Answer
Vertices: $(-2\sqrt 2,0),(2\sqrt 2,0)$
Foci: $(-\sqrt 6,0),(\sqrt 6,0)$
See graph
Work Step by Step
We are given the ellipse:
$4y^2+x^2=8$
Put the equation in standard form:
$\dfrac{4y^2}{8}+\dfrac{x^2}{8}=1$
$\dfrac{x^2}{8}+\dfrac{y^2}{2}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=8\Rightarrow a=\sqrt{8}=2\sqrt 2$
$b^2=2\Rightarrow b=\sqrt 2$
$c^2=a^2-b^2=8-2=6$
$c=\sqrt{6}$
The center is:
$(h,k)=(0,0)$
Find the vertices:
$(h-a,k)=(0-2\sqrt 2,0)=(-2\sqrt 2,0)$
$(h+a,k)=(0+2\sqrt 2,0)=(2\sqrt 2,0)$
Find the foci:
$(h-c,k)=(0-\sqrt 6,0)=(-\sqrt 6,0)$
$(h+c,k)=(0+\sqrt 6,0)=(\sqrt 6,0)$
Use the value of $b=\sqrt 2$ to find the two points above and below the center:
$(h,k-b)=(0,0-\sqrt 2)=(0,-\sqrt 2)$
$(h,k+b)=(0,0+\sqrt 2)=(0,\sqrt 2)$
Plot the center, the vertices, the two points above and below the center, and the foci. Then, graph the ellipse:
