196 



MR. CHARLES CHAMBERS ON THE 



by an envelope on the xz plane, the Cartesian equation to 

 which we shall presently show to be (/3), viz. that of the 

 hyperbola /3 (fig. i) . This hyperbola has its vertices in the 

 foci of the ellipse (a), and its foci, therefore, at the vertices 

 of the ellipse ; and its transverse axis is of the same mag- 

 nitude as that of the ellipse. 



Fig. I. 



,^-(j^- 



Case 3. When x^ alone is negative — that is, when y^ is 

 positive and greater than IP'. The y ordinates are tubular, 

 and the x ordinates circular. The envelope of the latter 

 has (7) for its Cartesian equation, and is an impossible 

 curve ; or, in other words, there is no intersection of the 

 circular ordinates, and therefore no envelope. 



Case 4. When both x^ and ?/* 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 dif- 

 ferent from that of every other compound ordinate ; hence 



