PARABOLA. 



greater than C H = H B = K L, it fo) 

 lows that K C is greater than K L, and 

 consequently that the point K falls with- 

 out the parabola : and as this holds of 

 every other point, except H, it follows 

 that K H is a tangent to the parabola in 

 H. Q. E. D. 



Prop. 3. Every right line, parallel to a 

 tangent, and terminated on each side by 

 the parabola, is bisected by the diameter 

 passing through the point of contact: 

 that is, it will be an ordinale to that diame- 

 ter. For let E e (fig. 3 and 4) termina- 

 ting in the parabola in the points E e, be 

 parallel to the tangent 1) K ; and let A D 

 be a diameter passing through the print 

 of contact D, and meeting E e in L ; then 

 shall E L = L e. 



Let A D meet the directrix in A, and 

 from the points E e t let perpendiculars 

 E F, ef, be diftwn to the directrix ; let C 

 A be drawn, meeting E e in G ; and on 

 the centre E, with the distance E C, let a 

 circle be described, meeting A C again in 

 H, and touching the directrix in F ; and 

 let D C be joined. Then because D A = 

 D C, and the angle A I) K = the angle 

 C D K, it follows (4. 1) that D K per- 

 pendicular to A C ; wherefore E e perpen- 

 dicular to A C, and C G = G H (3. 3) ; 

 so that e C = e H (4. 1), and a circle des- 

 cribed upon the centre e with the radius 

 e C, must pass through H; and because 

 e C = ef, it must likewise pass through 

 f. Now because F / is a tangent to both 

 these circles, and A H C cuts them, the 

 square A F = the rectangle C A H (36. 

 3) = the square A /; therefore A F = 

 A/, and F E, A L, and fe are parallel ; 

 and consequently L E = L e. Q E. D. 



Prop. 4. If from any point of a para- 

 bola, D, (fig. 5) a perpendicular, D H, be 

 drawn to a diameter B H, so as to be an 

 ordinate to it ; then shall the square of 

 the perpendicular, D H 1 , be equal to the 

 rectangle contained under the absciss H 

 F, and the parameter of the axis, or to 

 four times the rectangle II F B. 



1. When the diameter is the axis ; let 

 D H be perpendicular B C,join D C, and 

 draw D A perpendicular A B, and let F 

 be the vertex of the axis. Then, because 

 H B = D A = D C, it follows that H B* 

 = D O = ITU 1 -f H C 1 . Likewise, be- 

 cause B F = F C, H B 1 = 4 times the 

 rectangle H F C -f H C 1 (by 8. 2). 

 Wherefore D II 1 -f- H O = 4 times the 

 rectangle H F B -f H O ; and D H 1 = 

 4 times the rectangle H F B ; that is, D 

 H 1 = the rectangle contained under the 

 absciss H F, and the parameter of the 

 axis. 



2. When the diameter is not the axis : 

 let E N (fig. 3 and 4) be drawn perpendi- 

 cular to the diameter A I), and E L an or- 

 dinate to it ; and let D be the vertex of 

 the diameter. 



Then shall E N 1 = to the rectangle 

 contained under the absciss, L D, and the 

 parameter of the axis. For let D K be 

 drawn parallel to L E, and consequently 

 a tangent to the parabola in the point D ; 

 and let it meet the axis in K : let E F be 

 perpendicular A B the directrix ; and on 

 the centre E, with the radius E F, de- 

 scribe a circle, which will touch the di- 

 rectrix in F, and pass through the focus 

 Q : then join A C, which will meet the 

 circle again in II, and the right lines D K, 

 L E, in the points P G ; and, finally, let 

 L E meet the axis in O. 



Now since the angles C P K, C B A are 

 right, and the angle B C P common, the 

 triangles C B A, C P K are equiangular ; 

 and A C : C B (or C K : C P) : : O K : 

 G P ; and A C X G P = O K X C B. 

 Again, because C A = 2 C P, and C H 

 = 2CG, AH = 2GP; and consequent- 

 ly the rectangle C A H = C A X 2 G P 

 = O K x 2 C B. But, EN J = F A J = 

 rectangle C A II ; and consequently, 

 E N J = O K x 2 C B = the rectangle 

 contained under the absciss, L D, and the 

 parameter of the axis. Q. E. D. 



Hence, 1. The squares of the perpendi- 

 culars, drawn from any points of the para- 

 bola to any diameters, are to one another 

 as the abscissx intercepted between the 

 vertices of the diameters and the ordi- 

 nates applied to them from the same 

 points. 



2. The squares of the ordinates, ap- 

 plied to the same diameter, are to each 

 other as the abscissas between each of 

 them and the vertex of the diameter. 

 For let E L, Q R be ordinates to the same 

 diameter D N ; and let E N, Q S be per- 

 pendiculars to it. Then, on account of 

 the equiangular triangles E L N, Q R S, 

 E L 1 : Q R 1 : : E N 1 : Q S 1 : that is, as the 

 absciss D L to the absciss D R. 



Prop. 5 If from any point of a para- 

 bola E (fig. 3 and 4), an ordinate, E L, be 

 applied to the diameter A D ; then shall 

 the square of E L be equal to the rectan- 

 gle contained under the absciss D L, and 

 the latus rectum or parameter of that di- 

 ameter. 



For since Q R = D K, Q R 1 will be 

 equal to I) M* + M K ; but (by case 1. 

 of Prop. 4), D M 1 = 4 times the rectan- 

 gle M Q B ; and because M Q = Q K, 

 M K 1 = 4 M Q 1 : wherefore Q R 1 = 4 



