The given ellipse is $\frac{{(x+6)}^2}{16}+\frac{{(y+2)}^2}{25}=1$.

The given ellipse will have the same size and shape as the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$whose center is $(0,0)$.

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

Compare the equation $\frac{x^2}{16}+\frac{y^2}{25}=1$with the equation $\left(1\right)$to find $a^2$and $b^2$.

$\begin{array}{rcl}{a}^2& =& 16\\ b^2& =& 25\end{array}$

Since$25>16$ and $25$is in the role="math" $y^2$term, the major axis will be vertical.

The value is $a^2=16$.

This implies that $a=\pm 4$.

The coordinates of the vertices are $(-4,0)$and $(4,0)$.

The value is $b^2=25$.

This implies that $b=\pm 5$.

The coordinates of the endpoints of the major axis are $(0,5)$and $(0,-5)$.

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 $\frac{(x-{(-6))}^2}{16}+\frac{(y-{(-2))}^2}{25}=1$is in standard form and the center is $(-6,-2)$.

Now, translate the graph of localid="1645704555147" $\frac{x^2}{16}+\frac{y^2}{25}=1$six units to the left and then down$2$units.

The graph of a new ellipse is shown below: