according as that term is positive or negative. 130. The hyperbolic paraboloid —^ — — = 1cz. If the coeffi- cient of //" in (4), Art. 125, is negative, the locus is called a hyperbolic paraboloid. 1. The hyperbolic paraboloid is symmetrical with respect to the FZ-plane and the ZA-plane and the Z-axis. 2. It passes through the origin, but does not cut the axes elsewhere. 3. Its traces on the coordinate planes are respectively the -^-•h = 0, - = 2cz, -% = 2 cz, of which the first is a pair of intersecting lines and the other two are parabolas. 32(> NEW ANALYTIC GEOMETRY 4. The equation of the curve in which a plane parallel to the A' }'-plane, z = k, cuts the paraboloid is 2 U'ck = 1. The locus is a hyperbola. If c is positive, the transverse axis of the hyperbola is j)arallel to the A'- or F-axis accord- ing as k is positive or negative. If k increases from to oo, or decreases from to — oo, the plane recedes from the AF-plane and the axes of the hyperbolas increase indefinitely. Hence the surface recedes indefinitely from the AF-plane and the Z-axis. The surface has approximately the shape of a saddle. In like manner the sections I)arallel to the other coordinate planes are parabolas whose vertices recede from the AF-plane as their planes recede from the coordinate planes. The surface is said to " lie along the Z-axis." The loci of the equations ^ O 7 V ^ O — — Z 01/, —: :, = 2 ax, are hyperbolic paraboloids lying along the F-axis and the A'-axis respectively. A hyperbolic paraboloid also lies along the axis which corresponds to the first-degree term in its equation. A plane of symmetry of a quadric is called a principal plane. Each paraboloid has two principal planes ; each central quadric, three. Axes of symmetry are called principal axes. A parab- oloid possesses one such axis ; a central quadric, three. The existence of a center of symmetry for a central quadric explains the designation " central quadric." Plate II Elliptic Faraboloid Hyperbolic Paraboloid NONCENTRAL QUADRICS Hyperboloid of oue slict't Hyperbolic raiaboloid Ruled Quadrics QUADRIC SURFACES 327 PROBLEMS 1. Discuss and construct the following loci : (a) 2/2 + z2 = 4 a;. (e) 9z2 - 4x2 = 288 j^. (b) 1/2- z2 = 4x. (f) 10x2 + z^ = 64?/. (c) x2 - 4^2 = \6y. (g) ?/2 _ x2 = 10 2. (d) x~ + ?/ = 8 z. (h) I/' + 16 z2 + X = 0. 2. Reduce by transformation of coordinates each of the following to a standard form and determine the type of paraboloid it represents : {&) z = xy. (c) x2 + 2?/2_ 6x + 4?/ + 3z + 11 = 0. (b) z = x2 + xy + 2/2. (d) z2-3?/2- 4x + 2z — 6?/ + 1 = 0. 3. A point is equidistant from a fixed plane and a fixed point. Show that the locus is an elliptic paraboloid of revolution. 4. A point is equidistant from two nonintersecting perpendicular lines. Show that the locus is a hyperbolic paraboloid. 5. Prove that the parabolas obtained by cutting (a) an elliptic parabo- loid, and (b) a hyperbolic paraboloid by planes parallel to one of the principal planes, are all congruent. 6. Show analytically that any plane parallel to the axis along which (a) an elliptic paraboloid, and (b) a hyperbolic paraboloid lies, intersects the surface in a parabola. 131. Rectilinear generators. The equation of the hyperboloid of one sheet, Art. 127, may be written in the form As this equation is the result of eliminating k from the equa- tions of the system of lines a c \ hj'' a c k\ h the hyperboloid is a ruled surface. Equation (1) is also the re- sult of eliminating k from the equations of the system of lines a c \ b) a c k\ b and the hyperboloid may therefore be regarded in two ways as a ruled surface. 328 NEW ANALYTIC GEOMETRY In like manner the hyperbolic paraboloid contains the two systems of lines a k' and ah ^ a b k T^ese lines are called the rectilinear generators of these surfaces. Hence the Theorem. Tli.e liyperboloid of one sheet and the hyperbolic paraboloid have two systems of rectilinear generators, that is, they may be regarded in tivo ivays as ruled surfaces. The two systems of generators are shown in Plate 11. REVIEW PROBLEMS Name and draw the surfaces in detail all their characteristics xy = 0. xy = 1. xy-z. xy = z^. xy = z"^ + 1. xy = z^ + z. x'^ + y" — 0. X" + y" = 1. X- + y'^ = x. x2 + y'^ = z. x^ + ?/2 = z^. x^ + y~ = 2 xy. x + y = 0. x + y = 1. x + y = z. X + y = z^. x + y = xy. a;2 + 2/2 = z2 + 1, a;2 + y- = z^ — 1. x2 + 2/2 = 1 _ z2. 8. th g following groups, giving (a) x2 + 2/2 = z- + 2z. (b) x2 + 2/2=2:2-22. (c) x~ + y- = 2z- z2. (a) x2 + 2 2/2 + 3 z2 = 0. (b) x2 + 22/2 + 3z2 = i. (c) x2 + 2 2/2 + 3 z2 = 2 X. (d) x2 + 22/2 + 3z2 = 2x -1. (a) x2 + 2 2/2 - 3 z2 = 0. (b) x2 + 2y2_ 322=1. (c) x2 + 2 2/2 — 3z2 = 2a;. (d) x2 + 22/2-3z2 = 2x + 1. (e) a;2 + 2 7/2- 3^2 = 2x -1. (f) a;2 + 2 2/2 - 3 z2 = 2 X -2. (a) xy + yz + zx = 0. (b) z^ +yz + zx = 0. (c) z + yz -{■ zx = 0. (d) Z2 + X2 + zx = 0. (e) z2 + x^t+ xy = 0. (f) z^ + xy + X = 0, QUADRIC SURFACES 329 MISCELLANEOUS PROBLEMS 1. Construct the following surfaces and shade that part of the first intercepted by the second : (a) a;2 + 42/2 + 9z2 = 36, x^ + y^ + z^ = 16. (b) a;2 + 2/2 + 22 _ 64, x^ + y^ -8x = 0. (c) 4 X- + y" - 4 z = 0, x^ + 4i/^ - z^ = 0. 2. Construct the solids bounded by the surfaces (a) x^ + y^ = a^. z = mx, 2 = 0; (b) x2 + ?/2 = az, a;2 + ?/2 = 2 ax, 2 = 0. 3. Show that two rectilinear generators of (a) a hyperbolic paraboloid, and (b) a hyperboloid of one sheet, pass through each point of the surface. 4. If a plane passes through a rectilinear generator of a quadric, show that it will also pass through a second generator, and that these generators do not belong to the same system. 5. The equation of the hyperboloid of one sheet may be written in ?/2 2^ X" the form = 1 By treating this equation as In Art. 131, we obtain the equations of two systems of lines on the surface. Show that these systems of lines are identical with those already obtained. 6. Show that a quadric may, in general, be passed througii any nine points. 7. If a > 6 > c, what is the nature of the locus of x2 ?/2 2'- + ,, , + a--\ b--\ c2-\ if X > a2 ? if a2 > X > 62 ? if 62 > X > c2 ? if X < c2 ? 8. Show that the traces of the system of quadrics in Problem 7 are confocal conies. 9. Show that every rectilinear generator of the hyperbolic paraboloid x2 ?/2 X y — rr 2cz is parallel to one of the planes - ± = 0. 10. Prove that the projections of the rectilinear generators of (a) tlic hyperboloid of one sheet, (b) the hyperbolic paraboloid, on the principal planes are tangent to the traces of the surface on those planes. 11. A plane passed through the center and a generator of a hyper- boloid of one sheet intersects the surface in a second generator which is parallel to the first. 12. Show how to generate each of the central (juadrics by inoving an ellipse whose axes are variable. 