SOLUTION
Given the question on the question tab;
Explanation:
[tex]\frac{(y+2)^2}{16}-\frac{(x-3)^2}{9}=1[/tex][tex]h=3,k=-2,a=3,b=4[/tex][tex]The\text{ }standardform\text{ }is\frac{\text{ }\left(y+2\right)^2}{4^2}-\frac{\left(x - 3\right)^{2}}{3^{2}}=1.[/tex][tex]The\text{ }linear\text{ }eccentricity\text{ }is\text{ }c=\sqrt{b^{2} + a^{2}}=5.[/tex][tex]\begin{gathered} The\text{ first focus is:} \\ \left(h,k−c\right)=\left(3,−7\right). \\ The\text{ second focus is:} \\ \left(h,k+c\right)=\left(3,3\right) \end{gathered}[/tex]
Final answer:
