ANALY1K gMOMKTltY [Cii. Ml. 



K.XAM M tin- .-.juanon 



to det-nimi' the nature and portion of its locus. Turn ih axM through 

 an angle 0, i.., substitute for s ami //, respectively, x 1 cos - y 7 sin ami 

 a/ sin + y' co80; equation (5) then becomes 



sin'0 + 4sin0cos0) 



- 8cas0sin + 2cos0 sin0 - 4sin 2 + 4 co**0) 

 (48in<0 + cos'0 - 4sin0cos0) 

 - 18 cos + 26 8iii0) 



n0 + 26cos0)+64 = 0, . . (6) 



in which is to be so determined that the coefficient of x'y' shall be 



On placing this coefficient equal to zero, it is at once seen that tan 20=}, 



from which it follows (cf. exercise 3, Art. 16, second method) that 



sin 20= f and cos20 = |; 

 remembering that cos 2 = cos 2 - sin 3 = 2 cos 2 0-1 = 1-2 sin* 



it is easily deduced that sin = ^- and cos = 



V6 V5 



Substituting these values in equation (6), it becomes 

 5*" - 2V5z / + 14 Vr, ,/ + 64 = 0, 



which is the equation of a parabola whose vertex is at the point 



_L 63 \ 

 Vo' 14V5/' 



whose focus is at the point ( -, ), whose axis is parallel to the 



\/5 V5' , 4 



negative end of the y'-axis, and whose latus rectum is ^. All these 



V5 



results refer to the new axes ; the locus of the above equation is given in 

 Fig. 79, p. 178 (Art 108). 



EXERCISES 



1. For thn hyperbola in Fig. 121 find the 000rd$flBtM of flip confer 

 and of the foci, and also the equations of its axes and directrices, all 

 referred to the axes OX and OY. 



