If a fourth proportional be taken to

PN, PN+QN,, and PN-QN,,

i.e. if a length / be determined such that

PN: PN+QN, :: PN- QN, : 1,

the above equation may be written

i.e. AN is a fourth proportional to such length 1, NN, and PN. But this is really what has been done, for

i. e.

i.e.

and

That PN

2

=

Nx is the required length 7,

Na NN, PN: AN.

4AF. AN may be shewn thus: Join PA and let it meet the directrix in E. Join EF (F being the focus) and produce it to meet the diameter through P in L, while the diameter meets the directrix in M. Then since FA = AX, PL = PM=PF, for ML is parallel to FX, therefore the circle on ML as diameter goes through F, and therefore the angles MFL, MFE are both right angles and

also

2

EX. XM = FX2 = 4AF 2,

AN AX PN: EX by similar triangles,

:

PN2=4AF. AN, since AF= AX.

PROBLEM 47. (Fig. 43.) To draw a parabola, the axis AN, a point P on the curve and a tangent OT being given.

[The tangent must not be parallel to the axis, and the point must lie within the angle formed by the tangent and a symmetrical line on the other side of the axis.]

Draw the ordinate PN and let it meet the given tangent in O. Make NP, on the other side of the axis equal to PN, and P, will

by symmetry be a point on the curve.

Find a mean proportional

between OP and OP, (Prob. 5) and set off its length OE on OP

from O towards the axis.

Draw through E a parallel to the axis meeting the tangent in Q. Q is the point of contact of such tangent. Draw QN,, the ordinate of Q, and the vertex, A, will bisect N,T, the subtangent of Q (Prob. 36). The problem is therefore again reduced to Prob. 37.

Proof. That the diameter through Q, the point of contact of the given tangent, meets OP in E such that OE3 = OP. OP1, may be shewn thus. Let PAP, be a parabola and OQ a tangent at Q. Take any point a on the given tangent, and draw any two chords as abc, ab'P, and let q and q, be the vertices of the corresponding diameters, and let the diameter through q meet be in v: through a draw ad parallel to qv meeting the parabola in d, and draw du parallel to be meeting its diameter in u.

i.e. the ratio of the rectangles depends only on the positions of q and is independent of the position of the point a.

and 91'

If the lines abc, ab'P move parallel to themselves until they become the tangents at q and q1, we shall then obtain, if these tangents intersect in t1,

but the tangent aQ may be regarded as a chord cutting the parabola in two coincident points, and therefore if the tangent at q meet aQ in t and vq meet it in m

Also if Qk is the diameter at Q meeting ac in k, by similar triangles

PROBLEM 48. (Fig. 44.) To draw a parabola, the axis UN and two tangents PT, QT being given.

[The point 7 must not be on the axis.]

If from the point U in which either of the tangents (as QT) cuts the axis, a line UR be drawn making the same angle with the axis as QT but on the opposite side of it, this will, by symmetry, be a third tangent to the curve. Let it meet the other tangent (PT) in V. Describe a circle through the three points T, U, V (Prob. 20), cutting the axis in F. F will be the focus of

the required parabola, and FU will be the distance from F of the point of contact of either of the tangents QU, RU. With

Fig.44.

N

centre F and radius FU describe an arc cutting UR in R, with centre R and the same radius describe a circle, and the directrix will touch this circle and is of course perpendicular to the axis. The problem is therefore reduced to Prob. 36.

Proof. The fact that the circle through the points of intersection of three tangents is a locus of the focus is generally true, and is not confined to the case of two tangents meeting on the axis. For draw any tangent pab meeting the parabola in P, the two given tangents in a and b and the axis in c, and let Tb meet the axis in t. It has been shewn (Prob. 42) that the angle abt is equal to either of the angles pFb, PFb, also the angle Fpc = the angle Fcp the angle bct,

.. the remaining angle Fba of the triangle Fpb,

i. e. if two tangents intersect in b the angle which either makes with Fb is equal to the angle which the other makes with the axis.

Similarly, since QT, PT intersect in T, the angle FTa is equal

to the angle Ftb, i.e. btc,

... angle Fba = angle FTa,

or a circle goes round aFTb.

(Euc. III. 27.)

PROBLEM 49. (Fig. 45.) To draw a parabola, two tangents AT, BT, and their points of contact A and B being given.

First method. Divide AT, BT into any (the same) number of equal parts; the lines joining opposite points on the two tan

gents, (i.e. supposing each divided into 8 parts, the lines joining 1 on AT to 7' on BT, 2 on AT to 6' on BT, and so on,) will be tangents to the curve, which can easily be drawn to touch them all. Or points on the curve may be found successively thus. Bisect AT, BT in the points 4, 4'. The line joining these points is a tangent to the curve at its centre point, i. e. bisect 4, 4' in P and P is a point on the curve. Similarly the line joining the point of bisection (6) of 44 and the point of bisection of (m) 4P