I’^ARABOLA. 
to it ; tlien shall the square of the perpen- 
dicular, DH% be equal to the rectangle 
contained under the absciss HF, and the 
parameter of the axis, or to four times the 
rectangle H F B. 
1. Wlien 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 tliat H B* 
= D = D -f H Ch Likewise, be- 
cause B F = F C, H B^ = 4 times the 
rectangle H F C -|- H (by 8. 2). Where- 
fore D H = 4 times the rectangle 
H F B HC^ ; and D = 4 times the 
rectangle H F B ; that is, Dff = the rect- 
angle 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 D, and E L an or- 
dinate to it ; and let D be the vertex of the 
diameter. 
Then shall E N“ = to the rectangle con- 
tained under the absciss, L D, and the para- 
meter 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 perpendi- 
cular A B the directrix ; and on the centre 
E, with the radius E F, describe a circle, 
which will touch the directrix in F, and 
pass through the focus C ; then join A C, 
which will meet the circle again in H, and 
the right lines D K, L E, in tlie 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 tri- 
angles C B A, C P K are equiangular; and 
AC:CBCorCK: CP)::OK:GP; and 
ACxGP = OKxCB. Again, because 
C A = 2 C P, and C H = 2 C G, AH = 
2 GP; and consequently the rectangle 
CAH = CA X SJGP = OK x 2 C B. 
But, EN^ =: F A* = rectangle C A H ; and 
consequently, EN^ =:OK x 2CB = 
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 abscissae intercepted between the 
vertices of the diameters and the ordinates 
applied to them from the same points. 
2. The squares of the ordinates, applied 
to the same diametqr, are to each other as 
the abscissae between each of tliem and the 
vertex of the diameter. For let £ L, Q R 
be ordinates to the same diameter D N ; 
and let E N, Q S be perpendiculars to it. 
Then, on account of the equiangular tri- 
angles E L N, Q R S, E : Q R2 : : E N* 
: Q : that is, as the absciss D L to the 
absciss D R. 
Prop. 5. If fi'om 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 rectangle 
contained under the absciss D L, and the 
latus rectum or parameter of that diameter. 
For since QR =: D K, Q R- will be 
equal to D M K^; but (by case 1. 
of Prop. 4), D = 4 times the rectangle 
M Q B ; and because M Q = Q K, M 
= 4 M : wherefore Q R^ = 4 times the 
rectangle M Q B -|- 4 M ; that is, to 4 
times the rectangle QMB. ButMQ = 
Q K = D R, and M B = D A ; wherefore. 
Q R^ = 4 times the rectangle R D A : and 
because Q R, E L are ordinates to the dia- 
meter A D, Q R^ (by cor. 2, of Prop. 4), : 
E L“ (: R D ; LD) : : 4 times the rect- 
angle RD A : 4 times the rectangle L D A. 
Therefore E =: 4 times the rectangle 
L D A, or the rectangle contained under 
the absciss L D, and the parameter of the 
diameter A D : and from this property, 
Apollonius called the curve a parabola. 
Q. E. D. 
Prop. 6. If from any point of a parabola, 
A, (fig. 6) there be drawn an ordinate, 
A C, to the diameter B C ; and a tangent 
to the parabola in A, meeting the diameter 
in D : then shall the segment of the dia- 
meter, CD, intercepted between the or- 
dinate and the tangent, be bisected in the 
vertex of the diameter B. For let B E be 
drawn parallel to AD, it will be an or- 
dinate to the diameter A E; and tlie ab- 
sciss B C will be equal to the absciss A E, 
or B D. Q. E. D. 
Hence, if A C be an ordinate to B C, 
and A D be drawn so as to make B D = 
D C, then is A D a tangent to the para- 
bola. Also the segment of the tangent, 
AD, intercepted between the diameter and 
point of contact, is bisected by a tangent 
B G, passing through the vertex of D C. 
“ To draw Tangents to the Parabola.’’ 
If the point of contact C be given : (fig. 7) 
draw the qrdinate C B, and produce the 
axis till A 'T be = AB ; then join T C, 
which will be the tangent. Or if the point 
be given in the axis produced; take AB 
= A T, and draw the ordinate B C, which 
will give C the point of contact; to which 
draw the line T C as beforp. If D be any 
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