Find the coordinates of the center, the foci, and the vertices, sketch the curve, and determine the eccentricity of the ellipse given by the equation
![\[ \frac{(x+1)^2}{16} + \frac{(y+2)^2}{25}= 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-33457ddd23bf94a7747ec1d27463fbb6_l3.png "Rendered by QuickLaTeX.com")
This is an equation for an ellipse in standard form with center at 
.
![\[ \frac{(x-x_0)^2}{b^2} + \frac{(y-y_0)^2}{a^2} = 1 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-39c48b7613f0db1ce1736ded8feed8e3_l3.png "Rendered by QuickLaTeX.com")
with 
and 
. We then know its foci are located 
and 
where
![\[ c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = 3. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-54f1e7428bd5e4edd51f45a13bd0a893_l3.png "Rendered by QuickLaTeX.com")
Therefore, the foci of this ellipse are 
and 
. Further, the vertices are given by 
and 
. Finally, we compute the eccentricity
![\[ c = ae \quad \implies \quad 3 = 5e \quad \implies \quad e = \frac{3}{5}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-821caea23416a26b9689605ac391090e_l3.png "Rendered by QuickLaTeX.com")