A tangent to a parabola parallel to a given line may be drawn by constructing the angle GFP twice the angle which the line makes with the axis, so determining the point of contact P.

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PROBLEM 37. (Fig. 37.) To draw a parabola, the vertex A, the axis AN and a point P on the curve being given.

This might be solved by first finding the focus and proceeding as in the last problem. It can however be solved independently

without using circular arcs, and the method is evidently applicable to the last problem after any one point on the curve has been found.

Draw the tangent at the vertex and a diameter through P meeting it in M. Divide MP into any number of equal parts (say four), and AM into the same number. Then diameters through the several points on AM will meet lines joining A to the corresponding points on MP (counting from A in the first case and from M in the second) in points of the curve as B, C, D. As the curve recedes from the axis the points found get more and more distant from each other (compare C to D and D to P), but, if desirable, points can be interpolated between any two points already found by subdividing the corresponding spaces on MP and

AM. In the figure points are thus interpolated between C and D and between D and P. The curve can be carried beyond P by carrying on the divisions on the two lines as in the figure.

The other half of the curve can be put in by symmetry.

The tangent at any point D can be drawn by drawing the ordinate DN, and making AT on the axis equal to AN, on the other side of the vertex; DT will be the tangent at B, as has already been shewn.

The focus F is found by drawing the normal at any point D, bisecting the sub-normal NG and setting off AF=\NG.

The construction for the curve depends on the fact that if a diameter be drawn through the centre point of any chord, the tangents at the extremities of the chord intersect on the diameter, and the curve cuts the diameter at the centre point between the chord and the intersection of the tangents. Thus AP is a chord, the diameter through 2 (on AM) will intersect it in its centre point V, 42 is the tangent at A and therefore the tangent at P will also pass through 2, and C, which bisects V2 since

C2 CV M2 : P2

will be a point on the curve.

Similarly B may be shewn to be on the curve, since it bisects the diameter between 1 and the centre point of the chord AC, and D may be shewn to be on the curve as bisecting the diameter between C3 the tangent at C, and the centre point of the chord СР.

PROBLEM 38. (Fig. 37.) To draw a parabola the focus F, the axis FN, and a point P on the curve being given.

The directrix and consequently the vertex can at once be determined by drawing PM parallel to the given axis, measuring along it a length PL equal to FP and from L dropping a perpendicular on the axis intersecting it in X. This perpendicular is, of course, the directrix, and the vertex bisects FX. The curve can then be drawn by either of the preceding methods.

PROBLEM 39. (Fig. 38.) To draw a curve formed of circular arcs approximating to a parabola the focus F, and vertex A being given.

The following method depends on the fact that in the parabola the sub-normal is constant and equal to twice AF. Fig.38,

equal to FA, and draw With 1 as centre de

Draw the axis AFN and on it take F1 any ordinates as BB1, CC1, DD1, &c. scribe an arc through A, extending as far as the centre ordinate between A and BB,, from L the foot of ordinate BB make L2 equal to twice AF, and with centre 2 and radius to the point where arc through A meets the centre ordinate between A and B describe an arc extending to half-way between B and C; from M the foot of ordinate CC, make M3 equal to twice AF and with centre 3 and radius to the point where arc through B has been stopped describe an arc extending to half way between C and D. Similarly from the foot of the ordinate DD, measure a distance on the axis equal to twice AF so determining the centre (4) for an arc through D, and continue the process for any number of successive ordinates. It will be seen that the centres are determined by measuring a constant distance from the foot of the successive ordinates

equal to the known constant length of the sub-normal in the parabola (p. 60), but that the radius of each arc depends entirely on the arc previously drawn, so that the curve must be commenced from the vertex. Each successive arc extends some distance on each side of the ordinate from which its centre is determined. It is convenient, though not essential, to commence with ordinates dividing AF into equal parts, and tolerably close together, and as the curve recedes from the vertex and cuts the ordinates more nearly at right angles the distance between them may be increased. Carefully drawn, the method gives a remarkably close approximation to the real form of the curve, as may be seen by comparing the distance of the point P in the figure from F with the distance NX, its perpendicular distance from the directrix. The half distance between the ordinates to which each successive arc has to extend, and which furnishes the starting point for the next arc can generally be estimated with quite sufficient accuracy by the eye. PROBLEM 40. (Fig. 39.) To draw a parabola, the focus F, and two points A and B on the curve being given.

With centre A and radius AF describe a circle CFD, and with centre B and radius BF describe a circle C,FD ̧. Draw common tangents CC, and DD, to the two circles. (Prob. 31.) These will be the directrices of two parabolas fulfilling the given conditions, and the curves may be drawn by any of the preceding methods.

The construction is obvious.

PROBLEM 41. (Fig. 40.) To draw a parabola, the focus F, a point A on the curve, and a tangent YT being given.

The point of contact of the tangent is not given, as this would be a fifth condition. With centre A and radius AF

describe a circle FM, and on FA as diameter describe a circle EFE,, the centre being C. From F drop a perpendicular FY on the given tangent, and from Y draw tangents YE, YE, to the given circle. Join FE, FE, and produce them to meet the larger circle in M, M1, then MX, M ̧X, drawn parallel to YE, YE, respectively will be the directrices of two parabolas fulfilling the given conditions.

Proof. It is known that the perpendicular from the focus on a tangent passes through the point of intersection of that