Full text: A treatise on algebraic plane curves

* Cf. Newton, vol. i, p. 357, and also Cramer, p. 54. 
45 
Chap. Ill REAL SINGULAR POINTS 
them ? It is well, since /x is probably a fraction, to make a more 
elaborate change of variable 
x — x' r y = y'x' s 
y'Pi =0. (11) 
We wish to retain those terms where has the lowest 
value, and momentarily reject the others. How can we give 
to r and s such values that 
rotp+rfp = ... = roit+sPi = ... = ra Q +sp q ; (12) 
Op < ... <oc t < ... < a ff , P p > ... > Pi > ... > P q (13) 
whereas in all the other terms the value is greater ? The true 
method was discovered by Newton.* 
Let us start in the north-east quadrant of a new coordinate 
plane, and mark every point with the exponents (a, /3), i.e. we 
take the exponents of every term in (x, y) actually appearing, 
and mark the corresponding point in the (a, /3) plane. The line 
from {oi p ,p p ) to (a g ,p a ) has the slope 
Pg~Pp _ r 
ix g — 0i p S 
The points («;,&) indicated in the inequality above lie on this 
line, whose equation is 
ra+s/3 = rotpH-SjSp. 
The terms we wish to reject say a m ,j6 m for which 
lie on the other side of this line from the origin. The method of 
procedure is, then, as follows. 
The origin in the (ex, j8) plane is not a marked point, but there 
is surely some marked point on the /3-axis as otherwise our 
original equation would be divisible by a power of x. Take the 
marked point on the /3-axis nearest the origin, and call it P v 
Let the half-line which starts from P 1 and goes through the 
origin rotate positively about P x till it passes through one or 
more marked points; P 2 shall be the most distant of these from 
P v Let the half-line through P 2 away from P x rotate positively 
about P 2 till it passes through at least one other marked point, 
the most distant being P 3 , etc. Continuing thus we get a broken
	        
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