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And eliminating z and c we find
(m a) sin.a (n

b) cos. a
(m – a) cos. a + (n - b) sin a --

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x cot. y.

p cot. g

:: substituting this value of cot. ' 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 I BP, and passes through the given point P'. Q. E.I.

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

- a tan, a t t. tan. O ..... a which, being independent of 2, determines the straight line BL when z= 0, or BL lies in the plane YAX.

y=6

Eliminating (2'-) cot. g from equation (d) there results, after proper reduction, y'= n + m cot. a

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

a cot. a.....

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.

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yd.r

Then, by the question, we have

ydx x :y :: :

dy dy

yder .. r: x y

:X

dy
dy
..

which being homogeneous, put y = ur, dor

yx-y?

x?

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origin of abscissæ A (Fig. 42). Let x = - z'; then y=1(-1)

+ lx'.

= l.e

*V=1 But 1(-1)=1. (cos.n't N - 1 sin. 7)

ET

-1, an imaginary quantity. :: the curve does not go beyond A. Again, let lx = – C, or x = e = AB; then y =

0

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

dy Since tan. 0 = C-1+ la

= 0, the dir (C+lx)" curve touches the line of abscissæ at A.

1

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 I CA and make PM = y, MV = x. Now, by similar A, we have

y : MA:: CB': CA'

:: b BB : a + AA'. But, by the question, BB' = AA'= AV' b= VM — MA' — b =

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a-6

x = cy +

-1

the former the general integral of (1), And (1-y)' = 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

1 = y +2 Va 6. Vy..... (2) there are but two infinite values of y for one of x, or the curve

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is a parabola.

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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'ICR, and putting CM'=', PM'=y, we shall easily obtain y=

x'

2 x' + y, and substituting in (2), we get y's = 12. (a - b) x' - (a−b)....

(3)

and I = bcat

2

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and the latus-rectum = 4Cv = 2. (a - b).

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

y'? = N 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 I line of abscissæ meeting the tangent in D. Required the nature of the curve, when AT O ADm; 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 AD" (a being a constant )required the curve touched by the several lines passing through T, D. Making A the origin of x, we have AT = subtangent

yd.o

dy and AD = AT.

dy

zdy
= y
dir

dx
ydx

xdy
x = a. (y

dx
:: dividing by ydx xdy
dr
= a. (y

xdy
dar

:)

dy

m-1

dy

1

1

1

and

dy
y
xdy

ta
dx

dx

(a) which, coming under Clairauts Formula, is integrable by differentiation. By that process, and proper reduction, we get

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which being substituted in equation (a) give respectively,

y = cx + a 1-m for the general solution, and

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(6)

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L m-1.2

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or y = 1

(c) a".(m-1) for the particular solution.

The equation (c) expresses the nature of the curve required. See

PP. 47, 48, 67.

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

Let PN = y, G'N' = y'
AN = x, AN' - x'.
Then, &
AN AN

NN' = x - - f.x

ydy = G'N' = subnormal =

dx :: making dx constant, d.x' = dx - dx f'r

and y'

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