Therefore, we have -
Tr: Ta :: Tp: Tn,
TR x TN = TA x TP,
but - TR=1/2 TP;
therefore - 1/2 TP x TN = TA x TP,
1/2 TN = TA.
Or: The snbtangent of a parabola is bisected by the vertex; or is equal to twice the abscissa.
And because of the similarity of the two triangles TRA and TP N, as above shown, we have -
NP=2AR, y = 2 A R.
Or: The ordinate equals twice the vertical tangent.
464. - Parabola: Normal and Subnormal. - The line PO (Fig. 305) perpendicular to P T, is the normal and NO, the part of the axis intercepted between the normal and the ordinate, is the subnormal. For the normal, from similar triangles, we have -
TN: NP::. TP: P0,
PO = NPx TP/TN
Or: The normal equals the rectangle of the ordinate and tangent, divided by twice the abscissa.
The subnormal equals half the parameter. For (181.) -
y 2 = 2 px,
In the similar triangles (Art. 443) OPN and PTN, we have -
No: Np:: Np: N T,
As shown in the previous article, N T - 2 A N; therefore -
Comparing equations (A.) and (B.), we have -
NO = FB.
Or: The subnormal of a parabola equals half the parameter, a constant quantity for the subnormal to all points of the curve.
465. - Parabola: Diameters. - In the parabola BAC (Fig. 306), PD, a diameter (a line parallel with the axis) to the point P, is in proportion to BDx D C, the rectangle of the two parts into which the base of the parabola is divided by the diameter.
This may be shown in the following manner:
For EA we have, taking the co-ordinates, for the point
C, (181.) -
y2 = 2 p x,
or - y2/2p = x
For NA we have, taking the co-ordinates to the point P, (181.) -
y2 = 2px,
or - y2/2p=x
Using these values (B.) and (C.) in (A.), we have -
If l be put for B C and n for D C, then -
NP=EC-DC=1/2 l - n
and - Dp = (1/2 l2) - (1/2 l - n )2/2p then (Art. 413) -
Dp = 1/4 l2 - (1/4 l2 - ln+n2)/2p
or (Art. 415) -
Dp = 1/4 l2 - 1/4 l2 + ln - n2/2p
DP = ln -n2/2p
DP = ln - n2/2p
DP - DCxBD/2p
Now, since 2 p, the parameter, is constant, we have D P, the diameter, in proportion to D C x B D, the two parts of the base.
Putting d for the diameter, we have -
d = n (l - n)/2p. (183.)
Or: The diameter of a parabola equals the quotient obtained by dividing the rectangle - formed by the two parts into which the diameter divides the base - by the parameter.
It has been shown by writers on Conic Sections that a diameter, P J (Fig. 307), to any point P in a parabola bisects all chord lines, SG,DP, etc., drawn parallel with the tangent to the point P; the diameter being parallel with the axis of the parabola.
466. - Parabola: Elements. - From any given parabola, to find the axis, tangent, directrix, parameter and focus, draw any two parallel lines or chords, SG and D E (Fig. 307), and bisect them in H and J; through these points draw JP; then JP will be a diameter of the parabola - a line parallel with the axis. Perpendicular to P J draw the double ordinate PQ and bisect it in N; through Nand parallel with PJ draw TO, cutting the curve in A; then TO will be the axis. Make AT=AN, join T and P; then TP will be the tangent to the point P; from P draw PO perpendicular to P T; then P0 will be the normal, and NO the subnormal.
With NO for radius, from N as a centre, describe the quadrant O R; draw R C parallel with A O, cutting the curve in C; from C draw CB perpendicular to A O, cutting A O in F; then F will be the focus and C B the parameter. Make AL = AF; draw L M perpendicular to TO; then LM will be the directrix. Extend PJ to meet LM at M; join P and F; then, if the work has been properly performed, FP will equal MP.
467. - Parabola: Described Mechanically. - With NP (Fig. 308) a given base, and NA a given height, set perpendicularly to the base, extend NA beyond A, and make A T equal to NA; join T and P; from P perpendicularly to TP draw PO; bisect ON in R; make AL and A F each. equal to N R; through L, perpendicular to L 0, draw D E, the directrix.
Placing the square against the ruler and with its edge
JH coincident with the line MP, fasten to it a fine cord on the edge PE, and extend it from P to F, the focus, and secure it to a pin fixed in F. The cord FP will equal the edge MP. To describe the curve set the triangle J G H at MPE, slide it gently along the ruler towards V, keeping the edge JG in contact with the ruler, and, as the square is moved, keep the cord stretched tight, holding for this purpose a pencil, as at K, against the cord. Thus held, as the square is moved the pencil will describe the curve. That this operation will produce the true curve we have but to consider that at all points the line FK will equal KJ, which is the law of the curve (Art. 460).