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maxima; the latter a minimum; the position of the maximum value being found by taking 6 = 60°. Since in the Trisectrix the equation is

p=r.(1 + 2 cos. C) and here

é = r.(1 + cos. 6) the analogy between the curves is very evident. We may, in fact, derive the Trisectrix from the latter by constantly taking Mm = CN. They will intersect when 0 = 90° scil. in M'. The Trisectrix will pass through the centre C; for at A, CN becomes negatively = radius, and p= 0. When p = 2r, ø = Ab = 3r. Multiplying the two equations together, &c., the result p=

+ 7 4 + 12 cos. 0 + 9 cos. < 0 will represent the com

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2

2

pound curve.

For a complete discussion of the Trisectrix, see Trisection de l'Angle, par Azemar et Garnier, Paris, 1809.

The problem may be generalized, by requiring the locus of the intersection of the tangent and perpendicular let fall upon it, from any point given in position, of any curve whose polar equation is

r=f(p) r being the radius vector, and f (P) any function of the corresponding I upon the tangent.

First, let (A) the given point be also the pole of the given

curve.

Let P and Q (Fig. 21) be two points indefinitely near in the given curve BPQ, and corresponding tangents Pp, Qq. Also let p, q, be the intersections of these tanyents with the perpendiculars Ap, Aq, and suppose pqc the locus required.

Since P, Q, and :: p, y, are indefinitely near, p q being joined and produced will touch the locus in p, and p q will be a straight

7

line. Draw Ay I pq. Again, because the 2 at p, and g, are right angles, and Q and P may be considered as coincident, a semicircle may be described passing through A Ppq.

Hence 2 ypA = 2 pPA, and the 2 p and y are right k.
:: A Apy is similar to A ApP, and we have

Ay : Ap :: Ap : AP
Let Ay = *, Ap (= p) = R, and AP = r,
Then 7 =
pl p? pa

(a) the equation between

.(p) fle) the radius vector p, and the perpendicular upon the tangent of the required locus. Hence the locus may be found. Again, let pAB = 0. Then by similar A Ayp, qpr

Ay : py :: qr : pr
:. Ay: Apl :: gro : qr? + pr?
But gr = Aq.de = p.de
and pr = de

pa
..
pod6° + dpa

dp
p? +

d 02
pe

by (a)
si
dp?

f(d)
d02

p* d 02

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Sa

2

de Hence 0 =

(6)

v(.)+p the equation between the radius vector, and polar angle of the Locus.

Equations (a) and (6) will each determine the locus when its pole is coincident with that of the given curve.

But when its pole (A') does not coincide witb A, let pq', &c., be the required locus,

and put A'p = p, 2 p'A'B = 6',
the given distance AA' = a,

and the given ZAA'B = B. Draw AM I A'p'. Then: Aq, A'q are I Q7, and A'p, A'p are IPp, the L,'A'p' is evidently = 2 Ap,

.. dll' d 4.

Also p = p'A' = p M MA

= p'M + AA'. cos. AAM

= pt a . cos. (a 6) and substituting in equation (1) we get

d. a. COS. a

-0')

(6). V{f.p-a.cos. («— 6')}: -(-a.cos.z – 0")* by the integration of which we may find the locus wherever the given point, and the pole of the given curve, may be situated.

do' =

The formula (b), which will generally be most commodious in practice, may readily be exemplified by applying it to the solution of our problem.

Since RM (Fig. 20,) touches the O in R, the Z MRA, = _ RBA, and .:. the right-angled A ARM, ARB, are similar.

Hence AM. AB = AR, or making (as before) AC = 1, and AM = p = p, and AR = 1,

2p = p?

::r = N2p = 2p = f (p) Heuce by substituting in equation (6), the negative sign being taken, because dp and do have different signs,

de

√ 2q - p
Let 0 = 0. Then p = 2, and C = vers. 2.
:: 0 = vers. - 1 (2 - P)
.. 2 = vers. 0 =1
iig=+ cos. 6, the same as before determined.

vers.

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Ex. 2. Let the given curve be the common parabola referred to its focus. The latus rectum being called (1) it is known that

1

.-=pe,

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:: by substitution in equat. (b) we get, (measuring 9 from the vertex)

1

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16 p2

12

1
Put u =

substitute, &c.
4 p
- du

1
and 0 =

= cos. -1 utc=cos.
N1-22
1

= cos. 0
4p

1
and
P

. sec. 0, which is the equation of a straight

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line I at the vertex to the line joining the focus and vertex of the parabola.

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Ex. 3. Let the given curve be the common cycloid; required the locus of the intersection of the tangents and perpendiculars upon them from the centre of the generating circle.

Since PT (Fig. 19,) touches the cycloid, it is parallel to AB (PM being I AO, &c. &c.)

Then Op being I AQ is also I TP. Let AOD = 0, AO = 1, Op, the radius vector of the locus = p.

Then CG being joined and produced to G since ABG = R.L, CG is parallel Op, and G is a R. Z. Hence, the A PBG, MBC, are similar, and

:: Dp = BG : BP :: BM : BC
But BP = arc AB (by property of cycloid)

= L AOB = 20
BM = OB. sin. 20 = sin. 20

CA
and BC = OD X = 20D = 2 cos. 9

OA

20. sin. 20 :: Dp =

- 20. sin. A

2 cos. A Hence p = OD + Dp = cog. 8 + 20. sin. O, the equation of the locus.

After due attention to the above discussion (which will be useful hereafter in the theory of the Aberration of Light), the student may proceed to find the Locus of the intersections of the tangents, and perpendiculars of the first locus of a given curve ; the locus of the intersections of tangents and perpendiculars of the second locus, and so on to any number of Loci. He will thus obtain many curious and interesting results.

28. To trace and construct the curve whose equation is,

.r3 + bor?

y? =

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In the equation

y = $x.

6+x... (a) let x = 0,c, and – b. The cor

C-T

responding values of y are£0, = 00, 0. Hence, if from A (Fig. 22,) the origin of abscissæ, we take AC = c AB = -b, and draw CD, CD | BC, the curve will pass through A, and B, and CD, Cd will be asymptotes to the infinite branches AE, Ae. Also for all values of x > c, or > (- 6) (the sign not being considered), the values of y are imaginary. Again, putting 0 = inclination the tangent to the axis, we have

dy
tan. 0 =

b+c
= + + +
dx

(c x)}.76+x Let x = 0, and - b, and the corresponding values of tan. 6

= 0, respectively

2

C

are

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