the pairs K, G and K, G, give parabolas which while touching BC, DB, do not touch CD.

There are in general two solutions.

Proof. It is shewn at the end of Chap. IV. among the harmonic properties of conics, that the three diagonals of a complete quadrilateral circumscribing a conic form a self-conjugate triangle. It is easily proved analytically that every parabola touches the line at infinity, i.e. has one tangent situated at an altogether infinite distance. Now BE and DE meet CD, CB respectively in infinitely distant points, pass, that is, through the points in which this infinitely distant tangent meets CB and CD, they are therefore diagonals of the circumscribing quadrilateral formed by the three given and the infinitely distant tangent, and its third diagonal must be the line CK since this meets BD in infinitely distant points. E is therefore the pole of the line CK, and conversely the polar of K passes through E.

But a straight line drawn through any point is divided harmonically by the point, the curve and the polar of the point (see end of Chap. IV.), therefore A, must be a point on the

curve.

1

PROBLEM 56. (Fig. 52.) To draw a parabola to pass through four given points A, B, C, D.

[The points must not lie at the angles of a parallelogram, and must be so situated, that being joined in pairs, the two points of each pair are both on the same side, or on opposite sides of the point of intersection of the joining line*.]

Join BC, AD to meet in E. Through C draw CK parallel to AB meeting AD in K. Take a mean proportional EG between ED and EK (Prob. 5) and CG will be the direction of the axis of the required parabola. The Problem is therefore reduced to Prob. 51.

Since the distance EG may be set off on either side of E as EG,, the line CG, will be the direction of the axis of a second parabola fulfilling the given conditions.

1

* Puckle's Conic Sections. Fourth Edition, Art. 313, Ex. 1.

EG: EC2 :: ED. EA: EC.EB,

a relation "which is known to hold in the parabola. Geom. Conics, 3rd Ed., Art. 213.)

(Besant's

PROBLEM 57. (Fig. 53.) To draw a parabola to touch four given lines AB, BC, CD, DA, no two of which are parallel.

Let CD meet AB in E and AD meet BC in G.

The circle circumscribing the triangle formed by any three of the lines will be a locus of the focus (Prob. 48), which may therefore be determined as the intersection of the circles circumscribing any two of such triangles. In the figure, the circles. circumscribing BCE and ABG are drawn. They intersect in F,

the focus. The tangent at the vertex can be at once determined, by dropping perpendiculars from F on any two of the given

Fig,53,

E

B

D

G

Y

1

1.

tangents as FY, FY, perpendiculars on AB, BC; Y and Y, are points on the tangent at the vertex. (Notes to Problem 36.)

PROBLEM 58. (Fig. 54.) To determine the centre of curvature at any point P of a given parabola.

[A circle can be drawn through any three points of a curve, but cannot in general be drawn through a greater number taken arbitrarily. If a circle be drawn through three points of a curve and the outside points be conceived to gradually move up to the centre one, the circle in the limiting position it assumes when the points approach indefinitely near to each other so as ultimately to coincide, is called the circle of curvature at the point, and its centre is called the centre of curvaturę. The circle is said to pass through three consecutive points of the curve, and obviously has closer contact with it at the point than any other circle can

90

CIRCLE OF CURVATURE.

have, since it is not possible to draw a circle through four consecutive points. The centre of curvature will necessarily lie on the normal at the given point, and any circle having its centre on the normal and passing through the point really passes through two consecutive points of the curve, since curve and circle have a common tangent.]

F is the focus, PT the tangent, and PG the normal at the point P of the given parabola.

Join PF and produce it to K, making FK equal to FP. Draw KO perpendicular to PK to intersect the normal at P in 0. 0 will be the centre of curvature at P.

If the circle of curvature cuts the parabola again in Q, it will be found that PQ, the common chord, makes the same angle with the axis as PT, the tangent, does, and that

PQ = 4PT.

The focal chord FR of the circle of curvature is known to be in length equal to 4FP, and it is on this known value of the focal chord that the construction depends.

The chord (PV) of the circle of curvature through P parallel

to the axis is also equal to 4FP, since this chord and PR are equally inclined to the tangent at P.

The length PO of the radius of curvature may also be determined by taking a fourth proportional to FY, FP and 2FP, where FY is the perpendicular from F on the tangent at P.

The locus of the centre of curvature of any curve is called the Evolute of that curve; and the original curve, when considered with respect to its evolute, is called an Involute. The chaindotted curve in Fig. 54 is the evolute of the portion of the parabola lying above the axis.

Normals to the curve are tangents to the evolute; and since the focal radius of curvature at the vertex = 2. AF, the evolute must touch the axis at a point = 2. AF from A.

If the ordinate of the point of intersection of the curve and evolute be drawn meeting the axis in N, it will be found that

AN8. AF twice the latus rectum.

The evolute of the parabola is a curve known as the semicubical parabola.

PROBLEM 59. To draw a parabola to touch two given circles, the axis being the line joining the centres.

Let C be the centre of the larger circle, c that of the smaller, R and r their radii. Determine a fourth proportional to 20c, R+r, and R-r. From C towards c set off on Cc a length CN equal to this fourth proportional, i.e. a length such that

CN : R-r :: R+r : 20c.

Draw NP perpendicular to Cc meeting the circle in P, and P will be the required point of contact of the curve. The problem therefore reduces to Prob. 47, the given point being also the point of contact of the given tangent.