This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

460. - Parabola: Definitions - The parabola is one of the most interesting of the curves derived from the sections of a cone. The several curves thus produced are as follows: When cut parallel with its base the outline is a circle; when the plane passes obliquely through the cone, it is an ellipse; when the plane is parallel with the axis, but not in the axis, it is a hyperbola; while that which is produced by a plane cutting it parallel with one side of the cone is a parabola.

Fig. 304.

Let the lines L M and L N (Fig. 304) be at right angles; draw CFB parallel with LM; make LQ = LF; draw QB parallel with LF; then FB = B Q. Now let the line AL move from FL, but remain parallel with it, and as it moves let it gradually increase in length in such manner that the point A shall constantly be equally distant from the line LM and from the point F. Then A BP, the curve described by the point A will be a semi-parabola. For example, the lines FB and B Q are equal; the lines FP and PM are equal, and so of lines similarly drawn from any point in the curve A BP. Let PN be drawn parallel with LM; then for the point P, A N is, the abscissa and NP its ordinate (see Art. 452).

The double ordinate C B drawn through F, the focus, is the parameter. A F is the focal distance. A is the vertex of the curve. The line L M is the directrix.

461. - Parabola: Equation to the Curve. - In Fig. 304 FPN is a right-angled triangle, therefore -

but - FP=MP = LN=AN+AL;

and - FN = A N- A F.

Therefore -

or -

y2 = (x+1/2p)2 - (x-1/2p)2

p being- put for the distance LF= FB (see Art. 452). As in Arts. 412 and 413, we have -

(x+1/2 p)2 | = | x2 | + | px | + | 1/4p 2 |

(x-1/2p)2 | = | x2 | - | px | + | 1/4 p2 |

y2 | = | 2 px |

(181.)

by subtraction. This is the usual equation to the parabola, in which we have the rule: The square of the ordinate equals the rectangle of the corresponding abscissa with the parameter.

From (181.) we have -

x: y:: y: 2p,

or: The parameter is a third proportional to the abscissa and its corresponding ordinate.

462. - Parabola: Tangent. - From M, any point in the directrix, draw a line to F, the focus (Fig. 305); bisect MF in R, and through R draw U T perpendicular to MF, then the line T U will be a tangent to the curve. For, draw MD perpendicular to L V, and from P, the point of its intersection with the line T U, draw a line to F, the focus; then, because RP is a perpendicular from the middle of MF, MPF is an isosceles triangle, and therefore the lines MP and FP are equal, or the point P is equidistant from the focus and from the directrix, and therefore is a point in the curve.

To show that the line T U touches the curve but does not pass through it, take U, any point in the line T U, other than the point P; join U to M and to F. Then, since U is a point in the line T U, M UF, for reasons above given, is an isosceles triangle; from Udraw U V perpendicular to L V. Now, if the point U be also in the curve, the lines UV and UF, by the law of the curve, must be equal; but UF, as before shown, is equal to UM, a line evidently longer than UV; therefore, it is evident that the point U is not in the curve. A similar absurd result will be reached if any other point than the point U in the line U T be assigned, excepting the point P. Therefore the line T P touches the curve in only one point, P; hence it is a tangent.

Fig. 305.

Parallel with L V, from A, draw A S, the vertical tangent. Now A S bisects MF or intersects it in the point R. For the two right-angled triangles FL M and FA R are homologous; and because FA = A L, by construction, therefore FR = RM.

Or: The vertical tangent bisects all lines which can be drawn from the focus to the directrix.

The lines PF and FT are equal; for the lines MP and NT being parallel, therefore the alternate angles MPT and N TP are equal (Art. 345); and because the line P T bisects M F, the base of an isosceles triangle, therefore the angles MP T and FP T are equal. We thus have the two angles N TP and FP T each equal to the angle MPT; therefore the two angles N TP and FP T are equal to each other; hence the triangle PF T is an isosceles triangle, having the points T and P equidistant from F, the focus.

Also because the line MF is perpendicular to P T, therefore the line MF bisects the tangent PT in the point R. And because TR = R P, therefore, comparing triangles TRF and TPO, TF= FO.

The opposite angles MPT and U PD made by the two intersecting lines U T and MD (Art. 344) are equal, and since the angles MP T and FP T are equal, as before shown, therefore the angles FP Zand UPD are equal.

It is because these two angles are equal, that, in reflectors, rays of light and heat proceeding from F, the focus, are reflected from the parabolic surface in lines parallel with the axis.

For an equation expressing the value of the tangent, we have -

t2 = (2x)2+y2,

(182.)

Or: The tangent to a parabola equals the square root of the sum of four times the square of the abscissa added to the square of the ordinate.

463. - Parabola: Subtangent. - The line TN (Fig. 305), the portion of the axis intercepted between T, the point of intersection of the tangent, and N, the foot of a perpendicular to the axis from P, the point of contact, is the snbtangent. The subtangent is bisected by the vertex, or TA = A N. For, the two triangles TRA and TPN are homologous; and, as shown in the last article, the line MF bisects PT in R; or TR = R P.

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