352 Mr. A. Cayley on the Stenographic Projection 



and that thence 



the equation is 



It is now very easy to trace the curve. We see first that the 

 curve is symmetrical with respect to the axes, and that it meets 

 the axis of y in four imaginary points, but the axis of x in four 

 real points, the coordinates whereof are 



x=±{>/l + c 2 ±c), 



so that the two points on the same side of the centre are the 

 images one of the other in regard to the circle radius unity. 

 Moreover the curve touches the lines 



y= +#tan/3 

 at their intersections with the circle. By developing in regard 

 to y, the equation becomes 



2/ 4 + 2 (^-l + 2c 2 cot 2 /3) 2 / 2 -f-(^ 2 -l) 2 -4cV = 0; 

 and putting 



*=5±(Vl+?±(fl) J 



the last term vanishes, and the equation gives y 2 = 0, or 

 2/ 2 = 2(l-^ 2 -2c 2 cot 2 /3) 



= 4(-c 2 + c^T+^-c 2 cot 2 /3) 



= 4c( — ccosec 2 /3+ \/l + c 2 ), 

 the upper sign corresponding to the exterior values 



±x= V\ + c* + c, 

 and the lower sign to the interior values 



±x= a/1 + c 2 — c. 



In the former case the values of y are imaginary ; in the latter 

 case they are real if 



V'l-j-c^ccosec 2 /?, 

 or, what is the same thing, if 



sin 2 /3> * ; 



Vl + c* 



that is, if, for a given value of c, /3 is sufficiently great, but other- 

 wise they are imaginary. 



5 

 If, as in the annexed figures, c= ^o 



