91Ó°ÊÓ

The given ellipse is $\frac{{(x+3)}^2}{4}+\frac{{(y-5)}^2}{16}=1$.

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

Compare the given equation with standard form and conclude that $\frac{{(x+3)}^2}{4}+\frac{{(y-5)}^2}{16}=1$is in standard form and the center is$(-3,5)$.

Compare the equation $\left(1\right)$with $\frac{{(x+3)}^2}{4}+\frac{{(y-5)}^2}{16}=1$to find $a^2$and width="18">b2.

$\begin{array}{rcl}{a}^2& =& 4\\ b^2& =& 16\end{array}$

Since $16>4$and $16$is in the$y^2$term, the major axis will be vertical.

The value is $a^2=4$.

This implies that$a=Â\pm 2$.

This follows that the minor axis is horizontal and the distance from the center to the vertices is $2$.

The value is $b^2=16$.

This implies that $b=Â\pm 4$.

This follows that the distance from the center to the endpoints of the major axis is localid="1645686523387" $4$.

The graph of an ellipse is shown below: