Answer
See graph
Work Step by Step
We are given the equation:
$y=-\sqrt{16-4x^2}$
The graph of the above equation is half of the graph of the ellipse:
$y^2=(-\sqrt{16-4x^2})^2$
$y^2=16-4x^2$
$4x^2+y^2=16$
$\dfrac{4x^2}{16}+\dfrac{y^2}{16}=1$
$\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$
Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$:
$h=0$
$k=0$
$a^2=16\Rightarrow a=4$
$b^2=4\Rightarrow b=2$
Determine the foci:
$c^2=a^2-b^2$
$c^2=4^2-2^2$
$c^2=12$
$c=\pm\sqrt{12}=\pm 2\sqrt 3$
The foci are:
$F_1(h,k-c)=(0,-2\sqrt 3)$
$F_2(h,k+c)=(0,2\sqrt 3)$
Draw the equation $y=-\sqrt{16-4x^2}$, which is the half of ellipse with negative $y$.
