Question 1027926
The hyperbola would have the standard form of {{{y^2/36 - x^2/b^2 = 1}}}, and we have to find b.

==> {{{y^2 = (36/b^2)(b^2+x^2)}}}
==> y = ±{{{(6/b)x}}} (asymptotically) as x goes to {{{infinity}}}.

==> {{{6/b = 3/4}}} ==> b = 8, and the standard form would be {{{y^2/36 - x^2/64 = 1}}}.