Answer
Center: $(-5,4)$
Foci: $(-5-2\sqrt 3,4),(-5+2\sqrt 3,4)$
Vertices: $(-9,4),(-1,4)$
See graph
Work Step by Step
We are given the ellipse:
$(x+5)^2+4(y-4)^2=16$
Put the equation in standard form:
$\dfrac{(x+5)^2}{16}+\dfrac{4(y-4)^2}{16}=1$
$\dfrac{(x+5)^2}{16}+\dfrac{(y-4)^2}{4}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Identify $h,k,a,b$:
$h=-5$
$k=4$
$a^2=16\Rightarrow a=\sqrt{16}=4$
$b^2=4\Rightarrow b=\sqrt 4=2$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=16-4$
$c^2=12$
$c=\sqrt {12}$
$c=2\sqrt 3$
Determine the center:
$(h,k)=(-5,4)$
Determine the foci:
$(h-c,k)=(-5-2\sqrt 3,4)$
$(h+c,k)=(-5+2\sqrt 3,4)$
Determine the vertices:
$(h-a,k)=(-5-4,4)=(-9,4)$
$(h+a,k)=(-5+4,4)=(-1,4)$
Determine the $y$-intercepts:
$(h,k-b)=(-5,4-2)=(-5,2)$
$(h,k+b)=(-5,4+2)=(-5,6)$
Graph the ellipse:
