the calculus of functions. 
4 1 9 
Required the nature of the curve ADE such that taking any 
abscissa AB and corresponding ordinate DB : if the abscissa 
AC be taken, equal to DB and the ordinate EC be drawn, 
then the rectangle under the two ordinates shall be equal to 
the square of the first abscissa let the equation of the curve 
be y = $ oc then AB = oc D B = y = \J/<r AC = DB, 
and EC = 4/ ( AC )=%(/( DB ) = i|y = 
the given condition is therefore 
•vf / 8 oc x \|/ x — a ? 2 
taking the usual substitution of oc = <p" f f (p oc it becomes 
<P /* (p X X f (p X =^ a 
putting oc for oc we have 
<P /* <37 X $ f0C = (<£ OC)* 
Assume <p oc = % (oc,foc,f\v) 
the equation then becomes 
% {f*x>f 3 x> = [ % {#,/r,/v}]‘ 
If now /be determined from the equation 
f z oc — x 
we have/ 4 ac =fcc, and the equation becomes identical, 
hence <p‘ and/ being found, we may determine -p. 
One of the simplest curves which satisfies the equation is 
a line of the third order, the 69th in Newton's arrangement, 
its form is 
3 H 2 
