Question #fdc59

1 Answer
May 6, 2015

Write the equation in standard form as #(x-1/3)^2/1^2 - (y-3)^2/6^2#=1. The transverse axis of this hyperbola is parallel to x axis and its center is (1/3,3). a=1 , b= 6
Since #c^2= a^2+b^2#, c= #sqrt37#

The vertices are 'a' units from the center, hence one vertex would be #(4/3,3)# and the other would be #(-2/3,3)#.

Foci are c units from the center, hence one would be
#sqrt37+1/3#,3 and the other would be -#sqrt37 +1/3#,3

Asymptotes would be y= #3+6(x-1/3) and y= 3-6(x-1/3) #, that is y= 6x+1 and y= -6x+5