Segment of a circle.-BCD (Fig. 7) is the segment. Its area is best found by subtracting the triangle ABD from the sector ABCD.

The Ellipse. This is the figure produced when a circular cylinder is cut across in any direction not perpendicular to its axis. It is not of great importance to us, so we shall not examine its properties. There are several methods of constructing an ellipse, of which we will give one, which is simple, and assists the student to remember the value of the area.

To construct an ellipse with given axes.

AB and CD are the given axes, O the centre. The lengths of the semi-axes are a and b respectively, a being the major or greater.

On AB, with centre O, describe a circle. Divide AB into a number of parts, and draw ordinates of the circle as shown. Reduce each ordinate in the proportion of b to a. Then

D

Fig. 8.

B

through the tops of the reduced ordinates draw a curve, which is the required ellipse.

The method of reduction is shown for the ordinate cd. The inner circle with radius is described, d is joined to O cutting the inner circle in e, and ef is drawn parallel to AB, cutting cơ in f, then ƒ is one of the points.

For cf:cd::ba (similar triangles).

Only one quarter need be drawn in this manner, the others are quite symmetrical.

Now each ordinate of the circle is diminished in the ratio b/a, hence evidently so is the whole area.

Hence one rule will suffice both for ellipse and circle as follows.

RULE. The area of an ellipse is π/4 times that of the circumscribing rectangle. The circle is then a particular case, where b=a, and the rectangle becomes a square.

The Parabola. This curve is the most important we have to deal with, its construction enabling us to solve graphically many problems.

To construct a parabola of given height on a given base. First case.

Let AB be the given base; and D the aper be directly over C, the middle point of AB.

The parabola will then

be symmetrical, CD being the given height.

Complete the rectangle AEFB.

Divide DE into any number of equal parts (in the figure we take five), and EA into the same number.

Number the divisions 1', 2', etc., from D, and 1, 2, etc., from E as shown.

Join D to 1, 2, 3, and 4.

Through 1' draw a vertical cutting DI in I; through 2' a vertical cutting D2 in II; and so on, obtaining the points I, II, III, IV. Then the curve is drawn through A, the above four points, and D.

Similarly for the other side.

DC is called the axis, D the apex; and the nature of the curve is such that the distance of any point in it from DE varies as the square of its distance from DC. It will be seen that the construction fulfils this condition.

For example, IV is twice as far from DC as II is, but is 4, i.e. 22, times as far from DE.

The parabola may also be required to have its greatest height at some point not directly over C.

AB is again the base, and D the given highest point, at a height h say over M.

Join CD, and complete the parallelogram AF.

The construction now proceeds exactly as in the preceding case, the ordinates being drawn parallel to CD. The two sides are now unsymmetrical, CD is not the axis, nor is D the apex. CD is, however, an axis.

The curve has an important property, which we shall require to use hereafter, viz. The tangent at any point bisects the abscissa of that point.

Thus take for example IV, in either figure, then the tangent at IV passes through 2', i.e. bisects D4'.

Another method of constructing the curve, showing

yet another property, is illustrated in Fig. II.

This method is of use in cases where the actual height in inches is not fixed. AB is divided into ten parts, and they are numbered from each end as shown. Then at each point set up an ordinate representing the product of the two numbers at the point; either vertically, as in the figure, or all parallel in any direction.

The curve then passes through the tops of the ordinates.

The property which leads to this method of construction is, in words—

The ordinate is proportional to the rectangle contained by the segments into which it divides the base. For the area we have

RULE. The area of a segment of a parabola is of that of the circumscribing parallelogram; i.e.of the product of base and perpendicular height.

The circumscribing parallelogram means the parallelogram AEFB in Figs. 9 or 10.

The Hyperbola.-This curve together with the two preceding constitute the conic sections, which are treated in Analytical Geometry. For our purposes we select only those properties which are directly useful; and we define the curve by those properties, and not as it is defined in geometry.

For our purposes then we define the curve thus:

The hyperbola is the curve traced by a point which moves so that the product of its distance from two rectangular axes is constant.

This is strictly a special case, and is the rectangular hyperbola. It is, however, the only one we concern ourselves with.

Construction of the hyperbola.

Let OX, OY be the axes, and A one known position of the moving point. Through A draw lines AN and AM, parallel to OX and OY respectively. Produce AN as far as necessary. (The curve has no limits.)

Mark off along MX distances M1, 12, etc., equal or

not, so far as the curve is required to extend, and draw the ordinates II, 22, etc. Join now the top points 1, 2, etc., to O, cutting MA in 1', 2′, etc.

Through ' draw 'I parallel to OX, cutting 11 in I; through 2′, 2′II cutting 22 in II; and so on for III, IV, V, etc.

Then I, II, etc., are points on the required curve, and we draw it through them.

By Euclid we easily prove that the rectangles, OI, OII, etc., are equal to OA.

Area. The particular area required is that bounded by the curve, two ordinates, and OX; such as ABCD in Fig. 13.

The log. is the Napierian or natural

log., or from its present property the

hyperbolic log., and is 2.3026 times the ordinary tabular log. The number 2.3026 being known as the modulus.

Any Curve.

To find the area between the irregular curve CD, the ordinates AC and BD, and the base AB; we proceed as follows: