the axis take AD = *:::: at right angles to it draw Dc and Note. This method, of making p = 1, has the obvious advantage of requiring only one parabola for any number of biquadratics, the necessary variation being made in the radius of the circle. Cor. 1. When De represents a negative quantity, the ordinates on the same side of the axis with c represent the negative roots of the equation; and the contrary. Cor. 2. If the circle touch the parabola, two roots of the equation are equal; if it cut it only in two points, or touch it in one, two roots are impossible ; and if the circle fall wholly within the parabola, all the roots are impossible. cubic equation, with any given parabola, whose half parameter is AB (see the preceding figure): from the point B take, in the axis, (forward is the equation have —q, but backward if q be positive) the line BD = # ; then raise the perpendicular DC E #. and from c describe a circle passing through the vertex A of the parabola; the ordinates PM, &c. drawn from the points of intersection of the circle and parabola, will be the roots required. PROBLEM IV. To construct an equation of any order by means of a locus of the same degree as the equation proposed, and a right line. As the general method is the same in all equations, let it be one of the 5th degree, as a" — br'+acr"— a dr”+a'ez —af - 0. Let the last term afbe transposed; and, takin one of the linear divisors, % V. T." of the last term, make it equal :to z, for example, and divide the equation by a'; then will _ x5–br"--acra—a"dra-Ha'er a? On the indefinite line Bo describe the curve of this equation, BMDRLFC, by the method taught in prob. 2, sect. 1, of this chapter, taking the values of r from the fixed point B. The ordinates PM, SR, &c. will be equal to z; and therefore, from the point B draw the right line RA = f; parallel to the ordinates PM, sR, and through the point A draw the indefinite right line KC both ways, and parallel to Ba. From the points in which it cuts the curve, let fall the perpendiculars MP, Rs, cq; they will determine the abscissas, BP, Bs, Ba, which are the roots of the equation proposed. Those from A towards Q are positive, and those lying the contrary way are negative. If the right line Ac touch the curve in any point, the corresponding abscissa z will denote two equal roots; and if it do not meet the curve at all, all the roots will be imaginary. If the sign of the last term, af, had been positive, then we must have made z = —f, and therefore must have taken BA = —f, that is, below the point P, or on the negative side. ExERCISES. Ex. 1. Let it be proposed to divide a given arc of a circle into three equal parts. Suppose the radius of the circle to be represented by r, the sine of the given arc by a, the unknown sine of its third part by r, and let the known arc be 3a, and of course the Then substituting for sin u its value r, and for cos’ u its value r"—a", and arranging all the terms according to the powers of z, we shall have r°–3ror--aro-0, a cubic equation of the form r" — pro-- q = 0, with the condition that Hip">{q”; that is to say, it is a cubic equation falling under the irreducible case, and its three roots are represented by the sines of the three arcs u, u + 120°, and u + 240°. Now, this cubic may evidently be constructed by the rule in prob. 3, cor. 3. But the trisection of an arc may also be effected by means of an equilateral hyperbola, in the following manner. Let the arc to be trisected be Ab. In the circle ABC draw the semi - - E diameter AD, and to AD as a diame- B ter, and to the vertex A, draw the H equilateral hyperbola AE to which the right line AB (the chord of the A arc to be trisected) shall be a tan- A G D gent in the point. A ; then the arc AF, included within this hyperbola, is one third of the arc AB. For, draw the chord of the arc AF, bisect AD at G, so that G will be the centre of the hyperbola, join DF, and draw GH parallel to it, cutting the chords AB, AF, in 1 and K. Then, the hyperbola being equilateral, or having its transverse and conjugate equal to one another, it follows from Def. 16 Conic Sections, that every diameter is equal to its parameter, and from cor. theor. 2 Hyperbola, that GK . Ki = Ak”, or that GK : AK :: Ak : K1 ; therefore the triangles GKA, Aki are similar, and the angle KAI = AGK, which is manifestly –ADF. Now the angle ADF at the centre of the circle being equal to KAI or FAB ; and the former angle at the centre being measured by the arc AF, while the latter at the circumference is measured by half FB; it follows that AF = #FB, or = }AB, as it ought to be. Mr. Lardner, in his Analytical Geometry, gives the following elegant solution of this problem. -“Let A be the given angle. By trigonometry cos” # A-3 cos A– ; cos A=0; which, by supplying the radius r, and representing cos #A by r, becomes 4x”-3r*r-r” cos A=0; which, multiplied by r, gives 4x"—3rox”—r” cos A‘a’ =0. “Let the equation of one of the curves be, 22* = ry, and the other by substitution will be, 2y”—3ry—2 cos A:a::=0. The former is the equation of a parabola, the axis of which is the axis of y, the origin in the vertex, and the principal parameter equal to #r. “The latter is also a parabola, the equation of its axis is y = 3r; the co-ordinates of its vertex are y=}r, z= 9r2 - - - 16 cos A* and its principal parameter is cos A. “These parabolas being described, their points of intersection give the roots of the equation. The intersection at the origin gives the root a = 0, which was introduced by the multiplication by ar. . “The equation having more than one real root, it might appear that there were more values than one for the third of the given angle. But upon examining the process, it will be seen that the question really solved was not to find an angle equal to the third of a given angle, but to find the cosine of an angle which is the third of an angle whose cosine is given. Since, then, the arcs and in general all arcs which come under the general formula, 2m r + A have the same cosine, the question really solved is to find the cosine of the third of any of these arcs. And here again another apparent difficulty arises. If the number of arcs involved in the question be unlimited, shall there not be an unlimited number of values for the cosine of the third parts of these ? To account for this it - - - 2in A ehould be considered that in general the arc = * + 3 must have the same cosine as some one of the three arcs, for the number; must be either of these forms n, n + #, or n + 4, where n is an integer. If it have the form n, that is, if 3 measures m, then “And hence it follows that the cos &: + A), whatever be the value of m, must be equal to one or other of the quanti. ties. cos. 4A, cos. # (27—A), cos. # (4r—A), which correspond to the three roots of the cubic equation already found.” Er. 2. Given the side of a cube, to find the side of another of double capacity. Let the side of the given cube be a, and that of a double one y, then 2a"=y', or, by putting 2a=b, it will be a2b–y”: there are therefore to be found two mean proportionals between the side of the cube and twice that side, and the first of those mean proportionals will be the side of the double cube. Now these may be readily found by means of two parabolas : thus: Let the right lines AR, As, be joined at right angles; and a parabola AMH be described about the axis AR, with the parametera; and another parabola AMI about the axis As,with the parameter b; cutting the former in M. Then AP=z, PM==y, Vol II. - 20 |