Answer
See graph
Foci: $(0,-2\sqrt{5}),(0,2\sqrt{5})$
Work Step by Step
We are given the hyperbola:
$y^2-4x^2=16$
Bring the equation to the standard form:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
$\dfrac{y^2}{16}-\dfrac{4x^2}{16}=1$
$\dfrac{y^2}{16}-\dfrac{x^2}{4}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=16\Rightarrow a=\sqrt{16}=4$
$b^2=4\Rightarrow b=\sqrt {4}=2$
$c^2=a^2+b^2$
$c^2=16+4$
$c^2=20$
$c=\sqrt{20}$
$c=2\sqrt 5$
The centre of the hyperbola is:
$(h,k)=(0,0)$
Determine the coordinates of the vertices:
$(h,k-a)=(0,0-4)=(0,-4)$
$(h,k+a)=(0,0+4)=(0,4)$
Determine the coordinates of the foci:
$(h,k-c)=(0,0-2\sqrt{5})=(0,-2\sqrt{5})$
$(h,k+c)=(0,0+2\sqrt{5})=(0,2\sqrt{5})$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y-0=\pm\dfrac{4}{2}(x-0)$
$y=\pm 2x$
Graph the hyperbola:
