(p2 sin.

or p2

+ p2 cos.2 0) = a2 (p2 cos.” ◊ — p2 sin.2 0)

a2 (cos.20 sin.2 0)

a2. cos. 20.√cos. 20, the equation required.



The loci of the corresponding values of x and y, 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 x", the dimension of each term will


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

( 2/2 ) ?


Hence, after dividing by the coefficient of





the terms according to the powers of, the equation will be re


duced to the form,






У "

( 2 ) ̄ + ^ ( 4 ) + ^ ( 2 ) * + ( ~ ) + A = 0.







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Let the roots of this equation be denoted by a1, ɑ2, α.....ɑmi then, by the general theory of equations, we have

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Ex. To find what lines are defined by

· 2xy2 + x3 = 0, we have

and 1-√5, we have


α 2 =0,....

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and ordering



or y = a1 x, y = A2 X,.... y = ax, which are equations to straight lines. Consequently the corresponding loci of x and are straight lines. Q.E.D.


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y + 1 = 0, whose roots being 1, 1+√5

1+ √‍ 3

y = x,y=



x, and y =

X, which define three


straight lines issuing from the origin of abscissæ, and inclined to

the line of abscissæ at the angles 45°, tan.-1 1 + √ 5, and


-√3 respectively.


But y

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.

(MQ — MV) tan. Q.

For VtVQ. tan. Q = = y x tan. Q,

and AT (AV + VQ) tan. Q.

Hence AT × Vt




= y + (2a x} tan. Q.

and tan. Q=

(2ax-x) b being the axis-minor,


y2 + 2 (a−x) y tan. Q — (2ax-a2) tan.2 Q.




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.. by substitution, we get

AT x Vt =

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62 {2ax − x2 + 2(a — x)2 — (a — x)3}



= b2 which is constant.


Conversely, if AT x Vt be constant, and b', AT, AV, then the curve to which Tt is perpetually a tangent, is an ellipse.

Vt being

By the preceding problem, we have

AT × Vt = y2 + 2. (a — x) y tan. Q — (2ax − x2) tan.'Q

by supposition,

y2 + 2. (a

x)y tan. Q-(2ax − x2) tan.'Q = b2 ..... (a)

Now, supposing Tt to receive a minute change of position,


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 dtan. Q-2. (2ax 2. (2ax - x2) tan. Q. d tan. Q = 0.

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

y=(ax) c± √ b2 + c2a2

for the general solution, which defines all the lines Tt, 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,

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y2 =

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the particular solution, defining the locus of the intersections of Tt, Tt,&c., which is, therefore, an ellipse whose semi-axes are a and b.

dy And =

If for tan. Q we put its equivalent dy, equation (a) will become



– x)y! — – do

y2+ 2 (a


- (2ax — x2). = b2,



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

2 (ax) y

(2arx) 0







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dx 2ax x2

√2ax x2




which give the same results as before.


To find the curve which cuts off a third part from 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
and SA

PP' = P'T

AA' — A'T.

Join SP, PP, PT, and from C the middle of ST draw CGL ST. Then it is easily proved that chord SP chord PP' = chord PT, PG = GP' and .. SP = 2PG, and that SA = 2AC. y, and (since the curve evidently passes

Let ACc, PN through A) AN = x.

Then SP2 = (2PG)2 = (2NC)2 = also SN2 + PN3.

But 4NC

(2c — x)2 = 4c2

And SN
.. y2 = PN2 = 12cx + 3x2 = 3 (4cx + x2).

Put a = 2c

4 (c+x)2 = 4c


+ 8cx + 4x2

4cx + x2



and = 3, 12c2

Then y = (2ax + x2)=


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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.


Find the equation to a straight line any how disposed 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,

AX, AY, AZ, and

be referred to the rectangular co-ordinates suppose it to meet the plane YAX in B. Draw from P, any point in the line BL, Pp 1 plane YAX, and join Bp. Draw, also pm, pn AY, AX, and through B draw bC, aD parallel to pm, pn, respectively.

Then, putting Pp = z, pm = An = x, pn = Am = y, ≤ PBp = inclination of BL to the plane YAX ≈y, and ≤ 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 2, and y = Am = Ab + bm = b + BD = b + Bp, sin. « = b + sin. a. cot. y × 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 + z. cos. a. cot. yi
3 = 6


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', ', by the preceding determination, we get

x' = a' + z'. cos. a'. cot. y'?
ybz. sin. a'. cot.y'S


in which it remains for us to obtain the values of a', b', cos. a', sin. a, cot. ', in terms of the known constants, a, b, cos. a, sin. a. cot. y'.

Now, since B'Q is BP, if we make Qg plane YAX, and join B'q, B'q will be Bp. Hence the inclination of BQ to the plane ZAX = B'sC = a + 90° a'. Therefore, sin. a' = sin. (a + 90°) = cos. a, and cos. a' cos. (a + 90)° =

sin. a; and by substitution, equations (b) becomes x' = a' z'. sin. a. cot. y


Y = b' + z'. cos. a. cot. 'S

x = m

and y'

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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. y'

n = b'+r cos. a. cot. y'

Hence, deducing a' and b', and substituting in (c) we get


r) sin. a. cot. y'



n + (z'

r) cos. a. cot. 'S

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

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azcos. a. cot. y = m and bz' sin. a. cot. y =


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(z) sin. a. cot. y'l n + (z' — r) coș. a. cot. y'

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