2 s which is the ift Theorem, and gives II - 12 = 0222}{, the Analogy Analogy 131D-:y::D-y:z. Viz. CA: SA: TA: AP 10-yZ 14 yy Dy — yz = 1 Dz 14C 15 yy - Dy-yz+ DD–Dz+izz={DD+izz 15 w? 16y-D z =VDD-122 That is, 17 y=įD+ EVIDD+izz, which is the 2d Theor. Q. E. D. The Geometrical Performance of these two Theorems is very easy, as by the following Figure. 1. Suppose the Point B in the Ellipfis Periphery were given, and it were requir'd to find the Point P, &c. Make IC Radius, and upon the common Center C describe the Semi-circle I ds, and join the Points Card d with a Right Line; then bisect that Line (by Prob. 2, Pag. 287) and mark the Point where the bisecting Line would cross the Transverse, as at | e. Upon that Point | e, with the Radius Ce (or Cd) describe another Semi-circle, producing the Transverse Diameter to its Periphery, and it will assign the Point P. For if D=TS, y=AS, z=AP, as before. CA: A:: A: AP T P N Therefore the Point P is truly found. Consequently, if a Right Line be drawn through those Points B and P, it will be the Taxgent requir’d, according to the firft Theorem. And 41 6 2. The Converse of this is as easy, to wit, if the Point P be given, thence to find the Point B in the Ellipfis Periphery. Thus, circumscribe half the Ellipsis with the Semi-circle TIS, as before; and bifect the Distance between the points and P, as at e, viz. Let Ce=eP. Then making Ce Radius, upon the point c, describe the Semi-circle CdP; and from the Point where the two Semi-circles intersect or cross each other, as at d, draw the Right Line d A perpendicular to the Tranfverse TS, TS, and it will assign the Point of Contact B in the Ellips Periphery, through which the Tangent must pass. But the Practical Method of drawing Tangents to any assign'd Point in the Ellipfis Periphery may (without finding the aforesaid Point P) be easily deduced from the following Property of Tangents drawn to a Circle, which is this. B If to any Radius of a Circle, as CB, there be drawn a Tangent Line (as H K) to touch the Radius at the Point B ; the two ,4ngles, which the Tangent makes with the Radius, will always be two Right Angles (16, 17, 18, 19 Euclid 3.) that is, < HBC = SCBK = 90°. C In like Manner the two Angles, made between the Tangent and the two Lines drawn from the Foci of any Ellipsis to the point of Contact, will always be equal, but not Right Angles, save only at the two Ends of the Transverse Diameter. These being well consider'd, and compar’d with what hath been said in Page 366, it must needs be easy to understand the following Way of drawing Tangents to any align’d Point in the Ellipfis Periphery ; which is thus : Having by the transverse and conjugate Diameters found the two Foci f and F, by Sećt. 3. from them draw two Right Lines to meet each other in the asign’d Point h of Contact, as fb and Fb (or f B H K and FB) in the annex'd Figure. Next set off (viz. make) bd=b F(or BD =BF) and join the Points Fd (or T FD) with a Right Line. f F Then, I say, if a Right Line be drawn through the Point of Contaet H b (or B) parallel to d F, or D F, B it will be the Tangent requir'd. For it is plain, that as the sf NH=> FNK when the Tangent is parallel to the Transverse Diameter, even fo is the =fbb=sFBk, (and <fBH=FBK) and will be every where so, as the point of Contact b (or B) and its Tangent is carry'd about the Ellipsis Periphery with the Lines f 6 F (or f BF). S D B В CH A P. III. Concerning the Chief Properties of every Parabola. NOTE, in every Parabola, the intercepted Diameter, or that Part of its Axis, which is between the Vertex and