scale, and draw m'N parallel to kg cutting MN in N. N will be a point on the tangent at P.

The lines corresponding to m'N will of course be parallel for all points on the curve, so that the points k and g need only be found once.

A parallel to kg through the point 6 (the quadrisection of CA) cutting 47 in T determines AT and CT, the tangents at A and C.

Ovals of Cassini.

When a point moves in a plane so that the product of its distances from two fixed points in the plane is constant, it traces out one of Cassini's ovals. The fixed points are called the foci. The equation of the curve is therefore rr, k2, where r and r are the distances of any point on the curve from the foci and k is a constant.

=

Corresponding to any given foci an infinite number of ovals may of course be drawn by varying k

PROBLEM 154. To describe an oval of Cassini, the foci F and F and the constant k of the curve being given (Fig. 156).

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1

1

Draw a line through F and F, and bisect FF, in C: through C draw BCB, perpendicular to FF,, and with F as centre and

radius=k describe an arc cutting BCB, in B and B. B and B will evidently be points on the curve.

Draw FK perpendicular to FF, and make FK=k, and on CF make CA and CA, each = CK. A and A, will be points on the

With centre F and any radius greater than FA and less than FA, describe an arc dD cutting FA in d. Through K draw Kd, perpendicular to dK and cutting FF, in d. A circle described with centre F, and radius Fd, will cut the arc dD in D, a point

on the curve.

Evidently by symmetry D,, the intersection of arcs of the same radii as the above but struck from the opposite foci as centres, will also be on the curve, and so also will be the intersections on the other side of AA1. Similarly any number of points may be found.

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1

An alternative method may be adopted as soon as two points such as A and D, not very far apart, and the two corresponding points A, and D1 are found. If two series of terms in geometrical progression are found, FA and FD being successive terms of the one and FA, and FD, successive terms of the other (Problem 8), circles struck with the corresponding terms of each as radii and with the opposite foci as centres intersect in points of the curve, the radii increasing from the one focus and diminishing from the other. This is shewn in the figure, and this construction moreover enables at once any number of ovals to be drawn, the intersection of any two circles of opposite series being taken as a starting point, and the successive intersections giving succeeding points. The second curve drawn in the figure is an example of this.

It may be noticed that a circular arc with centre at the focus coincides very closely with the oval at the vertices A and 4.

To draw the tangent at any point P.

The angle FPG which the normal at any point P makes with the focal chord FP is equal to the angle which the other focal

chord FP makes with the chord CP drawn from P to the centre.

The Cissoid of Diocles.

This curve, named after Diocles, a Greek mathematician, who is supposed to have lived about the sixth century of our era, was invented by him for the purpose of constructing the solution of the problem of finding two mean proportionals. The curve is generated in the following manner:

In the diameter ACB of the circle ADBE (fig. 157) make AN- BM, and draw MQ and NR perpendicular to AB, and let MQ meet the circle in Q, then AQ and NR intersect in a point on the curve, i.e. the locus of this intersection is the Cissoid.

RN QM JAM.MB

AN AM

=

AM

since AQB is a

By similar triangles right angle; or if we call RN=y, AN=x, and the radius of the circle a,

which is the equation to the curve referred to rectangular axes with A as origin and AB as axis of x.

PROBLEM 155. To describe the Cissoid corresponding to a circle of given diameter (Fig. 157).

Of course the above description is really a construction for the curve, since by it any number of points can be determined. The curve may also be described by continuous motion thus:

Draw a diameter AB of the circle, and the tangent at B. If A is a point on the curve, this tangent will be an asymptote. Through C, the centre of the circle, draw a parallel to the tangent at B of indefinite length, and make 40 on CA produced equal to AC. Cut a piece of paper to a right angle as abc, and on one side of it mark off from the angle the points d, c, making bd = dc=AC, the radius of the given circle. If the paper be now placed so that the edge ba passes through O, and the point c is always on ECD, the point d will be on the curve, and by moving it

the positions of any number of points can easily be marked off The curve is evidently symmetrical about AB,

on the paper.

there is a cusp at A, and D and E, the extremities of the diameter perpendicular to AB, are points on the curve.

To draw the tangent at any point P.

From P, with radius AC, mark off L on the diameter ECD. Through L draw LG parallel to AB, and through O draw OG parallel to PL, meeting LG in G. G will be a point on the normal at P, and the tangent is therefore perpendicular to PG.

It may be noted that the area included between the curve and the asymptote is three times the area of the generating circle.

The problem of finding two mean proportionals between two given quantities a and b is, to find two quantities m and n such that

or that

m2

an and n2 = mb,

m3 = ab and n3 = ab2.

By means of the cissoid corresponding to the circle, the radius of which is equal to a, the smaller of the given quantities a and b, the first term m can easily be found thus: