GIVEN THREE TANGENTS AND DIRECTION OF AXIS.

82 the points of contact of TU and TV. The problem is therefore reduced to Prob. 49, or it may be completed by utilising other known properties of the curve already demonstrated, e. g.—making the angle TQF equal to angle TQM, QF is a locus of the focus ; similarly QF (the angle TQ,F being made equal to angle TQ,M1) is a second locus, and F, the focus, is therefore the intersection of QF, Q1F.

Again, the circle round UT, TV, VU is known to be a locus of the focus (Prob. 48), and the angle UFQ is known to be equal to the angle TUV. Prob. 42. Therefore, if on UQ a segment of a circle be described containing an angle equal to the angle TUV (Prob. 30), the intersection of this segment with the above circle will determine F. Any number of tangents to the curve between Q and Q, can be at once drawn without previously determining the focus by measuring the length mm, anywhere on MM, between M and M1 and from the extremities drawing perpendiculars to MM, to meet TQ, TQ1. Any pair of such points being joined will of course give a tangent to the curve.

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Proof. That the projected length on MM, of the portion of any tangent intercepted between TQ, TQ, is constant may be shewn thus. Let R be the point of contact of UV and let the diameter through R meet MM, in t. Draw Qn, Q1n, parallel to UV meeting tR in n and n1. Then UR = Qn (Prob. 49) and therefore tm = 1⁄2tM. Similarly tm1 = 1⁄2 tM ̧.

Therefore mm1 =

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MM,= constant, since MM, is the projection

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of the chord of contact of two fixed tangents.

PROBLEM 53. (Fig. 49.) To draw a parabola, two points A, B on the curve and two tangents TL, TM being given.

[The tangents must not be parallel and the points must not be on opposite sides of either tangent.]

Draw a line through A and B meeting the given tangents in L and M. Take LC on LM a mean proportional between LA and LB (Prob. 5), and MD on ML a mean proportional between MB and MA. Bisect CD in E. TE will be the direction of the axis

of a parabola fulfilling the required conditions and CQ, DQ, drawn parallel to TE to meet the given tangents will determine Q and Q,,

their points of contact. The problem therefore reduces to Prob. 49, or may be completed similarly to the preceding. Since LC and MD may be set off on either side of L and M, as LC ̧, MD ̧ in the figure, the point of bisection E, of C,D, determines TE, the direction of the axis of a second parabola fulfilling the required conditions. Further, either CD or CD ̧ may also be taken as the segment to be bisected, and there are consequently four solutions. The proof depends entirely on the property of the parabola already referred to in Prob. 47.

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GIVEN THREE POINTS AND A TANGENT.

PROBLEM 54. (Fig. 50.) To draw a parabola, three points A, B, C', and a tangent LM being given.

[The points must all be on the same side of the tangent.]

Join two pairs of the given points as AB, BC and let the joining lines cut the given tangent in L and M. On LB take LD a mean

and on MB take ME Then by the property

proportional between LA and LB (Prob. 5), a mean proportional between MC and MB. of the parabola already referred to (Prob. 47) a line through D parallel to the axis of a parabola through A and B and touching LM, will pass through the point of contact of LM with such parabola; and a line through E parallel to the axis of a parabola through B and C and touching LM will pass through the point of contact of LM with such parabola. Hence the line joining DE will be parallel to the axis of a parabola which can be described through AB and C to touch the given line, and its intersection with LM will determine the point of contact of such parabola.

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Since LD, ME can be set off on either side of L and M (as LD1, ME1), similarly the line joining D, and E, will be parallel to the axis of a second parabola fulfilling the conditions of the problem; its point of contact being P: and similarly DE, and DE will determine the direction of the axes of two more such parabolas. The line DE, determines P, as the point of contact.

Hence there are four solutions, and the problem in either case is reduced to Prob. 51. In the fig. two of the four parabolas are drawn, viz. those whose axes are parallel to D,E, and DE, respectively; the necessary construction in each case being indicated.

It might be considered at first sight that if a mean proportional were taken between the segments NA, NC of the line joining AC, the third pair of the given points, cutting the given tangent in N, two additional points would be obtained which, being joined to either D, D1, E or E,, would give the directions of axes of additional parabolas. This however is not so, since it will be found that the points thus obtained coincide with the intersections of ED, ED1, and of DE, ED, respectively, and therefore no more solutions than the four already mentioned are obtainable.

PROBLEM 55. (Fig. 51.) To draw a parabola, a point A on the curve and three tangents BC, CD, DB being given.

[No two of the tangents must be parallel, and the given point must not lie within the triangle formed by the tangents, nor so that any one tangent lies between it and either of the remaining tangents.]

Let C be the vertex of the triangle formed by the tangents, which cannot be reached from the given point without crossing BD. Through B draw BE parallel to CD and through D draw DE parallel to CB, meeting BE in E. Through C draw CK parallel to BD and join EA meeting CK in K, CB in L, and BD in M.

First let lie between E and K; complete the harmonic range KAEA ̧, i.e. find a point A, beyond E on KL such that

KA: KA, AE: EA,

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(Prob. 12.)

[Through A, K draw Aa, Ka any two lines intersecting in a, produce a to a, making Aa, Aa. Join a, E and produce it to aA = meet Ka in b. Draw bД, parallel to aA and it will intersect KA in the required point.]

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Then A1 will be a point on the curve and the problem reduces to Prob. 53.

Second, let the given point lie beyond E as A,, then, completing the harmonic range AEAK (Prob. 11), A will be a

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GIVEN THREE TANGENTS AND A POINT.

second point on the curve and the problem again reduces to Prob. 53.

In completing the figure, one of the tangents employed should be the one situated as BD in the figure, because it is necessary to take a mean proportional between the segments of the chord AA, included between the tangent and the curve, i.e. to take a mean proportional between MA and MA,: but it will be found that ME, MK are each equal to such mean proportional, and therefore E and K can be at once used without any further construction. If CB is the second tangent made use of, a mean proportional LG or LG, must be determined between LA, LA, (Prob. 5), and two of the four parabolas which can be constructed by means of pairs of the points K, E, G, G, to pass through A and A, and to touch BL, BD will also touch CD. There is an ambiguity as to which particular pairs of points must be selected, but this can easily be settled by trial in any given case. In the fig. it will be found that the pairs E, G and E, G, are those required, and that

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