MATHEMATICS. PRACTICAL GEOMETRY. 



[THB PARABOLA. 



IBT METHOD. Divide any line A B (Fig. G8) repre- 

 senting the trans- Fig. n. 

 vane diameter, into 

 three equal part* by 

 the points C D. 

 With tho centres 

 anil D and the equal 

 radii C A and D H, 

 describe two circles. 

 At the points E and 

 F, where they in- 

 tersect, as centres, 

 and the equal radii 

 AD or B C, de- 

 scribe parts of other two circles joining the circum- 

 ferenoes of the two first circles (the exact point* of junc- 

 tion of the two circlet may be got by joining tte centres of 

 both circlet, at shown by the dotted Una), which will com- 

 plete the oval required. 



I'ND METHOD. It is often required to draw a flat or 



long-shaped oval, in 

 which case divide 

 the line AB (Fig. 

 69) into four equal 

 parts by the points 

 C, D, G, and with 

 the centres and 

 G and radii equal 

 B to CA and G B, 

 describe two cir- 

 cles ; also with the 

 same centres and 

 radii equal to any 

 two parts of the 

 line AB, such as 

 A D, describe arcs 

 intersecting at E and F, with which, as centres and 

 radii equal to three parts, of A B, such as A G, describe 

 portions of other two circles joining the circumferences 

 of the first circles, which will complete the oval 

 required. 



SRD METHOD. To draw an egg-shaped oval 

 Bisect the diameter A B in Fig. 70. 



the point D (Fig. 70), with 

 which, as a centre, and D A 

 or D B as a radius, describe 

 a circle cutting the diameter 

 EF in the point G: with 

 the centres A and B and 

 radii equal to AB describe 

 two arcs, AH and BH'; 

 join AG and BG, and pro- 

 duce them until they cut 

 the arcs AH and BH' in 

 the points H and H', and 

 with the centre G and radius 

 G H, describe a part of a 

 circle touching the two latter 

 drawn arcs, which will com- 

 plete the oval. 



THE PARABOLA. 



(1.) To show that P N ! = 4A S, Aft. 



Let Dd (Fig. 71) be the directrix, 8 the focus ; draw 

 DAN perpendicular to Dd Fig. 71. 



through S ; this line is called 

 the axis of the parabola. 

 Draw P N perpendicular to 

 D N. and Yd perpendicular 

 to P N. Now S P - Pd. 

 Hence if AS AD, A is 

 point in the parabola, and is 

 called its vertex. 



Now P 8 =. Pd (by Def . 2) 

 -DN-DA+ AN-A8 

 + AN. 



Also P S - P N + 



A N - A S) 



.'. PN'-4A8, AN. 



Con. (1.) Hence A S. is a third proportional to A N, 

 and tho half of I 1 N. 



Cos. (2.) Also if P, be any other point, and P, N, be 

 parallel to P N, we have P, N. : PN : : 4 A S. AN, 

 : 4 A S. A N : : A N, : A N. 



N.B. If P, N, bound the curve it is sometimes 

 called the base, and AN. the height of the curve. 

 Also it is plain that if P N be produced, so that pN = 

 N P, then pN 1 = 4 A S. A N, and therefore p is a point 

 in the parabola ; which is therefore symmetrical about 

 A N, ; the line Pp is called a double ordinate. 



(2.) If ABCD (Fig. 72) be any rectangle, and if 

 D C and C B are each 

 divided into n equal 

 parts, and let Dp, = p 

 of the equal parts into 

 which CD is divided, 

 and Bp 2 = p of the 

 parts into which CB 

 is divided ; join Ap, 

 meeting p a q (parallel 

 to A B) in "P ; then P 

 is a point in the para- 

 bola which has for its base C B and height B A. 



For, draw PN perpendicular to AB, and 

 parallel to PN. 



n e . 72. 



^CB., An, 



1- A B and PN=Bj>,. 



A, PN Pii 



Also, T-iTj -=j - 



'AM An, 



'AN 

 And PN 

 PN 



CBn t 

 ABp, 



.Pi. 



CB. 



CB 1 



AN AB 

 .'. PN a :CB:: AN 



AB. 



.'. P is a point in the specified parabola by corollary 

 to last article. The second of the following practical 

 methods is founded on the above principles. The first and 

 third depend entirely on the definition of the parabola. 



To describe a parabola, any absciss (that it, a portion of, as 

 Cl C2, <fcc. ) of the axis and the corresponding ordinate 

 being given ; or, the half width of t!ie base (A D or 

 D B), and the height of the curve (C D) being given. 

 IST METHOD. Bisect D B (Fig. 73) in the point E ; 



join C E, and from E draw E F perpendicular to C E, 



and meeting C D pro- 

 duced in F ; make 



C O in DC produced 



and OF, each equal 



to DF, F! will be 



the focus of tho re- 

 quired parabola. Take 



any number of points, 



1, 2, 3, 4 (&c.), in 



C D ; through them 



draw double ordi- 



nates, or lines per- 

 pendicular to the axis 



CD; then with the 



centre F,, and radii OF,, Ol, O2, O3, O4 (O <fec.), 



describe arcs cutting the ordinates ; the curve drawn 



through the points of intersection will ba the parabola 



required. 



2ND METHOD. A B (Fig. 74) being the width at the 



base, and C D the height 



of the curve, as before, 



construct the parallelo- 

 gram A B E F, divide 



DA, DBand AF, BE 



respectively into the 



same number of equal 



parts in the points 1 , 2, 



3, 4; from the points 



of division in A F 



and B E draw lines 



A 



Fig. 74. 



C 



P 



x 



\ 



\ 



o s 



to the point C, and from the 

 points of division in D A and D B draw lines perpen- 



