The given hyperbola is a vertical hyperbola, with centre (h,k) =(1,-2) , and vertices are (1,-6),(1,2).
So we have h =1, k =-2, and k+a =2
That gives a = 4
And the equation of asymptotes are
[tex] y=-\frac{4}{3}x-\frac{2}{3},\frac{4}{3}x- \frac{2}{3}=\pm\frac{4}{3}(x-1)}+2 [/tex]
Therefore
[tex] \frac{a}{b} =\frac{4}{3} [/tex]
Substituting the value of a and solving for b
[tex] b=3 [/tex]
Required equation is
[tex] \frac{(y+2)^2}{9} -\frac{(x-1)^2}{16} =1 [/tex]
