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(p® sin.6 + pe cos. 6) = a(* cos. 0 – pe sin.()
or p = a (cos.0

= a.cos. 20.
:: p= a. cos. 20, the equation required.

sin.: 0)

55. The loci of the corresponding values of x and y, in an homogeneous equation, are straight lines.

Let the sum of the indices in each term of the equation determining the degree of its homogeneity be m. Then if we divide each term of the equation by r", the dimension of each term will

obviously be zero, i.e., each term will be of the form

X

( 3 ) ? Hence, after dividing by the coefficient of ( 1 )" and ordering

the terms according to the powers of y, the equation will be reduced to the form, Y +A, Y

+A,

Let the roots of this equation be denoted by a 1, 0g, a ....ami then, by the general theory of equations, we have y y y

y

- am) = 0

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or y = 0, X, y = 2, X,.... y = an X, which are equations to straight lines. Consequently the corresponding loci of x and y are straight lines. Q.E.D.

2

Ex. To find what lines are defined by y3 2.xyl + x = 0, we have y -2 y +1= 0, whose roots being 1, 1+5

2 and 1-15, we have

2

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2

y = x, y=1+v5 æ, and y = 1–7 5 x, which define three

2 straight lines issuing from the origin of abscissæ, and inclined to the line of abscissæ at the angles 45°, tan.-- 1+ 5, and tan.-- 1 - 5

15 respectively.

2

56. The rectangle under the two segments of the tangents at the extremities of the axis major of an ellipse, cut off by any other tangent, is constant.

Let APM (Fig. 39) be an ellipse whose axis-major AV = 2a. Vt, AT and Tt being tangents, we have AT > Vt=const. For Vt = VQ. tan. Q = (MQ – MV) tan. Q. =y

– x tan. Q, and AT = (AV + VQ) tan. Q.

= y + (2a – x) tan. Q. Hence ATX Vt = y2 + 2(a—x) y tan. Q (2ax -2%) tan.” Q.

62

(2ax – xo) 6 being the axis-minor,

But y =

a-C

6
and tan. Q=

dy
dx
a 2ar

.:*
:: by substitution, we get
AT X Vt =

62

{2ax – x2 + 2 (a -- x) – (a – x)"} a

= 6' which is constant.

57.

Conversely, if AT x Vt be constant, and = b', AT,
Vt being 1 AV, then the curve to which Tt is perpetually a tan-
gent, is an ellipse.
By the preceding problein, we have
AT * Vt = y2 + 2. (a – x) y tan. Q - (2ax — xo) tan."Q

And :: by supposition,
y' + 2. (a -- x)y tan. Q - (2ax - xo) tan.'Q = b .....

? (a)

Now, supposing Tt to receive a minute change of position, y and x will remain the same, whilst tan. Q will undergo a variation. Differentiating, therefore, on that hypothesis, we get 2. (a-x) y x d tan. Q - 2. (2ax - x*) tan. Q. d tan. Q=0.

i. d tan. Q=0 Or (a – x)y= (200 – **) tan. Q.

i. tan. Q = const. = c(1) And tan. Q=

(ax) y 2ax x2

Substituting the value of (1) in equation (a) we get, after solving the resulting quadratic,

y= (a - x)c+ b + c'a' for the general solution, which defines all the lines Tt, , &c., which can be made by giving to c all possible values, and by substituting for tan. Q in (a) its value in equation (2), we get, after proper reduction,

62

2ax

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the particular solution, defining the locus of the intersections of Tt, T't',&c., which is, therefore, an ellipse whose semi-axes are a and b.

If for tan. Q we put its equivalent

dy

equation (a) will become

dx
dy
2ax — )

dy?

= b?, dx

dxa

And differentiating, all the terms will be destroyed by opposition of signs, except

doy dyd'y 2(a – x) y

- 2

(2ar - x) = 0 dir

dx?

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rom

58. To find the curve which cuts off a third part all circular arcs described

upon

the same chord.

Let SPT (Fig. 40) be any one of these arcs described upon the chord ST.

Suppose SP = PP = P'T

and SA = AA' = AT. Join SP, PP, PT, and from the middle of ST draw CGI ST. Then it is easily proved that chord SP = chord PP = chord PT, PG = GP' and - SP = 2PG, and that SA = 2AC.

Let AC =C, PN = y, and since the curve evidently passes through A) AN = x. Then SP = (2PG) = (2NC) = also SN + PN”. But 4NC = 4 (c+x) = 4c + 8cx + 4x2 And SN = (2c - x) = 4c – 4cx + ** :: y = PN? = 12cx + 3x = 3 (4c2 + xo).

Put a = 20

62
and = 3, ..63 = 12c

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to an hyperbola, whose semi-axes are 2c and 2c. 3.

That the Hyperbola is the curve required, is also evident from the property of that curve with regard to its directrix. If S be the focus, and CG the directrix ; then SP: PG in a constant ratio, viz. 2:1. Consequently AP is an hyperbola.

59. Find the equation to a straight line any how disposed in space, and thence deduce the equation to another passing through a given point, at right angles to the former.

Let the straight line B, (Fig. 41) whose equation is required, be referred to the rectangular co-ordinates AX, AY, AZ, and suppose it to meet the plane YAX in B. Draw from P, any point in the line BL, Pp I plane YAX, and join Bp. Draw, also pm, pn 1 AY, AX, and through B draw 6C, aD parallel to pm, pn, respectively.

Then, putting Pp = z, pm= An= x, pn = Am = y, 2 PBp = inclination of BL to the plane YAX = y, and 2 pBC = its

inclination to the plane ZAX = a, and the co-ordinates Bb, Ba, of the given point B, a and b, we have x = An = Aa + an = a + BC = a + Bp cos. a = a + cos. a. cot. y X z,

and

y = Am = Ab + bm = b + BD = b + Bp, sin. a = b + sin. a. cot. y X z; or the two equations, which evidently determine the position of BL in space (since they do any point of it, P) are x = a + 2. cos. a. cot. r?

(a) y = b + z. sin. a. cot. S Again, let another straight line B'Q, passing through the given point P. meet BP at right-angles in Q, and the plane YAX in B'. Then, calling its co-ordinates a', y', z', by the preceding determination, we get

X' = a + z. cos. a'. cot.g
a ''?

(6) y = b + z'. sin. a'. coty's in which it remains for us to obtain the values of a, b, cos. a', sin. cé, cot. Ý, in terms of the known constants, a, b, cos. a, sin. a. cot. a'.

Now, since B'Q is I BP, if we make Qq I plane YAX, and join B'q, B'q will be I Bp. Hence the inclination of BQ to the plane ZAX = B'sC = a + 90o = a'. Therefore, sin. á = sin. (a + 90°) = cos. a, and cos. a' = cos. (c + 90)° sin. «; and by substitution, equations (6) becomes x = a z'. sin. a. cot. ♡

(c) y : = b' + z. cos. a. cot. v's Again, supposing the co-ordinates of the given point P to be m, n, r, we have, by equations (c) m = a

r sin, a. cot. a' n = b' + r cos. a. cot. g'? Hence, deducing a' and b', and substituting in (c) we get (z r) sin. a. cot. a'2

(d). n + (z' r) cos. a. cot. g's Lastly, since the point Q is common to both lines, at that point we have, z' = x, y' = y, and z' = z. Hence, by (a) and (d),

a + z' cos. a. cot. g = m – (z -- ») sin. a. cot. i'? and b + z' sin. a. cot. y = n + (z' – ) coş.a. cot. y'I

x = m and y

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