that Ordinate which limits its Length, as Sa or S A, &c. is callid an Abfcifla. Se&t. I. The Plain or Figure of every Parabola is proportion'd by its Ordinates and Abscillæ, as in the following Theorem. As any one Abscisa: is to the Square of its Semi-ordinate Theorem. :: fo is any other Abscissa : to the Square of its Semi dinate. S bA B A А Demontration. Let the following Figure HVG represent a Right Cone cut into two Parts by the Right Line SÀ, parallel to its Side V H. Then the Plain of that Section, viz. Bb Sb B will be a Parabsla, by Sect. 4. Page 364, wherein let us suppose S A to be its Axis, and bab, B A B to be Ordinates rightly apply'd to that Axis. Again, imagine the Cone to be cut by the Right Line hg parallel to its Base HG. Then will bg be the Diameter of a Cir cle, by Sect. 2. Page 363. and a Sag like to A SAG. Therefore lils Sa: ag :: SA: AG V S SHA=ha, because SA BA=AGXHA By Lem. Oba = ag x ha ŚP. 363 5 Saxo BA=SAX Oba 5 { By Axiom 5. Sa: o ba :: SA: O BA 6, Analogy B These I 2 x bal}|{ By Axiom ht Vide Page 194. These Proportions being prov'd to be the common Property of overy Parabola, all that is farther requir'd about that Section, or Figure, may from thence easily be deduced, Sect. 2. To find the Latus Kedum or Right Parameter of any Parabola. The Latus Rectum of a Parabola hath the same Ratio or Pro. portion to any Abscisa, and its Semi-ordinate, as the Latus Rectum of any Ellipfis hath to its Transverse and Conjugate Diameters, and may be found by this Theorem. Theorem. { Sa 4 As any Abscisla : is in Proportion to its Semi-ordinate :: fo is that Semi-ordinate : 10 the Latus Rectum. Let L= the Latus Rectum. Then, I Sa:ba::ba: L2 where ever the Points a, and And : 2 SA: BA::BA:LS? A, are taken in the Axis. Ова 3 =L: Or Sa XL= Oba Ова. 2 '.' =L: Or SAXL=OBA Sa ОВА, O sa 3=4 5 Per Axiom 5. SA 5 X 6 Sa X OBA=SAXO ba, which gives this Analogy 71 Sa: Oba::SA: OB A, the same as at the 7th Step of the last Process; therefore L (thus found) is the true Latus Re&tum, by which all the Ordinates may be regulated and found, according to its Definition in Section 4, Page 364. For by the third Step Sa XL= ba, and by the 4th Step SAXL=0 BA. Consequently v SaXL=ba and ✓ SAXL=BA; and so for any other Ordinate. ba Or if the Ordinates are given, to find their Abscisæ ; then it will be, L:ba::ba: Sa, and L:BA:: BA:SA, &c. Çonfequently = sa, and and Ba= 84, &c. From the Consideration of these Proportions, it will be easy to conceive how to find the Latus Rectum Geometrically, thus : Join S Join the vertical Point S of the Axis, and either extream Point of any Ordinate, as B (or b) with a Right Line, viz. SB (or Sb) and bifect that Line (by Problem. 2. Page 287.) marking the point where the bisecting Line doth intersect or cross the Axis, as at E (or e) and with the Radius S E (or Se) upon the Point bi E for e) describe a Circle ; (as in the E annex'd Figure) then will the Distance between the Ordinate and that Point where the Circle's Periphery cuts the Axis, viz. AR (or a r) be the true Latus ReElum required. For SA:BA::BA: A R, and Sa: ba:: ba: or, by Theor. 13. therefore AR=L. And ar=L, by the ift and 2d Steps above. R ConseEtary. From these Proportions of finding the Latus Rectum, it will be easy to deduce and demonstrate this following Theorem. As the Latus Rectum : Is to the Sum of any two SemiCheorem. ordinates :: fo is the Difference of those two Semi-ordi nates : to the Difference of their Abscise. Demantration. . s Ова OBA- Oba E B D 3 x L 41SA - SaxL= OBA-ba Which gives L:BA+ ba :: BA-ba;SA-Sa First 1 2 I |