[tex]\frac{y^{2}}{36}-\frac{x^{2}}{28}=1[/tex]
And the value of h=0
And the value of k is 0
And the value of a:2√7
And the value of b is 6
1) Since we were told that the vertices are (0,6) and (0,-6) and one focus is at (0,-8) we can write out the following equation to find the hyperbola:
[tex]\begin{gathered} h=0 \\ (k+8)^2=a^2+b^2 \\ \left(k-6\right)^2=b^2 \\ \left(k+6\right)^2=b^2 \\ (k+8)^2=a^2+\left(k-6\right)^2,h=0,k=0 \\ 64=a^2+36\Rightarrow64-36=a^2\Rightarrow a^2=28,a=\sqrt{28},a=2\sqrt{7} \\ \left(k-6\right)^2=b^2\Rightarrow b^2=36\operatorname{\Rightarrow}b=6 \end{gathered}[/tex]
Note that to find a, and b we set h=0, k=0
2) Now, we can set the standard equation for this hyperbola:
[tex]\begin{gathered} \frac{y^2}{a^2}-\frac{x^2}{b^2}=1 \\ \\ \frac{y^2}{6^2}-\frac{x^2}{(2\sqrt{7)}^2}=1 \\ \\ \frac{y^{2}}{6^{2}}-\frac{x^2}{(28)}=1 \\ \frac{y^2}{36}-\frac{x^2}{28}=1 \\ \end{gathered}[/tex]