13. Show how to generate each of the paraboloids by moving a parabola CHAPTER XX EMPIRICAL EQUATIONS 132. A problem quite distinct from any thus far treated in this text arises when it is required to find tlte (yjiidtlon of a curre which shall pass throur/h a series of emplriaxlly gloen j^olnts. That is, we suppose that certain values of the vr.riable and of the function are known from an actual experiment, and the cor- responding points are plotted on cross-section paper. A smooth curve is then drawn to " fit " these points, and an equation for this curve is required. The general treatment of this important problem is beyond the scope of an elementary text, and the following sections are concerned with simple cases only. 133. Straight-line law. If the curve suggested by the plotted points is a straight line, assume the law (1) y = mx + h, and determine the values of m and b from the observed data. The straight line representing the required law will not neces- sarily pass thnnujh all the points plotted, for experimental work is subject to error. It is suflBcient if the line fits the points within the limits of accuracy of the experiment. In general, the straight line may be drawn through two of the plotted points, and m and h may be calculated from their coordinates. 330 1 t "^ *w-. ^^ . V 1 ^ "v \^ rS I \ [ \ ' 1 ■ 1 '- 1 ]_ 1 \ ' \ \ - _ -^ 1 N M 1 1 — LU — 1 11 M M M 1 — >. EMPIRICAL EQUATIONS 331 EXAMPLE In an experiment with a pulley, the effort, E lb., required to raise a load of Wlb. was found to be as follows : w 10 20 30 40 50 60 70 80 00 100 E n 61- 11 9 10* 12i 13| 15 16i Find a straight-line law to fit these data. Solution. Plotting the points as in the figure, it is seen that the straight line drawn through (30, 6^) and (100, 16^) fits the observed data very- well. To find its equation, substitute these values in the equation (2) E = jnW+b. E This gives Q,\ = 30 m + 6, 20 16^=100?n + 6. Solving for m and 6, we find m = /j^g, ^ 15 b = -Y-- Substituting in (2), =| (3) E = ^S'-oW+Y, ^1*^ the required equation. For the purpose of calcu- lation we write (3) in the form (4) £■ = 0.146 n"+ 1.86, — ~ bHi — ' '" 40 60 80 100 W Load W in lbs. keeping three decimal places in the coefficient of W in order to secure three-figure accuracy. We now test (4) by comparing the observed values and calculated values. w 10 20 30 40 50 60 70 80 90 100 E, observed 3.25 4.87 6.25 7.50 9 10.5 12.2 13.7 15 16.5 E, calculated 3.32 4.78 6.24 7.70 9.16 10.6 12.1 13.5 15 16.5 The formula (4) may be used for calculating values of E for values of W intermediate between 10 and 100, and not given in the table. For example, the effort required to raise a load of 25 lb. is E = 0.146 X 25 -f 1.86 = 5.51 lb. 332 NEW ANALYTIC GEOMETRY PROBLEMS The following data treated in the same way will yield laws represented by the formula y = mx + b. 1. F is the volume in cubic centimeters of a certain quantity of gas at the temperature i° C, the pressure being constant. Find the law connecting V and t. t 27 33 40 55 68 V 109.9 112.0 114.7 120.1 125. Ans. r= 100 + 0.367 <. 2. V is the volume of a certain quantity of mercury at a temperature of $°C. Find the law connecting V and 0. e°c. 18 36 60 72 90 F(cc.) 100.32 100.65 101.07 101.30 101.61 Ans. F= 100 + 0.018(9. 