Question

Can you please help me with 37

Answer 1

The equation of the ellipse in the question is given as

`((y-3)^2)/(9)-((x-3)^2)/(9)=1`

Concept:

The general formula of a hyperbola is

`\begin{gathered} ((y-h)^2)/(b^2)-((x-k)^2)/(a^2)=1 \\ \text{Where } \\ h,k\text{ are the centres} \end{gathered}`

From the above equation by comparing coeficient, we will have

`\begin{gathered} a^2=9 \\ a=3 \\ b^2=9 \\ b=3 \\ (a,b)\Rightarrow(3,3) \\ h=3,k=3 \end{gathered}`

The linear eccentricity c, will be

`\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}`

The vertices of the hyperbola will be calculate using the formula below

`\begin{gathered} (h,k-b)\Rightarrow(3,3-3)\Rightarrow(3,0) \\ (h,h+b)\Rightarrow(3,3+3)\Rightarrow(3,6) \end{gathered}`

The co-vertex are calculated using the formula below

`\begin{gathered} (h-a,k)\Rightarrow(3-3,3)\Rightarrow(0,3) \\ (h+a,k)\Rightarrow(3+3,3)\Rightarrow(6,3) \end{gathered}`

The foci of the hyperbola will be calculated using the formula below

`\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}`

The eccentricity, e is

`e=(c)/(b)=\frac{3\sqrt[]{2}}{3}=\sqrt[]{2}`

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