y1 — q32 + ry — s = 0.

Assume also p = 1; then comparing the terms of the two equations, it will be, 2a-1=q,

9+1

or a = -2b=r, or b == ; a2 + b2 — n2 = − s, or

, 2

9

n2 = a2 + b2 + s, and consequently n = √(a2 + b2 + s). Therefore describe a parabola whose parameter is 1, and in at right angles to it draw DC and

the axis take AD =

---

9+1
2

r; from the centre C, with the radius √(a2+b2 + s), describe the circle M'M'GM, cutting the parabola in the points M, M', M", M""; then the ordinates PM, P'M', P"M", P"M", will be the roots required.

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 DC 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.

-

Cor:3. If a + b2 n2, or the circle pass through the point A, the last term of the equation, i. e. (a2 + b2— n2)p2=0; and therefore y1 ± (2pa + p2)y2± 2bp3y = 0, or y3± (2pa + p2)y ± 2bp2 = 0. This cubic equation may be made to coincide with any proposed cubic, wanting its second term, and the ordinates PM, P"M", P"M' are its roots. Thus, if the cubic be expressed generally by y3±qy±s=0. By comparing the terms of this and the preceding equation, we shall have 2pa + p2 = ± 9, and ± 26p2 = ± s, or Fap F and b So that, to construct a 2p' cubic equation, with any given parabola, whose half parameter is AB (see the preceding figure): from the point B take, in the axis, (forward if the equation have -q, but backward if be positive) the line BD ⇒ ; then raise the perpendicular

9

±

9

2p

DC = 2p2, and from ç 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

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 x3-bx++ acx3 — á2dx2 + a3ex -af0. Let the last term afbe transposed; and, taking one of the linear divisors, f, of the last term, make it equal

to z, for example, and divide the equation by at; then will

z=

x5 bx4 + acx3- a2dx4 + a3ex

a4

On the indefinite line вQ describe the curve of this equation, BMDRLFC, by the method taught in prob. 2, sect. 1, of this chapter, taking the values of x from the fixed point B. The ordinates PM, SR, &c, will be equal to z; and therefore, from the point в draw the right line BA =f, parallel to the ordinates PM, SR, and through the point A draw the indefinite right line KC both ways, and parallel to BQ. From the points in which it cuts the curve, let fall the perpendiculars MP, RS, ca; they will determine the abscissas BP, BS, BQ, which are the roots of the equation proposed. Those from A towards a are positive, and those lying the contrary way are negative.

If the right line AC touch the curve in any point, the corresponding abscissa x 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 è, 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 x, and let the known arc be 3u, and of course the required arc be u. Then, by equa. VIII, IX, chap. iii, we shall have

་

Putting, in the first of these equations, for sin 3u its given value a, and for sin 2u, cos 2u, their values given in the two other equations, there will arise

3 sin u

cos2 u. sin3 u

r

Then substituting for sin u its value x, and for cos2 u its value r2x2, and arranging all the terms according to the powers of x, we shall have

x3 — r2x + žar2 = 0,

a cubic equation of the form x3- px + q = 0, with the condition that p3 > 4q2; 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 semidiameter AD, and to AD as a diameter, and to the vertex A, draw the equilateral hyperbola AE to which the right line AB (the chord of the

H

E

arc to be trisected) shall be a tangent 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 I 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 = AK2, or that GK: AK:: AK: KI; therefore the triangles GKA, AKI are similar, and the angle KAI=AGK, which is manifestly 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.

Ex. 2.

= ADF.

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 oney, then 2a3=y3, or, by putting 2a=b, it will be a2b=y3: there are therefore to be found two mean proportionals be

tween

tween 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 AмH be described about the axis AR, with the parameter a; and another parabola AMI about the axis As, with the parameter b; P cutting the former in M. Then AP=x, PMy, are the two mean proportionals,

R

of which y is the side of the double cube required.

=

M

H

For, in the parabola AMH the equation is y2 = ax, and in the parabola AMI it is r2 = by. Consequently a: y::y: x, and y: x:x : b. Whence yx ab; or, by substitution, yby ab, or, by squaring, ybab2; or lastly, y3 = a2b 2a3, as it ought to be.

Note. For other exercises of the construction of equations, take some of the examples at the end of chap. viii.

GENERAL SCHOLIUM.

On the Construction of Geometrical Problems. Problems in Plane Geometry are solved either by means of the modern or algebraical analysis, or of the ancient or geometrical analysis. Of the former, some specimens are given in the Application of Algebra to Geometry, in the first volume of this Course. Of the latter, we here present a few examples, premising a brief account of this kind of analysis.

Geometrical analysis is the way by which we proceed from the thing demanded, granted for the moment, till we have connected it by a series of consequences with something anteriorly known, or placed it among the number of principles known to be true.

Analysis may be distinguished into two kinds. In the one, which is named by Pappus contemplative, it is-proposed to ascertain the truth or the falsehood of a proposition advanced; the other is referred to the solution of problems, or to the investigation of unknown truths. In the first we assume as true, or as previously existing, the subject of the proposition advanced, and proceed by the consequences of the hypothesis to something known; and if the result be thus found true, the proposition advanced is likewise true. The direct demonstration is afterwards formed, by taking up again, in an inverted order, the several parts of the analysis. If the consequence at which we arrive in the last place is found false,

we

we thence conclude that the proposition analysed is also false. When a problem is under consideration, we first suppose it resolved, and then pursue the consequences thence derived till we come to something known. If the ultimate result thus obtained be comprised in what the geometers call data, the question proposed may be resolved: the demonstration (or rather the construction), is also constituted by taking the parts of the analysis in an inverted order. The impossibility of the last result of the analysis, will prove evidently, in this case as well as in the former, that of the thing required.

In illustration of these remarks take the following examples.

Ex. 1. It is required to draw, in a given segment of a circle, from the extremes of the base A and B, two lines AC, BC, meeting at a point c in the circumference, such that they shall have to each other a given ratio, viz, that of м to N. Analysis. Suppose that the thing is affected, that is to say, that AC: CB:: M: N, and let the base AB of the segment be cut in the same ratio in the point E. Then EC, being drawn, will bisect the angle ACB (by th. 83 Geom.); consequently, if the circle

A

E

be completed, and CE be produced to meet it in F, the remaining circumference will also be bisected in F, or have FA FB, because those arcs are the double measures of equal angles: therefore the point F, as well as E, being given, the point c is also given.

Construction. Let the given base of the segment AB be cut in the point E in the assigned ratio of M to N, and complete the circle; bisect the remaining circumference in г; join FE, and produce it till it meet the circumference in c: then drawing CA, CB, the thing is done.

Demonstration. Since the arc FA the arc FB, the angle ACF = angle BCF, by theor. 49 Geom.; therefore AC: CB :: AE: EB, by th. 83. But AE: EB:: M: N, by construction; therefore AC: CB:: M: N. Q. E. D.

Ex. 2. From a given circle to cut off an arc such, that the sum of m times the sine, and n times the versed sine, may be equal to a given line.

Ana Suppose it done, and that AEE'B is the given circle, BEE the required arc, ED its sine, BD its versed sine; in DA (produced if necessary) take BP an nth part of the given sum; join PE, and produce it to meet BF to AB, or to ED, in the point F. Then, since m. ED+n. BDn. BP = n. PD + n. BD;

AL

D

consequently