Answer :
Solution:
The standard equation of a hyperbola is expressed as
[tex]\begin{gathered} \frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1\text{ ---- equation 1} \\ \text{where} \\ (h,\text{ k) is the coordinate of its center} \\ a\text{ is the axis} \\ b\text{ is the conjugate axis} \end{gathered}[/tex]Given the equation of the hyperbola to be
[tex]y^2-x^2=81\text{ ---- equation 2}[/tex]Express equation 2 in a similar form as equation 1.
Thus,
[tex]\begin{gathered} y^2-x^2=81 \\ \text{divide both sides of the equation by 81} \\ \frac{y^2-x^2}{81}=\frac{81}{81} \\ \Rightarrow\frac{y^2}{9^2}-\frac{x^2}{9^2}=1\text{ ---- equation 3} \end{gathered}[/tex]In comparison with equation 1, we can conclude that
[tex]\begin{gathered} a=9 \\ b=9 \end{gathered}[/tex]Vertices of the hyperbola:
The vertices of the hyperbola are expressed as
[tex]\mleft(h,k+a\mright),\: \mleft(h,k-a\mright)[/tex]where
[tex]h=0[/tex]The vertices of the hyperbola are evaluated to be
[tex]\begin{gathered} (h,\: k+a)\Rightarrow(0,\text{ 0+9)=(0, 9)} \\ (h,\: k-a)\Rightarrow(0,0-9)=(0,-9) \end{gathered}[/tex]Hence, the vertices of the hyperbola are
[tex](0,9),\text{ (0, -9)}[/tex]Foci of the hyperbola:
The foci of the hyperbola are expressed as
[tex]\begin{gathered} \mleft(h,k+c\mright),\: \mleft(h,k-c\mright) \\ \text{where c }\mathrm{\: }is\: the\text{ distance from the center (h,k) to a focus} \\ c\text{ is evaluated as } \\ c=\sqrt{a^2+b^2} \end{gathered}[/tex]Evaluating c gives
[tex]\begin{gathered} c=\sqrt{a^2+b^2} \\ =\sqrt[]{9^2+9^2} \\ =\sqrt[]{81+81} \\ =\sqrt[]{162} \\ c=9\sqrt[]{2} \end{gathered}[/tex]Thus, the foci are evaluated as
[tex]\begin{gathered} (h,k+c)\Rightarrow(0,\text{ 0+9}\sqrt[]{2})=(0,9\sqrt[]{2}) \\ (h,k-c)\Rightarrow(0,\text{ 0-9}\sqrt[]{2})=(0,-9\sqrt[]{2)} \end{gathered}[/tex]Hence, the foci of the hyperbola are
[tex]\mleft(0,\: 9\sqrt{2}\mright),\: \mleft(0,\: -9\sqrt{2}\mright)[/tex]