468. - Parabola: Described from Points. - With given base, NP (Fig. 309), and given height, A N, to find the points D, F, M, etc., and describe the curve. Make A T equal to A N (Art. 462); join T and P; perpendicular to TP draw

Fig. 309.

PO; make A B equal to twice NO; take G, any point in the axis A 0, and bisect B G in J; on J as a centre describe the semi-circle B C G cutting A L, a perpendicular to B 0 in C; on A C and A G complete the rectangle A CDG. Then D is a point in the curve. Take H, another point in the axis; bisect B H in K; on K as a centre describe the semi-circle B E H cutting A L in E; this by E F and HF, gives F, another point in the curve; in like manner procure M, and as many other points as may be desired. This simple and accurate method of obtaining points in the curve depends upon two well-established equations; one, the equation to the parabola, and the other, the equation to the circle. The line G D, an ordinate in the parabola, is equal to A C, an ordinate in the circle B CG; AG, the abscissa of the parabola, is also the abscissa of the circle; in which we have (Art. 443) -

Ag: Ac:: Ac: A B,

x: y:: y: a - x, y2 = x(a - x).

For the parabola, we have (181.) -

y2 = 2p x.

Comparing these two equations, we have -

x (a - x) = 2 p x,

a - x = 2 p,

or -

BG-AG=2p.

By construction A B equals 2 NO, or twice the subnormal; the subnormal (Art. 464) equals half the parameter. Hence, twice the subnormal equals the parameter - equals 2 p. Therefore, the method shown in Fig. 309 is correct.

469. - Parabola: Described from Arcs. - Let NP (Fig. 310) be the given base and A N the given height of the parabola. Make A T(Art. 462) equal A N Join TtoP; draw PO perpendicular to PT; bisect N O in R; make A L and A F each equal to NR; then L M, drawn perpendicular to TO, will be the directrix. Parallel to LM draw the lines BD, CE, etc., at discretion. Then with the distance B L for radius, and on F as a centre, mark the line B D with an arc; the intersection of the arc and the line will be a point in the curve (Art. 460). Again, with CL for radius and on F as a centre, mark the line CE with an arc; this gives another point in the curve. In like manner, mark each horizontal line from F as a centre by a radius equal to the perpendicular distance between that line and L M, the directrix. Then a curve traced through the points of intersection thus obtained will be the required parabola.

Fig. 310.

470. - Parabola: Described from Ordinates. - With a given base, NP(Fig. 311), and height, A N, a. parabola may be drawn through points J, H, G, etc., which are the extremities of the ordinates B J, C H, D G, etc.; the lengths of the ordinates being computed from the equation to the curve, (181.)-

y2 = 2px.

For any given parabola, in base and height, the value of p may be had by dividing both members of the equation by 2 x; by which we have -

(A.)

from which, NPand A N being known,p may be computed. With the value of p, a constant quantity, determined, the equation is rendered practicable. For, taking the square root of each member of equation (181.), we have -

(B.)

which by computation will produce the value of y, for every assigned value of x, as A B, A C, A D, etc.

Fig. 311.

As an example: let it be required to compute the ordi-nates in a parabola in which the base, NP, equals 8 feet, and the height, A N, equals 10 feet. With these values equation (A.) as above becomes -

p = NP2/2AN=82/2x10=64/20=3.2; 2p = 6.4.

Then, with this value in (B.) as above, we have, for each ordinate -

In order to assign values to x, let A N be divided into any number of parts at B, C, D, etc., say, for convenience in this example, in ten equal parts; then each part will equal one foot, and we shall have the consecutive values of x = 1, 2, 3, 4, etc., to 10, and the corresponding values of y will be as follows. When -

With these values of y, respectively, set on the corresponding horizontal lines B J, CH, DG, E S, etc., points in the curve J H, G, S, etc., are obtained, through which the curve may be drawn. The decimals above shown are the decimals of a foot; they may be changed to inches and decimals of an inch by multiplying each by 12. For example: 12x0.5297 = 6.3564 equals 6 inches and the decimal 0.3564 of an inch, which equals nearly 3/8 of an inch.

Near the top of the curve, owing to its rapid change in direction and to the approximation of the direction of the curve to a parallel with the direction of the ordinates, it is desirable to obtain points in the curve more frequent than those obtained by dividing the axis into equal parts.

Instead, therefore, of dividing the axis into equal parts, it is better to divide it into parts made gradually smaller toward the apex of the curve - or, to obtain points for this part of the curve as shown in the following article.

471. - Parabola: Described from Diameters. - Let EC (Fig. 312) be the given base and A E the given height, placed perpendicularly to E C. Divide E C in several parts at pleasure, and from the points of division erect perpendiculars to E C. The problem is to compute the length of these diameters, as D P, and thereby obtain points in the curve, as at P. For this purpose we have equation (183.), which gives the length of the diameters, and in which n equals D C (Fig. 312), l equals twice E C, and p equals half the parameter of the curve. The value of p is given in equation (A.), (Art 470), in which y equals EC (Fig. 312), and; x equals A E. Substituting: these symbols in equation (A.), we have -

Fig. 312.

where b = E C, the base, and h = A E, the height. For p, substituting this, its value, in equation (183.), we have -

d = n(l-n)/2p = n(l-n)/2.b2/2h

t = hn.(2b-n)/b2. (184.)

As an example: let it be required in a parabola in which the base equals 12 feet and the height 8 feet, to compute the length of several diameters, and through their extremities describe the curve. Then h will equal 8, and b 12.

If the base be divided into 6 equal parts, as in Fig. 312, each part will equal 2 feet. Then we have -

h/b2=8/122=8/144=1/18, and -

d = h/b2 n (2b - n),

d= n (24-n)/18.

In this equation, substituting the consecutive values of n, we have, when -

 n = 0, d = 0X 24 = 0 18 n = = 2, d = 2 X 22 =2.444 18 n = 4, d = 4X20 = 4.444 18 n = : 6, d= 6x 18 = 6. 18 n = 8, d = 8x 16 = 7. 111 18 n = 10, d = 10x 14 = 7.777 18 n = 12, d = 12 X 12 18 = 8.0

The several diameters, as P D, in Fig. 312, may now be made equal respectively to these computed values of d, and the curve traced through their extremities.

472. - Parabola: Area. - From (181.), the equation to the parabola, and by the aid of the calculus, it has been shown that the area of a parabola is equal to two thirds of the circumscribing rectangle. For example: if the height, A E (Fig. 312), equals 8 feet, and EC, the base, equals 12 feet, then the area of the part included within the figure APCEA equals 2/8 of 8x 12 = 2/3 x 96 = 64 feet; or, it is equal to 2/8 of the rectangle A B CE.