3. S is the weight of sodium nitrate dissolved by 100 g. of water at the temperature (°C. Find the law connecting S and t. s G8.8 72.9 87.5 102 t -6 20 40 An». 6' = 73.0 + 0. 73 i. 4. The following are corresponding values of the speed and induced volts in an arc-light dynamo. Find the law connecting volts and revolu- tions per miinite. Rev. per minute, n 200 320 495 621 744 Volts induced, v 165 270 410 525 625 EMPIRICAL EQUATIONS 333 5. S is the weight of potassium bromide which will dissolve in 100 s;. of water at the temperature t° C. Find the law connecting ,S and t. t 20 40 60 80 s 53.4 64.6 74.6 84.7 93.6 6. Find the equation of the straight lines that best fit the following data. 0.5 0.31 0.82 1.5 1.29 1.85 2.5 2.51 3.02 t 5 10 15 20 25 30 T 15 20 24.4 28.4 32 35.2 38.2 2.50 0.64 400 0.80 500 600 0.91 0.99 750 1.12 800 1.15 900 1.22 (a) (b) (c) (d) 134. Laws reduced to straight-line laws. By suitable treat- ment of the given data many laws can be transformed into a linear relation. Some cases of this kind of frequent occurrence will now be given. 1. The law V = a -\- bx\ When the points plotted suggest a vertical parabola with its vertex on the ?/-axis, the assumed equation will have the above form. If now we set x^ = t, and plot the values of t and y, these values satisfy the relation y = a-\- ht, that is, a straight-line law. w 3 ' 13 23 33 4:] F f 1 13 91 n 334 NEW ANALYTIC GEOMETRY EXAMPLE An experiment to determine the coasting resistance R in pounds per ton of a motor wagon for the speed V miles per hour gave the following data : V 2i 5 '2 10 12i 15 R 40 40 42 45 50 55 G3 Plotting the points, the curve suggested (Fig. 1) appears to be a parabola with the equation (1) r J 00 lb r ^ 60 f e 4 i*- •• 25 , _ _ _ _ _ _ _ R = a + 6F2. 5 10 15 20 Speed V miles per Lour Fig. 1 250 t Fig. 2 To check this, calculate the values of V^ (table. Art. 3), set V^ = t, and retabulate the data thus : n=v^) H 25 56i 100 156i 225 R 40 40 42 45 50 55 63 Plotting these points (Fig. 2), it appears that they are fitted by a straight line. By the preceding section we find the equation of this line to be B = 39.3 + .107^. Hence the required law is (2) iJ = 39.3 + . 107 F2. EMPIRICAL EQUATIONS 335 PROBLEMS The following data satisfy laws of the form y = a + bx^. Determine the values of a and b. X 19 25 31 38 44 y 1900 3230 4900 7330 9780 Ans. y = 5x- + 102. s 1 4 3 8 1. 3 1 H H 2 p 2 2| H 5 n lOf 13| 221 3. V 10 20 30 40 50 R 7 9.1 14.5 20 29 d 3 a 1 2 5 8 f $ 1 S 663 1178 1841 2651 3608 4712 2. The law y = ax". Taking logarithms, we have (3) log i/ = \og a + n log x, that is, the logarithms of the given data satisfy a straight-line law. Hence in this case tabulate the logarithms of the given data, determine the straight-line law to lit them,* compare with (3) to find a and n, and substitute in y = ax". This law, as the study of the problems on page 337 will show, has wide application. ^ Logarithmic squared-paper is of great convciiiencu in tills I'ase. 386 NEW ANALYTIC GEOMETRY EXAMPLE The following data satisfy a law of the form y — ax". Find the values of a and ?;. , y X 4 7 11 15 21 y 28.G 79.4 182 318 589 Solution. Tabulating the values of log r and log y (table, p. 4), logjr 0.602 0.845 1.041 1.176 1.322 logy 1.456 1.900 2.26 2.50 2.77 Plotting x' = loga;, ?/' = log ?