.. substituting this value of cot. y' in (d), we ultimately obtain two equations between x', y' and z' and known constants, determining the exact position of the straight line B'Q, which is BP, and passes through the given point P.

Q. E. I.

Eliminating z cot. y from equation (a), we get

which, being independent of z, determines the straight line BL when z = 0, or BL lies in the plane YAX.

Eliminating (z-r) cot. ' from equation (d) there results, after proper reduction,

which defines the straight line B'Q when z = 0, and z' = 0, or both BP, B'Q lie in the plane YAX.

60.

Required the curve whose nature is such that its abscissa, ordinate, subtangent, and distance between the origin of abscissæ, and the point of intersection of the tangent and axis, are in continued proportion.

y¤—y2, which being homogeneous, put y = ux,

x2

F

origin of abscissæ A (Fig. 42). Let xx'; then y = 7(— 1)

BC, which is .. an asymptote; also for every value of x > ē we have a real value of y, and when x = ∞, y = 0, .. the curve has another branch D'A' to which BC, BD are asymptotes.

61. Let C (Fig. 43) be a right angle, and from A, B given in position, let two points move with equal velocities; and let A', B′ be any contemporaneous position of those points; required the curve to which A'B', or it produced, is perpetually a tangent?

Let CA > CB = a, and CB = b. Then it is evident, that CA' will touch the curve in V, (AV being taken = BC,) and it will therefore be convex to CA'.

Let. B'A' touch the curve in P, and draw PM CA and make PM y, MV = x.

Now, by similar ▲, we have

y: MA:: CB': CA'

:: b + BB' : a 4+ AA'.

But, by the question, BB' AA' AV' b VM - MA'-b

--

a

dx

in equation (1), the

dy

x = cy + - the former the general integral of (1),

C- 1

And (x − y)2 = 4 (a - b) y

and the latter its particular solution.

Now, the latter being of the second degree, shews the curve to be a conic section. Also, since

there are but two infinite values of y for one of x, or the curve

a parabola.

To find the vertex, and latus-rectum, we proceed as follows: Let BV' AC; then the point A' will be in C, and CV' will touch the curve in V'. Also CV CV'. Hence, by a well known property of the parabola, C is the intersection of the axis and the directrix. Making. CR the axis, and drawing PM'LCR, and putting CM' x', PM'y', we shall easily obtain y=. x-y

√2

Hence, the equation to the curve, reckoning the abscissæ from ༧, is

y'2 = √2. (a - b) x".

The problem may be generalized, without increasing, materially, the difficulty of its solution, by supposing the angle C any whatever, and the points A', B' to move according to any given law. For problems involving other particular solutions, see pp. 47, 48, and the next problem.

62. Let A (Fig. p. 205 of the Problems) be the origin of abscissæ, PT any tangent to the curve, and AD be drawn 1 line of abscissæ meeting the tangent in D. Required the nature of the curve, when AT ∞ AD"; or the enunciation may be stated, If from A, a given point in the given straight line AT, AD be drawn at right-angles to AT, and AT be always taken a x ADTM (a being a constant)required the curve touched by the several lines passing through T, D.

Making A the origin of x, we have

which, coming under Clairaut's Formula, is integrable by dif

ferentiation. By that

process, and proper

reduction, we get

which being substituted in equation (a) give respectively,

The equation (c) expresses the nature of the curve required. pp. 47, 48, 67.

See

63. Required the nature of a curve G'P, (Fig. 44,) such, that if, PN being the ordinate, and PG the normal, we bring the ▲ PNG into the position P'N'G' determinable from that of PNG by making NN' Some given function of the abscissa, denoted by NN'= f.x, then G' shall be a point in the curve, and G'P' the normal at that point.

ydy

dx f'x