The equations of the hyperbola are
[tex]y=\pm\frac{b}{a}(x-h)+k[/tex](h, k) are the center of it
(h + a, k), (h - a, k) are the vertices
(h + c, k), (h - c, k) are the foci
From the graph, we can see the vertices are (-3, 3) and (3, 3), then
[tex]\begin{gathered} h+a=3\rightarrow(1) \\ k=3 \end{gathered}[/tex]The foci are (-5, 3) and (5, 3), then
[tex]h+c=5\rightarrow(2)[/tex]From the graph, we can see the center is the midpoint of the line joining the foci, then the center is (0, 3), then
h = 0
k = 3
Substitute the value of h in (1) and (2) to find a and c
[tex]\begin{gathered} 0+a=3 \\ a=3 \\ 0+c=5 \\ c=5 \end{gathered}[/tex]Use the relation
[tex]\begin{gathered} c^2=a^2+b^2 \\ b^2=c^2-a^2 \\ b^2=5^2-3^2 \\ b^2=25-9=16 \\ b=\pm4 \end{gathered}[/tex]Now, substitute the values of a, b, h, k in the equations of the asymptote above
[tex]\begin{gathered} y=\frac{4}{3}(x-0)+3 \\ y=\frac{4}{3}x+3 \\ y-3=\frac{4}{3}x\rightarrow(1st) \end{gathered}[/tex][tex]\begin{gathered} y=-\frac{4}{3}x+3 \\ y-3=-\frac{4}{3}x \end{gathered}[/tex]The correct answer is A (1st equation of asymptote)
