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
![\[ \frac{(x-2)^2}{16} + \frac{(y+3)^2}{9} = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3c5dbf6b274d846b5b33eeed8e6c0e3d_l3.png "Rendered by QuickLaTeX.com")
This is an equation for an ellipse in standard form with center at 
.
![\[ \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-23507852673110437d3f09a0c04c25e5_l3.png "Rendered by QuickLaTeX.com")
with 
and 
. We then know its foci are located 
and 
where
![\[ c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5d52ddd5b0fcc6e9d91f76f5d3b7438d_l3.png "Rendered by QuickLaTeX.com")
Therefore, the foci of this ellipse are 
and 
. Further, the vertices are given 
and 
. Finally, we can compute the eccentricity
![\[ c = |a| e \quad \implies \quad \sqrt{7} = 4e \quad \implies \quad e = \frac{\sqrt{7}}{4}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f76f121eacc7e085336aa955c04e52dd_l3.png "Rendered by QuickLaTeX.com")