Question 1028772
The transverse axis should be parallel to the y-axis, and the center is at the point (1,2), the midpoint of the vertices (1,6) and (1,-2).

The standard form is then given by {{{(y-2)^2/a^2 - (x-1)^2/b^2 = 1}}}

<==> y-2 = +/- {{{(a/b)sqrt(b^2+(x-1)^2)}}}

For very large positive/negative x-values, 

y-2 = +/-{{{(a/b)(x-1))}}}.  (Suppress {{{b^2}}}.
==> y = +/-{{{(a/b)(x-1))}}} + 2.

y = +/-{{{(4/3)(x-1))}}} + 2.

Hence let a = 4 and b = 3, and the standard form is  {{{(y-2)^2/16 - (x-1)^2/9 = 1}}}