Since the x-coordinate of the vertices are constant, we can conclude it is a vertical hyperbola. The equation for a vertical hyperbola is given by:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1[/tex]Where:
[tex]\begin{gathered} F=(h,k\pm c) \\ V=(h,k\pm a) \end{gathered}[/tex]so:
[tex]\begin{gathered} k+c=12 \\ so: \\ k=12-c \\ k-c=8 \\ 12-c-c=8 \\ 12-2c=8 \\ c=2 \\ k=10 \end{gathered}[/tex][tex]\begin{gathered} k-a=9 \\ a=k-9 \\ a=1 \\ \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} b=\sqrt{2^2-1^2} \\ b=\sqrt{3} \end{gathered}[/tex]So, the equation is:
[tex]\frac{(y-10)^2}{1}-\frac{(x-4)^2}{3}=1[/tex]
