The given ellipse is $\frac{{(x-5)}^2}{9}+\frac{{(y+4)}^2}{4}=1$.

The given ellipse will have the same size and shape as the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$whose center is $(0,0)$.

The standard form of an ellipse with center $(0,0)$is given by localid="1645700232210" $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\dots \left(1\right)$.

Compare the equation $\frac{x^2}{9}+\frac{y^2}{4}=1$with the equation $\left(1\right)$to find $a^2$and $b^2$.

$\begin{array}{rcl}{a}^2& =& 9\\ b^2& =& 4\end{array}$

Since $9>4$and $9$ is in the$x^2$ term, the major axis will be horizontal.

The value is $a^2=9$.

This implies that $a=\pm 3$.

The coordinates of the vertices are $(3,0)$and $(-3,0)$.

The value is $b^2=4$.

This implies that $b=\pm 2$.

The coordinates of the endpoints of the minor axis are$(0,2)$ and $(0,-2)$.

The graph of an ellipse is shown below:

The standard form of an ellipse with center $(h,k)$is given by $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$.

The equation $\frac{{(x-5)}^2}{9}+\frac{{(y-(-4\left)\right)}^2}{4}=1$is in standard form and the center is $(5,-4)$.

Now, translate the graph of $\frac{x^2}{9}+\frac{y^2}{4}=1$five units to the right and then down$4$units.

The graph of a new ellipse is shown below: