1921] An Interesting Maximal Case 71 



each other in either quadrant, since only one corner of the maximum 



rectangle inscribed in the ellipse lies in either quadrant. 



Since the ellipse and the rectangular hyperbola have a pair of 



common tangents at two corners of the maximum rectangle, it is 



clear that the critical values may be arrived at by solving the two 



ab 

 equations simultaneously. Thus substituting y = -— in the equation 



2x 



ot the ellipse we have 



a*b2 

 b^x^ + ~ — = a-b^, 



or 



1. e. 



giving 



4x* — 4a2x2 + a^ = 0, 



(2x2 _ a2)2 = 0, 



a 



and accordingly 



ab 



2+ -^ ^2 



At this point, or at an earlier point, in the investigation, the in- 

 quiry may be raised why both rectangular hyperbolas: 



ab 



xy = + Y 



and 



ab 



xy = — — 

 ^ 2 



do not enter into the problem. 



Although the question has virtually been answered in the course of 



the treatment in Method V, in the rejection of the solution <^ = 



— a — b 



135°, it may be pointed out that while x = ~~rK-> y ~ ~~/^ 



formally satisfy the equation 



ab 

 xy = — - 



