3 
Case 4. When both x 2 and y 2 are negative : the x and y 
ordinates are both circular, and each pair being compounded 
(in the manner of the compounding of moments of rotation) 
forms a single circular ordinate, the plane of which is 
different from that of every other compound ordinate ; 
hence, there can be no intersection of compound ordinates, 
and these can have no envelope. 
The three equations (a), (j3), (7) possess this remarkable 
reciprocal property : if any one of them be taken as the 
equation to be represented, it is itself the Cartesian equa- 
tion of the corresponding curve when x 2 and y 2 are both 
positive, and the two others are the Cartesian equations of 
the respective envelopes when either x 3 or y 3 is alone 
negative : and thus the ellipse (a) and the hyperbola (j3) 
are the representative curves of all the three equations. 
The three similar equations (a'), (j3'), (7'), in which the 
only difference from the former three is in the reversed 
sign of unity on the right-hand sides— possess, of course, 
the same reciprocal property; but here the impossible 
curve is on the xy plane, that on the xz plane is the con- 
jugate hyperbola to the former one, and is on the same 
plane with it and has therefore the same asymptotes, whilst 
the curve on the yz plane is an ellipse whose semi-axes are 
a and (a 2 — 6 2 )*. 
When a = b, that is, (a) is a circle, the three other real 
curves coincide with the 0 axes. 
To find the Cartesian equation to the envelope of the 
circular ordinates in the case No. 2, in which y 2 alone is 
negative, we have for the positive square of the type radius 
of the circles %{x 2 - a 2 ), and for the equation to a type circle 
CL 
+ ? =- 2 (x 2 - a 2 ) • • * * • * r • » » • • • (i) 
CL 
where x is the variable parameter : hence, taking the first 
derivative, 
