Sidebilder
PDF
ePub

The equation a’ (y – a)' = (r – a) (2 a x -- z) belongs to the same curve referred to a different origin.

a2 + a? Ex. Trace the locus of the equation y = xe

a tra 316. AM, fig. 1, is a tangent to a circle ACQ, M Q an ordinate to the abscissa 'A M; M P is taken a mean proportional between A M and MQ;

a required the locus of P.

[merged small][merged small][merged small][subsumed][merged small][ocr errors][ocr errors][merged small][subsumed][merged small][merged small][merged small][merged small]
[ocr errors]
[ocr errors]
[ocr errors]

x

[ocr errors]

Let AM = x, and M P = y, be the rectangular co-ordinates of P, and let the radius of the circle b,

then the square on MP = the rectangle A M, M Q. To find MQ, we have the equation to the circle

(y y') + (x - x')' = r,

- ' x
or ys - 2 b y + x = 0, since d' = 0, and y' =r= b,

:: MQ — 6 IN 62 - x?,

:: M P or y = {b x + x 763 – }. Since 1% - x is < 69, there are four values of y to each positive value of x < b, and no value of y to a negative ; hence if A B = b, fig. 1, the straight line C B C' perpendicular to A B is a limit to the curve, and when x = 6, the ordinate to the curve is equal to the extreme ordinate of the circle, that is, to the tangent B C.

Between x = 0, and x = b, we have four values of y, which give the two dotted ovals of fig. (1).

To make the question more general we shall suppose the line A B to be a chord of the circle, figs. (2) (3) (4).

Then if b and a are the co-ordinates to the centre of the circle, and A the origin, the equation to the curve will be

y= + v{bx Fx16% + 2 a x x2}, and we have four cases depending on the values of b and a; hence we have four curves of different forms, yet partaking of the same character and generation.

Case (1). a = 0, fig. (1) already discussed.
Case (2). a and b positive, fig. (2). A E= a + a + bo.
Case (3). b = 0, fig. (3).
Case (4). b negative, fig. (4), the equation is
y=+/{-bx 5 x 7b+ + 2 a r — x?}.

M

There are two values of y for x positive, and < 2 a; but four values for x negative, and <v a + 62 – a, that is, < A E.

The gradual transition of one curve to another is apparent, but that the same problem should produce such very different curves as (2) and (4) requires some explanation.

In fig. (1) P and P are determined by mean proportionals between A M and M Q, and also between A M and MQ'. Moreover P may be in QM produced as well as in MQ, thus we have the double oval, fig. (1.) 011 the left of A the abscissas A M will be negative, and the ordinates MQ positive; hence no possible mean proportional can exist, or no part of the curve can be on the left of A.

In fig. (2) A M and MQ determine the points P and P'; but A M and M Q' give only an imaginary locus.

Fig. (3) requires no comment.

In fig. (4) the reasoning on fig. (2) will explain the positive side of A; on the left of A the abscissa and both ordinates are negative ; therefore two mean proportionals can be found, or four points in the curve for each abscissa.

Such curves may be invented at pleasure, by taking the parabola or other curves for the base instead of the circle.

Ex. To find the locus of the equation yö + 2 a x ye a 28 = 0.

317. To find a point P', such that drawing straight lines to two given points S and H, we may have the rectangle SP, H P constant.

Join the points S and H, and bisect S H in C ; let C be the origin of rectangular axes, S H = 2 a, CM = x, MP = y and let the rectangle SP, HP, = ab. Then since S M = a + x, and H M= a - X; we have {+ (a + x)' } {y* + (a - x)} = a bo,

? ,
or (y' + i + a + 2 a x) (y + y + a* - 2 a x)=a* 69,
* a

' b
or, {y+ + a'}" - 4 a ta' b;
hence y = v{-(a + ) + a v 62 + 4x2}.

a2
Let y = 0, : x = IVa (a £6) (1).
Let x = 0, y = va (6

(2).

[ocr errors]
[ocr errors]
[ocr errors]

ВІ

1. Let a be less than b.

Then from (1) we have the points A and A', and from (2) we have the points B and B'

a

Also hy comparing the values of y in the original equation and in equation (2) we shall find that M P is greater than C B as long as x is greater than N 2 a (2 a - b); thus the form of the curve must be like that of the figure A PB A' B' A.

As b increases, the oval becomes flatter at the top, and takes the form of the outer curves.

2. Let a = b, then we have the dotted curve passing through C; also since the equation becomes (r2 + y2):= 2 a? (x2 - y2) the locus is in this case the lemniscata of Bernouilli.

