Answer
Center: $ \quad (-5,4)$
Foci: $ \quad (-5-2\sqrt{3},4), \quad (-5+2\sqrt{3},4)$
Vertices: $ \quad (-9,4), \quad (-1,4)$
Work Step by Step
Divide with 16 to obtain the standard form
$\displaystyle \begin{aligned}\frac{(x+5)^{2}}{16}+\frac{4(y-4)^{2}}{16}&=1\displaystyle \\\frac{(x+5)^{2}}{16}+\frac{(y-4)^{2}}{4}&=1\displaystyle \end{aligned}$
1. $ \quad $The form of the equation is (See table 3)
Major axis horizontal (parallel to the x-axis)$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h+c, k)}&{(h+a,k)}&{\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1}\\
{(h-c,k)}&{(h-a,k)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\\ \end{array}$
2. $ \quad $Center = $(h,k)=(-5,4)$,
3.$ \quad a=4, b=2.$
Find $c: $
$c^{2}=a^{2}-b^{2}=16-4=12$
$c=\sqrt{12}=2\sqrt{3}\approx 3.464$
Vertices: $ \quad (h\pm a,k)=(-5\pm 4,4)$
$(-9,4), \quad (-1,4)$
Foci: $ \quad (h\pm c,k) = (-5\pm 2\sqrt{3},4)$
