7. The same things being given to find the curve, by another method by points. Fig. 8.
Draw DE parallel to CB, and BE parallel to CD ; divide BC and BE proportionally at g and f, make CI equal to CD ; draw I g h and f h D, and the point h will be in the curve. In the same manner may other p‘,ints in, 72 be found. Perhaps to find the points m, n, it would be easier to some to divide the lines BC and BE each into the same number of equal parts.
8. The same ti.ings being still giA en to describe the reprosentation with a compass, Fin. 9.
Take hail Lie conjugate CD, which apply on the se mitransverse from A to E ; divide EC into three equal parts, and set one towards A from E to F ; make CC equal to CF; mith the distance FG describe an equila teral triangle 111G ; produce 1W to I, and HG to K from II, with the distance HD, describe an arc IK; from G, with the distance GK, describe an arc KB, and with the radius IF describe an arc IA ; and AIDKB will be the curve of the scmiellipse required.
9. To describe the curve of a parabola, having the double ordinate AB, and abscissa CD given in position, Fig. 10.
Draw AH parallel to CD, and HD to AC ; divide AC and AH in the same proportion at e and f ; draw e g parallel to CD, and draw f g D, the point g is in the curve of a parabola. In the same manner all other points h, i are to be found.
10. To describe the curve of a hyperbola, having a double ordinate AB, and a diameter DE, given in posi tion, Fig. 11.
Draw AF parallel to CD, and FD to AB ; divide AC and AF, the former in f, and the latter in g, in the same proportion; draw f h E and g h D, and the point is will be in the curve. In the same manner all other points k are to be found.
11. Upon a given straight line AB, to describe any re gular polygon, Fig. 15.
Produce the side AB to H, on AH, as a diameter, describe the semicircle ACH ; divide the semicircumfe. rence FICA into as many equal parts as the polygon is to have sides ; draw BC through the second division ; bisect AB at K, and BC at L ; draw KI at right angles to AB, and LI at right angles to BC ; from 1, with the distance IA, I13, or IC, describe the circle ABCD, &c.
to A, which will contain the side AB the number of times required.
12. To cut off the angles of a square ABCD, so as to form an octagon, Fig. 16.
Draw the diagonals AC and BD, intersecting each other at E; through E, draw FG and III parallel to the sides ; make El, EG, EH, EF, each equal to any half diagonal Ell ; join IF, FH, MG, and GI, cutting the sides of the square at I', Q, R, K, N, 0 ; join KL, I,M, MN, NO, OP, PQ, QR, RK and KLMNOPQR will be the octagon required.
13. To inscribe an octagon in a square ABCD, hav ing four of its angles in the middle of the sides, Fig. 17.
Draw the diagonals AC and BD, cutting each other at E ; through E draw FK and UM parallel to the sides, rutting the sides of the square in the middle at I', K, ; make EC, El, EL, EN, each equal to the half side of the ;quire; join FG, GII, III, IK, KL, LM, MN, NV, FG, and l'0111KLMN will be the octagon re quired.
14. To cut off the angles of an oblong ABCD, so as to form an elliptic octagon, Fig. 18.
Draw the diagonals AC and BD, cutting each other at E ; draw FG and III parallel to the sides ; make FU and AK cach equal to FA; and join UK and AU; make UM equal to UF ; draw MI, parallel to FK, cutting AD at I. ; make FT equal to FL, and OP and C Q each equal to FL ; draw LN and QR parallel to DB ; l'O and TS parallel to AC; then will LNOPQRST be tile elliptic octagon required.
15. In a given oblong ABCD to inscribe an elliptic oc tagon, Fig. 19.
Draw the diagonals AC and BD, cutting each other at E ; make IK equal to 113, and join KB, cutting FG at L; make EM equal to LB, and EN equal to KB ; OG, GR, RH, HQ, QF, EP, PI, 10, and 01PFQHRGO will be the elliptic octagon required.
16. Through a given point K, to draw the circumfe rence of an ellipse concentric with a given ellipse ABC DEFGH, Fig. 20.
Let I be the centre; take any number of points A,