(13.) Let PT touch the curve in P, and meet the axis produced in T; draw PO at right angles to PT, and let it cut the axis in O. PT is called the tangent, TN the subtangent, PO the normal, and NO the subnormal, to the point P.

IV.

On the Properties of the Parabola.

PROPERTY 1.

(14.) The Latus-rectum BC is equal to 4 AS.

Draw BD (Fig. in page 14.) parallel to LZ, then SB =BD=SL. But since SAAL, SL is equal to 2AS; hence SB=2AS, and 2SB or BC=4AS.

PROPERTY 2.

(15.) The tangent PT bisects the angle MPS.

curve,

that it may

Take Pp so small a part of the be considered as coinciding with the tangent, and consequently as a right line. Join Sp, and draw pm parallel to AZ; let fall po, pn, perpendiculars upon SP, PM.

The figure Mnpm is a parallelogram, .. nM=pm; and since po is at right angles to SP, it may be considered as a small circular arc described with radius Sp, .. So-Sp. Also SP=PM, and Sp=pm.

Now Po SP - So SP-Sp, and Pn-PM-n M-SP-pm,

=SP-Sp,)

.. Po=Pn.

In the small right-angled triangles Ppo, Ppn, we have therefore Pp common, and Po=Pn, .. (47. 1.) po=pn; having their three sides equal, the angle p Po must be equal to the pPn; hence, since pT may be considered as the continuation of the line Pp, PT bisects the angle MPS.

(16.) COR. Since the angle MPS continually increases as P moves towards A, and at A becomes equal to two right angles, the tangent at A must be perpendicular to

the axis.

PROPERTY 3.

If PT meets the axis produced in T, then SP=ST, and TN=2AN.

(17.) Since PM is parallel to TZ, the angle MPT= alternate angle STP; but (by Prop. 2.) MPT= ▲ SPT, :. 4 STP=L SPT, and consequently SP=

ST.

(17.) Now

(18.) Now (7.) SP= AN+AS;
and ST TA+ AS.

Hence, since SP=ST, we have AN+AS=TA+AS, ..TA=AN, or TN=2AN.

PROPERTY 4.

(19.) The square of the ordinate (PN')=latus-rectum x abscissa (BC× AN).

By Art. 7. y=4ax, or PN'=4ASX AN; but by Prop. 1. BC=4AS, .. PN'=BC× AN.

(20.) COR. Hence BC= ; and BC=

PN
AN

PROPERTY 5.

(21.) The subnormal NO= BC.

PN

2AN

Since TPO is a right-angled triangle, (Euc. B. VI.

Prop. 8.) NO: PN :: PN: TN, .. NO =

PN'

PN'

; but

TN

by Prop. 3, TN=2AN, .. NO=; ; hence (19.) NO

2AN

= BC.

PROPERTY 6.

(22.) The square of the ordinate (QV3)=4SP× PV. (Fig. in p. 16.)

Produce VQ to H; draw EQ, GV parallel to PN, and QD parallel to AZ; then the figures PTHV, PNGV will be parallelograms, and TH=PV=NG; .. HN+ NG=HN+TH, or HG=TN.

Now (19) EQ=4AS× AE=4axx+y-z; and by similar triangles, HEQ, TNP, we have

HE: EQ' :: TN2 : PN3,

i. e. 2x-x: 4axx+y-x: 4x2 : 4ax,

Again, by sim. ▲', HGV, QDV, we have

HG GV or PN' :: QD': DV',

:

But QV=QD'+DV'

=4xy+4ay=4.x+a.y=4SPX PV.

The same demonstration, with a very slight alteration, is applicable to the case when P and Q are on opposites of A.

PROPERTY 7.

(23.) If QV is produced to meet the curve

qV=QV. (Fig. in page 18.)

in 9, then

Draw qe at right angles to AZ, cutting PW in k; then He=HG+ Ge, and Ae=AG+ Ge; if therefore Ge or Vkz, we have He=2x+x, and Ae=x+y+z. In this case, the sign of x is changed throughout; reasoning therefore as in the last Property, we should have z' or Vk=4xy.

By sim. A', HGV, Vkq, HG2 : GV or PN1 :: Vk2 : kq', i. e. 4x : 4 ax :: 4xy: kq3.

but Vq' = Vk2+kq'

=4xy+4ay=4.x+a.y=4SP × PV.

(24.) COR. Since QV and Vq' are each equal to 4SPX PV, it follows that Vq=QV, and consequently Vq=QV. Hence all lines drawn parallel to the tangent PT, and terminated both ways by the curve, are bisected by the diameter PW.

PROPERTY 8.

(25.) The Parameter bc is equal to 4SP.

Let bc and PW intersect each other in g, then (24) cg=gb; ..cg={bc, and cg2=4bc3.