/, it ap- pears that x' and y' satisfy a straight-line law. The equation is found to be - - - - - - - - v- - - \r\- / 1 5 / n } J I I \ / 1 J 1 f / 1 1 1 1 i (4) y'= .364 + 1.82 x'. x' = log X Hence, comparing with (3), n = 1.82, log a = .364. Then a = 2.3, and the required law is (6) ?/ = 2.3x1-82. Ans. PROBLEMS 1. Find a law of the form y = ax» for the following data: X 2000 4000 6000 8000 10,000 y 2869 8700 16,660 26,370 37,660 2. Find a law of the form p = od" to fit the following data : V 4 4.5 5 5.5 6 7 p 110 97.1 86.8 78.4 71.5 60.7 EMPIRICAL EQUATIONS 337 3. The time, t seconds, that it took for water to flow through a tri- angular notch, under a pressure of h feet, until the same quantity was in each case discharged, was found by experiment to be as follows : h .043 .057 .077 .094 .100 t 1260 540 275 170 135 Find the law. 4. The indicated horse power I required to drive a vessel with a displacement of D tons at a ten-knot speed is given by the following data. Find the law connecting I and D. D 1720 2300 3200 4100 I 655 789 1000 1164 5. u is the volume in cubic feet of 1 lb. of saturated steam at a pressure of p lb. per square inch. Find the law of the form p«« = const, connect- ing p and u. u 26.43 22.40 19.08 16.32 14.04 p 14.7 17.53 20.80 24.54 28.83 6. F is the force between two magnetic poles at a distance of d centi- meters. Find the law connecting F and d. dcm. 1.2 1.9 2.3 3.2 4.5 F dynes 4.44 1.77 1.21 0.625 0.316 7. 1) is the diameter in inches of wrought-iron shafting required to transmit H horse power when running 70 revolutions per minute. Find a fornnila. H 10 20 30 40 50 60 70 80 D 2.11 2.67 3.04 3.36 3.61 3.82 4.02 4.22 338 NEW ANALYTIC GEOMETRY 8. Q is the quantity of water in cubic feet per second flowing tlirougli a riglit-angled isosceles notcli when the surface of quiet water is H feet above the bottom of the notch. Find the law. H 1 2 3 4 Q 2.63 15 41 84.4 9. A certain ship draws h feet of water and displaces V cubic feet. AYhat is the law connecting h and F? Draft ft 18 13 11 9.5 Displacement V 107,200 65,800 51,200 41,100 135. Miscellaneous laws. In many experiments the analytic form of the law is known in advance. If, however, such fact is unknown, and if the preceding methods fail, the points deter- mined by the data should in any case be plotted, and then the shape of the required curve may suggest an equation to be tried. The following problems furnish a variety in this regard. PROBLEMS 1. The curve suggested in each of the following is a vertical parabola y = a -\- bx -\- cx^ . The values of a, 5, and c may be found from three pairs of values of the data. Determine the law in each case. (a) (b) The resistance, R ohms, of a wire at t° C is given by the follow- ing table : X 1 2 3 4 5 6 7 y 25 41 55 67 77 85 91 t 5 10 15 20 25 R 25 25.49 25.98 26.48 26.99 27.51 EMPIRICAL EQUATIONS 339 (c) (d) X 0.5 1 1.5 2 2.5 3 y 5.4 6.3 6.6 6.1 5.0 3.2 0.6 u 20 40 60 80 100 V 2U0 253 215 176 136 94 2. The points may be fitted by a branch of an equihiteral hyperbola whose equation is ^ In this case plot y and - — t^ thus transforming into a straight-line law. Find the law after this manner for the following data : (a) (b) X 4 5 6 7 8 y 4410 3530 2940 2520 2210 (c) A 1.06 2.46 2.97 3.45 3.96 4.97 5.97 V 50.25 48.7 47.9 47.5 46.8 45.7 45 S 10 11 12 13 14 W 8370 48S0 - 1970 -490 - 2600 3. In some cases a branch of the rectangular hyperbola {(12), p. 184), xy=bx-\-ay a b will fit the points. Dividing through by xy, this becomes 1 = - H 11 ^ ^'^ Hence if - = h, - = v, are plotted, u and v will satisfy a straight-line law. X y Show tliat this is the case in the following and find the law. (a) X 1 10 20 30 40 50 60 70 80 y 12.8 17.1 20 22.2 23.1 23.8 23.8 24.2 340 NEW ANALYTIC GEOMETRY (b) s 1 2 3 4 5 6 7 8 t 2.05 3.23 3.95 4.49 4.87 5.20 5.40 5.60 4. Compound-interest law (p. 103). If the law sought is of the form taking logarithms, y = ae"^, logy = loga + fcx loge, where log e = 0.434, as usual. Hence log y and x satisfy a linear relation, and we accordingly plot x and log y, determine the straight line, the values of a and k, and substitute. Proceed in this manner in the following : (a) X 3.45 10.85 19.30 28.8 40.1 53.75 y 19.9 18.9 16.9 14.9 12.9 10.9 8.9 Hint Plot z and log y. (b) h 886 2753 4763 6942 10,693 P 30 29 27 25 23 20 Hint. Plot h and log p. (c) t 10 27.4 42.1 s 61.5 62.1 66.3 70.3 X 2.1 5.6 9.3 11.5 y 20 18.92 17.34 15.8 14.96 136. The problem under discussion requires for thorough solution the method of least squares, and for an exposition of this theory the student is referred to treatises on that subject. INDEX Abscissa, 10 Algebraic curve, 44 Amplitude, 108 Anchor ring, 306 Angle,eccentric,215 ; vectorial, 119 Arch, parabolic, 158 Area of ellipse, 175 Asymptotes, 51, 170 Auxiliary circle, 164 Axis, conjugate, 167; major, 161.; minor, 161 ; transverse, 167 Axis of parabola, 153 Cardioid, 221 Catenary, 113 Center, instantaneous, 218 Center of conic, 100, 167 Central conic, 186 Central quadric, 318 Circle, point-, 92 Cissoid of Diodes, 54, 210, 220 Cocked hat, 55 Colatitude, 313 Compound interest curve, 103 Compound interest law, 340 Conchoid of Nicomedes, 221 Confocal conies, 189 Conicoid, 316 Conjugate diameters, 229 Coordinates, oblique, 11 Cubical parabola, 46 Curtate cycloid, 216 Cycloid, 208, 212 Degenerate ellipse, 179 Degenerate hyperbola, 179 Degenerate parabola, 179 Degenerate quadric, 318 Director circle, 225 Directrix, 153, 186, 295 Discriminant of the equation of a circle, 93 Eccentricity, 162, 168 Ellipse, point, 165 Epicycloid, 217 Exponential curves, 102 Focal radii of conies, 187 Eocus, 153, 160, 107, 186 Folium of Descartes, 209 Four-leaved rose, 126, 223 Helix, 301 Hyperbolic spiral, 132 Hypocycloid, 217; of four cusps, 210, 213 ; of three cusps, 206 Intercepts, 46 Involute of a circle, 216 Latus rectum, 154, 161, 168 Lemniscate of Bernoulli, 55, 122, 225 LimaQon of Pascal, 55, 222, 225 Lituus, 132 Logarithmic curves, 102 Longitude, 313 341 342 NEW ANALYTIC GEOMETRY IMaximum value of a function, 136 Minimum value of a function, 136 Octant, 232 Ordinate, 10 Parabola, cubical, 46; semicubical, 205 Parabolic spiral, 223 Parameter, 8-4 Period of sine curves, 107 Point-circle, 92 Point of contact, 191 Polar axis, 119 Pole, 119 Principal axes, 326 Principal planes, 319, 326 Probability curve, 105 Prolate cycloid, 216 Radian, 2 Radius vector, 119 Reciprocal spiral, 132 Rose, three-leaved, 125, 126 ; four- leaved, 126,223; eight-leaved, 126 Semicubical parabola, 205 Spiral, hyperbolic or reciprocal, 132 ; logarithmic or equiangular, 132, 133 ; parabolic, 223 Spiral of Archimedes, 132 Strophoid, 54, 226 Symmetry, 43 System of logarithms, common, 101 ; natural, 101 Torus, 306 Traces of a surface, 257 Triangle problems, 90 Vertex of a conic, 153 Whispering gallery, 203 Witch of Agnesi, 219 I UC SOUTHERN REGIONAL LIBRARY FACILITY AA 001 318 401 5 I A V*^'