78 Mr Spurge, On the curves of constant intensity of [Jan. 28, 

 The general equation being 



. * • 2 



sin 29 

 we have for the radii of the points of contact the equation 



smr = + T = + l, 

 substituting k for sin 20. 



If the point of contact is on the nth. oval we must take the 

 solution 



r 2 = 2n — 1 o ■ 



Thus the distance of the point of contact of the tangents to the 

 nth oval of intensity k is 



r = s /(2n-l)\. 



Now this radius is independent of k but contains n. Therefore 

 the tangents to all nth ovals have their points of contact on a 



circle centre the origin and radius \J 2n — l ■= . 



Proposition VII. All ovals of the same intensity have the 



same area. 



See Glazebrook, loc. cit. 



Proposition VIII. Let the tangents be drawn from the origin to 

 ovals of intensity k. The area contained by the tangents and 

 the two parts of any two consecutive ovals that they intercept 

 will be always the same. 



Consider the area between the parts intercepted by the tangents 

 to the nth. and n + 1th ovals of intensity k. 



In Fig. 1, PP' is the nth oval, QQ' the w + lth oval and 

 PP'Q'Q is the element bounded by the curves and lines through 

 the origin. OEE' OFF' the common tangents, see Prop. V. Just 

 as in the last proposition 



Area EE'F'F= if (0 Q 2 - OP 2 ) dO. 



