Solution:
The equation of the ellipse in the question is given as
[tex]\frac{(y-3)^2}{9}-\frac{(x-3)^2}{9}=1[/tex]
Concept:
The general formula of a hyperbola is
[tex]\begin{gathered} \frac{(y-h)^2}{b^2}-\frac{(x-k)^2}{a^2}=1 \\ \text{Where } \\ h,k\text{ are the centres} \end{gathered}[/tex]
From the above equation by comparing coeficient, we will have
[tex]\begin{gathered} a^2=9 \\ a=3 \\ b^2=9 \\ b=3 \\ (a,b)\Rightarrow(3,3) \\ h=3,k=3 \end{gathered}[/tex]
The linear eccentricity c, will be
[tex]\begin{gathered} c=\sqrt[]{a^2+b^2} \\ c=\sqrt[]{3^2+3^2} \\ c=\sqrt[]{9+9} \\ c=\sqrt[]{18} \\ c=3\sqrt[]{2} \end{gathered}[/tex]
The vertices of the hyperbola will be calculate using the formula below
[tex]\begin{gathered} (h,k-b)\Rightarrow(3,3-3)\Rightarrow(3,0) \\ (h,h+b)\Rightarrow(3,3+3)\Rightarrow(3,6) \end{gathered}[/tex]
The co-vertex are calculated using the formula below
[tex]\begin{gathered} (h-a,k)\Rightarrow(3-3,3)\Rightarrow(0,3) \\ (h+a,k)\Rightarrow(3+3,3)\Rightarrow(6,3) \end{gathered}[/tex]
The foci of the hyperbola will be calculated using the formula below
[tex]\begin{gathered} \mleft(h,k-c\mright)\Rightarrow(3,3-3\sqrt[]{2)} \\ \mleft(h,k+c\mright)\Rightarrow\Rightarrow\Rightarrow\mleft(3,3+3\sqrt[]{2}\mright) \end{gathered}[/tex]
The eccentricity, e is
[tex]e=\frac{c}{b}=\frac{3\sqrt[]{2}}{3}=\sqrt[]{2}[/tex]
Hence,
The sketch of the graph is given below as
The foci is represented with the two red dots on the graph
The vertices are represented above by the two coordinates
