Answer
See below:
Work Step by Step
The larger number is under the expression involving $y$, so the major axis is vertical and parallel to the $y$ -axis.
Compare equation (II) $\frac{{{\left( x-3 \right)}^{2}}}{9}+\frac{{{\left( y+3 \right)}^{2}}}{36}=1$ with the standard form $\frac{{{\left( x-h \right)}^{2}}}{{{b}^{2}}}+\frac{{{\left( y-k \right)}^{2}}}{{{a}^{2}}}=1$ ,
$h=3$, $k=-3$
$\begin{align}
& {{a}^{2}}=36\Rightarrow a=\pm 6 \\
& {{b}^{2}}=9\Rightarrow b=\pm 3
\end{align}$
So, the center of the ellipse is given by $\left( h,k \right)=\left( 3,-3 \right)$.
For a vertical major axis with center $\left( 3,-3 \right)$, the vertices are given by points
$\begin{align}
& \left( 3,-3+6 \right)=\left( 3,3 \right) \\
& \left( 3,-3-6 \right)=\left( 3,-9 \right) \\
\end{align}$
End-points of the minor axis are given by the points
$\begin{align}
& \left( 3+3,-3 \right)=\left( 6,-3 \right) \\
& \left( 3-3,-3 \right)=\left( 0,-3 \right) \\
\end{align}$
The coordinates of the foci of an ellipse are given by $\left( h+c,k-c \right)$
We know that,
$\begin{align}
& {{c}^{2}}={{a}^{2}}-{{b}^{2}} \\
& =36-9 \\
& =27 \\
& c=\pm 3\sqrt{3}
\end{align}$
For a vertical major axis with center $\left( 3,-3 \right)$, the foci are at points
$\left( 3,-3+3\sqrt{3} \right)$ And $\left( 3,-3-3\sqrt{3} \right)$
Consider the equation of an ellipse:
$4{{x}^{2}}+{{y}^{2}}-24x+6y+9=0$ (I)
Now, we will convert the equation in standard form as follows:
$\begin{align}
& 4{{x}^{2}}+{{y}^{2}}-24x+6y+9=0 \\
& 4{{x}^{2}}-24x+{{y}^{2}}+6y+9=0 \\
& 4\left( {{x}^{2}}-6x \right)+\left( {{y}^{2}}+6y \right)+9=0 \\
& 4\left( {{x}^{2}}-6x+9 \right)-36+\left( {{y}^{2}}+6y+9 \right)-9+9=0
\end{align}$
It can be further simplified as:
$\begin{align}
& 4{{\left( x-3 \right)}^{2}}+{{\left( y+3 \right)}^{2}}-36-9+9=0 \\
& 4{{\left( x-3 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=36 \\
& \frac{{{\left( x-3 \right)}^{2}}}{9}+\frac{{{\left( y+3 \right)}^{2}}}{36}=1
\end{align}$
So one will have the equation of the ellipse as:
$\frac{{{\left( x-3 \right)}^{2}}}{9}+\frac{{{\left( y+3 \right)}^{2}}}{36}=1$ (II)
Therefore, the foci of the ellipse are $\left( 3,-3+3\sqrt{3} \right)$ and $\left( 3,-3-3\sqrt{3} \right)$.
