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144. To draw a tangent at any point of the parabola.Let P (fig. 117) be the given point, draw the ordinate PM perpendicular to the axis A X. Take A T on the axis produced outside the curve, equal to A M; then T P is the tangent at the point P.
AX; bisect the angle SPK by the line PT, which is the tangent required. PT also bisects the line SK, PK being equal to PS.
145. From a given point on the axis outside the parabola to draw a tangent.—Let T be the given point (fig. 117); take A M equal to A T, and draw MP at right angles to A M, meeting the curve at P; then PT is a tangent to the curve.
146. To draw a normal at any given point on the parabola.—Let P (fig. 117) be the given point, A X the axis, S the focus. Join SP, and draw S Y parallel to the axis AX; bisect the angle SPY by the line PN, which is the normal to the curve at P, and is at right angles to the tangent PT.
Let BKC be the directrix, perpendicular to the axis, A B being equal to A S. Draw K P parallel to the axis, and join SK. Then the normal is found by drawing PN parallel to SK, PK being equal to PS.
The normal can also be drawn without finding the focus, by first drawing the tangent at P (144), and then drawing PN perpendicular to PT.
Let MP be the ordinate at P, at right angles to the axis ; measure M N, equal to SB, or to twice SA, S being the focus ; then N is the point where the normal cuts the axis, and N P is the normal required.
147. To apply the parabola to form the entasis of a column.-Let A B (fig. 120) be the given height of the column, A C the difference between its radii at top and bottom. Draw B P at right angles to A B, and divide A B and B P into the same number of equal parts. Draw the
lines Af, A g, &c., cutting the lines a l, 6 m, &c., perpendicular to A B, in the points l, m, &c. These will be
points on the required curve, through Fig. 120.
which it can be drawn by bending a Bf I h i k P
ruler or lath. The curve thus drawn is part of a parabola (142), having A for its vertex.
148. To find the focus of a given parabola.—Let A X (fig. 117) be the axis, A the vertex, P a given point on the parabola. Draw PM perpendicular to the axis, and take A T, equal to A M. Draw the tangent TP, and also the line P Y parallel to A X.
Draw the normal P N at right angles to PT. Draw PS, making with PN the angle SPN, equal to the angle NPY; then if PS cuts the axis at S that point is the required focus.
Let Q be a point (fig. 117) upon the curve, draw XQ perpendicular to A X, and produce it to V, making X V equal to twice X A. Join A V, and let the line A V cut the curve at the point L. Drop the perpendicular LS upon A S, and S will be the focus of
the parabola. 149. To apply the parabola to form the outline of a Tudor arch.—Let ABC (fig. 121) be the springing line of the arch, A M or A N the half span, MP or N Q its height at the vertex. Through P or Q draw the parabola A PQ,
having A for its vertex (142); then A P or A Q will form one-half of a flat-pointed or Tudor arch; and by repeating the curve on the opposite side of the centre line MP or N Q we obtain the other half PB or QC of the arch. The joints of the voussoirs must all be in the directions of the normals (146) to the curve. Take M V, equal to twice
Fig. 121. .
MA, join A V, cutting the curve at b, draw 6 S perpendicular to A M, then S is the focus of the parabola (148).
150. To approximate to the contour of the parabola by arcs of circles.—Find any number of points, a, b, &c. (fig. 121) on the curve by one of the methods described above (142). Draw the normal at a (146), cutting the axis at the point p, and from p as a centre, with p A or pa as radius, describe the arc A a.
Draw the normal at b, meeting the normal a p in the point q, and from q as a centre, with qa or qb as radius, describe the arc a b. Proceeding in this way, we can draw a number of arcs of circles gradually increasing in radius, which will approximate to the form of a parabola ; and the larger the number of points taken, the nearer will the resulting curve approach to the contour of a parabola.
151. Application of the parabola to mouldings, architectural ornaments, and other purposes.—The parabola can be made available to form the outline of a classical ovolo, as Q A P (fig. 122), A X being the axis. Drawing A B at
right angles to the axis, and B P parallel thereto, we can find any number of points on the curve by the previous method (142).
The parabola can also be used for a variety of ornamental purposes where curved lines are required, as in a leaf (fig. 123), of which C D is the axis, DT the tangent at D perpendicular to CD. Then DAC, D A'C are two similar arcs of the parabola, having A X, A'X' for the axes making equal angles with CD, and D T for their common