3. Let a be greater than b.

Then from (1) we have two values of X, and from (2) an impossible value of y; hence the curve must consist of the two small oval figures round S and H.

As b decreases, the little ovals decrease; and when b = 0, we have the points S and H themselves for the locus.

These curves are called the ovals of Cassini, that celebrated astronomer having imagined that the path of a planet was a curve like the exterior one in the above figure.

The equation (y + 2%) = b ye + ax*, found in art. (123), gives a figure like that in case 1.

318. There are some cases in which it is useful to introduce a third variable; for example, if the equation be y + xy + 2 y + x = 0, it requires the solution of an equation of three or four dimensions, in order to find corresponding values of x and y; to avoid this difficulty, assume x = UY,

.. y* + u® ya + 2 y: - u y3 = 0, or, y tu y + 2

US = 0,

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

816

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

u 2

u 2 and ru.

3 U2 +1

u +1 from these equations we can find a series of corresponding values for x and y.

3
2
2

4
1
11

1*
170

*7
0
2

0

1}
1


1 }
2
1 }

2 en
3
2 }

7 7
4

317
&c.
&c.

&c.

[ocr errors]
[ocr errors]
[ocr errors]

1

[ocr errors]

14

&

Y

[ocr errors]

Also when y=0, x = 0, hence the curve passes through A. Let A X, A Y be the axes ; along the axis of y take values equal to those in the table for y; and from the points thus determined draw lines equal to the corresponding values in the table for x (these are the dotted lines in the figure); by this method we obtain a number of points in the curve sufficient to determine its course.

This example is taken from the “ Analyse des Lignes Courbes, by G. Cramer. Geneva. 1750,” a work which will be found extremely useful in the study of algebraical curves.

319. To trace the curve whose equation is yö – 5 a x* y' + x = 0.

[ocr errors]
[subsumed][subsumed][subsumed][ocr errors]

Let x be very small :: 25 being exceedingly small may be omitted, and the equation becomes y' = 5 a x y*, or y: = 5 a x®, which is the equation to a semi-cubical parabola P A P' fig. (1.); and if y be very

Y small, we have = 5 a y, which gives the parabola Q AQ'; hence near the origin the curve assumes the forms of the two parabolic branches. Again when x is infinitely great, ze may be neglected in comparison with 25 and the equation becomes y' = - moy :: = – x; hence for x positive, we have an infinite branch in the angle X A y, and for v negative an infinite branch in the angle Y A. x. To find the asymptote : y' =

+ 5 a x' y?,

a ya (

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

Therefore the equation to the asymptote is y + x = a; this being drawn and the branches A P', AQ produced towards it, we have nearly a correct idea of the curve.

If the equation be yo 5 a' re y + x = 0, the curve will be traced in the same manner, tig. (2).

If the equation be y® a' x y* + 20 = 0, we have fig. (3);
And the equation yo - a' z* y* – 20 = 0 will give fig. (4).

z )
Ex. Find the locus of the equation – 4 a' x y - * = 0.

For the above method of tracing curves of this species, see a treatise on the Differential Calculus, by Professor Miller. Cambridge, 1832.

320. B C is a straight line of given length (2 b), having its extremities always in the circumferences of two equal circles, to find the locus of the middle point P of the line B C.

Let the line joining the centres 0,0' of the circles be the axis of z, and let the origin of rectangular axes be at A, the bisecting point of 0 0'.

Let
x y be the co-ordinates of B.

C.
X Y

P.

a' y

.

[ocr errors]
[ocr errors]
[ocr errors]

.

[ocr errors][merged small]
[ocr errors]

AO=AO' = a,

OB = O' C = C,
the equation to B is 'y' + (x – a) =

(1)
to C is y'! + (x' + a?)! = c* (2)
also (y - y) + (x − x)

m')! = 4.62
2 Y = y + y' (4)
2 X

= x + x From these five equations we must eliminate the four quantities y, k, y' and x'; from (1) and (2)

ya y'? + 22
y'? + x - x - 2 a (x + x) = 0,

X' ?
or (y y') Y + (x - x') X – 2 a X = 0

=

(6), from (4) and (5) y + y + 3 + x + 2 yy' + 2 xx=4Y + 4 X”, from (3) y + y2 + m + / - 2 yy' – 2 x d' = 46',

x from (1) and (2) 2 y + 2 y'? + 2 x + 2 x'! – 4 a (x - ')= 40 - 4 a', '

% :. by substitution 4 a (x - 2) = 4 (Y + X* + * + a' – c'),

')

.

[ocr errors]

2

2

[ocr errors]
[ocr errors]
« ForrigeFortsett »