Nature of its Equation, is more fimple than the Circle; and higher Equations may be conftructed by the Help of more compounded Curves; befides, a Geometrical Conftruction, rightly manag'd, fhews at once, as well the Number and, nearly, Quantity of the real Roots, as their Signs, viz. whether they be affirmative or negative: But feeing the Parabola and more compounded Curves cannot be defcribed but by Points, and the uncertain Motion of the Hand, the Antients hardly admitted them into their Geometry, and would fcarce allow that to be Geometrically effected, which could not be described by the Help of the Rule and Compaffes. Let b AB A AD=12. H Suppofe 2 a BF BE HD DG. 42aa(v:EBq+BFq:)=EF=a√2. 51:b—a:x√2 (=√: FCq+CGq:)=FG. By the Problem 6m (3) · n (6) :: a √ 2 ⋅ b — a √2. b mana. a. 72ey 4C. 6 +7. 8yy + zey + ee=aa+4c. 4 a. 92+1 =b a. 92.10 ee + 2ey+yy=bb — 2 ba+aa. 810. 11 aa+4cbb-2 ba+aa. II, 12 2 b a= bb 122b. 13 a 6 - 7 = 5. b 9, 13. 14 e + y = ; b + 20. b 6-7.15 ee2ey+yyaa4c. 4cc bb 15, 13. 16ee2ey+y y=4bb¬60 + 66 c 16 w. 2. 17e-y=+ √ : 200 - 6c+ 400 2 C 66: CC 4CC bb PROB. III. (Apol. Perg.) Two Points A, B in a Plane being given; and, knowing it poffible to be done; fo to defcribe in the fame Plane a Circle D, D, that to every Point of its Circumference D, the ftraight AEP B Inquifition. Suppofe it done, and C the Center of the Circle: Then, because the Points A, B are given, the Line AB (which fuppofe 7) is given; and because of m to n, its Point E, this being one of the Points D; therefore AE and BE are given. Put AE=b(4); then m(8) · n (6)::b(4) 12 m b(3)=BE. Suppofing then from D on A B produc'd, if need be, a 1 Dp drawn, and taking (on oppofite Sides of AB) ADA, and BD B4, the As on oppofite Sides of AB will be fimilar and equal; and therefore Dpd one ftraight Line bifected in p, and at Ls to AB in which therefore is the Center C, and the Diameter ECF, which muft lie towards the Side B, not A, otherwife B F would be AF, that is here BD AD. = Now fuppofe EC=r? And Epx; then Ap= b+x, and Bp = m bex; then (by the Property of a Circle) 2x x x = pDq: And (by 47. 1 Eucl. Elem.) 2rx-xx+bb + 2 b x + x x = AD2 = 2rx + 2b x Now (by Hypothefis) mn:: AD. BD; Confequently, mmnn::ADqBDq; that is :: 2 rx+2x + b b - 2 r x n n ·6x+ bb; wherefore 2 mmrx 2 nm bx + mm nnbb = 2nnrx + 2 n n b x + n nbb: mm r m nb + nnb; Confequently, by dividing each Part by nb 6×4 72 7 2212 you'll have r = = 12.) If m be , then r will be infinite; and (fince by how much the greater the Radius of a Circle is, by fo much the nearer its Circumference approaches to a ftraight Line) the Circumference dED will become an endless ftraight Line, perpendicular to AB, and bifecting it in E. 1. Suppofe it done, and through C draw CG || EB, and CF AB. 2. Thro' D likewife draw DL || EB, and DK || A B. 3. In the ABCF you have the Ls B and F, and the Side CB given; And therefore the Sides BF and CF are easily found by Trigonometrical Calculation; wherefore, fuppofe them known; And put CF = b (= 5) and BF = d (= √75.) 4. Put BC=c(10) and therefore BD = 2c. Then 4. 6 E. El 5LD=BK = 2d, and DKLB = 2 b. Suppofe 6 KE=a=? 7FEd + a. CE=FE+ CF = dd + 2 da + aa + bb. 4. 6 E. El. 9a (KE) · 2 b (KD) : : 2 d (LD) · 4db 10 AG=AL+LG=" 47. 1E.E. 11 DAC 4 db .. -LA. α 4 db + ba a +b= a DAGOGC = 1 6 dd b b + 8 d b b a + b b a a + dd. Now, Now, by Suppofition, the Square-Root of the 8th Step is the Square-Root of the 11th the 8th Step is the IIth. 3 13 a + 2d 14 a3 — 8 db b = 0; confeq. a = √ 8ddbb (= 3 V200 75 <=12.00 +.) A Tree AF (2co) Foot high, ftanding upon the Side of a Hill, was by a tempef tuous Wind broke in a Point C. The upper Part of it CF fell fo as to become CD: And the Distance from D its Top, to A its Root, was found to be c(95) Foot: And a horizontal Line A B being drawn, till it cut the Part CD of the Tree in B, was found to be d (40) Foot. It is required to tell how many Foot long the ftanding Part A C is. In order to folve this Problem, I fhall fuppofe the Tree AF to be perpendicular to the Horizon; And then, in the ▲ ACD, I have DCCA, AD c, Angle CAB Right, and ABd given; to find CA a = ? Let fall AE 1 DC, fo is DC divided into two Segments DE and EC. |